
Velocity and acceleration Newton's laws and momentum Gravity Energy Circular motion Friction Spin Pressure Buoyancy History
Waves Music Overtones Resonance
Electromagnetism Electric force Magnetic force Electric field Magnetic field Batteries Resistance
Atoms Chemistry Particles Nuclei Fusion Fission
Aerodynamic drag Pendulum Thermodynamics Balance Spin II
The fundamental units are the meter, second, and kilogram, and all other units are derived from these.
Quantity Units Length Meter Time Second Mass Kilogram Velocity = Length / Time Meter / second Acceleration = Velocity/ Time Meter / second^{2} Momentum = Mass * Velocity Kilogram meter / second Force = Mass * Acceleration Kilogram meter / second^{2} = Newtons Energy = Force * Length Kilogram meter^{2} / second^{2} = Joules = ½ *Mass * Velocity^{2} Area = Length^{2} Meter^{2} Volume = Length^{3} Meter^{3} Pressure = Force / Area Kilogram / meter / second^{2} = Pascals = Joules/meter^{3} Density = Mass / Volume Kilogram / meter^{3} Frequency = 1 / Time Second^{1} = Hertz Angular momentum= Mass * Velocity * Length Kilogram meter^{2} / second = Joule seconds
Electric charge is also a fundamental unit and it is measured in Coulombs, but it won't appear in the laws of mechanics.
Linear variables Spin variables Position = X Angle = $\theta $ = X / R Velocity = V Angular velocity = $\omega $ = V / R Acceleration= A Angular accel. = $\alpha $ = A / R Mass = M Moment of inertia = I = M * R^{2} Force = F Torque = $\Gamma $ = F * R Momentum = Q Angular momentum = L = Q * R Energy = E Time = T Radius = R Force Momentum Energy Centripetal Constant Constant acceleration velocity acceleration Linear motion: F = M A Q = M V = F T E = .5 M V^{2} = F X A = V^{2} / R X = V T V = A T Spin: $\Gamma $ = I $\alpha $ L = I $\omega $ = Q R E = .5 I $\omega 2$ = $\Gamma $ $\theta $ A = $\omega 2$ R $\theta $ = $\omega $ T $\alpha $ = $\omega $ T Energy = .5 M V^{2} + .5 I $\omega 2$ + M g Height F_{grav} =  M g =  G M_{0} M / R^{2}
If an object starts at X=0 and moves with constant velocity,
Time = T (seconds) Velocity = V (meters/second) Distance traveled = X = V T (meters)
In the previous case the acceleration is 0.
If an object starts at rest with X=0 and V=0 and moves with constant acceleration,
Time = T (seconds) Acceleration = A (meters/second^{2}) Final velocity = V = A T (meters/second) Average velocity = V_{a} = .5 V (meters/second) Distance traveled = X = V_{a} T = .5 V T = .5 A T^{2} (meters)All of these equations contain the variable T. We can solve for T to obtain an equation in terms of (X, A, V).
V^{2} = 2 A XSim: Position, velocity, and acceleration
There are four variables (X, V, A, T) and four equations, and each equation contains
three of the variables.
At T=0, X=0 and V=0.
Equations Variables in Variable not the equation in the equation V = A T V A T X X = .5 A T^{2} X A T V X = .5 V T X V T A V^{2} = 2 A X X V A T
The figure shows the position of a ball at regular time intervals and the green arrow shows the direction of the acceleration.
Top row Zero acceleration (constant velocity) Second row Positive acceleration Third row Negative acceleration (deceleration) Fourth row Freefall in gravityIn the language of calculus,
Time = T Position = X = = $\int $ V dT Velocity = V = ∂X/∂T = $\int $ A dT Acceleration = A = ∂V/∂TExamples of position, velocity, and acceleration.
Sim: Position, velocity, and acceleration #2
Mass = M Acceleration = A Force = F = M A (Newton's law)
For an object falling in gravity, the acceleration doesn't depend on mass and the acceleration is the same everywhere on the surface of the Earth.
Mass = M Gravity constant = g = 9.8 m/s^{2} Gravity acceleration = A Gravity force = F = M g (Law of gravity) = M A (Newton's law)Cancelling the "M's", the acceleration experienced by the object is
A = gIf gravity is the only force involved, then all objects experience the same gravitational acceleration.
We can distinguish between gravitational mass and inertial mass.
M_{grav} = Gravitational mass M_{inertial} = Inertial mass F = M_{grav} g Gravitational mass causes gravitational force F = M_{inertial} A Inertial mass governs the response to forceFor all known forms of matter,
M_{grav} = M_{inertial}
For this example we set g=10 m/s^{2} and assume there is no air drag. If an object starts at rest and falls under gravity, the distance fallen is
Time Velocity Average Distance Acceleration (s) (m/s) velocity fallen (m/s^{2}) (m/s) (m) 0 0 0 0 10 1 10 5 5 10 2 20 10 20 10 3 30 15 45 10 4 40 20 80 10 Distance fallen = ½ * Acceleration * Time^{2}
g = 9.8 m/s^{2} PoundAsMass = .4535 kg = Pound interpreted as mass PoundAsForce = 4.448 Newtons = Pound interpreted as force = The force exerted by .4535 kg in Earth's gravity = .4535 kg * 9.8 m/s^{2} PoundAsForce = PoundAsMass * g
Mass = M Velocity = V Momentum = Q = M V
Suppose an object undergoes a constant force for time T.
Impulse = F T = M A T = M V = Momentum
Suppose 2 objects both start with X=0, V=0, and T=0, and that they exert a constant repelling force on each other.
Object 1 accelerates to the right (positive force) and object 2 accelerates to the left (negative force).
Mass Acceleration Force Velocity after time T Momentum after time T Object 1 M_{1} A_{1} F_{1} = M_{1} A_{1} V_{1} = A_{1} T Q_{1} = M_{1} V_{1} = F_{1} T Object 2 M_{2} A_{2} F_{2} = M_{2} A_{2} V_{2} = A_{2} T Q_{2} = M_{2} V_{2} = F_{2} TThe forces and momenta are equal and opposite.
F_{2} = F_{1} Q_{1} = Q_{1} Total momentum = Q_{1} + Q_{2} = 0The total momentum is constant in time. This is the principle of "conservation of momentum".
Conservation of momentum is equivalent to the fact that forces are equal and opposite.
Equal and opposite forces imply conservation of momentum.
Conservation of momentum implies equal and opposite forces.
Suppose an object starts from rest at X=0 and experiences a constant force.
X = Distance traveled F = Force F X = M A X = .5 M V^{2} (using V^{2} = 2 A X)We can define a kinetic energy, which is equal to the force times the distance.
Kinetic energy = F X = .5 M V^{2}Newton's law implies conservation of energy. For example, suppose an object starts at rest at height X and falls in Earth's gravity until it reaches the ground.
Initial height = X Mass = M Gravity constant = g = 9.8 meters/second^{2} Velocity upon reaching ground = V Time to reach the ground = T Kinetic energy upon reaching ground = E_{k} = .5 M V^{2} Gravity energy when released = E_{g} = M A X Total energy = E_{t} = E_{k} + E_{g} E_{g} = E_{k} because .5 M V^{2} = M A XThe gravitational energy at the start of the fall is converted to kinetic energy at the end of the fall so that the total energy is constant.
[d/dT] E_{t} = [∂/∂T] E_{g} + [∂/∂T] E_{k} = M A V + M A V = 0
Mass of Earth = M = 5.972e24 kg Radius of Earth = R = 6371 km Gravity constant = G = 6.67⋅10^{11} Newton m^{2}/kg^{2} Test mass = m Force on test mass = F = G M m / R^{2} = g m Acceleration = g = G M / R^{2} = 9.8 m/s^{2}
The simpliest case of a collision is two billiard balls colliding headon and with equal speeds.
We henceforth use dimensionless units.
Mass = M Time = T Velocity = V Position = X = X T Momentum = Q = M V Energy = E = .5 M V^{2} Mass Initial Final Initial Final Initial Final velocity velocity Momentum momentum energy energy Ball 1 1 +1 1 +1 1 1/2 1/2 Ball 2 1 1 +1 1 +1 1/2 1/2 Total 2 n/a n/a 0 0 1 1Momentum is always conserved. In this example the total initial momentum is equal to the total final momentum.
Energy is either conserved or some energy is lost to heat. In this example energy is conserved.
If energy is lost to heat then the rebound velocity is less than the initial velocity. If the rebound velocity is "K" then
Mass Initial Final Initial Final Initial Final velocity velocity Momentum momentum energy energy Ball 1 1 +1 K +1 K .5 .5 K^{2} Ball 2 1 1 +K 1 +K .5 .5 K^{2} Total 2 n/a n/a 0 0 1 K^{2}
We can define a collision coefficient "K" as
K^{2} = Collision coefficient = FinalEnergy / InitialEnergyIf you know the value of K for a collision then you can solve it by writing down down equations for momentum and energy conservation.
The easiest case is if you are in the frame of the center of mass so that the total momentum is zero.
Mass Initial Final Initial Final Initial Final velocity velocity Momentum momentum energy energy Object 1 M_{1} V_{1i} V_{1f} M_{1} V_{1i} M_{1} V_{1f} .5 M_{1} V_{1i}^{2} .5 M_{1} V_{1f}^{2} Object 2 M_{2} V_{2i} V_{2f} M_{2} V_{2i} M_{2} V_{2f} .5 M_{2} V_{2i}^{2} .5 M_{2} V_{2f}^{2} Total momentum = M_{1} V_{1i} + M_{2} V_{2i} = M_{1} V_{1f} + M_{2} V_{2f} = 0If energy is conserved then
V_{1f} = V_{1i} V_{2f} = V_{2i}If energy is not conserved then
V_{1f} = V_{1i} K V_{2f} = V_{2i} Kwhere K is the collision coefficient.
Power = Energy / Time = Force * Distance / Time = Force * VelocitySuppose you climb a flight of stairs.
Height of a flight of stairs = X = 4 meters Typical time to climb a flight of stairs = T = 2 seconds Mass of a typical human = M = 75 kg Gravitational energy gained = E = MgX = 3000 Joules Power delivered in climbing the stairs = P = E/T = 1500 Watts
Often a problem can be simplified with a strategic choice of reference frame. For example, for an object in free fall the center of mass follows a parabola regardless of its angular momentum.
The motion of an object can be described as the motion of the center of mass plus an angular momentum vector. In the above figure the red dot is the center of mass.
Practice being in the frame of the sword, or your opponent, or the center of mass between you and your opponent.
There is a sequence of reference frames: The Earth, the tip of the spine (the atlas vertebra), the tip of the index finger, and the tip of the sword. They should be programmed hierarchically in this order. For example, you should be able to move the atlas vertebra while preserving the reference frame of the Earth (maintain balnance), you should be able to move the index finger while preserving the frame of the atlas vertebra, etc.
If two objects are placed on a seesaw, the center of mass is the position of the fulcrum.
Distance from the left ball to the fulcrum = a = 1 Distance from the right ball to the fulcrum = b = 20 Mass of the left ball = M_{1} = 100 Mass of the right ball = M_{2} = 5 M_{1} a = M_{2} b
The equations of constant acceleration usually assume an initial velocity and position of 0. If not, then
Time = T Initial position = X_{i} Final position = X Initial velocity = V_{i} Final velocity = V Acceleration = A X(T) = X_{i} + V_{i} T + 1/2 A T^{2} V(T) = V_{i} + A T A(T) = AIf X_{i} = V_{i} = 0 then the equations reduce to
X(T) = .5 A T^{2} V(T) = A T A(T) = AFor example, suppose a car starts at X=5, has an initial speed of 20 meters/second, and decelerates uniformly at a rate of 10 meters/second^{2}.
X(T) = 5 + 20 T  5 T^{2}
The "center of mass" (COM) and "velocity of mass" (VOM) are defined as
Mass Position Velocity Object 1 M_{1} X_{1} V_{1} Object 2 M_{2} X_{2} V_{2} Center of mass M_{c} X_{c} V_{c} Total mass = M_{c} = M_{1} + M_{2} Center of mass = X_{c} = (X_{1} M_{1} + X_{2} M_{2}) / M_{c} Velocity of mass = V_{c} = (V_{1} M_{1} + V_{2} M_{2}) / M_{c}If object 1 and 2 are balanced on a seesaw then the center of mass is at the fulcrum.
(X_{c}  X_{1}) M_{1} = (X_{2}  X_{c}) M_{2}If two objects exert a force on each other then the trajectories are
X_{1} = X_{1i} + V_{1i} T + .5 A_{1} T^{2} X_{2} = X_{2i} + V_{2i} T + .5 A_{2} T^{2} X_{c} = X_{c} + V_{c} TThe center of mass moves with constant velocity.
Displacement of the spring = X Spring constant = K Force on the spring = F = K X (Hooke's law)
For a spring, the force is proportional to displacement.
A spring contains compression energy if compressed and tension energy if stretched.
Energy of the spring = E = $\int $ F dX = $\int $ K X dX = .5 K X^{2}
Suppose a football is kicked vertically upward with a velocity of 20 m/s.
Initial vertical velocity = V = 20 m/s Maximum height reached = X = .5 V^{2} / g = 20 m Time to reach max height = T = V/g = 2 sThe "Hang time" is the total time the football spends in the air. The upward and downward parts of the trajectory each take 2 seconds, for a hang time of 4 seconds. The downward trajectory is the mirror image of the upward trajectory.
We could write the Y trajectory as
Y(T) = 20 T  5 T^{2} V_{y}(T) = 20  10 T A(T) = 10
Suppose a punt is kicked with a horizontal velocity of 20 m/s and a vertical velocity of 20 m/s.
The horizontal motion corresponds to constant velocity and the vertical motion corresponds to constant acceleration.
Time = T Horizontal velocity = V_{x} = 20 m/s (constant) Vertical velocity = V_{y} = 20  10 T Football X coordinate = X = V_{x} T Football Y coordinate = Y = 20 T  5 T^{2}We can eliminate T from the trajectories and express Y in terms of X.
Y = X  X^{2} / 80The ball follows a parabolic trajectory. In general, the trajectory of any object moving under gravity is a parabola.
The ball hits the ground when X=80 and Y=0.
Suppose a ball is pitched horizontally, with no initial vertical velocity. Suppose a second ball is dropped with zero initial velocity from the same release point as the pitch. Both balls hit the ground at the same time.
Distance from the pitcher's plate to home plate = 18.0 m (measurement) Distance from the pitcher's plate to the ball release point = 2.0 m (estimate) Distance between the ball release point and home plate = 16.0 m (estimate) Height of the pitcher's mound = .25 m (measurement) Height of the ball release point above the pitcher's mound = 1.25 m (estimate) Height of the ball release point above the field = 1.50 m (estimate) Speed of a typical fastball = 43 m/s (96 mph) Ball travel time from the release point to home plate = .372 s Vertical distance the ball drops before reaching home plate = .69 m = .5 g T^{2}
A spinning ball curves, a fact that was first observed by Newton while playing tennis. This is called the "Magnus force".
A ball with topspin curves downward and a ball with backspin curves upward.
A fastball is thrown with maximum speed, which puts backspin on the ball, giving it an upward force. This frce works against gravity and makes the ball easier to hit. A curveball has topspin, which works with gravity and makes the ball harder to hit. The topsin comes with a sacrifice in speed.
mph Fastball 95 Split finger fastball 90 Curveball 85 Knuckleball 60A split finger fastball is thrown with wide fingers to decrease the backspin. A knuckleball is thrown with minimal spin so that it curves in multiple directions on its way to the plate. A knuckleball curves because of airflow around the seams.
Radius of the circle = R Velocity = V Centripetal acceleraton = A = V^{2} / R
If artificial gravity is generated by spinning a spaceship, then according to en.wikipedia.org/wiki/Artificial_gravity, the spin period has to be at least 30 seconds for the inhabitants to not get dizzy.
Spin period = T = 2$\pi $ R / V = 30 s Centripetal acceleration = A = V^{2} / R = 10 m/s^{2} Spin radius of the spaceship = R = T^{2} A / (2$\pi $)^{2} = 228 meters Tangential velocity of the spaceship = V = (A R)^{1/2} = 48 m/s
Suppose an satellite is on a circular orbit around a central object.
Graviational constant = G
Mass of central object = M
Mass of satellite = m
Distance of satellite from central object = R
Velocity of satellite = V
Gravity force = F = G M m / R^{2}
Gravity potential energy = E = G M m / R = $\int $ F ∂R
The velocity of a circular orbit is obtained by setting gravitational force equal to
centripetal force.
G M m / R = m V^{2} / RThe escape velocity is obtained by setting gravitational energy equal to kinetic energy.
G M m / R = 1/2 m V^{2} Circular orbit velocity = (G M / R)$1/2$ Escape velocity = $\surd 2\; (G\; M\; /\; R)$ 1/2$Escape\; velocity\; =$ \surd 2\; *\; Circular\; orbit\; velocity\; Escape\; velocity\; Circular\; orbit\; velocity\; (km/s)\; (km/s)\; Earth\; 11.2\; 7.9\; Mars\; 5.0\; 3.6\; Moon\; 2.4\; 1.7$$
For an object on a circular orbit,
Gravitational energy = 2 * Kinetic energyThe relationship between the kinetic and gravitational energy doesn't depend on R. If a satellite inspirals toward a central object, the gain in kinetic energy is always half the loss in gravitational energy.
The total energy is negative.
Total energy = Gravitational energy + Kinetic energy = ½ * Gravitational energy = ½ G M m / R Angular momentum = m V R = m (G M R)^{1/2}As R decreases, both energy and angular momentum decrease. In order for a satellite to inspiral it has to give energy and angular momentum to another object.
Density = D Radius = R Volume = Υ = (4/3) π R^{3} Mass = M = D Υ Acceleration at surface = A = G M / R^{2} = (4/3) π G D R Orbit speed at surface = V = (G M / R)^{1/2} = [(4/3) π G D]^{½} R Orbit time at surface = T = 2 π R / V = [(16/3) π^{3} G D]^{½}Acceleration is proportional to R
Density Radius Gravity g/cm^{2} (Earth=1) m/s^{2} Earth 5.52 1.00 9.8 Venus 5.20 .95 8.87 Uranus 1.27 3.97 8.69 Mars 3.95 .53 3.71 Mercury 5.60 .38 3.7 Moon 3.35 .27 1.62 Titan 1.88 .40 1.35 Ceres 2.08 .074 .27
F_{contact} = Contact force between the object and a surface (usually gravity) F_{friction} = Maximum friction force transverse to the surface of contact. C = Coefficient of friction, usually with a magnitude of ~ 1.0. F_{friction} = C F_{contact}The larger the contact force the larger the maximum friction force.
Coefficient of friction Ice .05 Tires 1When two surfaces first come together there is an instant of large surface force, which allows for a large friction force.
Mass = M Contact force between the car and the road = F_{contact} = M g Maximum friction force that the road can provide = F_{friction} = C F_{contact} Maximum acceleration that friction can provide = A = F_{friction} / M = C F_{contact} / M = C g M / M = C gThis clip shows the magnitude and direction of the acceleration while a Formula1 car navigates a racetrack. Formula1 lap
Villeneuve vs. Arnoux At 0:49 Arnoux breaks before he hits the turn.
For maximum cornering acceleration, the same equations apply as for the maximum drag racing acceleration. It doesn't matter in which direction the acceleration is.
Maximum cornering acceleration = C g
Suppose an object with mass m rests on a ramp inclined by an angle theta. The gravitational force on the object is
F = m gThe force between the object and the surface is equal to the component of the gravitational force perpendicular to the surface.
F_{contact} = F_{grav} * cos($\theta $)
The force of gravity parallel to the ramp surface is
F_{ramp} = F_{grav} sin($\theta $)
Th maximum friction force that the ramp can exert is
F_{friction} = C F_{contact}This is balanced by the gravitational force along the ramp
F_{friction} = F_{ramp} F_{grav} sin($\theta $) = C F_{grav} cos($\theta $) C = tan($\theta $)This is a handy way to measure the coefficient of friction. Tilt the ramp until the object slides and measure the angle.
$\theta $ = Angle in radians (dimensionless) X = Arc distance around the circle in meters (the red line in the figure) R = Radius of the circle in meters X = $\theta $ R
Pi is defined as the ratio of the circumference to the diameter.
Full circle = 360 degrees = 2 $\pi $ radians
1 radian = 57.3 degrees
1 degree = .0175 radians
Radius = R Angle = $\theta $ X coordinate = X = R cos($\theta $) Y coordinate = Y = R sin($\theta $)
T = Time in seconds $\theta $ = Angle (dimensionless) $\omega $ = Angular velocity in radians/second = 1.75 in the animated figure
$\omega $ = Angular frequency F = Spin frequency in Hertz or 1/second $\omega $ = 2 Pi F 1 Hertz = 1 revolution/second = 2$\pi $ radians/second
T = Period in seconds F = Frequency in Hertz or 1/second F T = 1
If an object is at the edge of a record then the position is the arc length around the circumference.
Time = T Radius = R = .15 meters (for a vinyl record) Angular velocity = $\omega $ = 3.49 radians/second (= 33.33 revolutions per minute) Velocity of the outer edge V = $\omega $ R = .523 meters
Suppose a billiard ball rolls across a table with a speed of 2 m/s.
Ball velocity = V = 2 meters/second Ball radius = R = .03 meters Angular frequency = $\omega $ = V/R = 67 radians/second Ball spin frequency in Hertz = F = $\omega $/(2$\pi $)= 10.6 HertzA point on the edge of the ball is moving at Velocity=0 when it is in contact with the ground and it is moving at Velocity=2V when it is at the opposite point from the ground.
Ball velocity = V Edge velocity = V_{edge} Spin parameter = Z = V_{edge} / V = 2 π R F / V Spin frequency = F = Z V / (2 π R)For a rolling ball, Z=1. For curveballs thrown in the air, Z is typically less than 1. If Z=.5 then the following table shows typical speeds and spin rates for various balls.
Radius Speed Spin (mm) (m/s) (1/s) Ping pong 20 20 80 Golf 21.5 80 296 Tennis 33.5 50 119 Baseball 37.2 40 86 Soccer 110 40 29
For a mass moving around a circle, we can describe the motion in terms of either a position or an angle.
R = Radius of the circle X = Position of the mass on the circumference of the circle $\theta $ = Angle pointing to X X = $\theta $ REvery linear quantity has a corresponding spin quantity, and every linear equation has a crresponding spin equation. By changing variables from X to $\theta $ we can translate a linear equation into a spin equation.
Linear variables Spin variables Position = X Angle = $\theta $ = X / R Velocity = V Angular velocity = $\omega $ = V / R Acceleration= A Angular accel. = $\alpha $ = A / R Mass = M Moment of inertia = I = M * R^{2} Force = F Torque = $\Gamma $ = F * R Momentum = Q Angular momentum = L = Q * R Energy = E Time = T Radius = R Force Momentum Energy Centripetal Constant Constant acceleration velocity acceleration Linear motion: F = M A Q = M V = F T E = .5 M V^{2} = F X A = V^{2} / R X = V T V = A T Spin: $\Gamma $ = I $\alpha $ L = I $\omega $ = F T R E = .5 I $\omega 2$ = $\Gamma $ $\theta $ A = $\omega 2$ R $\theta $ = $\omega $ T $\alpha $ = $\omega $ T Derivation: Force Momentum Energy Energy Constant Constant velocity acceleration F = M A Q = M V E = .5 M V^{2} E = F X X = V T V = A T F R = M A R Q R = M V R E = .5 M R^{2} (V/R)^{2} E = F R X/R X/R = (V/R) T V/R = (A/R) T $\Gamma $ = M R^{2} A/R L = M R^{2} V/R E = .5 I $\omega 2$ E = $\Gamma $ $\omega $ $\theta $ = $\omega $ T $\omega $ = $\alpha $ T $\Gamma $ = I $\alpha $ L = I $\omega $
Angle = $\theta $ = X / R Angular velocity = $\omega $ = V / R Angular accel. = $\alpha $ = A / R Constant velocity: X = V T Constant spin: $\theta $ = $\omega $ T Constant acceleration: V = A T Constant angular accel.: $\omega $ = $\alpha $ T
Newton's law for linear motion: F = M A Newton's law for circular motion: $\Gamma $ = I $\alpha $ = F R Momentum: Q = M V Angular momentum: L = I $\omega $ = M V R Energy: E = .5 M V^{2} Angular energy: L = .5 I $\omega 2$
Angular momentum is conserved. In the following figure, the angular momentum is constant and V*R is constant.
The direction of the angular momentum vector is given by the right hand rule. The Earth's angular momentum points to the north pole and it is constant throughout the orbit.
The angular momentum vector is the cross product of the position and momentum vectors. Denoting the cross product by "x" and the angle between R and Q by Theta,
L = R x Q = R Q sin(Theta)
The moment of inertia of an object depends on its mass and how far the mass is from the axis of rotation.
I / (M R^{2}) Point mass 1 Tetherball Ring 1 Hula hoop Solid disk 1/2 Pizza Spherical shell 2/3 Tennis ball, soccer ball Solid sphere 2/5 Billiard ball Pole 1/12 Grasped at the center Sword 1/3 Grasped at the end
If the object has a simple shape then we can calculate the moment of inertia using the formulae above. If the shape is not simple then we often assume it is a solid sphere. For example, the moment of inertia of Chuck Norris as a solid sphere is
M = Mass of Chuck = 100 kg R = Radius of Chuck = .25 meters (estimate) I = Moment of inertia of Chuck = .4 * 100 * .25^{2} = 2.5 kg m^{2}
Kinetic energy = .5 M V^{2}
Gravitational energy = M g Height (In a constant gravitational field)
Gravitational energy =  G M_{1} M_{2} / R
Rotational energy = .5 I $\omega 2$
Energy of matter = M C^{2} C = Speed of light = 3.00e8 m/s
Conservation arises from multiplying F=MA by various quantities.
Force = Mass * Acceleration Force * Time = Momentum Force * Distance = Energy Force * Radius = Torque Force * Time * Radius = Angular Momentum
For a rolling ball,
Velocity = V Radius = R Mass = M Angular frequency = $\omega $ = V/R Moment of inertia = I = .4 M R^{2} Kinetic energy = E_{k} = .5 M V^{2} Spin energy = E_{s} = .5 I $\omega 2$ = .2 M V^{2} Total energy = E_{t} = E_{k} + E_{s} = 1.4 E_{k} Spin energy / Kinetic energy = E_{s} / E_{k} = .4 Total energy / Kinetic energy = E_{s} / E_{k} = 1.4
Suppose a bowling ball is launched so that it initially slides along the floor with zero spin. The friction force decreases the ball velocity. It also exerts a torque on the ball that increases the spin angular velocity.
Ball initial velocity = V_{0} = 10 m/s Mass of the ball = M = 7.26 kg (=16 pounds, which is the maximum) Radius of the ball = R Friction coefficient = C Gravitational force = F_{g} = M g Friction force = F_{f} = C M g (directed opposite to the ball's velocity) Ball acceleration = A = F_{f} / M = C g (The friction force decelerates the ball) Torque on ball = $\Gamma $ = R C M g Moment of inertia = I = .4 M R^{2} Angular acceleration = $\alpha $ = $\Gamma $ / I = 2.5 C g / R Time sliding = T Rolling angular velocity = $\omega $ = $\alpha $ T = 2.5 C g T / R Ball velocity when rolling = V = Vo + A T = Vo  C g T = Vo  .4 C g $\omega $ R / C / g = Vo  .4 $\omega $ R = Vo  .4 VSolving for V,
V = V_{0} / 1.4 = 7.1 m/sThe time it takes to being rolling smoothly is
T = (2/7) V_{0} / C / g = .29 / CFor surfaces with C=1, it usually takes less than half a second for a ball to begin rolling smoothly.
A bowling lane is covered in a layer of oil and has a friction coefficient of C=.08, giving T=3.6, giving the ball plenty of time to slide before it starts rolling. This allows one to use sidespin. While the ball is still sliding, sidespin can deliver a sideways force. Once the ball starts rolling the sidespin is lost.
If there are no torques on an object then the angular momentum is conserved.
In free fall you can't change your angular momentum but you can change your orientation.
In the absence of external torques, a rigid object can't change its orientation axis and a deformable object can. Cats change their orientation axis by generating internal torques and by varying their moment of inertia.
A cat can change its orientation using either a 2axis strategy or a 3axis strategy. Each can work indepenently and the cat uses a combination of both. The 3axis strategy is depicted in the figure above an the 2axis strategy is as follows:
A can can right itself with the following 2step procedure. The first step is to compact the arms and extend the legs, turning the upper torso one direction and the legs the opposite direction. Because the upper torso has a smaller moment of inertia it rotates farther than the legs. The second step is to extend the arms and compact the legs and perform an opposite set of rotatons as step 1. The sum of steps 1 and step 2 produces a net change in orientation.
The more deformable you are the more precise internal torques you can generate.
Bruce Lee: "Be like water"
Fumio, from the film "Fist of Legend": "if you learn to be fluid, to adapt, you will be unbeatable."
Paul Atreides, from the film "Dune" during the duel with FeydRautha Harkonnen: "I will bend like a reed in the wind."
Most of the elements of the breathing cycle and axis cycle are determined by conservation of momentum.
The cat's first move is to maximize its moment of inertia to slow down its rotation.
Jupiter's spin makes it oblate.
For a spinning 3D object,
Torque: 3D vector Angular acceleration: 3D vector Moment of inertia: 3x3 matrix Torque = MomentOfInertia * AngularAccelerationIf the axis of rotation passes through the object's center of mass then the moment of inertia matrix has a tridiagonal form.
A typical set of parameters for a racing bike is
Velocity = V = 20 m/s (World record=22.9 m/s) Power = P = 2560 Watts (Typical power required to move at 20 m/s, measured experimentally) Force on ground = F = P/V = 128 Newtons
We assume a high gear, with 53 teeth on the front gear and 11 teeth on the rear gear.
Number of links in the front gear = N_{f} = 53 Number of links in the rear gear = N_{r} = 11 Length of one link of a bicycle chain = L = .0127 m = .5 inches Radius of the front gear = R_{f} = N_{f} L / (2 π) = .107 m Radius of the rear gear = R_{r} = N_{r} L / (2 π) = .0222 mTorque balance:
Ground force * Wheel radius = Chain force * Rear gear radius
Pedal force * Pedal radius = Chain force * Front gear radius
Chain force = Ground force * Wheel radius / Rear gear radius
= 128 * .311 / .0222
= 1793 Newtons
Pedal force = Ground force * Wheel radius / Pedal radius * Front gear radius / Rear gear radius
= Ground force * Wheel radius / Pedal radius * Front gear teeth / Rear gear teeth
= 128 * .311 / .17 * 53 / 11
= 1128 Newtons
Radius Force Torque Gear
(m) (N) (Nm) teeth
Pedal crank .170 1128 191.9 
Front gear .107 1793 191.9 53
Rear gear .0222 1793 39.8 11
Rear wheel .311 128 39.8 
Wheel frequency = Velocity / (Radius * 2Pi)
= 20 / (.311 * 2$\pi $)
= 10.2 Hertz
Pedal frequency = Wheel frequency * Rear gear teeth / Front gear teeth
= 10.2 * 53 / 11
= 2.12 Hertz
= 127 revolutions per minute
Humans can pedal effectively in the range from 60 rpm to 120 rpm. Gears allow one to
choose the pedal frequency. There is also a maximum pedal force of around 1200 Newtons.
When going fast the goal of gears is to slow down the pedals.
When one is climbing a hill the goal of gears is to speed up the pedals so that you don't have to use as much force on the pedals.
Pedal period * Rear gear teeth = Wheel period * Front gear teeth Pedal radius / Pedal velocity * Front gear teeth = Wheel radius / Wheel velocity * Front gear teeth Pedal force = Power / Pedal velocity = Power / Wheel velocity * Wheel radius / Pedal radius * Front gear teeth / Rear gear teeth = Power / Wheel velocity * .311 / .17 * Front gear teeth / Rear gear teeth = Power / Wheel velocity * 1.83 * (Front gear teeth / Rear gear teeth) = Power / Wheel velocity * 1.83 * Gear ratio Gear ratio = Front gear teeth / Rear gear teethFor a given power and wheel velocity, the pedal force can be adjusted by adjusting the gear ratio.
Suppose a bike is going uphill at large power and low velocity.
Power = 1000 Watts Velocity = 3 m/s Front gear teeth = 34 (Typical for the lowest gear) Rear gear teeth = 24 (Typical for the lowest gear) Pedal force = Power / Wheel velocity * 1.83 * Front gear teeth / Rear gear teeth = 1000 / 3 * 1.83 * 34 / 24 = 864 Newtons = 88 kg equivalent forceThis is a practical force. If you used the high gear,
Pedal force = Power / Wheel velocity * 1.83 * Front gear teeth / Rear gear teeth = 1000 / 3 * 1.83 * 53 / 11 = 2939 Newtons = 300 kg equivalent forceThis force is impractically high.
Surface area = A Force = F Pressure = P = F / A (Pascals or Newtons/meter^{2} or Joules/meter^{3})
Mass of the Earth's atmosphere = M = 5.15e18 kg Surface area of the Earth = A = 5.10e14 m^2 Gravitational constant = g = 9.8 m/s^2 Pressure of Earth's atmosphere = P = M g / A = 101000 Pascals = 15 pounds/inch^{2} = 1 BarOne bar is defined as the Earth's mean atmospheric pressure at sea level
Height Pressure Density (km) (Bar) (kg/m^{3}) Sea level 0 1.00 1.225 Denver 1.6 .82 1.05 One mile Everest 8.8 .31 .48 Airbus A380 13.1 .16 .26 F22 Raptor 19.8 .056 .091 SR71 Blackbird 25.9 .022 .034 Space station 400 .000009 .000016
Properties of atmospheres:
Density Pressure S Gravity (kg/m^{2}) (Bar) (tons/m^{2}) (m/s^{2}) Venus 67 92.1 1050 8.87 Titan 5.3 1.46 109 1.35 Earth 1.22 1 10.3 9.78 Mars .020 .0063 .54 3.71 Mass of atmosphere above one meter^{2} of surface = M = 10.3 tons for the Earth P = M gYou don't need a pressure suit on Titan. You can use the kind of gear arctic scuba divers use. Also, the gravity is so weak and the atmosphere is so thick that humanpowered flight is easy. Titan will be a good place for the X games.
Kelvin Celsius Fahrenheit Absolute zero 0 273.2 459.7 Water freezing point 273.2 0 32 Room temperature 294 21 70 Water boiling point 373.2 100 212 Kelvin Absolute zero 0 Helium boiling point 4.2 Hydrogen boiling point 20.3 Pluto 44 Nitrogen boiling point 77.4 Oxygen boiling point 90.2 Hottest superconductor 135 Mercury barium calcium copper oxide Mars 210 H2O melting point 273.15 0 Celcius = 32 Fahrenheit Room temperature 293 20 Celcius = 68 Fahrenheit H2O boiling point 373.15 100 Celcius = 212 Fahrenheit Venus 740 Wood fire 1170 Iron melting point 1811 Bunsen burner 1830 Tungsten melting point 3683 Highest melting point among metals Earth's core 5650 Innercore boundary Sun's surface 5780 Solar core 13.6 million Helium4 fusion 200 million Carbon12 fusion 230 million
P = Pressure T = Temperature Vol = Volume E = Kinetic energy of gas molecules within the volume e = Kinetic energy per volume of gas molecules in Joules/meter^{3} = E / Vol Mol = Number of moles of gas molecules in the volumeIdeal gas law:
P = 2/3 e Form used in physics P Vol = 8.3 Mol T Form used in chemistryPressure has units of energy density, where the energy corresponds to kinetic energy of gas molecules.
1660 Boyle law P Vol = Constant at fixed T 1802 Charles law T Vol = Constant at fixed P 1802 GayLussac law T P = Constant at fixed Vol 1811 Avogadro law Vol / N = Constant at fixed T and P 1834 Clapeyron law P Vol / T = Constant combined ideal gas law
Distance below the surface = X Density of water = D = 1000 kg/m^{3} Mass of water above 1 meter^{2} of surface = M = D X Force from the water above 1 meter^{2} of surface = F = D X g Pressure at depth X relative to the surface = P = D X gAt a depth of 10 meters,
P = 1000 * 10 * 10 = 100000 Pascals = 1 Bar
In the 2014 AFC championship football game the Patriot's footballs were found to be underinflated. The Patriots claimed this was because the balls were inflated at warm temperature and used at cold temperature.
"Gauge pressure" is the pressure difference between the inside and outside of the football. The rules state that the gauge pressure for a football should be between 12.5 and 13.5 psi.
When the ball is moved from warm temperature to cold temperature, the external atmospheric pressure doesn't change and the pressure inside the football decreases.
The game was played at 4 Celsius and the gauge pressure was measured to be 11 psi. If we assume the footballs were inflated at warm temperature at 12.5 psi then we can solve for the inflation temperature.
Atmospheric pressure = P_{atm} = 15 psi Pressure of the cold football during the game = P_{cold} = P_{atm} + 11 psi = 15 + 11.0 psi = 26.0 psi Pressure of the warm football at inflation = P_{warm} = P_{atm} + 12.5 psi = 15 + 12.5 psi = 27.5 psi Temperature of the cold football = T_{cold} = 277 Kelvin = 4 Celsius Temperature of the warm football = T_{warm}The number of gas molecules inside the football is constant and we assume that the volume of the football doesn't change. Using the ideal gas law for constant volume and molecule number,
Pressure = Constant * Temperature P_{cold} = Constant * T_{cold} P_{warm} = Constant * T_{warm} T_{warm} = T_{cold} * P_{warm} / P_{cold} = 277 * 27.5 / 26 = 293 Kelvin = 20 Celsius
Gas pressure arises from kinetic energy of gas molecules, and the average kinetic energy per molecule is proportional to the temperature.
For air at sea level and room temperature,
P = Pressure = 101325 Pascals D = Density = 1.22 kg/m^{3} $\gamma $ = Adiabatic constant = 7/5 for air V_{therm}= Thermal speed = 544 meters/second V_{sound}= Sound speed = 343 meters/second at 20 Celsius = ($\gamma $ P / D)^{1/2} = ($\gamma $/3)^{1/2} V_{therm} = .63 V_{therm} M = Average mass of an air molecule = 4.78e26 kg n = Number of molecules per volume = 2.55e25 meter^{3} = D / M k = Boltzmann constant = 1.38e23 Joules/Kelvin T = Gas temperature = 293 Kelvin (Room temperature, or 20 Celcius) E = Ave kinetic energy per molecule = 5.96e21 Joules = .5 M V_{therm}^{2} = 1.5 k T = e / n e = Kinetic energy per volume = 152000 Joules/meter^{3} = 1.5 P Mol = Moles of molecules in 1 meter^{3} = 42.34 = n / 6.022e23 Avo = Avogadro number = 6.022e23 molecules H = Heat capacity = 1004 Joules/kg/Kelvin (calculated in the thermodynamics section)The characteristic thermal speed of a gas molecule is defined in terms of the mean energy per molecule. The Boltzmann constant relates the average kinetic energy to the temperature.
E = .5 M V_{therm}^{2} = 1.5 k TThe ideal gas law can be written as:
P = 2/3 e (Physics form) = 8.3 Mol T / Vol (Chemistry form)Writing the pressure as an energy density allows one to connect pressure with molecular kinetic energy.
The following table estimates the average mass per air molecule. We have neglected the argon molecules.
Atmosphere oxygen fraction = .21 Atmosphere nitrogen fraction = .78 Atmosphere argon fraction = .01 Mass of a nitrogen molecule = 28 Atomic mass units Mass of an oxygen molecule = 32 Atomic mass units Mass of one Atomic mass unit = 1.66e27 kg Average molecule mass = Oxygen fraction * Oxygen mass + Nitrogen fraction * Nitrogen mass = 28.8 Atomic mass units = 4.78e26 kg
Ice heat capacity = 2110 J/kg/K At 10 Celsius Water heat capacity = 4200 J/kg/K At 20 Celsius Steam heat capacity = 2080 J/kg/K At 100 Celsius Air heat capacity = 1004 J/kg/K Melting energy of water at 0 Celsius = 2501000 J/kg Vaporization energy of water at 100 Celsius = 2257000 J/kg Energy required to raise the temperature of 1 kg of H2O from 40 Celsius to 140 Celsius = Energy to raise the temperature of ice from 40 C to 0 C + Energy to turn ice to water (at 0 C) + Energy to raise the temperature of water from 0 C to 100 C + Energy to turn the water from a liquid to steam (at 100 C) + Energy to raise the temperature of steam from 100 C to 140 C = 2110 * 40 + 2501000 + 4200 * 100 + 2257000 + 1004 * 40 = 5302560 Joules
Archimedes was commissioned by the king to develop a method to measure the volume of an irregular object, such as a crown. The king wanted to measure the crown's density to determine if it was made of pure gold.
In the animation above, the crown and the cylinder have equal masses and densities and they displace equal volumes of water. This is "Archimedes' principle". A submerged mass displaces an equal mass of water.
Inventions of Archimedes:
For a ship floating in water,
Gravitational acceleration = g = 9.8 m/s^{2} Density of water = D_{water} = 1000 kg/m^{3} Mass of a ship = M_{ship} Mass of water displaced by the ship = M_{water} = M_{ship} (Archimedes' principle) Gravity force on ship = F_{grav} = M_{ship} g Buoyancy force on ship = F_{buoy} = M_{water} g Volume of water displaced by the ship= Υ_{water} = M_{water} / D_{water} M_{grav} = F_{buoy} M_{ship} = M_{water}The mass of water displaced is the same for a floating and a sunk ship.
For ice floating in water,
Gravitational acceleration = g = 9.8 m/s^{2} Density of ice = D_{ice} = 920 kg/m^{3} Density of water = D_{water} = 1000 kg/m^{3} Volume of the iceberg = Υ_{ice} Volume of water displaced by iceberg = Υ_{water} Mass of the iceberg = M_{ice} = D_{ice} Υ_{ice} Mass of water displaced by iceberg = M_{water} = D_{water} Υ_{water} Gravity force on the iceberg = F_{grav} = M_{ice} g Buoyant force on the iceberg = F_{buoy} = M_{water} g Fraction of iceberg above surface = f = (Υ_{ice}  Υ_{water}) / Υ_{ice} F_{grav} = F_{buoy} (Principle of bouyancy) M_{ice} = M_{water} (Principle of Archimedes) f = (D_{ice}  D_{water}) / D_{ice} = (M_{ice}/Υ_{ice}  M_{water}/Υ_{water}) / (M_{ice}/Υ_{ice}) = 1  Υ_{ice} / Υ_{water} = .08
We estimate the number of helium balloons required to lift a person.
Gravitational acceleration = g = 9.8 m/s^{2} Density of air = D_{air} = 1.22 kg/m^{3} Density of helium = D_{helium} = .179 kg/m^{3} Radius of one balloon = R_{balloon} = .2 m Volume of one balloon = Υ_{balloon} = .0335 m^{3} Volume of helium in all balloons = Υ_{helium} = 83 m^{3} Mass of helium in one balloon = M_{balloon} = .00600 kg = D_{helium} Υ_{balloon} Mass of helium in all balloons = M_{helium} = 14.6 kg = D_{helium} Υ_{helium} Mass of air displaced by the balloons= M_{air} = 101.3 kg = D_{air} Υ_{helium} Mass of payload = M_{payload} = 80 kg (Typical person) Gravity force on payload & balloons = F_{grav} = 927 N = (M_{payload} + M_{helium}) g Buoyant force on a helium balloon = F_{buoy} = 927 N = F_{air} (Principle of buoyancy) Number of balloons required = Z = 2477 = Υ_{helium} / Υ_{balloon} F_{buoy} = F_{air} (Principle of buoyancy) F_{grav} = F_{buoy} (Balance of gravity and buoyancy) (M_{payload} + M_{helium}) g = M_{air} g (M_{payload} + D_{helium} Υ_{helium}) g = D_{air} Υ_{helium} g Υ_{helium} = M_{payload} / (D_{air}  D_{helium}) = 1.04 M_{payload} = 83 m^{3}The volume of helium doesn't depend on gravity.
Helium is more expensive and more dense than hydrogen, but it is not flamable.
kg/m^{3} Hydrogen .0899 Helium .179 Air (hot) 1.12 320 Kelvin Air (room temp) 1.22 293 Kelvin Ice 920 Water 1000
For a hot air balloon, the volume is fixed and the pressures on the inside and outside are equal For example,
Inside temperature = T_{in} = 320 Kelvin Outside temperature = T_{out} = 293 Kelvin Inside density = D_{in} = 1.12 kg/m^{3} Outside density = D_{out} = 1.22 kg/m^{3} D_{in} T_{in} = D_{out} T_{out}
Jacques Charles made the first hot air balloon flight in 1783.
1585 Simon Stevin introduces decimal numbers to Europe. (For example, writing 1/8 as 0.125) 1586 Simon Stevin drops objects of varying mass from a church tower to demonstrate that they accelerate uniformly. 1604 Galileo publishes a mathematical description of acceleration. 1614 Logarithms invented by John Napier, making possible precise calculations of equal tuning ratios. Stevin's calculations were mathematically sound but the frequencies couldn't be calculated with precision until logarithms were developed. 1637 Cartesian geometry published by Fermat and Descartes. This was the crucial development that triggered an explosion of mathematics and opened the way for the calculus. 1676 Leibniz defines kinetic energy and notes that it is conserved in many mechanical processes 1684 Leibniz publishes the calculus 1687 Newton publishes the Principia Mathematica, which contained t hecalculus, the laws of motion (F=MA), and a proof that planets orbit as ellipses. 1776 Smeaton publishes a paper on experiments related to power, work, momentum, and kinetic energy, supporting the principle of conservation of energy 1798 Thompson performs measurements of the frictional heat generated in boring cannons and develops the idea that heat is a form of kinetic energy 1802 GayLussac publishes Charles's law. For a gas at constant pressure, Temperature * Volume = Constant 1819 Dulong and Petit find that the heat capacity of a crystal is proportional to the number of atoms 1824 Carnot analyzes the efficiency of steam engines; he develops the notion of a reversible process and, in postulating that no such thing exists in nature, lays the foundation for the second law of thermodynamics, and initiating the science of thermodynamics 1831 Melloni demonstrates that infrared radiation can be reflected, refracted, and polarised in the same way as light 1834 Clapeyron combines Boyle's Law, Charles's Law, and GayLussac's Law to produce a Combined Gas Law. Pressure * Volume = Constant * Temperature 1842 Mayer calculates the equivalence between heat and kinetic energy
The properties of a wave are
Frequency = F (seconds^{1} Wavelength = W (meters) Wavespeed = V (meters/second) Period = T (seconds) = The time it takes for one wavelength to pass byWave equations:
F W = V F T = 1
Length of a train car = W = 10 meters (The wavelength) Speed of the train = V = 20 meters/second (The wavespeed) Frequency = F = 2 Hertz (Number of train cars passing by per second) Period = T = .5 seconds (the time it takes for one train car to pass by)
Speed of sound at sea level = V = 340 meters/second Frequency of a violin A string = F = 440 Hertz Wavelength of a sound wave = W = .77 meters = V/F Wave period = T = .0023 seconds
A wave on a string moves at constant speed and reflects at the boundaries.
For a violin Astring,
Frequency = F = 440 Hertz Length = L = .32 meters Time for one round trip of the wave = T = .0023 s = 2 L / V = 1/F Speed of the wave on the string = V = 688 m/s = F / (2L)
If two notes are played at the same time then we hear the sum of the waveforms.
If two notes are played such that the frequency of the high note is twice that of the low note then this is an octave. The wavelength of the high note is half that of the low note.
Color Frequency Wavelength Orange 220 Hertz 1 Red 440 Hertz 1/2Because the red and orange waves match up after a distance of 1 the blue note is periodic. This makes it easy for your ear to process.
If we double both frequencies then it also sounds like an octave. The shape of the blue wave is preserved.
Color Frequency Wavelength Orange 440 Hertz 1/2 Red 880 Hertz 1/4When two simultaneous pitches are played our ear is sensitive to the frequency ratio. For both of the above octaves the ratio of the high frequency to the low frequency is 2.
440 / 220 = 2 880 / 440 = 2If we are talking about frequency ratios and not absolute frequencies then for simplicity we can set the lower frequency to 1.
Frequency Normalized frequency 220 1 440 2 880 4
If two notes are out of tune they produce dissonant beat frequencies.
F1 = Frequency of note #1 F2 = Frequency of note #2 Fb = Beat frequencyIf F1 and F2 are played together the beat frequency is
Fb = F2  F1For the beats to not be noticeable, Fb has to be less than one Hertz. On the E string there is little margin for error. Vibrato is often used to cover up the beat frequencies.
The more out of tune the note, the more pronounced the beat frequencies. In the first figure, the notes are in tune and no beat frequencies are produced.
If you play an octave out of tune you also get beat frequencies.
Violins, violas, and cellos are tuned in fifths. String basses, guitars, and bass guitars are tuned in fourths. Pianos are tuned with equal tuning.
Hertz Violin E 660 = 440*1.5 Violin A 440 Violin D 293 = 440/1.5 Violin G 196 = 440/1.5^2 Viola A 440 Same as a violin A Viola D 293 Viola G 196 Viola C 130 Cello A 220 One octave below a viola A Cello D 147 Cello G 98 Cello C 65 String bass G 98 = 55 * 1.5^2 String bass D 73 = 55 * 1.5 String bass A 55 3 octaves below a violin A String bass E 41 = 55 / 1.5 Guitar E 326 Guitar B 244 Guitar G 196 Guitar D 147 Guitar A 110 2 octaves below a violin A Guitar E 82When an orchestra tunes, the concertmaster plays an A and then everyone tunes their A strings. Then the other strings are tuned in fifths starting from the A.
A bass guitar is tuned like a string bass.
The viola is the largest instrument for which one can comfortably play an octave, for example by playing a D on the Cstring with the first finger and a D on the Gstring with the fourth finger. Cellists have to shift to reach the D on the Gstring.
According to legend Bach used a supersized viola, the "Viola Pomposa"
In a reed instrument, a puff of air enters the pipe, which closes the
reed because of the Bernoulli effect. A pressure pulse travels to the other
and and back and when it returns it opens the reed, allowing another puff of
air to enter the pipe and repeat the cycle.
Green dots indicate the frequencies of open strings.
An orchestral bass and a bass guitar have the same string tunings.
The range of organs is variable and typically extends beyond the piano in both the high and low directions.
If a wave is linear then it propagates without distortion.
If a wave is linear then waves add linearly and oppositelytraveling waves pass through each other without distortion.
If two waves are added they can interfere constructively or destructively, depending on the phase between them.
If a speaker system has 2 speakers you can easily sense the interference by moving around the room. There will be loud spots and quiet spots.
The more speakers, the less noticeable the interference.
Noisecancelling headphones use the speakers to generate sound that cancels incoming sound.
Two waves traveling in opposite directions create a standing wave.
Waves on a string simulation at phet.colorado.edu
Whan a wave on a string encounters an endpoint it reflects with the waveform
preserved and the amplitude reversed.
When an string is played it creates a set of standing waves.
L = Length of a string V = Speed of a wave on the string N = An integer in the set {1, 2, 3, 4, ...} W = Wavelength of an overtone = 2 L / N F = Frequency of the overtone = V/W = V N / (2L) N = 1 corresponds to the fundamental tone N = 2 is one octave above the fundamental N = 3 is one octave plus one fifth above the fundamental.Audio: overtones
For example, the overtones of an Astring with a frequency of 440 Hertz are
Overtone Frequency Note 1 440 A 2 880 A 3 1320 E 4 1760 A 5 2200 C# 6 2640 E 7 3080 G 8 3520 A
Overtone simulation at phet.colorado.edu
In the left frame the pipe is open at the left and closed at the right.
In the right frame the pipe is reversed, with the left end closed and the right
end open. Both are "halfopen pipes".
An oboe and a clarinet are halfopen pipes.
L = Length of the pipe ~ .6 meters for an oboe V = Speed of sound N = An odd integer having values of {1, 3, 5, 7, ...} W = Wavelength of the overtone = 4 L / N F = Frequency of the overtone = V / W = V N / (4L)The overtones have N = {1, 3, 5, 7, etc}
A cantilever has the same overtones as a halfopen pipe.
A flute and a bassoon are pipes that are open at both ends and the overtones are plotted in the figure above. In this case the overtones have twice the frequency as those for a halfopen pipe.
L = Length of the pipe V = Speed of sound N = An odd integer having values of {1, 3, 5, 7, ...} W = Wavelength of the overtone = 2 L / N F = Frequency of the overtone = V / W = V N / (2L)
A string has the same overtones as a closed pipe.
A closed pipe doesn't produce much sound. There are no instruments that are closed pipes. A muted wind or bass instrument can be like a closed pipe.
Modes 1 through 5 for a closed pipe.
An instrument of length L has overtones with frequency
Frequency = Z * Wavespeed / (2 * Length)Z corresponds to the white numbers in the figure above.
An oboe is a halfopen pipe (open at one end), a flute is an open pipe (open at both ends), and a string behaves like a pipe that is closed at both ends.
If a violin, an oboe, and a flute are all playing a note with 440 Hertz then the overtones are
Violin 440, 2*440, 3*440, 4*440, ... Oboe 440, 3*440, 5*440, 7*440, ... Flute 440, 3*440, 5*440, 7*440, ...
The fundamental mode is at the upper left. The number underneath each mode is the frequency relative to the fundamental mode. The frequencies are not integer ratios.
In general, overtones of a 1D resonator are integer multiples of the fundamental frequency and overtones of a 2D resonator are not.
Wikipedia: Virations of a circular membrane
In 1787 Chladni published observations of resonances of vibrating plates.
He used a violin bow to generate a frequency tuned to a resonance of the plate
and the sand collects wherever the vibration amplitude is zero.
A "formant" is a vocal resonance. Vowels can be identified by their characteristic mode frequencies.
The whispering gallery in St. Paul's Cathedral has the same modes as a circular drum.
Whispering gallery waves were discovered by Lord Rayleigh in 1878 while he was in St. Paul's Cathedral.
Overtones are ubiquitous in vibrating systems. They are usually referred to as "normal modes".
You can increase the pitch by pulling the string sideways. This increases the string tension, which increases the wavespeed and hence the frequency.
If you are playing a note on a guitar using a fret, you can change the frequency of the note by bending the string behind the fret.
Tension = Tension of a string D = Mass per meter of the string V = Speed of a wave on the string = (Tension/D)^(1/2) L = Length of the string T = Wave period of a string (seconds) = 2 L / V F = Frequency of a string = 1/T = V / (2L)
The vibration of the string depends on where it is plucked. Plucking the string close to the bridge enhances the overtones relative to the fundamental frequency.
A bow produces a sequence of plucks at the fundamental frequency of the string.
As a sound waves travels back and forth along the clarinet it forces the reed to
vibrate with the same frequency.
In a brass instrument your lips take the function of a reed.
P = Pressure V = Fluid velocity H = Height g = Gravity = 9.8 meters/second^2 D = Fluid densityThe bernoulli principle was published in 1738. For a steady flow, the value of "B" is constant along the flow.
B = P + .5 D V^2 + D g HIf the flow speeds up the pressure goes down and vice versa.
A wing slows the air underneath it, inreasing the pressure and generating lift.
In the right panel, air on the top of the wing is at increased speed and
reduced pressure, causing condensation of water vapor.
Lift incrases with wing angle, unless the angle is large enough for the airflowto stall.
A turbofan compresses the incoming airflow so that it can be combusted with fuel.
In a reed instrument, a puff of air enters the pipe, which closes the reed because of the Bernoulli effect. A pressure pulse travels to the other and and back and when it returns it opens the reed, allowing another puff of air to enter the pipe and repeat the cycle.
The vocal tract is around 17 cm long. For a halfopen pipe this corresponds to
a resonant frequency of
Resonant frequency = WaveSpeed / (4 * Length) = 340 / (4*.17) = 500 HertzOne has little control over the length of the vocal pipe but one can change the shape, which is how vowels are formed. Each of the two vocal chords functions like a string under tension. Changes in muscle tension change the frequency of the vibration.
Male vocal chords tend to be longer than female vocal chords, giving males a lower pitch. Male vocal chords range from 1.75 to 2.5 cm and female vocal chords range from 1.25 to 1.75 cm.
When air passes through the vocal chords the Bernoulli effect closes them. Further air pressure reopens the vocal chords and the cycle repeats.
The airflow has a triangleshaped waveform, which because of its sharp edges generates abundant overtones.
Audio file: Creating a triangle wave by
adding harmonics.
Lung pressure (Pascals) Passive exhalation 100 Singing 1000 Fortissimo singing 4000Atmospheric pressure is 101000 Pascals.
For a lung volume of 2 liters, 4000 Pascals corresponds to an energy of 8 Joules.
Singers, wind, and brass musicians train to deliver a continuous stable exhalation. String musicians train locking their ribcage in preparation for delivering a sharp impulse.
Suppose a microphone samples a wave at fixed time intervals. The white curve is the wave and the orange dots are the microphone samplings.
F = Wave frequency Fmic = Sampling frequency of the microphone Fny = Nyquist frequency = Minimum frequency to detect a wave of frequency F = 2 FIn the above figure the sampling frequency is equal to the Nyquist frequency, or Fmic = 2 F. This is the minimum sampling frequency required to detect the wave.
This figure shows sampling for Fmic/F = {1, 2, 4, 8, 16}. In the left panel the wave and samplings are depicted and in the right panel only the samplings are depicted.
The top row corresponds to Fmic=F, and the wave cannot be detected at this sampling frequency.
The second row corresponds to Fmic=2F, which is the Nyquist frequency. This frequency is high enough to detect the wave but accuracy is poor.
For each successive row the value of Fmic/F is increased by a factor of 2. The larger the value of Fmic/F, the more accurately the wave can be detected.
Human hearing has a frequency limit of 20000 Hertz, which corresponds to a Nyquist frequency of 40000 Hertz. If you want to sample the highest frequencies accurately then you need a frequency of at least 80000 Hertz.
Overtones can generate highfrequency content in a recording, which is why the sampling frequency needs to be high.
Wikipedia: Harmonic oscillator Q factor Resonance Resonance
A force can stretch or compresses a spring.
A spring oscillates at a frequency determined by K and M.
Frequency = Squareroot(K/M) / (2 Pi)
T = Time X = Displacement of the spring when a force is applied K = Spring constant M = Mass of the object attached to the spring Force = Force on the spring =  K X (Hooke's law)Solving the differential equation:
Force = M * Acceleration  K X = M * X''This equation has the solution
X = sin(2 Pi F T)where
F = SquareRoot(K/M) / (2 Pi)Wikipedia: Hooke's law
After a string is plucked the amplitude of the oscillations decreases with time.
The larger the damping the faster the amplitude decays.
T = Time for one oscillation of the string Tdamp= Characteristic timescale for vibrations to damp q = "Quality" parameter of the string = Characteristic number of oscillations required for the string to damp = Tdamp / TIn the above figure,
q = Tdamp / T = 4The smaller the damping the larger the value of q. For most instruments, q > 100.
The above figure uses the equation for a damped vibrating string.
t = Time X(t) = Position of the string as a function of time T = Time for the string to undergo one oscillation if there is no damping q = Quality parameter, defined below Typically q>>1 F = Frequency of the string if there is no damping = 1/T Fd = Frequency of string oscillations if there is damping = F Z Z = [1  1/(4 Pi^2 q^2)]^(1/2) ~ 1 if q>>1A damped vibrating string follows a function of the form: (derived in the appendix)
X = exp(t/(Tq)) * cos(Zt/T)The consine part generates the oscillations and the exponential part reflects the decay of the amplitude as a function of time.
For large q, the oscillations have a timescale of T and the damping has a timescale of T*q. This can be used to measure the value of q.
q = (Timescale for damping) / (Time of one oscillation)For example, you can record the waveform of a vibrating string and measure the oscillation period and the decay rate.
If you shake a spring at the same frequency as the oscillation frequency then a large amplitude can result. Similarly, a swing can gain a large amplitude from small impulses if the impulses are timed with the swing period.
Suppose a violin Astring is tuned to 440 Hertz and a synthesizer produces a
frequency that is close to 440 Hertz. If the synthesizer is close enough to
440 Hertz then the Astring rings, and if the synthesizer is far from
440 Hertz then the string doesn't ring.
This is a plot of the strength of the resonance as a function of the synthesizer frequency. The synthesizer frequency corresponds to the horizontal axis and the violin string has a frequency of 440 Hertz. The vertical axis corresponds to the strength of the vibration of the Astring.
A resonance has a characteristic width. The synthesizer frequency has to be within this width to excite the resonance. In the above plot the width of the resonance is around 3 Hertz.
F = Frequency of the resonator = 440 Hertz f = Frequency of the synthesizer Fwidth = Characteristic frequency width for resonance If fF < Fwidth then the resonator vibrates If fF > Fwidth then the resonator doesn't vibrateResonance simulation at phet.colorado.edu
Wind can make a string vibrate (The von Karman vortex).
The Tacoma Narrows bridge collapse was caused by wind exciting resonances in the bridge.
The larger the value of q, the stronger the resonance. The following plot shows resonance curves for various values of q.
If q>>1 then
Amplitude of the resonance = Constant * q
You can break a wine glass by singing at the same pitch as the glass's resonanant frequency. The more "ringy" the glass the stronger the resonance and the easier it is to break.
The width of the resonance decreases with q. In the following plot the peak amplitude of the resonance curve has been set equal to 1 for each curve. As q increases the width of the resonance decreases.
T = Time for one oscillation of the string Td = Characteristic timescale for vibrations to damp q = Characteristic number of oscillations required for the string to damp = Td / T F = Frequency of the resonator = 1/T f = Frequency of the synthesizer Fwidth = Characteristic frequency width for resonance (derived in appendix) = F / (2 Pi q) If fF < Fwidth then the resonator vibrates If fF > Fwidth then the resonator doesn't vibrateIf q>>1 then
Width of the resonance = F / (2 Pi q)Overtones can also excite a resonance. For example, if you play an "A" on the Gstring of a violin then the Astring vibrates. The open Astring is one octave above the "A" on the Gstring and this is one of the overtones of the Gstring.
The strings on an electric guitar are less damped than the strings on an acoustic guitar. An acoustic guitar loses energy as it generates sound while an electric guitar is designed to minimize damping. The resonances on an electric guitar are stronger than for an acoustic guitar.
Oscillators that are mechanically connected can transfer energy back and forth between them.
Suppose you measure the frequency of a wave by counting the number of crests and dividing by the time.
T = Time over which the measurement is made N = Number of crests occurring in a time T F = N/T dF = Uncertainty in the frequency measurement = 1/TSuppose the number of crests can only be measured with an uncertainty of +1. The uncertainty in the frequency is dF = 1/T. The more time you have to observe a wave the more precisely you can measure the frequency.
The equation for the uncertainty in a frequency measurement is
dF T >= 1
Pressure = P (Pascals or Newtons/meter^{2} or Joules/meter^{3}) Temperature = T (Kelvin) Volume = Vol (meters^{3}) Total gas kinetic energy = E (Joules) Kinetic energy per volume = e = E/Vol (Joules/meter^{3}) Number of gas molecules = N Mass of a gas molecule = M Gas molecules per volume = n = N / Vol Gas density = D = N M / Vol Avogadro number = Avo= 6.022⋅10^{23} moles^{1} Moles of gas molecules = Mol= N / Avo Boltzmann constant = k = 1.38⋅10^{23} Joules/Kelvin Gas constant = R = k Avo = 8.31 Joules/Kelvin/mole Gas molecule thermal speed = V_{th} Mean kinetic energy / gas molecule= ε = E / n = ½ M V_{th}^{2} (Definition of the mean thermal speed)Gas pressure arises from the kinetic energy of gas molecules and has units of energy/volume.
P = ^{2}⁄_{3} e Form used in physics = R Mol T / Vol Form used in chemistry = k N T / Vol = ^{1}⁄_{3} N M V_{th}^{2}/ Vol = ^{1}⁄_{3} D V_{th}^{2} = k T D / MGas simulation at phet.colorado.edu
1660 Boyle law P Vol = Constant at fixed T 1802 Charles law T Vol = Constant at fixed P 1802 GayLussac law T P = Constant at fixed Vol 1811 Avogadro law Vol / N = Constant at fixed T and P 1834 Clapeyron law P Vol / T = Constant combined ideal gas law
For a system in thermodynamic equilibrium each degree of freedom has a mean energy of ½ k T. This is the definition of temperature.
Molecule mass = M Thermal speed = V_{th} Boltzmann constant = k = 1.38⋅10^{23} Joules/Kelvin Molecule mean kinetic energy = εA gas molecule moving in N dimensions has N degrees of freedom. In 3D the mean energy of a gas molecule is
ε = ^{3}⁄_{2} k T = ½ M V^{2}_{th}
The sound speed is proportional to the thermal speed of gas molecules. The thermal speed of a gas molecule is defined in terms of the mean energy per molecule.
Adiabatic constant = γ = 5/3 for monatomic molecules such as helium, neon, krypton, argon, and xenon = 7/5 for diatomic molecules such as H_{2}, O_{2}, and N_{2} = 7/5 for air, which is 21% O_{2}, 78% N_{2}, and 1% Ar ≈ 1.31 for a triatomic gas such as CO_{2} Pressure = P Density = D Sound speed = V_{sound} Mean thermal speed = V_{th} K.E. per molecule = ε = ½ M V_{th}^{2} V^{2}_{sound} = γ P / D = ^{1}⁄_{3} γ V^{2}_{th}The sound speed depends on temperature and not on density or pressure.
For air, γ = 7/5 and
V_{sound} = .68 V_{th}These laws are derived in the appendix.
We can change the sound speed by using a gas with a different value of M.
M in atomic mass units Helium atom 4 Neon atom 20 Nitrogen molecule 28 Oxygen molecule 32 Argon atom 40 Krypton atom 84 Xenon atom 131A helium atom has a smaller mass than a nitrogen molecule and hence has a higher sound speed. This is why the pitch of your voice increases if you inhale helium. Inhaling xenon makes you sound like Darth Vader. Then you pass out because Xenon is an anaesthetic.
In a gas, some of the energy is in motion of the molecule and some is in rotations and vibrations. This determines the adiabatic constant.
1635 Gassendi measures the speed of sound to be 478 m/s with 25% error. 1660 Viviani and Borelli produce the first accurate measurement of the speed of sound, giving a value of 350 m/s. 1660 Hooke's law published. The force on a spring is proportional to the change in length. 1662 Boyle discovers that for air at fixed temperature, Pressure * Volume = Constant 1687 Newton publishes the Principia Mathematica, which contains the first analytic calculation of the speed of sound. The calculated value was 290 m/s.Newton's calculation was correct if one assumes that a gas behaves like Boyle's law and Hooke's law.
The fact that Newton's calculation differed from the measured speed is due to the fact that air consists of diatomic molecules (nitrogen and oxygen). This was the first solid clue for the existence of atoms, and it also contained a clue for quantum mechanics.
In Newton's time it was not known that changing the volume of a gas changes its temperature, which modifies the relationship between density and pressure. This was discovered by Charles in 1802 (Charles' law).
Melt Boil Solid Liquid Gas Mass Sound speed (K) (K) density density density (AMU) at 20 C g/cm^{3} g/cm^{3} g/cm^{3} (m/s) He .95 4.2 .125 .000179 4.00 1007 Ne 24.6 27.1 1.21 .000900 20.18 Ar 83.8 87.3 1.40 .00178 39.95 319 Kr 115.8 119.9 2.41 .00375 83.80 221 Xe 161.4 165.1 2.94 .00589 131.29 178 H2 14 20 .070 .000090 2.02 1270 N2 63 77 .81 .00125 28.01 349 O2 54 90 1.14 .00143 32.00 326 Air .0013 29.2 344 79% N_{2}, 21% O_{2}, 1% Ar H2O 273 373 .917 1.00 .00080 18.02 CO2 n/a 195 1.56 n/a .00198 44.00 267 CH4 91 112 .42 .00070 16.04 446 CH5OH 159 352 .79 .00152 34.07 AlcoholGas density is for 0 Celsius and 1 Bar. Liquid density is for the boiling point, except for water, which is for 4 Celsius.
Carbon dioxide doesn't have a liquid state at standard temperature and pressure. It sublimes directly from a solid to a vapor.
M = Mass of a gas molecule V = Thermal speed E = Mean energy of a gas molecule = 1/2 M V^2 H = Characteristic height of an atmosphere g = Gravitational accelerationSuppose a molecule at the surface of the Earth is moving upward with speed V and suppose it doesn't collide with other air molecules. It will reach a height of
M H g = 1/2 M V^2This height H is the characteristic height of an atmosphere.
Pressure of air at sea level = 1 Bar Pressure of air in Denver = .85 Bar One mile high Pressure of air at Mount Everest = 1/4 Bar 10 km highThe density of the atmosphere scales as
Density ~ (Density At Sea Level) * exp(E/E0)where E is the gravitational potential energy of a gas molecule and E0 is the characteristic thermal energy given by
E0 = M H g = 1/2 M V^2Expressed in terms of altitude h,
Density ~ Density At Sea Level * exp(h/H)For oxygen,
E0 = 3/2 * Boltzmann_Constant * TemperatureE0 is the same for all molecules regardless of mass, and H depends on the molecule's mass. H scales as
H ~ Mass^1
S = Escape speed T = Temperature B = Boltzmann constant = 1.38e23 Joules/Kelvin g = Planet gravity at the surface M = Mass of heavy molecule m = Mass of light molecule V = Thermal speed of heavy molecule v = Thermal speed of light molecule E = Mean energy of heavy molecule e = Mean energy of light molecule H = Characteristic height of heavy molecule h = Characteristic height of light molecule = E / (M g) = e / (m g) Z = Energy of heavy molecule / escape energy z = Energy of light molecule / escape energy = .5 M V^2 / .5 M S^2 = .5 m v^2 / .5 m S^2 = V^2 / S^2 = v^2 / S^2 For an ideal gas, all molecules have the same mean kinetic energy. E = e = 1.5 B T .5 M V^2 = .5 m v^2 = 1.5 B TThe light molecules tend to move faster than the heavy ones. This is why your voice increases in pitch when you breathe helium. Breathing a heavy gas such as Xenon makes you sound like Darth Vader.
For an object to have an atmosphere, the thermal energy must be much less than the escape energy.
V^2 << S^2 <> Z << 1 Escape Atmos Temp H2 N2 Z Z speed density (K) km/s km/s (H2) (N2) km/s (kg/m^3) Jupiter 59.5 112 1.18 .45 .00039 .000056 Saturn 35.5 84 1.02 .39 .00083 .00012 Neptune 23.5 55 .83 .31 .0012 .00018 Uranus 21.3 53 .81 .31 .0014 .00021 Earth 11.2 1.2 287 1.89 .71 .028 .0041 Venus 10.4 67 735 3.02 1.14 .084 .012 Mars 5.03 .020 210 1.61 .61 .103 .015 Titan 2.64 5.3 94 1.08 .41 .167 .024 Europa 2.02 0 102 1.12 .42 .31 .044 Moon 2.38 0 390 2.20 .83 .85 .12 Ceres .51 0 168 1.44 .55 8.0 1.14Even if an object has enough gravity to capture an atmosphere, it can still lose it to the solar wind. Also, the upper atmosphere tends to be hotter than at the surface, increasing the loss rate.
The threshold for capturing an atmosphere appears to be around Z = 1/25, or
Thermal Speed < 1/5 Escape speed
When an object collapses by gravity, its temperature increases such that
Thermal speed of molecules ~ Escape speedIn the gas simulation at phet.colorado.edu, you can move the wall and watch the gas change temperature.
For an ideal gas,
3 * Boltzmann_Constant * Temperature ~ MassOfMolecules * Escape_Speed^2For the sun, what is the temperature of a proton moving at the escape speed? This sets the scale of the temperature of the core of the sun. The minimum temperature for hydrogen fusion is 4 million Kelvin.
The Earth's core is composed chiefly of iron. What is the temperature of an iron atom moving at the Earth's escape speed?
Escape speed (km/s) Core composition Sun 618. Protons, electrons, helium Earth 11.2 Iron Mars 5.03 Iron Moon 2.38 Iron Ceres .51 Iron
We first derive the law for a 1D gas and then extend it to 3D.
Suppose a gas molecule bounces back and forth between two walls separated by a distance L.
M = Mass of molecule V = Speed of the molecule L = Space between the wallsWith each collision, the momentum change = 2 M V
Time between collisions = 2 L / V
The average force on a wall is
Force = Change in momentum / Time between collisions = M V^2 / LSuppose a gas molecule is in a cube of volume L^3 and a molecule bounces back and forth between two opposite walls (never touching the other four walls). The pressure on these walls is
Pressure = Force / Area = M V^2 / L^3 = M V^2 / Volume Pressure * Volume = M V^2This is the ideal gas law in one dimension. For a molecule moving in 3D,
Velocity^2 = (Velocity in X direction)^2 + (Velocity in Y direction)^2 + (Velocity in Z direction)^2 Characteristic thermal speed in 3D = 3 * Characteristic thermal speed in 1D.To produce the 3D ideal gas law, replace V^2 with 1/3 V^2 in the 1D equation.
Pressure * Volume = 1/3 M V^2 Where V is the characteristic thermal speed of the gasThis is the pressure for a gas with one molecule. If there are n molecules,
Pressure Volume = n 1/3 M V^2 Ideal gas law in 3DIf a gas consists of molecules with a mix of speeds, the thermal speed is defined as
Kinetic dnergy density of gas molecules = E = (n / Volume) 1/2 M V^2Using this, the ideal gas law can be written as
Pressure = 2/3 E = 1/3 Density V^2 = 8.3 Moles Temperature / VolumeThe last form comes from the law of thermodynamics:
M V^2 = 3 B T
A typical globular cluster consists of millions of stars. If you measure the total gravitational and kinetic energy of the stars, you will find that
Total gravitational energy = 2 * Total kinetic energyjust like for a single satellite on a circular orbit.
Suppose a system consists of a set of objects interacting by a potential. If the system has reached a longterm equilibrium then the above statement about energies is true, no matter how chaotic the orbits of the objects. This is the "Virial theorem". It also applies if additional forces are involved. For example, the protons in the sun interact by both gravity and collisions and the virial theorem holds.
Gravitational energy of the sun = 2 * Kinetic energy of protons in the sun
Because of Hooke's law, springs oscillate with a constant frequency. X = Displacement of a spring V = Velocity of the spring A = Acceleration of the spring F = Force on the spring M = Spring mass Q = Spring constant q = (K/M)^(1/2) t = time T = Spring oscillation periodHooke's law and Newton's law:
F =  Q X = M A A =  (Q/M) X =  q^2 XThis equation is solved with
X = sin(q t) V = q cos(q t) A = q^2 sin(q t) =  q^2 XThe oscillation period of the spring is
T = 2 Pi / q = 2 Pi (M/Q)^(1/2)According to Boyle's law, a gas functions like a spring and hence a gas oscillates like a spring. An oscillation in a gas is a sound wave.
For a gas,
P = Pressure dP = Change in pressure Vol = Volume dVol= Change in volumeIf you change the volume of a gas according to Boyle's law,
P Vol = Constant P dVol + Vol dP = 0 dP =  (P/Vol) dVolThe change in pressure is proportional to the change in volume. This is equivalent to Hooke's law, where pressure takes the role of force and the change in volume takes the role of displacement of the spring. This is the mechanism behind sound waves.
In Boyle's law, the change in volume is assumed to be slow so the gas
has time to equilibrate temperature with its surroundings. In this case
the temperature is constant as the volume changes and the change is
"isothermal".
P Vol = ConstantIf the change in volume is fast then the walls do work on the molecules, changing their temperature. If there isn't enough time to equilibrate temperature with the surroundings then the change is "adiabatic". You can see this in action with the "Gas" simulation at phet.colorado.edu. Moving the wall changes the thermal speed of molecules and hence the temperature.
If a gas consists of pointlike particles then
Vol = Volume of the gas Ek = Total kinetic energy of gas molecules within the volume E = Total energy of gas molecules within the volume = Kinetic energy plus the energy from molecular rotation and vibration dE = Change in energy as the volume changes P = Pressure dP = Change in pressure as the volume changes D = Density C = Speed of sound in the gas d = Number of degrees of freedom of a gas molecule = 3 for a monotomic gas such as Helium = 5 for a diatomic gas such as nitrogen G = Adiabatic constant = 1 + 2/d = 5/3 for a monatomic gas = 7/5 for a diatomic gas k = Boltzmann constant T = TemperatureThe ideal gas law is
P Vol = (2/3) Ek (Derived in www.jaymaron.com/gas/gas.html)This law is equivalent to the formula that appears in chemistry.
P Vol = Moles R TFor a gas in thermal equilibrium each degree of freedom has a mean energy of .5 k T. For a gas of pointlike particles (monotomic) there are three degrees of freedom, one each for motion in the X, Y, and Z direction. In this case d=3. The mean kinetic energy of each gas molecule is 3 * (.5 k T). The total mean energy of each gas molecule is also 3 * (.5 k T).
For a diatomic gas there are also two rotational degrees of freedom. In this case d=5.
In general,
Ek = 3 * (.5 k T) E = d * (.5 k T) Ek = (3/d) EIf you change the volume of a gas adiabatically, the walls change the kinetic and rotational energy of the gas molecules.
dE = P dVolThe ideal gas law in terms of E instead of Ek is
P Vol = (2/d) E dP = (2/d) (dE/Vol  E dVol/Vol^2) = (2/d) [P dVol/Vol  (d/2) P dVol/Vol] = (1+2/d) P dVol/Vol =  G P dVol/VolThis equation determines the speed of sound in a gas.
C^2 = G P / DFor air,
P = 1.01e5 Newtons/meter^2 D = 1.2 kg/meter^3Newton assumed G=1 from Boyle's law and calculated the speed of sound in air to be
C = 290 m/sThe correct value for air is G=7/5, which gives a sound speed of
C = 343 m/swhich is in accord with the measurement.
For a gas, G can be measured by measuring the sound speed. The results are
Helium 5/3 Monatomic molecule Argon 5/3 Monatonic molecule Air 7/5 4/5 Nitrogen and 1/5 Oxygen Oxygen 7/5 Diatomic molecule Nitrogen 7/5 Diatomic moleculeThe fact that G is not equal to 1 was the first solid evidence for the existence of atoms and it also contained a clue for quantum mechanics. If a gas is a continuum (like Hooke's law) it has G=1 and if it consists of pointlike particles (monatonic) it has G=5/3. This explains helium and argon but not nitrogen and oxygen. Nitrogen and oxygen are diatomic molecules and their rotational degrees of freedom change Gamma.
Kinetic degrees Rotational degrees Gamma of freedom of freedom Monatonic gas 3 0 5/3 Diatomic gas T < 1000 K 3 2 7/5 Diatomic gas, T > 1000 K 3 3 4/3Quantum mechanics freezes out one of the rotation modes at low temperature. Without quantum mechanics, diatomic molecules would have Gamma=4/3 at room temperature.
The fact that Gamma=7/5 for air was a clue for the existence of both atoms, molecules, and quantum mechanics.
For dark energy,
E = Energy dE = Change in energy e = Energy density Vol= Volume P = PressureThe volume expands as the universe expands.
As a substance expands it does work on its surroundings according to its pressure.
dE =  P dVolFor dark energy, the energy density "e" is constant in space and so
dE = e dVolHence,
P =  eDark energy has a negative pressure, which means that it behaves differently from a continuum and from particles.
Dark matter consists of pointlike particles but they rarely interact with other particles and so they exert no pressure.
For gases, the density at boiling point is used. Size data
Copper atoms stack like cannonballs. We can calculate the atom size by assuming the atoms are shaped like either cubes or spheres. For copper atoms,
Density = D = 8900 kg/m^{3} Atomic mass unit= M_{0} = 1.661⋅10^{27} kg Atomic mass = M_{A} = 63.55 Atomic mass units Mass = M = M_{A}⋅M_{0} = 9.785⋅10^{26} kg Number density = N = D / M = 9.096⋅10^{28} atoms/m^{3} Cube volume = Υ_{cube}= 1 / N = 1.099⋅10^{29} m^{3} Volume/atom if the atoms are cubes Cube length = L = Υ^{1/3}_{cube} = 2.22⋅10^{10} m Side length of the cube Sphere fraction = f = π/(3√2) = .7405 Fraction of volume occupied by spheres in a stack o spheres Sphere volume = Υ_{sph}= Υ_{cube} f = 8.14⋅10^{30} m^{3} = ^{4}⁄_{3}πR^{3} Volume/atom if the atoms are spheres Sphere radius = R = 1.25⋅10^{10} m
Dot size = Density Color = Shear Modulus, an indicator of the element's strength. Blue: The element is a liquid at room temperature Red: Weak White: StrongShear data Density data
The "de Broglie wavelength" of a particle is
Particle momentum = Q Planck constant = h = 6.62*10^34 Joule seconds Particle wavelength = W = h/Q (de Broglie formula)The Bohr hypothesis states that for an electron orbiting a proton, the number of electron wavelengths is an integer. This sets the characteristic size of a hydrogen atom.
Orbit circumference = C = N W where N is a positive integer N Orbital 1 S 2 P 3 D 4 F Electron mass = m = 9.11*10^{31} kg Electron velocity = V Electron momentum = Q = m V Electron charge = e = 1.60*10^{19} Coulombs Coulomb constant = K = 9.0*10^{9} Newtons meters^{} / Coulombs^{2} Electric force = F_{e} = K e^{2} / R^{2} Centripetal force = F_{c} = M V^{2} / R Orbit radius = R = N h^{2} / (4 π^{2} K e^{2} m) = N * 5.29e11 meters Electron energy = E =  .5 K e^{2} / (R N^{2}) = N^{2} 2.18e18 Joules = N^{2} 13.6 electron Volts (Ionization energy)For an electron on a circular orbit,
F_{e} = F_{c}
Charges of the same sign repel and charges of opposite sign attract.
Charge 1 Charge 2 Electric Force + + Repel   Repel +  Attract  + Attract Charge = Q (Coulombs) 1 Proton = 1.602e19 Coulombs Distance between charges = R Mass of the charges = M Gravity constant = G = 6.67e11 Newton m^{2} / kg^{2} Electric constant = K = 8.99e9 Newton m^{2} / Coulomb^{2} Gravity force = F = G M_{1} M_{2} / R^{2} = M_{2} g Electric force = F = K Q_{1} Q_{2} / R^{2} = Q_{2} E Gravity field from M_{1} = g = G M_{1} / R^{2} Electric field from Q_{1} = E = K Q_{1} / R^{2} Gravity voltage = H g (H = Height, g = Gravitational acceleration) Electric voltage = H E (H = Distance parallel to the electric field) Gravity energy = G M_{1} M_{2} / R Electric energy = K Q_{1} Q_{2} / R
A charge generates an electric field. The electric field points away from positive charges and toward negative charges.
A moving charge is an "electric current". In an electric circuit, a battery moves electrons through a wire.
Charge = Q Time = T Electric current = I = Q / T (Coulombs/second)The current from a positive charge moving to the right is equivalent to that from a negative charge moving to the left.
Moving charges and currents exert forces on each other. Parallel currents attract and antiparallel currents repel.
Charge = Q Velocity of the charges = V Current = I Length of a wire = L Distance between the charges = R Electric force constant = K_{e} = 8.988e9 N m^{2}/C^{2} Magnetic force constant = K_{m} = 2e7 = K_{e}/C^{2} Electric force between charges = F_{e} = K_{e} Q_{1} Q_{2} / R^{2} Magnetic force between charges = F_{m} = K_{m} V^{2} Q_{1} Q_{2} / R^{2} = (V^{2}/C^{2}) F_{e} Magnetic force between currents = F_{m} = K_{m} I_{1} I_{2} Z / R Magnetic force / Electric force = V^{2} / C^{2}The magnetic force is always less than the electric force.
The electric force can be interpreted as an electric field, and the magnetic force can be interpreted as a magnetic field. Both interpretations produce the same force.
Radial distance = R (Distance perpendicular to the velocity of the charge) Magnetic field from charge Q_{1} = B = K_{m} V Q_{1} / R^{2} Magnetic field from current I_{1} = B = K_{m} I_{1} / R Magnetic force on charge Q_{2} = F_{m} = Q_{2} V B = K_{m} V^{2} Q_{1} Q_{2} / R^{2} Magnetic force on current I_{2} = F_{m} = I_{2} Z B = K_{m} I_{1} I_{2} Z / R
The direction of the magnetic force on a positive charge is given by the right hand rule. The force on a negative charge is in the opposite direction (the left hand rule).
We use the above symbols to depict vectors in the Z direction. The vector on the left points into the plane and the vector on the right points out of the plane.
The direction of the force is the cross product "×" of V and B. The direction is given by the "right hand rule".
Magnetic field = B Magnetic force on a charge = F = Q V × B Magnetic force on a current = F = 2e7 I × B
Quantity MKS units CGS units Conversion factor _{} Mass M kg gram .001 Wire length Z meter cm .01 Radial distance from wire R meter cm .01 Time T second second 1 Force F Newton dyne 100000 Charge Q Coulomb Franklin 3.336e10 Velocity of a charge V meter/second cm/s .01 Speed of light C 2.999e8 meter/second cm/s 100 Energy E Joule erg e7 Electric current I Ampere = Coulomb/s Franklin/s 3.336e10 Electric potential V Volt Statvolt 299.79 Electric field E Volt/meter StatVolt/cm 29979 Magnetic field B Tesla Gauss 10000 Capacitance C Farad cm 1.11e12 Inductance L Henry s^{2}/cm 9e11 Electric force constant K_{e} = 8.988e9 N m^{2}/C^{2} K_{e} = 1 dyne cm^{2} / Franklin^{2} Magnetic force constant K_{m} = 2e7 = K_{e}/C^{2} K_{m} = 1/C^{2} Vacuum permittivity ε = 8.854e12 F/m =1/4/π/K_{e} Vacuum permeability μ = 4 π e7 Vs/A/m =2 π K_{m} Proton charge Q_{pro} = 1.602e19 Coulomb Q_{pro}= 4.803e10 Franklin Electric field from a charge E = K_{e} Q / R^{2} E = Q / R^{2} Electric force on a charge F = Q E F = Q E Electric force between charges F = K_{e} Q Q / R^{2} F = Q Q / R^{2} Magnetic field of moving charge B = K_{m} V Q / R^{2} B = (V/C) Q / R^{2} Magnetic field around a wire B = K_{m} I / R B = (V/C) I / R Magnetic force on a charge F = Q V B F = (V/C) Q B Magnetic force on a wire F = K_{m} B Z F = I B z Magnetic force between charges F = K_{m} V^{2} Q_{1} Q_{2} / R^{2} F = (V/C)^{2} Q Q / R^{2} Magnetic force between wires F = K_{m} I_{1} I_{2} Z / R F = I_{1} I_{2} Z / R Energy of a capacitor E = .5 C V^{2} Field energy per volume Z = (8 π K_{e})^{1} (E^{2} + B^{2}/C^{2}) Z = .5 (E^{2} + B^{2}/C^{2})
Speed of light C Electric field E Electric field, time derivative E_{t} Magnetic field B Magnetic field, time derivative B_{t} Charge Q Charge density q Current density J MKS CGS K_{e}=8.988e9 K_{e}=1 K_{m}=2e7 K_{m}=2/C ∇˙E = 4 π K_{e} q ∇˙E = 4 π q ∇˙B = 0 ∇˙B = 0 ∇×E = B_{t} ∇×E = B_{t} / C ∇×B = 2 π K_{m} J + E_{t} / C^{2} ∇×B = 4 π J / C + E_{t} / C
Charge = Q Coulombs Voltage = V Volts Energy = E = VQ Joules Time = T seconds Current = I = Q/T Amperes Resistance = R = V/I Ohms Power = P = QV/T Watts = IV = V^{2}/R = I^{2}R Ohm's Law: V = IR
In a superconductor, electrons move without interference.
In a resistor, electrons collide with atoms and lose energy.
Resistance (Ohms) Copper wire .02 1 meter long and 1 mm in diameter 1 km power line .03 AA battery .1 Internal resistance Light bulb 200 Human 10000
Voltage = V Volts Capacitance = C Farads Total energy = E = ½ C V^{2} Joules Effective = E_{e} = ¼ C V^{2} JoulesNot all of the energy in a capacitor is harnessable because the voltage diminishes as the charge diminishes, hence the effective energy is less than the total energy.
A = Plate area Z = Plate spacing Ke = Electric force constant = 8.9876e9 N m^{2} / C^{2} Q = Max charge on the plate (Coulombs) Emax= Max electric field = 4 Pi Ke Q / A V = Voltage between plates = E Z = 4 Pi Ke Q Z / A En = Energy = .5 Q V = .5 A Z E^{2} / (4 π Ke) e = Energy/Volume = E / A Z = .5 E^{2} / (4 π Ke) q = Charge/Volume = Q / A / Z C = Capacitance = Q/V = (4 Pi Ke)A capacitor can be specified by two parameters:
The maximum electric field is equal to the max field for air times a dimensionless number characterizing the dielectric
Eair = Maximum electric field for air before electical breakdown Emax = Maximum electric field in the capacitor Rbohr= Bohr radius = Characteristic size of atoms = 5.2918e11 m = hbar^{2} / (ElectronMass*ElectronCharge^{2}*Ke) Ebohr= Bohr electric field = Field generated by a proton at a distance of 1 Bohr radius = 5.142e11 Volt/m Maximum energy density = .5 * 8.854e12 Emax^{2} Emax (MVolt/m) Energy density (Joule/kg) Al electrolyte capacitor 15.0 1000 Supercapacitor 90.2 36000 Bohr limit 510000 1.2e12 Capacitor with a Bohr electric field
A solenoid is a wire wound into a coil.
N = Number of wire loops Z = Length A = Area Mu = Magnetic constant = 4 π 10^{7} I = Current It = Current change/time F = Magnetic flux = N B A (Tesla meter^{2}) Ft = Flux change/time (Tesla meter^{2} / second) B = Magnetic field = Mu N I / Z V = Voltage = Ft = L It = N A Bt = Mu N^{2} A It / Z L = Inductance = Ft / It = Mu N^{2} A / Z (Henrys) E = Energy = .5 L I^{2}Hyperphysics: Inductor
White: High conductivity Red: Low conductivity
Electric Thermal Density Electric C/Ct Heat Heat Melt $/kg Young Tensile Poisson Brinell conduct conduct conduct/ cap cap number hardness (e7 A/V/m) (W/K/m) (g/cm^3) Density (AK/VW) (J/g/K) (J/cm^3K) (K) (GPa) (GPa) (GPa) Silver 6.30 429 10.49 .60 147 .235 2.47 1235 590 83 .17 .37 .024 Copper 5.96 401 8.96 .67 147 .385 3.21 1358 6 130 .21 .34 .87 Gold 4.52 318 19.30 .234 142 .129 2.49 1337 24000 78 .124 .44 .24 Aluminum 3.50 237 2.70 1.30 148 .897 2.42 933 2 70 .05 .35 .245 Beryllium 2.5 200 1.85 1.35 125 1.825 3.38 1560 850 287 .448 .032 .6 Magnesium 2.3 156 1.74 1.32 147 1.023 1.78 923 3 45 .22 .29 .26 Iridium 2.12 147 22.56 .094 144 .131 2.96 2917 13000 528 1.32 .26 1.67 Rhodium 2.0 150 12.41 .161 133 .243 3.02 2237 13000 275 .95 .26 1.1 Tungsten 1.89 173 19.25 .098 137 .132 2.54 3695 50 441 1.51 .28 2.57 Molybdenum 1.87 138 10.28 .182 136 .251 2896 24 330 .55 .31 1.5 Cobalt 1.7 100 8.90 .170 .421 1768 30 209 .76 .31 .7 Zinc 1.69 116 7.14 .388 693 2 108 .2 .25 .41 Nickel 1.4 90.9 8.91 .444 1728 15 Ruthenium 1.25 117 12.45 2607 5600 Cadmium 1.25 96.6 8.65 594 2 50 .078 .30 .20 Osmium 1.23 87.6 22.59 .130 3306 12000 Indium 1.19 81.8 7.31 430 750 11 .004 .45 .009 Iron 1.0 80.4 7.87 .449 1811 211 .35 .29 .49 Palladium .95 71.8 1828 Tin .83 66.8 505 22 47 .20 .36 .005 Chromium .79 93.9 .449 2180 Platinum .95 .133 2041 Tantalum .76 .140 3290 Gallium .74 303 Thorium .68 Niobium .55 53.7 2750 Rhenium .52 .137 3459 Vanadium .5 30.7 2183 Uranium .35 Titanium .25 21.9 .523 1941 Scandium .18 15.8 1814 Neodymium .156 1297 Mercury .10 8.30 .140 234 Manganese .062 7.81 1519 Germanium .00019 1211 Dimond iso 10 40000 Diamond e16 2320 .509 Tube 10 3500 Carbon nanotube. Electric conductivity = e16 laterally Tube bulk 200 Carbon nanotubes in bulk Graphene 10 5000 Graphite 2 400 .709 Natural graphite Al Nitride e11 180 Brass 1.5 120 Steel 45 Carbon steel Bronze .65 40 Steel Cr .15 20 Stainless steel (usually 10% chromium) Quartz (C) 12 Crystalline quartz. Thermal conductivity is anisotropic Quartz (F) e16 2 Fused quartz Granite 2.5 Marble 2.2 Ice 2 Concrete 1.5 Limestone 1.3 Soil 1 Glass e12 .85 Water e4 .6 Seawater 1 .6 Brick .5 Plastic .5 Wood .2 Wood (dry) .1 Plexiglass e14 .18 Rubber e13 .16 Snow .15 Paper .05 Plastic foam .03 Air 5e15 .025 Nitrogen .025 1.04 Oxygen .025 .92 Silica aerogel .01 Siemens: Amperes^2 Seconds^3 / kg / meters^2 = 1 Ohm^1For most metals,
Electric conductivity / Thermal conductivity ~ 140 J/g/K
Teslas Field generated by brain 10^{12} Wire carrying 1 Amp .00002 1 cm from the wire Earth magnetic field .0000305 at the equator Neodymium magnet 1.4 Magnetic resonance imaging machine 8 Large Hadron Collider magnets 8.3 Field for frog levitation 16 Strongest electromagnet 32.2 without using superconductors Strongest electromagnet 45 using superconductors Neutron star 10^{10} Magnetar neutron star 10^{14}
The critical electric field for electric breakdown for the following materials is:
MVolt/meter Air 3 Glass 12 Polystyrene 20 Rubber 20 Distilled water 68 Vacuum 30 Depends on electrode shape Diamond 2000
Relative permittivity is the factor by which the electric field between charges is decreased relative to vacuum. Relative permittivity is dimensionless. Large permittivity is desirable for capacitors.
Relative permittivity Vacuum 1 (Exact) Air 1.00059 Polyethylene 2.5 Sapphire 10 Concrete 4.5 Glass ~ 6 Rubber 7 Diamond ~ 8 Graphite ~12 Silicon 11.7 Water (0 C) 88 Water (20 C) 80 Water (100 C) 55 TiO2 ~ 150 SrTiO3 310 BaSrTiO3 500 Ba TiO3 ~ 5000 CaCuTiO3 250000
A ferromagnetic material amplifies a magnetic field by a factor called the "relative permeability".
Relative Magnetic Maximum Critical permeability moment frequency temperature (kHz) (K) Metglas 2714A 1000000 100 Rapidlycooled metal Iron 200000 2.2 1043 Iron + nickel 100000 Mumetal or permalloy Cobalt + iron 18000 Nickel 600 .606 627 Cobalt 250 1.72 1388 Carbon steel 100 Neodymium magnet 1.05 Manganese 1.001 Air 1.000 Superconductor 0 Dysprosium 10.2 88 Gadolinium 7.63 292 EuO 6.8 69 Y3Fe5O12 5.0 560 MnBi 3.52 630 MnAs 3.4 318 NiO + Fe 2.4 858 CrO2 2.03 386
Resistivity in 10^9 Ohm Meters
293 K 300 K 500 K Beryllium 35.6 37.6 99 Magnesium 43.9 45.1 78.6 Aluminum 26.5 27.33 49.9 Copper 16.78 17.25 30.9 Silver 15.87 16.29 28.7
Gauge Diameter Continuous 10 second 1 second 32 ms Resistance mm current current current current Ampere Ampere Ampere Ampere mOhm/meter 0 8.3 125 1900 16000 91000 .32 2 6.5 95 1300 10200 57000 .51 4 5.2 70 946 6400 36000 .82 6 4.1 55 668 4000 23000 1.30 12 2.0 20 235 1000 5600 5.2 18 1.02 10 83 250 1400 21.0 24 .51 3.5 29 62 348 84 30 .255 .86 10 15 86 339 36 .127 .18 4 10 22 1361 40 .080 1 1.5 8 3441
Particle Charge Mass Proton +1 1 Composed of 2 up quarks, 1 down quark, and gluons Neutron 0 1.0012 Composed of 1 up quark, 2 down quarks, and gluons Electron 1 .000544 Up quark +2/3 .0024 Down quark 1/3 .0048 Photon 0 0 Carries the electromagnetic force and binds electrons to the nucleus Gluon 0 0 Carries the strong force and binds quarks, protons, and neutronsCharge and mass are relative to the proton.
All of these particles are stable except for the neutron, which has a half life of 886 seconds.
Proton charge = 1.6022 Coulombs Proton mass = 1.673⋅10^{27} kg Electron mass = 9.11⋅10^{31} kg Hydrogen mass = Proton mass + Electron mass = 1.6739⋅10^{27} kg
An element has a fixed number of protons and a variable number of neutrons. Each neutron number corresponds to a different isotope. Naturallyoccuring elements tend to be a mix of isotopes.
Isotope Protons Neutrons Natural fraction Hydrogen1 1 0 .9998 Hydrogen2 1 1 .0002 Helium3 2 1 .000002 Helium4 2 2 .999998 Lithium6 3 3 .05 Lithium7 3 4 .95 Beryllium9 4 5 1 Boron10 5 5 .20 Boron11 5 6 .80 Carbon12 6 6 .989 Carbon13 6 7 .011Teaching simulation for isotopes at phet.colorado.edu
Alpha particle = Helium nucleus = 2 Protons and 2 Neutrons Beta particle = Electron Gamma ray = Photon Alpha decay: Uranium235 > Thorium231 + Alpha Beta decay: Neutron > Proton + Electron + Antineutrino (From the point of view of nuclei) Beta decay: Down quark > Up quark + Electron + Antineutrino (From the point of view of quarks)Beta decay is an example of the "weak force".
For a radioactive material,
Time = T Half life = T_{h} Original mass = M Mass remaining after time "T" = m = M exp(T/T_{h})
Suppose an element has a half life of 2 years.
Time Mass of element remaining (kg) 0 1 2 1/2 4 1/4 6 1/8 8 1/16
The weak force can convert a neutron into a proton, ejecting a highenergy electron.
From the point of view of nucleons: Neutron > Proton + electron + antineutrino From the point of view of quarks: Down quark > Up quark + electron + antineutrino
The unit of energy used for atoms, nuclei, and particle is the "electron Volt", which is the energy gained by an electron upon descending a potential of 1 Volt.
Electron Volt (eV) = 1 eV = 1.602e19 Joules Kilo electron Volt = 1 keV = 10^{3} eV Mega electron Volt = 1 MeV = 10^{6} eV Giga electron Vlt = 1 GeV = 10^{9} eV
Proton + Proton > Deuterium + Positron + NeutrinoHydrogen fusion requires a temperature of at least 4 million Kelvin, which requires an object with at least 0.08 solar masses. This is the minimum mass to be a star. The reactions in the fusion of hydrogen to helium are:
P + P > D + Positron + Neutrino + .42 MeV P + D > He3 + Photon + 5.49 MeV He3 + He3 > He4 + P + P + 12.86 MeV
As the core of a star star runs out of hydrogen it contracts and heats, and helium fusion begins when the temperature reaches 10 million Kelvin.
A heavy star continues to fuse elements until it reaches Iron56.
Beyond this, fusion absorbs energy rather than releasing it, triggering a
runaway core collapse that fuses elements up to Uranium. If the star explodes
as a supernova then these elements are ejected into interstellar space.
Star type Mass Luminosity Color Temp Lifetime Death Remnant Size of Output (solar (solar (Kelvin) (billions remnant masses) luminosities) of years) Brown Dwarf <0.08 1000 immortal Red Dwarf 0.1 .0001 red 2000 1000 red giant white dwarf Earthsize The Sun 1 1 white 5500 10 red giant white dwarf Earthsize light elements Blue star 10 10000 blue 10000 0.01 supernova neutron star Manhattan heavy elements Blue giant 20 100000 blue 20000 0.01 supernova black hole Central Park heavy elementsFate of stars, with mass in solar masses:
Mass < 9 > End as red giants and then turn white dwarf. 9 < Mass > End as supernova 9 < Mass < 20 > Remnant is a neutron star. 20 < Mass > Remnant is a black hole. 130 < Mass < 250 > Pairinstability supernova (if the star has low metallicity) 250 < Mass > Photodisintegration supernova, producing a black hole and relativistic jets.
A neutron triggers the fission of Uranium235 and plutonium239, releasing energy and more neutrons.
Fission releases neutrons that trigger more fission.
Two pieces of uranium, each with less than a critical mass, are brought together in a cannon barrel.
If the uranium is brought together too slowly, the bomb fizzles.
Plutonium is more difficult to detonate than uranium. Plutonium detonation requires a spherical implosion.
Blue elements are unstable with a half life much less than the age of the solar system.
The only elements heavier than Bismuth that can be found on the Earth are Thorium and Uranium, and these are the only elements that can be tapped for fission energy.
Natural Thorium is 100% Thorium232
Natural Uranium is .72% Uranium235 and 99.3% Uranium238.
Plutonium doesn't exist in nature.
Protons Neutrons Halflife Critical Isotope (10^6 yr) mass (kg) fraction Thorium232 90 142 14000  1.00 Absorbs neutron > U233 Uranium233 92 141 .160 16  Fission chain reaction Uranium235 92 143 700 52 .0072 Fission chain reaction Uranium238 92 146 4500  .9927 Absorbs neutron > Pu239 Plutonium238 94 144 .000088   Produces power from radioactive heat Plutonium239 94 145 .020 10  Fission chain reactionThe elements that can be used for fission energy are the ones with a critical mass. These are Uranium233, Uranium235, and Plutonium239. Uranium233 and Plutonium239 can be created in a breeder reactor.
Thorium232 + Neutron > Uranium233 Uranium238 + Neutron > Plutonium239The "Fission" simulation at phet.colorado.edu illustrates the concept of a chain reaction.
Natural uranium is composed of .7% Uranium235 and the rest is Uranium238. Uranium235 can be separated from U238 using centrifuges, calutrons, or gas diffusion chambers. Uranium235 is easy to detonate. A cannon and gunpowder gets it done.
Plutonium239 is difficult to detonate, requiring a perfect spherical implosion. This technology is beyond the reach of most rogue states.
Uranium233 cannot be used for a bomb and is hence not a proliferation risk.
Plutonium238 emits alpha particles, which can power a radioisotope thermoelectric generator (RTG). RTGs based on Plutonium238 generate 540 Watts/kg and are used to power spacecraft.
Creating Plutonium239 and Uranium233:
Uranium238 + Neutron > Plutonium239 Thorium232 + Neutron > Uranium233 Detail: Uranium238 + Neutron > Uranium239 Uranium239 > Neptunium239 + Electron + Antineutrino Halflife = 23 mins Neptunium239 > Plutonium239 + Electron + Antineutrino Halflife = 2.4 days Thorium232 + Neutron > Thorium233 Thorium233 > Protactinium233 + Electron + Antineutrino Halflife = 22 mins Protactinium233 > Uranium233 + Electron + Antineutrino Halflife =
A nuclear fusion bomb contains deuterium and lithium6 and the reaction is catalyzed by a neutron.
N + Li6 > He4 + T + 4.87 MeV T + D > He4 + N + 17.56 MeV Total energy released = 22.43 MeV Nucleons = 8 Energy / Nucleon = 22.434 / 8 = 2.80
1 ton of TNT 4*10^9 Joules 1 ton of gasoline 4*10^10 Joules North Korea fission device 0.5 kilotons TNT 10 kg uranium fission bomb 10 kilotons TNT 10 kg hydrogen fusion bomb 10 megatons TNT Tunguska asteroid strike 15 megatons TNT 50 meter asteroid Chixulub dinosaur extinction 100 trillion tons TNT 10 km asteroid
1885 Rontgen discovers Xrays 1899 Rutherford discovers alpha and beta rays 1903 Rutherford discovers gamma rays 1905 E=mc^2. Matter is equivalent to energy 1909 Nucleus discovered by the Rutherford scattering experiment 1932 Neutron discovered 1933 Nuclear fission chain reaction envisioned by Szilard 1934 Fermi bombards uranium with neutrons and creates Plutonium. First successful example of alchemy 1938 Fission discovered by Hahn and Meitner 1938 Bohr delivers news of fission to Princeton and Columbia 1939 Fermi constructs the first nuclear reactor in the basement of Columbia 1939 Szilard and Einstein write a letter to President Roosevelt advising him to consider nuclear fission 1942 Manhattan project started 19421945 German nuclear bomb project goes nowhere 1945 Two nuclear devices deployed by the United States
Fission Fusion U.S.A. 1945 1954 Germany Attempted fission in 1944 & failed Russia 1949 1953 Britain 1952 1957 France 1960 1968 China 1964 1967 India 1974 Uranium Israel 1979 ? Undeclared. Has both fission and fusion weapons South Africa 1980 Dismantled in 1991 Iran 1981 Osirak reactor to create Plutonium. Reactor destroyed by Israel Pakistan 1990 Centrifuge enrichment of Uranium. Tested in 1998 Built centrifuges from stolen designs Iraq 1993 Magnetic enrichment of Uranium. Dismantled after Gulf War 1 Iraq 2003 Alleged by the United States. Proved to be untrue. North Korea 2006 Obtained plutonium from a nuclear reactor. Detonation test fizzled Also acquired centrifuges from Pakistan Also attempting to purify Uranium with centrifuges Syria 2007 Nuclear reactor destroyed by Israel Iran 2009? Attempting centrifuge enrichment of Uranium. Libya  Attempted centrifuge enrichment of Uranium. Dismantled before completion. Cooperated in the investigation that identified Pakistan as the proliferator of Centrifuge designs. Libya 2010 Squabbling over nuclear material Libya 2011 Civil war
A tokamak fusion reactor uses magnetic fields to confine a hot plasma so that fusion can occur in the plasma.
The fusion reaction that occurs at the lowest temperature and has the highest reaction rate is
Deuterium + Tritium > Helium4 + Neutron + 17.590 MeVbut the neutrons it produces are a nuisance to the reactor.
A potential fix is to have "liquid walls" absorb the neutrons (imagine a waterfall of neutronabsorbing liquid lithium cascading down the walls of the reactor).
http://deadspin.com/megatronsbuttholetoremainclenched1797265016
Black: Carbon White: Hydrogen Red: Oxygen
The energy sources that can be used by vehicles are:
Energy/Mass Power/mass Energy/$ Rechargeable Charge Maximum charging MJoule/kg Watt/kg MJoule/$ time cycles Gasoline 45 60 Battery, aluminum 4.6 130 No Battery, lithium .8 1600 .010 Yes 1 hour 1000 Supercapacitor .016 8000 .00005 Yes Instant Infinite Aluminum capacitor .010 10000 .0001 Yes Instant Infinite
Energy/Mass Power/mass Density Energy/$ MJoule/kg Watt/kg MJoule/$ Antimatter 90000000000 Fusion bomb 250000000 Max for d+t fusion Fission bomb 83000000 Max for a uranium bomb Nuclear battery, Pu238 2265000 10 7.6 88 year half life Nuclear battery, Sr90 589000 10 59 29 year half life Hydrogen ( 0 carbons) 141.8 .07 Methane ( 1 carbon ) 55.5 .42 Natural gas Ethane ( 2 carbons) 51.9 .54 Propane ( 3 carbons) 50.4 .60 Butane ( 4 carbons) 49.5 .60 Octane ( 8 carbons) 47.8 .70 Kerosene (12 carbons) 46 .75 Diesel (16 carbons) 46 .77 Oil (36 carbons) 46 .8 Fat (20 carbons) 37 .9 1.0 9 Calories/gram Pure carbon 32.8 2.0 Coal 32 .8 Similar to pure carbon Ethanol 29 .79 .2 7 Calories/gram Wood 22 .6 Sugar 17 1.54 6 4 Calories/gram Protein 17 1.06 2 4 Calories/gram Plastic explosive 8.0 600000 1.91 HMX Smokeless powder 5.2 1.23 Modern gunpowder TNT 4.7 1.65 Black powder 2.6 1.7 Medieval gunpowder Phosphocreatine .137 Recharges ATP ATP .057 1.04 Adenosine triphosphate Battery, aluminumair 4.68 130 Battery, LiS 1.44 670 Battery, Liion .8 1600 .007 Battery, Lipolymer .6 4000 Battery, Alkaline .4 Battery, Lead acid .15 150 Lithium supercapacitor .054 15000 Supercapacitor .016 8000 .00005 Aluminum capacitor .010 10000 .0001 Spring .0003 Human 20 Solar cell 77 Gasoline engine 8000 Electric motor 8000 Jet engine 10000 Rocket engine 3200000The energy cost to convert water to hydrogen and oxygen is 13.16 MJ/kg. If hydrogen and oxygen are reacted to produce one kg of water, the energy produced is equivalent to a 1 kg mass moving at 5.13 km/s.
The characteristic distance a ball travels before air slows it down is the "Newton length". This distance can be estimated by setting the mass of the ball is equal to the mass of the air the ball passes through.
Mass of a soccer ball = M = .437 kg Ball radius = R = .110 meters Ball crosssectional area = A = .038 meters^{2} Ball density = D = 78.4 kg/meters^{3} Air density = d = 1.22 kg/meter^{3} (Air at sea level) Ball initial velocity = V Newton length = L Mass of air the ball passes through= m = A L d m = M L = M / (A d) = (4/3) R D / d = 9.6 metersThe depth of the penalty box is 16.45 meters (18 yards). Any shot taken outside the penalty box slows down substantially before reaching the goal.
Newton was also the first to observe the "Magnus effect", where spin causes a ball to curve.
The orange boxes depict the size of the court and the Newton length is the distance from the bottom of the court to the ball. Ball sizes are magnified by a factor of 20 relative to the court sizes.
Diameter Mass Drag Shot Drag/ Density Ball Max Spin (mm) (g) (m) (m) Shot (g/cm^{3}) speed speed (1/s) (m/s) (m/s) Ping pong 40 2.7 1.8 2.74 .64 .081 20 31.2 80 Squash 40 24 15.6 9.75 1.60 .716 Golf 43 46 25.9 200 .13 1.10 80 94.3 296 Badminton 54 5.1 1.8 13.4 .14 .062 Racquetball 57 40 12.8 12.22 1.0 .413 Billiards 59 163 48.7 2.7 18 1.52 Tennis 67 58 13.4 23.77 .56 .368 50 73.2 119 Baseball 74.5 146 27.3 19.4 1.4 .675 40 46.9 86 Whiffle 76 45 8.1 .196 Football 178 420 13.8 20 .67 .142 20 26.8 18 Rugby 191 435 12.4 20 .62 .119 Bowling 217 7260 160 18.29 8.8 1.36 Soccer 220 432 9.3 16.5 .56 .078 40 59 29 Basketball 239 624 11.4 7.24 1.57 .087 Cannonball 220 14000 945 1000 .94 7.9"Drag" is the Newton drag length and "Shot" is the typical distance of a shot, unless otherwise specified. "Density" is the density of the ball.
For a billiard ball, rolling friction is greater than air drag.
A bowling pin is 38 cm tall, 12 cm wide, and has a mass of 1.58 kg. A bowling ball has to be sufficiently massive to have a chance of knocking over 10 pins.
Mass of 10 bowling pins / Mass of bowling ball = 2.18
To estimate the distance a bullet travels before being slowed by drag,
Air density = D_{air} = .012 g/cm^{3} Water density = D_{water} = 1.0 g/cm^{3} Bullet density = D_{bullet} = 11.3 g/cm^{3} Bullet length = L_{bullet} = 2.0 cm Bullet distance in water = L_{water} ≈ L_{bullet} D_{bullet} / D_{water} ≈ 23 cm Bullet distance in air = L_{air} ≈ L_{bullet} D_{bullet} / D_{air} ≈ 185 meters
g/cm^{3} g/cm^{3} Air .00122 (Sea level) Silver 10.5 Wood .7 ± .5 Lead 11.3 Water 1.00 Uranium 19.1 Magnesium 1.74 Tungsten 19.2 Aluminum 2.70 Gold 19.3 Rock 2.6 ± .3 Osmium 22.6 (Densest element) Titanium 4.51 Steel 7.9 Copper 9.0
Cartridge Projectile Length Diameter Warhead Velocity (kg) (kg) (m) (m) (kg) (m/s) Massive Ordnance Penetrator  13608 6.2 .8 2404 PGU14, armor piercing .694 .395 .173 .030 1013 PGU13, explosive .681 .378 .173 .030 1020The GAU Avenger armorpiercing shell contains .30 kg of depleted uranium.
The massive ordnamce penetrator typically penetrates 61 meters of Earth.
The PGU13 and PGU14 are used by the A10 Warthog cannon.
The composition of natural uranium is .72% uranium235 and the rest is uranium238. Depleted uranium has less than .3% of uranium235.
The drag force on an object moving through a fluid is
Velocity = V Fluid density = D = 1.22 kg/m^{2} (Air at sea level) Crosssectional area = A Drag coefficient = C = 1 (typical value) Drag force = F = ½ C D A V^{2} Drag power = P = ½ C D A V^{3} = F V Terminal velocity = V_{t}"Terminal velocity" occurs when the drag force equals the gravitational force.
M g = ½ C D A V_{t}^{2}Suppose we want to estimate the parachute size required for a soft landing. Let a "soft landing" be the speed reached if you jump from a height of 2 meters, which is V_{t} = 6 m/s. If a skydiver has a mass of 100 kg then the area of the parachute required for this velocity is 46 meters^{2}, which corresponds to a parachute radius of 3.8 meters.
Drag coefficient Bicycle car .076 Velomobile Tesla Model 3 .21 2017 Toyota Prius .24 2016 Bullet .30 Typical car .33 Cars range from 1/4 to 1/2 Sphere .47 Typical truck .6 Formula1 car .9 The drag coeffient is high to give it downforce Bicycle + rider 1.0 Skier 1.0 Wire 1.2
Mach X15 6.7 Rocket Blackbird SR71 3.5 X2 Starbuster 3.2 MiG25 Foxbat 2.83 XB70 Valkyrie 3.0 MiG31 Foxhound 2.83 F15 Eagle 2.5 Aardvark F111 2.5 Bomber Sukhoi SU27 2.35 F22 Raptor 2.25 Fastest stealth aircraft
Fluid density = D Cross section = A Drag coef = C Drag force = F = ½ C A D V^{2} Drag power = P = ½ C A D V^{3} = K D V^{3} = F V Drag parameter = K = ½ C A Speed Density Drag force Drag power Drag (m/s) (kg/m^{3}) (kN) (kWatt) parameter Bike 10 1.22 .035 .305 .50 Bike 18 1.22 .103 1.78 .50 Bike, speed record 22.9 1.22 .160 3.66 .50 Bike, streamlined 38.7 1.22 .095 3.66 .104 Porche 911 94.4 1.22 7.00 661 1.29 LaFerrari 96.9 1.22 7.31 708 1.28 Lamborghini SV 97.2 1.22 5.75 559 1.00 Skydive, min speed 40 1.22 .75 30 .77 75 kg Skydive, max speed 124 1.22 .75 101 .087 75 kg Airbus A380, max 320 .28 1360 435200 94.9 F22 Raptor 740 .084 312 231000 6.8 SR71 Blackbird 1100 .038 302 332000 6.6 Sub, human power 4.1 1000 .434 1.78 .052 Blue Whale 13.9 1000 270 3750 2.8 150 tons, 25 Watts/kg Virginia nuclear sub 17.4 1000 1724 30000 11.4The drag coefficient is an assumption and the area is inferred from the drag coefficient.
For the skydiver, the minimum speed is for a maximum cross section (spread eagled) and the maximum speed is for a minimum cross section (dive).
Wiki: Energy efficiency in transportation
Airplanes fly at high altitude where the air is thin.
Altitude Air density (km) (kg/m^{3}) Sea level 0 1.22 Denver (1 mile) 1.6 .85 Mount Everest 9.0 .45 Airbus A380 13.1 .25 Commercial airplane cruising altitude F22 Raptor 19.8 .084 SR71 Blackbird 25.9 .038
m/s Mach Swim 2.39 Boat, human power 5.14 Aircraft, human power 12.3 Run 12.4 Boat, wind power 18.2 Bike 22.9 Car, solar power 24.7 Bike, streamlined 38.7 Land animal 33 Cheetah Bird, level flight 45 Whitethroated needletail Aircraft, electric 69 Helicopter 111 .33 Train, wheels 160 .54 Train, maglev 168 .57 Aircraft, propeller 242 .82 Rocket sled, manned 282 .96 Aircraft, manned 981 3.33 Rocket plane, manned 2016 6.83 Rocket sled 2868 9.7 Scramjet 5901 20Mach 1 = 295 m/s at high altitude.
Commercial airplanes fly at Mach .9 because the drag coefficient increases sharply at Mach 1.
The drag coefficient depends on speed.
Object length = L Velocity = V Fluid viscosity = Q (Pascal seconds) = 1.8⋅10^{5} for air = 1.0⋅10^{3} for water Reynolds number = R = V L / Q (A measure of the turbulent intensity)The drag coefficient of a sphere as a function of Reynolds number is:
Golf balls have dimples to generate turbulence in the airflow, which increases the Reynolds number and decrease the drag coefficient.
Reynolds Soccer Golf Baseball Tennis number 40000 .49 .48 .49 .6 45000 .50 .35 .50 50000 .50 .30 .50 60000 .50 .24 .50 90000 .50 .25 .50 110000 .50 .25 .32 240000 .49 .26 300000 .46 330000 .39 350000 .20 375000 .09 400000 .07 500000 .07 800000 .10 1000000 .12 .35 2000000 .15 4000000 .18 .30Data
If the cyclists are in single file then the lead rider has to use more power than the following riders. Cyclists take turns occupying the lead.
A "slingshot pass" is enabled by drafting. The trailing car drops back by a few lengths and then accelerates. The fact that he is in the leading car's slipstream means he has a higher top speed. As the trailing car approaches the lead car it moves the side and passes.
For an object experiencing drag,
Drag coefficient = C Velocity = V Fluid density = D Cross section = A Mass = M Drag number = Z = ½ C D A / M Drag acceleration = A = Z V^{2} Initial position = X_{0} = 0 Initial velocity = V_{0} Time = TThe drag differential equation and its solution are
A = Z V^{2} V = V_{0} / (V_{0} Z T + 1) X = ln(V_{0} Z T + 1) / Z
1672 Newton is the first to note the Magnus effect while observing tennis players at Cambridge College. 1742 Robins, a British mathematician and ballistics researcher, explains deviations in musket ball trajectories in terms of the Magnus effect. 1852 The German physicist Magnus describes the Magnus effect.For a spinning tennis ball,
Velocity = V = 55 m/s Swift groundstroke Radius = R = .067 m Area = Area = .0141 m^{2} Mass = M = .058 kg Spin number = S = W R / V = .25 Heavy topspin Spin rate = W = V / R = 205 Hz Air density = D_{air = 1.22 kg/m3 Ball density= Dball Drag coef = Cdrag = .5 For a sphere Spin coef = Cspin = 1 For a sphere and for S < .25 Drag force = Fdrag = ½ Cdrag Dair Area V2 = 13.0 Newtons Spin force = Fspin = ½ Cspin Dair Area V2 S = 6.5 Newtons Drag accel = Adrag = 224 m/s2 Spin accel = Aspin = 112 m/s2 Gravity = Fgrav = M g }For a rolling ball the spin number is S=1.
If the spin force equals the gravity force (F_{spin = Fgrav), }
V^{2} S C R^{1} D_{air}/D_{ball} = .0383
The drag force on an object moving through a fluid is
Velocity = V Fluid density = D = 1.22 kg/m^{2} (Air at sea level) Crosssectional area = A Drag coefficient = C Drag force = F = ½ C A D V^{2} Drag power = P = ½ C A D V^{3} = F V Drag parameter = K = C A"Terminal velocity" occurs when the drag force equals the gravitational force.
M g = ½ C D A V^{2}Suppose we want to estimate the parachute size required for a soft landing. Let a "soft landing" be the speed reached if you jump from a height of 2 meters, which is V_{t} = 6 m/s. If a skydiver has a mass of 100 kg then the area of the parachute required for this velocity is 46 meters^{2}, which corresponds to a parachute radius of 3.8 meters.
Drag coefficient Bicycle car .076 Velomobile Tesla Model 3 .21 2017 Toyota Prius .24 2016 Bullet .30 Typical car .33 Cars range from 1/4 to 1/2 Sphere .47 Typical truck .6 Formula1 car .9 The drag coeffient is high to give it downforce Bicycle + rider 1.0 Skier 1.0 Wire 1.2
1672 Newton is the first to note the Magnus effect while observing tennis players at Cambridge College. 1742 Robins, a British mathematician and ballistics researcher, explains deviations in musket ball trajectories in terms of the Magnus effect. 1852 The German physicist Magnus describes the Magnus effect.For a spinning tennis ball,
Velocity = V = 55 m/s Swift groundstroke Radius = R = .067 m Area = Area = .0141 m^{2} Mass = M = .058 kg Spin number = S = W R / V = .25 Heavy topspin Spin rate = W = V / R = 205 Hz Air density = D_{air = 1.22 kg/m3 Ball density= Dball Drag coef = Cdrag = .5 For a sphere Spin coef = Cspin = 1 For a sphere and for S < .25 Drag force = Fdrag = ½ Cdrag Dair Area V2 = 13.0 Newtons Spin force = Fspin = ½ Cspin Dair Area V2 S = 6.5 Newtons Drag accel = Adrag = 224 m/s2 Spin accel = Aspin = 112 m/s2 Gravity = Fgrav = M g }For a rolling ball the spin number is S=1.
If the spin force equals the gravity force (F_{spin = Fgrav), }
V^{2} S C R^{1} D_{air}/D_{ball} = .0383
Force of the wheel normal to ground = F_{normal} Rolling friction coefficient = C_{roll} Rolling friction force = F_{roll} = C_{roll} F_{normal }Typical car tires have a rolling drag coefficient of .01 and specialized tires can achieve lower values.
C_{roll} Railroad .00035 Steel wheels on steel rails Steel ball bearings on steel .00125 Racing bicycle tires .0025 8 bars of pressure Typical bicycle tires .004 18wheeler truck tires .005 Best car tires .0075 Typical car tires .01 Car tires on sand .3
Wheel diameter = D Wheel sinkage depth = Z Rolling coefficient = C_{roll} ≈ (Z/D)^{½}
For a typical car,
Car mass = M = 1200 kg Gravity constant = g = 9.8 m/s^{2} Tire rolling drag coeff = C_{r} =.0075 Rolling drag force = F_{r} = C_{r} M g = 88 Newtons Air drag coefficient = C_{a} = .25 Air density = D = 1.22 kg/meter^{3} Air drag crosssection = A = 2.0 m^{2} Car velocity = V = 17 m/s (City speed. 38 mph) Air drag force = F_{a} = ½C_{a}ADV^{2} = 88 Newtons Total drag force = F = F_{r} + F_{a} = 176 Newtons Drag speed = V_{d} = 17 m/s Speed for which air drag equals rolling drag Car electrical efficiency = Q = .80 Battery energy = E = 60 MJoules Work done from drag = EQ = F X = C_{r} M g [1 + (V/V_{d})^{2}] X Range = X = EQ/(C_{r}Mg)/[1+(V/V_{d})^{2}] = 272 kmThe range is determined by equating the work from drag with the energy delivered by the battery. E Q = F X.
The drag speed V_{d} is determined by setting F_{r} = F_{a}.
Drag speed = V_{d} = [C_{r} M g / (½ C_{a} D A)]^{½} = 4.01 [C_{r} M /(C_{a} A)]^{½} = 17.0 meters/second
m/s Mach Swim 2.39 Boat, human power 5.14 Aircraft, human power 12.3 Run 12.4 Boat, wind power 18.2 Bike 22.9 Car, solar power 24.7 Bike, streamlined 38.7 Land animal 33 Cheetah Bird, level flight 45 Whitethroated needletail Aircraft, electric 69 Helicopter 111 .33 Train, wheels 160 .54 Train, maglev 168 .57 Aircraft, propeller 242 .82 Rocket sled, manned 282 .96 Aircraft, manned 981 3.33 Rocket plane, manned 2016 6.83 Rocket sled 2868 9.7 Scramjet 5901 20Mach 1 = 295 m/s at high altitude.
Mach X15 6.7 Rocket Blackbird SR71 3.5 X2 Starbuster 3.2 MiG25 Foxbat 2.83 XB70 Valkyrie 3.0 MiG31 Foxhound 2.83 F15 Eagle 2.5 Aardvark F111 2.5 Bomber Sukhoi SU27 2.35 F22 Raptor 2.25 Fastest stealth aircraft
Speed Power Force Force Force Mass Drag Drag Area Drag Roll Year 100kph (total) (fluid) (roll) (data) (specs) coef coef time m/s kWatt kN kN kN ton m^{2} m^{2} m^{2} s eSkate 5.3 .11 .021 .013 .008 .08 .76 1.0 .01 eScooter Zoomair 7.2 .25 .035 .027 .008 .08 .85 1.0 .01 Bike 10 .30 .035 .030 .005 .10 .49 1.0 .005 Bike 18 1.78 .103 .098 .005 .10 .50 1.0 .005 eBike 250 Watt 8.9 .25 .028 .023 .005 .10 .48 1.0 .005 eBike 750 Watt 10.6 .75 .071 .066 .005 .10 .96 1.0 .005 eBike 1 kWatt 12.5 1.0 .080 .075 .005 .10 .79 1.0 .005 eBike 1.5 kWatt 15.3 1.5 .098 .093 .005 .10 .65 1.0 .005 eBike 3 kWatt 16.7 3.0 .180 .175 .005 .10 1.03 1.0 .005 eBike Stealth H 22.2 5.2 .234 .228 .006 .12 .76 1.0 .005 eBike Wolverine 29.2 7.0 .240 .234 .006 .12 .45 1.0 .005 Bike, record 22.9 3.66 .160 .155 .005 .10 .48 1.0 .005 Bike, steamline 38.7 3.66 .095 .090 .005 .10 .099 .11 .005 Loremo 27.8 45 1.62 1.58 .035 .47 3.39 .25 1.25 .20 .0075 2009 Mitsubishi MiEV 36.1 47 1.30 1.22 .081 1.08 1.53 .35 .0075 2011 Aptera 2 38.1 82 2.15 2.09 .062 .82 2.36 .19 1.27 .15 .0075 2011 Nissan Leaf SL 41.7 80 1.92 1.81 .114 1.52 1.71 .72 2.50 .29 .0075 2012 10.1 Volkswagen XL1 43.9* 55 1.25 1.19 .060 .80 1.01 .28 1.47 .19 .0075 2013 11.9 Chevrolet Volt 45.3 210 4.64 4.52 .121 1.61 3.61 .62 2.21 .28 .0075 2014 7.3 Saab 900 58.3 137 2.35 2.25 .100 1.34 1.09 .66 1.94 .34 .0075 1995 7.7 Tesla S P85 249+ 69.2* 568 8.21 8.06 .150 2.00 2.76 .58 2.40 .24 .0075 2012 3.0 BMW i8 69.4* 260 3.75 3.63 .116 1.54 1.24 .55 2.11 .26 .0075 2015 4.4 Nissan GTR 87.2 357 4.09 3.96 .130 1.74 .85 .56 2.09 .27 .0075 2008 3.4 Lamborghini Dia 90.3 362 4.01 3.89 .118 1.58 .78 .57 1.85 .31 .0075 1995 Porsche 918 94.4 661 7.00 6.88 .124 1.66 1.27 .29 .0075 LaFerrari 96.9 708 7.31 7.19 .119 1.58 1.26 .0075 Lamborghini SV 97.2 559 5.75 5.62 .130 1.73 .98 .0075 Bugatti Veyron 119.7 883 7.38 7.24 .142 1.89 .83 .74 .0075 2005 Hummer H2 242 .218 2.90 2.46 4.32 .57 .0075 2003 Formula 1 .053 .702 .9 .0075 2017 Bus (2 decks) 138 12.6 .005 2012 Subway (R160) 24.7 448 38.6 .0004 2006 Airbus A380 320 435200 1360 1360 21.8 F22 Raptor 740 231000 312 312 .93 Blackbird SR71 1100 332000 302 302 .41 Skydive, min 40 30 .75 .75 0 .075 .77 1.0 0 Skydive, max 124 101 .75 .75 0 .075 .080 1.0 0 Sub, human power 4.1 1.78 .434 .051 Blue Whale 13.9 3750 270 2.74 Sub, nuke 17.4 30000 1724 11.2 Virginia Class *: The top speed is electronically limited Drag (data) Drag parameter obtained from the power and top speed Data (specs) Drag parameter from Wikipedia Force (total) Total drag force = Fluid drag force + Roll drag force Force (fluid) Fluid drag force Force (roll) Roll drag force Area Cross section from Wikipedia Drag coef Drag coefficient from Wikipedia Roll coef Roll coefficient. Assume .0075 for cars and .005 for bikes. 100kph time Time to accelerate to 100 kphFor the skydiver, the minimum speed is for a maximum cross section (spread eagled) and the maximum speed is for a minimum cross section (dive).
Wiki: Energy efficiency in transportation
If the cyclists are in single file then the lead rider has to use more power than the following riders. Cyclists take turns occupying the lead.
A "slingshot pass" is enabled by drafting. The trailing car drops back by a few lengths and then accelerates. The fact that he is in the leading car's slipstream means he has a higher top speed. As the trailing car approaches the lead car it moves the side and passes.
0100kph 400m 400m Top Power Mass Top (s) (s) speed speed (kw) (kg) speed (kph) (kph) (m/s) Porche 918 2.2 9.8 233 340 661 1704 94.4 LaFerrari 2.4 9.7 240 349 708 1255 96.9 Bugatti Veyron 2.5 9.7 224 431 883 1888 119.7 Tesla S 2.6 10.9 198 249 568 2000 69.2 Lamborghini SV 2.6 10.4 218 350 559 1769 97.2 Porche 997 S 2.7 10.9 205 315 390 1570 87.5
If everything seems under control, you're just not going fast enough.  Mario Andretti
I will always be puzzled by the human predilection for piloting vehicles at unsafe velocities  Data
Car minimum mass = 702 kg Includes the driver and not the fuel Engine volume = 1.6 litres Turbocharged. 2 energy recovery systems allowed Energy recovery max power = 120 kWatts Energy recovery max energy = 2 Megajoules/lap Engine typical power = 670 kWatts = 900 horsepower Engine cylinders = 6 Engine max frequency = 15000 RPM Engine intake = 450 litres/second Fuel consumption = .75 litres/km Fuel maximum = 150 litres Forward gears = 8 Reverse gears = 1 Gear shift time = .05 seconds Lateral accelertion = 6 g's Formula1 1g downforce speed= 128 km/h Speed for which the downforce is 1 g Formula1 2g downforce speed= 190 km/h Speed for which the downforce is 2 g Indycar 1g downforce speed = 190 km/h Rear tire max width = 380 mm Front tire max width = 245 mm Tire life = 300 km Brake max temperature = 1000 Celsius Deceleration from 100 to 0 kph = 15 meters Deceleration from 200 to 0 kph = 65 meters (2.9 seconds) Time to 100 kph = 2.4 seconds Time to 200 kph = 4.4 seconds Time to 300 kph = 8.4 seconds Max forward acceleration = 1.45 g Max breaking acceleration = 6 g Max lateral acceleration = 6 g Drag at 250 kph = 1 g
1950 Formula1 begins. Safety precautions were nonexistent and death was considered an acceptable risk for winning races. 1958 Constructor's championship established 1958 First race won by a rearengine car. Within 2 years all cars had rear engines. 1966 Aerodynamic features are required to be immobile (no air brakes). 1977 First turbocharged car. 1978 The Lotus 79 is introduced, which used ground effect to accelerate air under the body of the car, generating downforce. It was also the first instance of computeraided design. It was unbeatable until the introduction of the Brabham Fancar. 1978 The Brabham "Fancar" is introduced, which used a fan to extract air from underneath the car and enhance downforce. It won the race decisively. The rules committee judged it legal for the rest of the season but the team diplomatically Wiki 1982 Active suspension introduced. 1983 Ground effect banned. The car underside must be flat. 1983 Cars with more than 4 wheels banned. 1989 Turbochargers banned. 1993 Continuously variable transmission banned before it ever appears. 1994 Electronic performanceenhancing technology banned, such as active suspension, traction control, launch control, antilock breaking, and 4wheel steering. (4wheel steering was never implemented) 1999 Flexible wings banned. 2001 Traction control allowed because it was unpoliceable. 2001 Beryllium alloys in chassis or engines banned. 2002 Team orders banned after Rubens Barrichello hands victory to Michael Schumacher at final corner of the Austrian Grand Prix. 2004 Automatic transmission banned. 2007 Tuned mass damper system banned. 2008 Traction control banned. All teams must use a standard electrontrol unit. 2009 Kinetic energy recovery systems allowed.
Place Points Place Points 1 25 6 8 2 18 7 6 3 15 8 4 4 12 9 2 5 10 10 1
Electric bikes are easy to make. All you have to do is replace a conventional wheel with an electric wheel and attach a battery pack. Electric wheels come in kits and you can make the battery pack yourself. Example configurations for various motor powers:
Power Max Range Motor Battery Battery speed cost cost energy kWatt mph miles $ $ MJoule .75 30 10 160 40 .5 1.5 35 20 240 60 1.2 3 45 40 570 100 1.8 6 55 80 1150 200 3.6The bikes have one electric wheel and one conventional wheel except for 6 kWatt bike, which has 2 electric wheels with 3 kWatt each.
Electric wheel prices are from Amazon.com.
Speed Power License mph kWatt required? Connecticut 30 1.5 Yes California 28 .75 No Massachusetts 25 .75 Yes Oregon 20 1.0 No Washington 20 1.0 No Pennsylvania 20 .75 No Delaware 20 .75 No Maryland 20 .5 No DC 20 ? No
Air density = D Velocity = V Wing area = A_{wing} Wing drag coefficient = C_{wing} Drag force on the wing = F_{drag} = ½ C_{Wing} A_{wing} D V^{2} C_{wing} F4 Phantom .021 (subsonic) Cessna 310 .027 Airbus A380 .027 Boeing 747 .031 F4 Phantom .044 (supersonic)
F_{lift} = Lift force (upward) F_{drag} = Drag force (rearward) Q_{lift} = Lifttodrag coefficient = F_{lift} / F_{drag} Q_{lift} U2 23 Highaltitude spy plane Albatross 20 Largest bird Gossamer 20 Gossamer albatross, humanpowered aircraft Hang glider 15 Tern 12 Herring Gull 10 Airbus A380 7.5 Concorde 7.1 Boeing 747 7 Cessna 150 7 Parachute 5 Sparrow 4 Wingsuit 2.5 Flying lemur ? Most capable gliding mammal. 2 kg max Flying squirrel 2.0
A glider is an airplane without an engine. The more efficient the glider, the smaller the glide angle. The minimum glide angle is determined by the wing lift/drag coefficient.
Wing lift/drag coefficient = Q_{lift} = F_{lift} / F_{drag} Glider horizontal velocity = V_{x} Glider vertical velocity = V_{z} Drag force = F_{drag} Gravitational force = F_{grav} Lift force = F_{lift} = F_{grav} Drag power = P_{drag} = F_{drag} V_{x} Power from gravit = P_{grav} = F_{grav} V_{z}If the glider descends at constant velocity,
P_{drag} = P_{grav}The goal of a glider is to maximize the glide ratio V_{x} / V_{z}.
V_{x} / V_{z} = (P_{drag} / F_{drag}) / (P_{grav} / F_{grav}) = F_{grav} / F_{drag} = Q_{lift}The glide ratio is equal to the lift coefficient Q_{lift}.
D = Air density A_{wing} = Wing area C_{wing} = Wing drag coefficient F_{drag} = Drag force on the wing = ½ C_{wing} D A_{wing} V^2 Q_{wing} = Wing lift coefficient = F_{lift} / F_{drag} F_{lift} = Lift force from the wing = Q_{wing} Fdrag M = Aircraft mass F_{eng} = Engine force F_{grav} = Gravity force = M g P_{drag} = Drag power = F_{drag} V = ½ C_{wing} D A_{wing} V^{3} V = Cruising speed Agility= Powertoweight ratio = P_{drag} / M = V g / Q (derived below)For flight at constant velocity,
F_{eng} = F_{drag} Horizontal force balance F_{lift} = F_{grav} Vertical force balance Agility = P_{drag} / M = F_{drag} V / M = F_{lift} V / M / Q = M g V / M / Q = V g / QWe can use this equation to solve for the minimum agility required to fly.
P_{drag} = M g V / Q = ½ C_{wing} D A_{wing} V^{3} Agility = g^{3/2} M^{½} Q^{3/2} (½ C D A)^{½}If we assume that mass scales as size cubed and wing area scales as size squared, then
A_{wing} ~ M^{2/3} Agility ~ g^{3/2} M^{1/6} Q^{3/2} C^{½} D^{½}
V_{cruise} V_{max} Mass Takeoff Ceiling Density Force Wing Len Wing Range m/s m/s ton ton km kg/m^{3} kN m^{2} m m km Cessna 150 42 56 .60 .73 4.3 .79 1.34 15 7.3 10.1 778 Boeing 747 254 274 178.1 377.8 11.0 .36 1128 525 70.6 64.4 14200 Boeing 7879 251 262 128.9 254.0 13.1 .26 640 360.5 62.8 60.1 14140 Airbus A380 243 262 276.8 575 13.1 .26 1360 845 72.2 79.8 15200 Concorde 599 605 78.7 190.5 18.3 .115 560 358.2 61.7 25.6 7223 F22 Raptor 544 740 19.7 38.0 19.8 .091 312 78.0 18.9 13.6 2960 U2 192 224 6.49 18.1 21.3 .071 84.5 92.9 19.2 31.4 10308 SR71 954 983 30.6 78.0 25.9 .034 302 170 32.7 16.9 5400Mach 1 = 298 m/s.
Altitude Density (km) (kg/m^{3}) Sea level 0 1.22 Cessna 150 3.0 .79 Boeing 747 11.0 .36 Airbus A380 13.1 .26 Concorde 18.3 .115 F22 Raptor 19.8 .091 U2 21.3 .071 SR71 Blackbird 25.9 .034
Cruise Max Ceiling Mass Cruise Motor Solar Cells Battery m/s m/s kW tons kw kW cells m^{2} tons kW Aquila 35.8 27.4 .40 5.0 .2 Solar Impulse 2 25.0 38.9 12 2.3 52 66 269.5 .633
The Loon balloon is 15 meters wide, 12 meters, tall, and .076 mm thick. The solar panels generate 100 Watts and the payload is 10 kg. It is too large to be selfpropelled and relies and buoyancy modulation and air currents to maneuver.
1961 Piggott accomplishes the first humanpowered flight, covering a distance of 650 meters. 1977 The "Gossomer Condor 2" flies 2172 meters in a figureeight and wins the Kremer Prize. It was built by Paul MacCready and piloted by amateur cyclist and hangglider pilot Bryan Allen. It cruised at 5.0 m/s with a power of 260 Watts. 1988 The MIT Daedalus 88 piloted by Kanellos Kanellopoulos flies from Crete to Santorini (115.11 km), setting the distance record, which still stands.Humanpowered helicopters can only reach a height of 3 meters and can only hover for 20 seconds.
Mass Power Agility (kg) (kW) (Watts/kg) Human 75 2500 33 BMW i8 1485 170 114 Cessna 150 600 75 125 Airbus A380 276000 49000 178 Formula1 car 642 619 964 SR71 30600 33000 1078 F22 Raptor 19700 33000 1675If you put a wing on a BMW i8, it would be able to go fast enough to take off.
X_{wing} = Length of the wing, from the fuselage to the tip Y_{wing} = Wing dimension in the direction of flight, measured along the point of attachment with the fuselage A_{wing} = Wing area R_{wing} = Wing aspect ratio = X_{wing} / Y_{wing} Q_{lift} = Wing liftdrag ratio Q_{Lift} R_{wing} Wing X_{wing} area (m^{2}) (m) U2 23 10.6 Highaltitude spy plane Albatros 20 1.7 Largest bird Gossamer 20 41.34 14.6 Gossamer albatross, humanpowered aircraft Airbus A380 7.5 7.5 845 36.3 Concorde 7.1 358.2 11.4 Boeing 747 7 7.9 525 29.3 Cessna 150 7 15 4.5 Wingsuit 2.5 1 2 1.0Q_{Lift} tends to be proportional to R_{wing}.
A wingtip creates a vortex as it moves. Birds fly in a "V" formation to use the updraft from their neighbor's wingtip vortex.
The minimum agility required to fly scales as
Agility ~ g^{3/2} M^{1/6} Q^{3/2} C_{wing}^{½} D^{½}We can normalize the Earth to 1 and estimate the minimum agility for other planets. For example,
MarsAgility / EarthAgility = (MarsGravity / EarthGravity)^{3/2} * (MarsDensity / EarthDensity)^{½} Gravity Atmosphere Agility Power/ Maximum density normalized mass mass for (m/s^2) (kg/m^3) to Earth (Watts/kg) flight (kg) Earth 9.78 1.22 1.0 400 20 Mars 3.8 .020 1.89 756 .44 Titan 1.35 5.3 .025 10 >1000000 Venus 8.87 67 .12 48 >1000000 Pandora 7.8 1.46 .65 261 265For the "Power/mass" column we assume that the power required for human flight is 400 Watts and estimate the power required for flight on other planets.
On Titan you can fly with a wingsuit. A creature as massive as a whale can fly.
"Pandora" is the fictional moon from the film "Avatar".
The largest flying birds on the Earth have a mass of 20 kg. We can use the agility scaling to estimate the maximum mass for flight on other planets.
Agility ~ g^{3/2} M^{1/6} Q^{3/2} C_{wing}^{½} D^{½} M ~ g^{9} D^{3}
The wing on a Formula1 car is an upsidedown aircraft wing that generates downforce, to help with friction.
M = Mass V = Velocity A = Acceleration (in any direction) C_{fri}= Friction coefficient C_{↓} = Wing coefficient for downforce F_{grav}= Gravitational force on the car = M g F_{↓} = Downforce from the wing = M g C_{↓} V^{2} F_{fri}= Maximum friction force = C_{fri} (F_{grav} + F_{↓}) = C_{fri} M g (1 + C_{↓} V^{2})A formula1 car generates 1 g of downforce at 50 m/s, hence C_{↓} = 1/50^{2}. At the top speed of 100 m/s the downforce is 4 g. The maximum accelerations incurred by the driver are of order 5 g.
The maximum cornering speed for a circle of radius R is:
F_{fri} = M V^{2}/R = M g C_{fri} (1 + V^{2}/C_{fri}^{2}) V^{2} = g R C_{fri} / (1  R/C_{fri}^{2})
The angle of attack is the angle of the plane's noze with respect to level fight. As the angle of attack increases the lift increases, with an accompanying increase in drag. If the angle of attack is too high then lift drops and the plane stalls.
Airplanes fly at high altitude where the air is thin.
Altitude Air density (km) (kg/m^{3}) Sea level 0 1.22 Denver (1 mile) 1.6 .85 Mount Everest 9.0 .45 Airbus A380 13.1 .25 Commercial airplane cruising altitude F22 Raptor 19.8 .084 SR71 Blackbird 25.9 .038
Speed Mass Takeoff Ceiling Thrust Range Cost Number Year Stealth Mach ton ton km kN km M$ SR71 Blackbird 3.3 30.6 78.0 25.9 302 5400 32 1966 MiG25 Foxbat 2.83 20.0 36.7 20.7 200.2 1730 1186 1970 MiG31 Foxhound 2.83 21.8 46.2 20.6 304 1450 519 1981 F22A Raptor 2.51 19.7 38.0 19.8 312 2960 150 195 2005 * F15 Eagle 2.5 12.7 30.8 20.0 211.4 4000 28 192 1976 F14 Tomcat 2.34 19.8 33.7 15.2 268 2960 712 1974 MiG29 Fulcrum 2.25 11.0 20.0 18.0 162.8 1430 29 1600 1982 Su35 2.25 18.4 34.5 18.0 284 3600 40 48 1988 F4 Phantom II 2.23 13.8 28.0 18.3 1500 5195 1958 Chengdu J10 2.2 9.8 19.3 18.0 130 1850 28 400 2005 F16 Falcon 2.0 8.6 19.2 15.2 127 1200 15 957 1978 Chengdu J7 2.0 5.3 9.1 17.5 64.7 850 2400 1966 Dassault Rafale 1.8 10.3 24.5 15.2 151.2 3700 79 152 2001 Euro Typhoon 1.75 11.0 23.5 19.8 180 2900 90 478 2003 F35A Lightning 1.61 13.2 31.8 15.2 191 2220 85 77 2006 * B52 .99 83.2 220 15.0 608 14080 84 744 1952 B2 Bomber .95 71.7 170.6 15.2 308 11100 740 21 1997 * A10C Warthog .83 11.3 23.0 13.7 80.6 1200 19 291 1972 Drone RQ180 ~15 18.3 ~2200 2015 * Drone X47B .95 6.4 20.2 12.2 3890 2 2011 * Carrier Drone Avenger .70 8.3 15.2 17.8 2900 12 3 2009 * Drone RQ4 .60 6.8 14.6 18.3 34 22800 131 42 1998 Drone Reaper .34 2.2 4.8 15.2 5.0 1852 17 163 2007 Drone RQ170 15 20 2007 * India HAL AMCA 2.5 14.0 36.0 18.0 250 2800 ? 0 2023 * India HAL FGFA 2.3 18.0 35.0 20.0 352 3500 ? 0 >2020 * Mitsubishi F3 2.25 9.7 ? ? 98.1 3200 ? 1 2024 * Chengdu J20 2.0 19.4 36.3 ? 359.8 ? 110 4 2018 * Sukhoi PAK FA 2.0 18.0 35.0 20.0 334 3500 50 6 2018 * Shenyang J31 1.8 17.6 25.0 ? 200 4000 ? 0 2018 * Mach 1 = 295 m/s5th generation fighters: F22, F35, X2, HAL AMCA, J20, J31, Sukhoi PAK FA
An aircraft moving at Mach 2 and turning with a radius of 1.2 km has a g force of 7 g's.
Mach Range Missile Warhead Year Engine km kg kg Russia R37 6 400 600 60 1989 Solid rocket Japan AAM4 5 100 224 ? 1999 Ramjet India Astra 4.5+ 110 154 15 2010 Solid rocket EU Meteor 4+ 200 185 ? 2012 Ramjet Russia R77PD 4 200 175 22.5 1994 Ramjet USA AIM120D 4 180 152 18 2008 Solid rocket Israel DerbyIR 4 100 118 23 Solid rocket Israel Rafael 4 50 118 23 1990 Solid rocket France MICA 4 50 112 12 1996 Solid rocket Israel Python 5 4 20 105 11 Solid rocket Russia K100 3.3 400 748 50 2010 Solid rocket UK ASRAAM 3+ 50 88 10 1998 Solid rocket Germany IRIST 3 25 87.4 2005 Solid rocket USA AIM9X 2.5+ 35 86 9 2003 Solid rocket USA Hellfire 1.3 8 49 9 1984 Solid rocket AGM114
Mach Range Missile Warhead Year Engine Stages Anti km kg kg missile USA SM3 15.2 2500 1500 0 2009 Solid rocket 4 * Israel Arrow 9 150 1300 150 2000 Solid rocket 2 USA THAAD 8.24 200 900 0 2008 Solid rocket * USA David 7.5 300 2016 Solid rocket * Russia S400 6.8 400 1835 180 2007 Solid rocket * India Prithvi 5 2000 5600 2006 Solid, liquid 2 * India AAD Ashwin 4.5 200 1200 0 2007 Solid rocket 1 Taiwan Sky Bow 2 4.5 150 1135 90 1998 Solid rocket China HQ9 4.2 200 1300 180 1997 Solid rocket 2 USA Patriot 3 4.1 35 700 90 2000 Solid rocket * China KS1 4.1 50 900 100 2006 Solid rocket * USA RIM174 3.5 460 1500 64 2013 Solid rocket 2 India Barak 8 2 100 275 60 1015 Solid rocket 2 Japan ChuSAM 570 73 2003 Solid rocket Korea KMSAM 40 400 2015 Solid rocket
Mach Range Missile Warhead Year Engine Launch km kg kg platform USA Tomahawk .84 2500 1600 450 1983 Turbofan Ground USA AGM129 .75 3700 1300 130 1990 Turbofan B52 Bomber USA AGM86 .73 2400 1430 1361 1980 Turbofan B52 Bomber
Speed Mass Payload Range Year mach tons tons km USA SR72 6 Future. Successor to the SR71 Blackbird USA HSSW 6 900 Future. High Speed Strike Weaspon USA HTV2 20 5500 17000 2 Test flights USA X41 8 450 Future USA X51 5.1 1.8 740 2013 Tested. 21 km altitude. Will become the HSSW Russia Object 4202 10 Tested India HSTDV 12 Future China Wu14 10 2014 7 tests. also called the DZZFThe SR72 has two engines: a ramjet for below Mach 3 and a ramjet/scramjet for above Mach 3. The engines share an intake and thrust nozzle.
Payload Paylod Range Mass Launch Year (tons) (Mtons) (km) (tons) USA Titan 2 9 15000 154 Silo 1962 Inactive USA Minuteman 3 .9 13000 35.3 Silo 1970 USA Trident 2 .95 11300 58.5 Sub 1987 USA Titan 3.75 10200 151.1 Silo 1959 Inactive USA Peacekeeper 3 9600 96.8 Silo 1983 Inactive Russia RS24 1.2 12000 49 Road 2007 Russia Voevoda 8.7 8 11000 211.4 Silo 1986 Russia Layner 11000 40 Sub 2011 Russia RS28 Sarmat 10 10000 >100 Silo 2020 Liquid rocket Russia Bulava .9 10000 36.8 Sub 2005 France M51.1 1 10000 52 Sub 2006 China DF5B 8 15000 183 Silo 2015 China DF5A 4 15000 183 Silo 1983 China JL2 6 12000 42 Sub 2001 China DF5 5 12000 183 Silo 1971 China DF31A 3 12000 42 Road China DF31 1 8000 42 Road 1999 China DF4 3.3 7000 82 Silo 1974 India Surya 15 16000 70 Road 2022 India AgniVI 10 12000 70 Road 2017 India AgniV 6 8000 50 Road 2012 India K4 2.5 3500 17 Sub 2016 Solid. Arihant nuclear sub India K15 ~6.5 750 1.0 Sub 2010 Solid. 2 stages. Arihant nuclear sub Israel Jericho 3 .75 11500 30 Road 2008 N. Kor. Taepodong2 6000 79.2 Pad 2006 Pakis. Shaheen 3 2750 Road 2015 Solid. 2 stages. Pakis. Shaheen 2 2000 25 Road 2014 Solid. 2 stages. Pakis. Ghauri 2 1.2 1800 17.8 Road Pakis. Ghauri 1 .7 1500 15.8 Road 2003 Liquid. 1 stage. Iran Shabab 3 1.0 1930 2003Payload in "tons" represents the mass of the payload.
Rotor radius = R Air density = D = 1.22 kg/meter^{3} Rotor tip speed = V Rotor lift force = F_{l} = D W R^{2} V^{2} Rotor drag force = F_{d} Rotor lift param = W = F_{l} D^{1} R^{2} V^{2} Rotor lift/drag = Q = F_{l} / F_{d} Rotor power = P = F_{d} V = F V / Q Rotor quality = q = Q W^{½} D^{½} = F_{l}^{3/2} P^{1} R^{1} Rotor force/power= Z = F_{l}/ P = Q / V = D^{½} W^{½} Q R F^{½} = q R F_{l}^{½}The physical parameters of a propeller are {R,Q,W,q}, with typical values of
Q = 5.5 W = .045 q = 1.29Most propellers have 2 blades and some have 3. If there are 4 or more blades then q declines.
A measurement of F_{l} and V determines W.
A measurement of P, F_{l}, and V determines Q.
A measurement of F_{l}, P, and r determines q.
Q and W are not independent. They are related to the blade aspect ratio.
Q ≈ Aspect ratio W ≈ Q^{½} q ≈ Q^{½}
A commonlyappearing quantity is the power/mass ratio, which is inversely proportional to the force/power ratio.
Mass = M Gravity = g Hover force = F = M g Hover power = P Force/Power ratio = Z = F/P Power/Mass ratio = p = P/M = g/Z
The larger the propeller radius the better, because the force/power ratio is proportional to radius. Also, increasing the radius decreases the tip speed, which is helpful for nice because noise scales as the tip speed to the fifth power. The only limit to propeller radius is mass. If the radius is too large then the mass is too large.
The smaller the number of propeller, the larger the propeller radius. One propeller is optimal but singlepropeller aircraft are difficult to control, and there is no failsafe if the rotor fails.
We specify a design using 2 large propellers (for power) and 2 small propellers (for stability and failsafe). The large propellers have a radius of 1.5 meters and the small propellers have a radius of 1.0 meters.
The large propellers are mounted forward and aft and the foward propeller tilts forward for horizontal flight. The small propellers are to the right and left.
We assume the total vehicle mass is 400 kg and we use the properties of propellers to calculate the power required to hover. We use a peak power that is comfortably larger than the hover power. The minimum battery mass is 1/4 the total vehicle mass.
Total aircraft mass = M = 400 kg (Includes passenger) Number of rotors = N = 2 Rotor radius = R = 1.5 meters Gravity constant = g = 9.8 meters/second^{2} Rotor force = F = Mg/N =1960 Newtons Rotor quality = q = 1.02 Air density = D = 1.22 kg/meter^{3} Rotor power = P_{r}=(qDR)^{1}F^{3/2}= 46.2 kWatts Hover power = P_{h}= N P_{r} = 92.4 kWatts Peak power = P = 150 kWatts Battery power/mass = p =1600 Watts/kg Battery energy/mass = e = .8 MJoules/kg Battery mass = m = P/p = 94 kg Battery energy = E = e m = 75 MJoules Hover time = T = E/P_{h} = 812 seconds = 14 minutesThe properties of propellers are discussed in the propeller section. The rotor tip speed is
Rotor lift/drag = Q = 5.5 Rotor tip speed = V = P Q / F = 130 m/sThe maximum horizontal speed is around 1/3 of the rotor tip speed. If we assume a horizontal speed of 40 meters/second then the range is 32 km.
For the large propellers,
Propeller radius = R = 1.5 meters Propeller mass parameter = C = 5 kg/meter^{3} Propeller mass = M = C R^{3} = 17 kgFor the motors on the large propellers,
Motor power = 60 kWatts Motor power/mass = 8 kWatts/kg Motor mass =7.5 kgThe masses of the components in kg is
2 large 1.5 meter rotors 32 2 small 1.0 meter rotors 10 2 motors fo rthe large rotors 16 2 motors for the small rotors 10 Battery 100 Cabin 50 Fuselage 50 Pilot 80 Cargo 20 Total 380
The flight time of a drone is determined by:
*) The battery energy/mass.
*) The power/mass required to hover.
*) The ratio of the battery mass to the drone mass.
Typical parameters for a drone are:
Drone mass = M = 1.0 kg Battery mass = M_{bat} = .5 kg (The battery is the most vital component) Battery energy = E = .38 MJoules Battery energy/mass= e_{bat}= E/M_{bat}= .75 MJoules/kg (Upper range for lithium batteries) Drone energy/mass = e = E/M = .38 MJoules/kg Drone power/mass = p = P/M = 60 Watts/kg (Practical minimum to hover. Independent of mass) Drone power = P = p M = 60 Watts (Power required to hover) Flight time = T = E/P = 6250 seconds = 104 minutesThe flight time in terms of component parameters is
T = (e_{bat}/p) * (M_{bat}/M)
One has to choose a wise balance for the masses of the motor, battery, fuselage, and payload. The properties of the electrical components are:
Energy/Mass Power/mass Energy/$ Power/$ $/Mass MJoule/kg kWatt/kg MJoule/$ kWatt/$ $/kg Electric motor  10.0  .062 160 Lithiumion battery .75 1.5 .009 .0142 106 Lithium supercapacitor .008 8 .0010 .09 90 Aluminum capacitor .0011 100If the battery and motor have equal power then the battery has a larger mass than the motor.
Mass of motor = M_{mot} Mass of battery = M_{bat} Power = P (Same for both the motor and the battery) Power/mass of motor = p_{mot} = P/M_{mot} = 8.0 kWatt/kg Power/mass of battery = p_{bat} = P/M_{bat} = 1.5 kWatt/kg Battery mass / Motor mass= R =M_{bat}/M_{mot} = p_{mot}/p_{bat} = 5.3The "sports prowess" of a drone is the drone power divided by the minimum hover power. To fly, this number must be larger than 1.
Drone mass = M_{dro} Motor mass = M_{mot} Motor power/mass = p_{mot} = 8000 Watts/kg Hover minimum power/mass = p_{hov} = 60 Watts/kg Drone power = P_{dro} = p_{mot} M_{mot} Hover minimum power = P_{hov} = p_{hov} M_{dro} Sports prowess = S = P_{dro}/P_{hov} = (p_{mot}/p_{hov}) * (M_{mot}/M_{dro}) = 80 M_{mot}/M_{dro}If S=1 then M_{mot}/M_{dro} = 1/80 and the motor constitutes a negligible fraction of the drone mass. One can afford to increase the motor mass to make a sports drone with S >> 1.
If the motor and battery generate equal power then the sports prowess is
S = (p_{bat}/p_{hov}) * (M_{bat}/M_{dro}) = 25 M_{bat}/M_{dro}If M_{bat}/M_{dro} = ½ then S=12.5, well above the minimum required to hover.
Suppose a drone has a mass of 1 kg. A squash racquet can have a mass of as little as .12 kg. The fuselage mass can be much less than this because a drone doesn't need to be as tough as a squash racquet, hence the fuselage mass is negligible compared to the drone mass. An example configuration is:
kg Battery .5 Motors .1 To match the battery and motor power, set motor mass / battery mass = 1/5 Rotors <.05 Fuselage .1 Camera .3 Drone total 1.0Supercapacitors can generate a larger power/mass than batteries and are useful for extreme bursts of power, however their energy density is low compared to batteries and so the burst is short. If the supercapacitor and battery have equal power then
Battery power/mass = p_{bat} = 1.5 kWatts/kg Supercapacitor power/mass = p_{sup} = 8.0 kWatts/kg Battery power = P Battery mass = M_{bat} = P / p_{bat} Supercapacitor mass = M_{sup} = P / p_{sup} Supercapacitor/Battery mass= R =M_{sup}/ M_{bat} = p_{bat}/p_{sup} = .19The supercapacitor is substantially ligher than the battery. By adding a lightweight supercapacitor you can double the power. Since drones already have abundant power, the added mass of the supercapacitor usually makes this not worth it.
If a battery and an aluminum capacitor have equal powers,
Aluminum capacitor mass / Battery mass = .015If a battery or supercapacitor is operating at full power then the time required to expend all the energy is
Mass = M Energy = E Power = P Energy/Mass = e = E/M Power/Mass = p = P/M Discharge time= T = E/P = e/p Energy/Mass Power/Mass Discharge time Mass MJoule/kg kWatt/kg seconds kg Lithium battery .75 1.5 500 1.0 Supercapacitor .008 8.0 1.0 .19 Aluminum capacitor .0011 100 .011 .015"Mass" is the mass required to provide equal power as a lithium battery of equal mass.
Max Mass Max Bombs Max Engine Range # Year speed mass alt Built kph ton ton ton km kWatt km UK Avro Lancaster 454 16.6 32.7 10.0 6.5 4x 954 4073 7377 1942 USA B29 Superfortress 574 33.8 60.6 9.0 9.7 4x1640 5230 3970 1944 Germany Heinkel He 177 565 16.8 32.0 7.2 8.0 2x2133 1540 1169 1942 UK Short Stirling 454 21.3 31.8 6.4 5.0 4x1025 3750 2371 1939 UK Handley Page Halifax 454 17.7 24.7 5.9 7.3 4x1205 3000 6176 1940 Germany FokkeWulf Condor 360 17.0 24.5 5.4 6.0 4x 895 3560 276 1937 Soviet Tupolev Tu2 528 7.6 11.8 3.8 9.0 2x1380 2020 2257 1942 USA B17 Flying Fortress 462 16.4 29.7 3.6 10.5 4x 895 3219 12731 1938 Japan Mitsubishi Ki67 537 8.6 13.8 1.6 9.5 2x1417 3800 767 1942 Soviet Petlyakov Pe2 580 5.9 8.9 1.6 8.8 2x 903 1160 11427 1941 Japan Yokosuka P1Y Ginga 547 7.3 13.5 1.0 9.4 2x1361 5370 1102 1944 Japan Mitsubishi G4M 428 6.7 12.9 1.0 8.5 2x1141 2852 2435 1941 Curtis LeMay: Flying fighters is fun. Flying bombers is important.
Max Climb Mass Max Bombs Max Engine Range # Year speed mass alt Built kph m/s ton ton ton km kWatt km USA P51 Black Widow 589 12.9 10.6 16.2 2.9 10.6 2x1680 982 706 1944 USA A20 Havoc 546 10.2 6.8 12.3 .9 7.2 2x1200 1690 7478 1941 USA F7F Tigercat 740 23 7.4 11.7 .9 12.3 2x1566 1900 364 1944 USA P38 Lightning 667 24.1 5.8 9.8 2.3 13.0 2x1193 10037 1941 UK Fairey Firefly 509 8.8 4.4 6.4 .9 8.5 1x1290 2090 1702 1943 UK Mosquito 668 14.5 6.5 11.0 1.8 11.0 2x1103 2400 7781 1941 UK Beaufighter 515 8.2 7.1 11.5 .3 5.8 2x1200 2816 5928 1940 UK Fairie Fulmar 438 3.2 4.6 .1 8.3 1x 970 1255 600 1940 UK Defiant 489 9.0 2.8 3.9 0 9.2 1x 768 749 1064 1939 Japan Dragon Slayer 540 11.7 4.0 5.5 0 10.0 2x 783 1701 1941 Ki45 Japan Flying Dragon 537 6.9 8.6 13.8 1.6 9.5 2x1417 3800 767 1942 Ki67 Japan J1N Moonlight 507 8.7 4.5 8.2 0 2x 840 2545 479 1942 Ger. Hornet 624 9.3 6.2 10.8 1.0 10.0 2x1287 2300 1189 1943 Ger. Flying Pencil 557 3.5 9.1 16.7 4.0 7.4 2x1287 2145 1925 1941 Do217 Ger. Heinkel He219 616 13.6 0 9.3 2x1324 1540 300 1943 Ger. Junkers Ju88 360 11.1 12.7 0 5.5 2x1044 1580 15183 1939 Ger. Me110 595 12.5 7.8 0 11.0 2x1085 900 6170 1937 SU Petlyakov Pe3 530 12.5 5.9 8.0 .7 9.1 2x 820 1500 360 1941 UK Gloster Meteor 965 35.6 4.8 7.1 .9 13.1 Jet 965 3947 1944 Ger. Me262 Swallow 900 ~25 3.8 7.1 1.0 11.5 Jet 1050 1430 1944 Ger. Heinkel He162 840 23.4 1.7 2.8 0 12.0 Jet 975 320 1945 Me262 Swallow jet = 2x 8.8 kNewtons Heinkel He162 jet = 1x 7.8 kNewtons Gloster Meteor jet = 2x16.0 kNewtons
Max Climb Mass Max Bombs Max Engine Range # Year speed mass alt Built kph m/s ton ton ton km kWatt km USA P39 Airacobra 626 19.3 3.0 3.8 .2 10.7 1x 894 840 9588 1941 USA P63 Kingcobra 660 12.7 3.1 4.9 .7 13.1 1x1340 725 3303 1943 USA F2A Buffalo 517 12.4 2.1 3.2 0 10.1 1x 890 1553 509 1939 USA P40 Warhawk 580 11.0 2.8 4.0 .9 8.8 1x 858 1100 13738 1939 USA P51 Mustang 703 16.3 3.5 5.5 .5 12.8 1x1111 2755 >15000 1942 USA F4F Wildcat 515 11.2 2.7 4.0 0 10.4 1x 900 1337 7885 1940 USA F6F Hellcat 629 17.8 4.2 7.0 1.8 11.4 1x1491 1520 12275 1943 USA F8F Bearcat 730 23.2 3.2 6.1 .5 12.4 1x1678 1778 1265 1945 USA P43 Lancer 573 13.0 2.7 3.8 0 11.0 1x 895 1046 272 1941 USA P47 Thunderbolt 713 16.2 4.5 7.9 1.1 13.1 1x1938 1290 15677 1942 USA F4U Corsair 717 22.1 4.2 5.6 1.8 12.6 1x1775 1617 12571 1942 Japan Zero 534 15.7 1.7 2.8 .3 10.0 1x 700 3104 10939 1940 Japan N1K Strong Wind 658 20.3 2.7 4.9 .5 10.8 1x1380 1716 1532 1943 Japan Ki84 "Gale" 686 18.3 2.7 4.2 .7 11.8 1x1522 2168 3514 1943 Japan Ki61 580 15.2 2.6 3.5 .5 11.6 1x 864 580 3078 1942 Japan Ki100 580 13.9 2.5 3.5 0 11.0 1x1120 2200 396 1945 Japan A5M 440 1.2 1.8 0 9.8 1x 585 1200 1094 1936 Japan A6M2 436 12.4 1.9 2.9 .1 10.0 1x 709 1782 327 1942 Japan J2M Thunderbolt 655 23.4 2.8 3.2 .1 11.4 1x1379 560 671 1942 Japan Ki27 470 15.3 1.1 1.8 .1 12.2 1x 485 627 3368 1937 Japan Ki43 530 1.9 2.9 .5 11.2 1x 858 1760 5919 1941 Japan Ki44 605 19.5 2.1 3.0 0 11.2 1x1133 1225 1942 UK Hawker Hurricane 547 14.1 2.6 4.0 .5 11.0 1x 883 965 14583 1943 UK Hawker Tempest 700 23.9 4.2 6.2 .9 11.1 1x1625 1190 1702 1944 UK Hawker Typhoon 663 13.6 4.0 6.0 .9 10.7 1x1685 821 3317 1941 UK Submarine Seafire 578 13.4 2.8 3.5 9.8 1x1182 825 2334 1942 UK Submarine Spitfire 595 13.2 2.3 3.0 0 11.1 1x1096 756 20351 1938 Ger. Fw190 685 17.0 3.5 4.8 .5 12.0 1x1287 835 >20000 1941 Ger. Bf109 640 17.0 2.2 3.4 .3 12.0 1x1085 850 34826 1937 SU MiG3 640 13.0 2.7 3.4 .2 12.0 1x 993 820 3172 1941 SU Yak1 592 15.4 2.4 2.9 0 10.0 1x 880 700 8700 1940 SU Yak3 655 18.5 2.1 2.7 0 10.7 1x 970 650 4848 1944 SU Yak7 571 12.0 2.4 2.9 0 9.5 1x 780 643 6399 1942 SU Yak9 672 16.7 2.5 3.2 0 10.6 1x1120 675 16769 1942 SU LaGG3 575 14.9 2.2 3.2 .2 9.7 1x 924 1000 6528 1941 SU La5 648 16.7 2.6 3.4 .2 11.0 1x1385 765 9920 1942 SU La7 661 15.7 3.3 .2 10.4 1x1230 665 5753 1944 SU Polykarpov I16 525 14.7 1.5 2.1 .5 14.7 1x 820 700 8644 1934
Class Speed Power Length Displace Planes # Year kph MWatt m kton built USA Essex 60.6 110 263 47 100 24 1942 USA Independence 58 75 190 11 33 9 1942 Japan Shokaku 63.9 120 257.5 32.1 72 2 1941 Japan Hiyo 47.2 42 219.3 24.2 53 3 1944 Japan Unryu 63 113 227.4 17.8 65 3 1944 Japan Chitose 53.5 42.4 192.5 15.5 30 2 1944 Japan Zuiho 52 39 205.5 11.4 30 2 1940
MJoules Speed Density C H N O /kg (km/s) (g/cm^{3}) Bombardier beetle .4 Hydroquinone + H_{2}O_{2} + protein catalyst Ammonium nitrate 2.0 2.55 1.12 0 4 2 3 Black powder 2.6 .6 1.65 Used before 1884 Smokeless powder 5.2 6.4 1.4 6 9 1 7 Used after 1884. Nitrocellulose TNT 4.7 6.9 1.65 7 5 3 6 Trinitrotoluene PETN 5.8 8.35 1.77 5 8 4 12 Dynamite 5.9 7.2 1.48 3 5 3 9 75% Nitroglycerine + stabilizer Composition 4 6.3 8.04 1.59 3 6 6 6 91% RDX. "Plastic explosive" PLX 6.5 1.14 1 3 1 2 95% CH_{3}NO_{2} + 5% C_{2}H_{4}(NH_{2})_{2} Nitroglycerine 7.2 8.1 1.59 3 5 3 9 Unstable RDX (Hexagen) 7.5 8.7 1.78 3 6 6 6 HMX (Octogen) 8.0 9.1 1.86 4 8 8 8 Dinitrodiazeno. 9.2 10.0 1.98 4 0 8 8 Octanitrocubane 11.2 10.6 1.95 8 0 8 16 Gasoline + Oxygen 10.4 8 18 0 13 Hydrogen + Oxygen 13.16 0 2 0 1 Uranium bomb 219000 Hydrogen bomb 10 mil Antimatter 90000 mil Speed: Detonation speed C: Carbon atoms H: Hydrogen atoms N: Nitrogen atoms O: Oxygen atoms
~808 Qing Xuzi publishes a formula resembling gunpower, consisting of 6 parts sulfur, 6 parts saltpeter, and 1 part birthwort herb (for carbon). ~850 Incendiary property of gunpower discovered 1132 "Fire lances" used in the siege of De'an, China 1220 alRammah of Syria publishes "Military Horsemanship and Ingenious War Devices", describes the purification of potassium nitrate by adding potassium carbonate with boiling water, to precipitate out magnesium carbonate and calcium carbonate. 1241 Mongols use firearms at the Battle of Mohi, Hungary 1338 Battle of Arnemuiden. First naval battle involving cannons. 1346 Cannons used in the Siege of Calais and the Battle of Crecy 1540 Biringuccio publishes "De la pirotechnia", giving recipes for gunpowder 1610 First flintlock rifle 1661 Boyle publishes "The Sceptical Chymist", a treatise on the distinction between chemistry and alchemy. It contains some of the earliest modern ideas of atoms, molecules, and chemical reaction, and marks the beginning of the history of modern chemistry. 1669 Phosphorus discovered 1774 Lavoisier appointed to develop the French gunpowder program. By 1788 French gunpowder was the best in the world. 1832 Braconnot synthesizes the first nitrocellulose (guncotton) 1846 Nitrocellulose published 1847 Sobrero discovers nitroglycerine 1862 LeConte publishes simple recipes for producing potassium nitrate. 1865 Abel develops a safe synthesis of nitrocellulose 1867 Nobel develops dynamite, the first explosive more powerful than black powder It uses diatomaceous earth to stabilize nitroglycerine 1884 Vieille invents smokeless gunpowder (nitrocellulose), which is 3 times more powerful than black powder and less of a nuisance on the battlefield. 1902 TNT first used in the military. TNT is much safer than dynamite 1930 RDX appears in military applications 1942 Napalm developed 1949 Discovery that HMX can be synthesized from RDX 1956 C4 explosive developed (based on RDX) 1999 Eaton and Zhang synthesize octanitrocubane and heptanitrocubane
Above 550 Celsius, potassium nitrate decomposes. 2 KNO_{3} ↔ 2 KNO_{2} + O_{2}.
Black powder = .75 KNO_{3} + .19 Carbon + .06 Sulfur 1 kg TNT equivalent = 4.184 MJ Fission bomb = 9.20e13 J = 22000 tons of TNT equivalent Fission bomb = 420 kg Fission bomb = 2.19e11 J/kg Fusion bomb maximum = 2.51e13 J/kg (Maximum theoretical efficiency) Fusion bomb practical = 1.0 e13 J/kg (Practical efficiency achieved in real bombs)
Potassium nitrate KNO_{3} 75% (Saltpeter) Charcoal C_{7}H_{4}O 15% Sulfur S 10% Oversimplified equation: 2 KNO_{3} + 3 C + S → K_{2}S + N_{2} + 3 CO_{2} Realistic equation: 6 KNO_{3} + C_{7}H_{4}O + 2 S → KCO_{3} + K_{2}SO_{4} + K_{2}S + 4 CO_{2} + 2 CO + 2 H_{2}O + 3 N_{2}Nitrite (NO_{3}) is the oxidizer and sulfur lowers the ignition temperature.
MJoules /kg Black powder 2.6 Smokeless powder 5.2 HMX (Octogen) 8.0 Gasoline + Oxygen 10.4 Hydrogen + Oxygen 13.16 Uranium bomb 219000 Hydrogen bomb 10 mil Antimatter 90000 mil Mass Energy Energy/Mass kg MJ MJ/kg MOAB 9800 46000 4.7 8500 kg of fuel
Form Ignition Density (Celsius) White 30 1.83 Red 240 1.88 Violet 300 2.36 Black 2.69Red phosphorus is formed by heating white phosphorus to 250 Celsius or by exposing it to sunlight. Violet phosphorus is formed by heating red phosphorus to 550 Celsius. Black phosphorus is formed by heating white phosphorus at a pressure of 12000 atmospheres. Black phosphorus is least reactive form and it is stable below 550 Celsius.
The safety match was invented in 1844 by Pasch. The match head cannot ignite by itself. Ignitition is achieved by striking it on a rough surface that contains red phosphorus. When the match is struck, potassium chlorate in the match head mixes with red phosphorus in the abrasive to produce a mixture that is easily ignited by friction. Antimony trisulfide is added to increase the burn rate.
Match head Fraction Striking surface Fraction Potassium chlorate KClO3 .50 Red phosphorus .5 Silicon filler Si .4 Abrasive .25 Sulfur S small Binder .16 Antimony3 trisulfide Sb2S3 small Neutralizer .05 Neutralizer small Carbon .04 Glue smallA "strike anywhere" match has phosphorus in the match head in the form of phosphorus sesquisulfide (P_{4}S_{3}) and doesn't need red phosphorus in the striking surface. P_{4}S_{3} has an ignition temperature of 100 Celsius.
Before the invention of iron, fires were started by striking flint (quartz) with pyrite to generate sparks. Flintlock rifles work by striking flint with iron. With the discovery of cerium, ferrocerium replaced iron and modern butane lighters use ferrocerium, which is still referred to as "flint".
Cerium .38 Ignition temperature of 165 Celsius Lanthanum .22 Iron .19 Neodymium2 .04 Praseodymium .04 Magnesium .04
Nitrous oxide is stored as a cryogenic liquid and injected along with gaoline into the combustion chamber. Upon heating to 300 Celsius the nitrous oxide decomposes into nitrogen and oxygen gas and releases energy. The oxygen fraction in this gas is higher than that in air (1/3 vs. .21) and the higher faction allows for more fuel to be consumed per cylinder firing.
Air density = .00122 g/cm^{3} Nitrous oxide gas density = .00198 g/cm^{3} Diesel density = .832 g/cm^{3} Gasoline density = .745 g/cm^{3} Diesel energy/mass = 43.1 MJoules/kg Gasoline energy/mass = 43.2 MJoules/kg Nitrous oxide boiling point = 88.5 Celsius Air oxygen fraction = .21 Nitrous oxide oxygen fraction= .33 Nitrous oxide decompose temp = 300 Celsius Nitrous oxide liquid pressure= 52.4 Bars Pressure required to liquefy N2O at room temperature
Hydroquinone and peroxide are stored in 2 separate compartments are pumped into the reaction chamber where they explode with the help of protein catalysts. The explosion vaporizes 1/5 of the liquid and expels the rest as a boiling drop of water, and the pquinone in the liquid damages the foe's eyes. The energy of expulsion pumps new material into the reaction chamber and the process repeats at a rate of 500 pulses per second and a total of 70 pulses. The beetle has enough ammunition for 20 barrages.
2 H_{2}O_{2} → 2 H_{2}O + O_{2} (with protein catalyst) C_{6}H_{4}(OH)_{2} → C_{6}H_{4}O_{2} + H_{2} (with protein catalyst) O_{2} + 2 H_{2} → 2 H_{2}O Firing rate = 500 pulses/second Number of pulses in one barrage = 70 Firing time = .14 seconds Number of barrages = 20
A turbojet engine compresses air before burning it to increase the flame speed and make it burn explosively. A ramjet engine moving supersonically doesn't need a turbine to achieve compression.
Airbus A350 compression ratio = 52 Air density at sea level = 1 bar Air density at 15 km altitude = .25 bar Air density in A350 engine = 13 barFrom the thermal flame theory of Mallard and Le Chatelier,
Temperature of burnt material = T_{b} Temperature of unburnt material = T_{u} Temperature of ignition = T_{i} Fuel density = D_{fuel} Oxygen density = D_{oxygen} Reaction coefficient = C Reaction rate = R = C D_{fuel} D_{oxygen} Thermal diffusivity = Q = 1.9⋅10^{5} m^{2}/s Flame speed = V V^{2 = Q C Dfuel Doxygen (Tb  Ti) / (Ti  Tu) }
If the pressure front moves supersonically then the front forms a discontinuous shock, where the pressure makes a sudden jump as the shock passes.
Metal powder is often included with explosives.
Energy/mass Energy/mass not including including oxygen oxygen (MJoule/kg) (MJoule/kg) Hydrogen 113.4 12.7 Gasoline 46.0 10.2 Beryllium 64.3 23.2 Aluminum 29.3 15.5 Magnesium 24.5 14.8 Carbon 12.0 3.3 Lithium 6.9 3.2 Iron 6.6 4.6 Copper 2.0 1.6
An oxygen candle is a mixture of sodium chlorate and iron powder, which when ignited smolders at 600 Celsius and produces oxygen at a rate of 6.5 manhours of oxygen per kilogram of mixture. Thermal decomposition releases the oxygen and the burning iron provides the heat. The products of the reaction are NaCl and iron oxide.
2000 System of hours, minutes, and seconds developed in Sumer 300 Water clock developed in Ancient Greece 100 Zhang Heng constructs a seismometer using pendulums that was capable of detecting the direction of the Earthquake. 1300 First mechanical clock deveoped. 1400 Springbased clocks developed. 1500 Pendulums are used for power, for machines such as saws, bellows, and pumps. 1582 Galileo finds that the period of a pendulum is independent of mass and oscillation angle, if the angle is small. 1636 Mersenne and Descartes find that the pendulum was not quite isochronous. Its period increased somewhat with its amplitude. 1656 Huygens builds the first pendulum clock, delivering a precision of 15 seconds per day. Previous devices had a precision of 15 minutes per day. Fron this point on pendulum clocks were the most accurate timekeeping devices until the development of the quartz oscillator was developed in 1921. 1657 Balance spring developed by Hooke and Huygens, making possible portable pocketwatches. 1658 Huygens publishes the result that pendulum rods expand when heated. This was the principal error in pendulum clocks. 1670 Previous to 1670 the verge escapement was used, which requires a large angle. The anchor escapement mechanism is developed in 1670, which allows for a smaller angle. This increased the precision because the oscillation period is independent of angle for small angles. 1673 Huygens publishes a treatise on pendulums. 1714 The British Parliament establishes the "Longitude Prize" for anyne who could find an accurate method for determing longitude at sea. At the time there was no clock that could measure time on a moving ship accurately enough to determine longitude. 1721 Methods are developed for compensating for thermal expansion error of a pendulum. 1726 Gridiron pendulum developed, improving precision to 1 second per day. 1772 Harrison builds a clock which James Cook used in his exploration of the Pacific. Cook's log is full of praise for the watch and the charts of the Pacific Ocean were remarkably accurate. 1772 Harrison gives one of his clocks to King George III, who personally tested it and found it to be accurate to 1/3 of one second per day. King George III advised Harrison to petition Parliament for the full Longitude Prize after threatening to appear in person to dress them down. 1851 Foucault shows that a pendulum can be used to measure the rotation period of the Earth. The penulum swings in a fixed frame and the Earth rotates with respect to this frame. In the Earth frame the pendulum appears to precess. 1921 Quartz electronic oscillator developed 1927 First quartz clocks developed, which were more precise than pendulum clocks. L = Length of the pendulum g = Gravity constant = 9.8 meters/second^{2} T = Period of the pendulum Z = Angle of maximum amplitude, in radians.If the angle Z is small (Z << 1) then the period of oscillation is
T = 2 Pi SquareRoot(L/g)As the angle increases the period of oscillation increases.
For a leg with a center of mass that is .5 meters below the hip joint the pendulum period is 1.4 seconds, similar to your walking candence. The pendulum frequency of your arms is slightly shorter.
Suppose an object starts at (X,Y) = (0,R) and moves with a speed V around the circle
X^{2} + Y^{2} = R^{2}Approximating the motion near (X,Y) = (0,R)
Y = (R^{2}  X^{2})^{1/2} = R (1  X^{2}/R^{2})^{1/2} ~ R (1  .5 X^{2}/R^{2} ~ R + .5 X^{2}/R X = V T Y = R + .5 V^{2} T^{2} / RThis has the form
Y = R + .5 A T^{2}where
A = V^{2}/RThis is the acceleration of an object moving with constant velocity around a circle.
Suppose a particle moves around a circle.
R = Radius of the circle $\omega $ = Angular frequency X = X position = R cos($\omega $ T) V_{x} = X velocity = R $\omega $ sin($\omega $ T) A_{x} = X acceleration = R $\omega $^{2} cos($\omega $ T) Y = Y position = R sin($\omega $ T) V_{y} = Y velocity = R $\omega $ cos($\omega $ T) A_{y} = Y acceleration = R $\omega $^{2} sin($\omega $ T) V^{2} = V_{x}^{2} + V_{x}^{2} = R^{2} K_{t}^{2} > V = R $\omega $ A^{2} = A_{x}^{2} + A_{y}^{2} = R^{2} K_{t}^{2} > A = R $\omega 2$
The pressure in a gas arises from kinetic energy of gas molecules.
Number of gas molecules = N Mass of a gas molecule = M Volume of the gas = Vol Number of gas molecules per volume = n = N / Vol Thermal speed of gas molecules = V_{th} Mean kinetic energy per gas molecule = E = .5 M V_{th}^{2} (Definition) Kinetic energy per volume = e = E / Vol Boltzmann constant = k = 1.38e23 Joules/Kelvin Density = D = N M / VolThe characteristic thermal speed of a gas molecule is defined in terms of the mean energy per molecule.
E = .5 M V_{th}^{2}The ideal gas law can be written in the following forms:
P = 2/3 e = 8.3 Mol T / Vol = k T N / Vol = 1/3 N M V_{th}^{2}/ Vol = 1/3 D V_{th}^{2} = k T D / M
For a system in thermodynamic equilibrium each degree of freedom has a mean energy of .5 k T. This is the definition of temperature.
A gas molecule moving in 3 dimensions has 3 degrees of freedom and so the mean kinetic energy is
E = 1.5 k T = .5 M V^{2}
Melt Boil Solid Liquid Gas Mass (K) (K) density density density (AMU) g/cm^3 g/cm^3 g/cm^3 O2 54 90 1.14 .00143 32.0 N2 63 77 .81 .00125 28.0 H2O 273 373 .917 1.00 .00080 18.0 CO2 n/s 195 1.56 n/a .00198 44.0 H2 14 20 .070 .000090 2.0 CH4 91 112 .42 .00070 16.0 CH5OH 159 352 .79 .00152 34.0 AlcoholGas density is for 20 Celsius and 1 Bar.
Carbon dioxide doesn't have a liquid state at standard temperature and pressure. It sublimes directly from a solid to a vapor.
M = Mass of a gas molecule V_{th}= Thermal speed H = Characteristic height of an atmosphere g = Gravitational accelerationSuppose a molecule at the surface of the Earth is moving upward with speed V and suppose it doesn't collide with other air molecules. It will reach a height of
M H g = .5 M V_{th}^{2}This height H is the characteristic height of an atmosphere.
The density of the atmosphere scales as
Density ~ Density At Sea Level * exp(E/E0)where E is the gravitational potential energy of a gas molecule and E is the characteristic thermal energy given by
E = M H g = 1/2 M V_{th}^{2}Expressed in terms of altitude h,
Density ~ Density At Sea Level exp(h/H)For oxygen,
E = 1.5 k TE is the same for all molecules regardless of mass, and H depends on the molecule's mass. H scales as
H ~ M^{1}
We first derive the law for a 1D gas and then extend it to 3D.
Suppose a gas molecule bounces back and forth between two walls separated by a distance L.
M = Mass of molecule V = Speed of the molecule L = Space between the wallsWith each collision, the momentum change = 2 M V
The average force on a wall is
Force = Change in momentum / Time between collisions = M V^2 / LSuppose a gas molecule is in a cube of volume L^3 and a molecule bounces back and forth between two opposite walls (never touching the other four walls). The pressure on these walls is
Pressure = Force / Area = M V^2 / L^3 = M V^2 / Volume Pressure Volume = M V^2This is the ideal gas law in one dimension. For a molecule moving in 3D,
Velocity^2 = (Velocity in X direction)^2 + (Velocity in Y direction)^2 + (Velocity in Z direction)^2Characteristic thermal speed in 3D = 3 * Characteristic thermal speed in 1D.
To produce the 3D ideal gas law, replace V^2 with 1/3 V^2 in the 1D equation. Pressure Volume = 1/3 M V^2 Where V is the characteristic thermal speed of the gasThis is the pressure for a gas with one molecule. If there are n molecules,
Pressure Volume = n 1/3 M V^2 Ideal gas law in 3DIf a gas consists of molecules with a mix of speeds, the thermal speed is defined as
Kinetic dnergy density of gas molecules = E = (n / Volume) 1/2 M V^2Using this, the ideal gas law can be written as
Pressure = 2/3 E = 1/3 Density V^2 = 8.3 Moles Temperature / VolumeThe last form comes from the law of thermodynamics: M V^2 = 3 B T
The "Balloons and Buoyancy" simulation at phet.colorado shows a gas with a mix of light and heavy molecules.
S = Escape speed T = Temperature B = Boltzmann constant = 1.38e23 Joules/Kelvin g = Planet gravity at the surface M = Mass of heavy molecule m = Mass of light molecule V = Thermal speed of heavy molecule v = Thermal speed of light molecule E = Mean energy of heavy molecule e = Mean energy of light molecule H = Characteristic height of heavy molecule h = Characteristic height of light molecule = E / (M g) = e / (m g) Z = Energy of heavy molecule / escape energy z = Energy of light molecule / escape energy = .5 M V^2 / .5 M S^2 = .5 m v^2 / .5 m S^2 = V^2 / S^2 = v^2 / S^2 For an ideal gas, all molecules have the same mean kinetic energy. E = e = 1.5 B T .5 M V^2 = .5 m v^2 = 1.5 B TThe light molecules tend to move faster than the heavy ones. This is why your voice increases in pitch when you breathe helium. Breathing a heavy gas such as Xenon makes you sound like Darth Vader.
For an object to have an atmosphere, the thermal energy must be much less than the escape energy.
V^2 << S^2 <> Z << 1 Escape Atmos Temp H2 N2 Z Z speed density (K) km/s km/s (H2) (N2) km/s (kg/m^3) Jupiter 59.5 112 1.18 .45 .00039 .000056 Saturn 35.5 84 1.02 .39 .00083 .00012 Neptune 23.5 55 .83 .31 .0012 .00018 Uranus 21.3 53 .81 .31 .0014 .00021 Earth 11.2 1.2 287 1.89 .71 .028 .0041 Venus 10.4 67 735 3.02 1.14 .084 .012 Mars 5.03 .020 210 1.61 .61 .103 .015 Titan 2.64 5.3 94 1.08 .41 .167 .024 Europa 2.02 0 102 1.12 .42 .31 .044 Moon 2.38 0 390 2.20 .83 .85 .12 Ceres .51 0 168 1.44 .55 8.0 1.14Even if an object has enough gravity to capture an atmosphere it can still lose it to the solar wind. Also, the upper atmosphere tends to be hotter than at the surface, increasing the loss rate.
Titan is the smallest object with a dense atmosphere, suggesting that the threshold for capturing an atmosphere is on the order of Z = 1/25, or
Thermal Speed < 1/5 Escape speed
When an object collapses by gravity, its temperature increases such that
Thermal speed of molecules ~ Escape speedIn the gas simulation at phet.colorado.edu, you can move the wall and watch the gas change temperature.
For an ideal gas,
3 * Boltzmann_Constant * Temperature ~ MassOfMolecules * Escape_Speed^2For the sun, what is the temperature of a proton moving at the escape speed? This sets the scale of the temperature of the core of the sun. The minimum temperature for hydrogen fusion is 4 million Kelvin.
The Earth's core is composed chiefly of iron. What is the temperature of an iron atom moving at the Earth's escape speed?
Escape speed (km/s) Core composition Sun 618. Protons, electrons, helium Earth 11.2 Iron Mars 5.03 Iron Moon 2.38 Iron Ceres .51 Iron
A typical globular cluster consists of millions of stars. If you measure the total gravitational and kinetic energy of the stars, you will find that
Total gravitational energy = 2 * Total kinetic energyjust like for a single satellite on a circular orbit.
Suppose a system consists of a set of objects interacting by a potential. If the system has reached a longterm equilibrium then the above statement about energies is true, no matter how chaotic the orbits of the objects. This is the "Virial theorem". It also applies if additional forces are involved. For example, the protons in the sun interact by both gravity and collisions and the virial theorem holds.
Gravitational energy of the sun = 2 * Kinetic energy of protons in the sun
Suppose an object undergoes periodic motion.
M = Mass of the object S = Amplitude of the oscillation t = Time T = Period of the oscillation F = Frequency of the oscillation S = Position of the object as a function of time = S sin(2 Pi t/T) V = Peak velocity of the object = 2 Pi S / T A = Peak acceleration of the object = 4 Pi^{2} S / T^{2} Force= Peak force during the oscillation = M A = 4 Pi^{2} M S / T^{2}The peak force and the period are related by
Force = 4 Pi^{2} M S / T^{2}The brain is good at measuring frequencies. Whenever possible, convert force measurements into frequency measurements.
If you shake a sword like a pendulum then we can translate force to torque and mass to moment of inertia.
R = Distance of the rotating object from the axis of rotation. I = Moment of inertia = M R^{2} Torque = 4 Pi^{2} I S / (R T^{2})
For a spinning top, the forces of gravity and the ground on the top generate a torque, which causes the spin to precess.
Precession of the spin axis from an external torque.
The gyroscope was invented by Serson in 1743 and it was used by Foucault in
1852 to measure the Earth's spin.
The Hubble telescope uses gyroscopes to orient itself. The gyroscopes periodically fail, requiring a servicing mission.
In the 1860s, the advent of electric motors made it possible for a gyroscope to spin indefinitely; this led to the first prototype heading indicators and the gyrocompass. The first functional gyrocompass was patented in 1904 by German inventor Hermann AnschutzKaempfe. The American Elmer Sperry followed with his own design later that year, and other nations soon realized the military importance of the invention  in an age in which naval prowess was the most significant measure of military power  and created their own gyroscope industries. The Sperry Gyroscope Company quickly expanded to provide aircraft and naval stabilizers as well.
Omega = Rotation speed in radians/second V = Velocity of the object A = Coriolis acceleration = 2 Omega V
For a central force, we can isolate the radial component of the motion. Using conservation of angular momentum we can write the radial force in terms of angular momentum and then convert it into an effective potential for centripetal acceleration.
For a satellite orbiting a central potential,
Gravity constant = G Mass of central object= M Mass of satellite = m Radius = R Distance of satellite from the center Tangential velocity = V Velocity transverse to the radius vector Angular momentum = L = m R V Centripetal accel = A_{C} = V^{2} R^{1} = L^{2} m^{2} R^{3} = ∂_{R} Φ_{C} Centripetal potential = Φ_{C} = .5 L^{2} m^{2} R^{2} Gravity acceleration = A_{G} = G M R^{2} = ∂_{R} Φ_{G} Gravity potential = Φ_{G} = G M R^{1} Total potential = Φ = Φ_{G} + Φ_{C} =  G M R^{1} + L^{2} m^{2} R^{2}
Suppose you rotate an airplane 90 degrees upward in the pitch direction and then roll it 90 degrees in the rightward direction, and then take note of its final position. Now start over and do the rotations in reverse order. The final orientation depends on order.
A rotation can be expressed as a matrix and matrix multiplication is noncommutative. In the following, "!=" stands for "not necessarily equal to".
A = Rotation matrix (3x3 matrix) Qo = Original orientation of an object (3D vector) Qf = Final orientation of an object after rotating using matrix "A". (3D vector) I = Identity matrix (diagonal elements equal to 1, offdiagonal elements equal to 0) Qf = A * Qo (Using "A" to rotate from "Qo" to "Qf") Qo = I * Qo (The identity matrix does not rotate an object) Ai = Inverse matrix of "A" B = 3x3 rotation matrix that is not necessarily the same as "A". Bi = Inverse matrix of "B" A * Ai = Ai * A = I B * Bi = Bi * B = I B * A (This stands for rotating by "A" and then rotating by "B") A * B (This stands for rotating by "B" and then rotating by "A") B * A != A * B (If there are 2 rotations then order matters) Ai * Bi * B * A = I (Doing 2 rotations & then unwinding them in order restores the original orientation) B * Ai * B * A != I (Doing 2 rotations & then unwinding them out of order does not necessarily restore the original orientation)
Commercial airplanes fly at Mach .9 because the drag coefficient increases sharply at Mach 1.
The drag coefficient depends on speed.
Object length = L Velocity = V Fluid viscosity = Q (Pascal seconds) = 1.8⋅10^{5} for air = 1.0⋅10^{3} for water Reynolds number = R = V L / Q (A measure of the turbulent intensity)The drag coefficient of a sphere as a function of Reynolds number is:
Golf balls have dimples to generate turbulence in the airflow, which increases the Reynolds number and decrease the drag coefficient.
Reynolds Soccer Golf Baseball Tennis number 40000 .49 .48 .49 .6 45000 .50 .35 .50 50000 .50 .30 .50 60000 .50 .24 .50 90000 .50 .25 .50 110000 .50 .25 .32 240000 .49 .26 300000 .46 330000 .39 350000 .20 375000 .09 400000 .07 500000 .07 800000 .10 1000000 .12 .35 2000000 .15 4000000 .18 .30Data
For an object experiencing drag,
Drag coefficient = C Velocity = V Fluid density = D Cross section = A Mass = M Drag number = Z = ½ C D A / M Drag acceleration = A = Z V^{2} Initial position = X_{0} = 0 Initial velocity = V_{0} Time = TThe drag differential equation and its solution are
A = Z V^{2} V = V_{0} / (V_{0} Z T + 1) X = ln(V_{0} Z T + 1) / Z
Electric quantities  Thermal quantities  Q = Charge Coulomb  Etherm= Thermal energy Joule I = Current Amperes  Itherm= Thermal current Watts E = Electric field Volts/meter  Etherm= Thermal field Kelvins/meter C = Electric conductivity Amperes/Volt/meter  Ctherm= Thermal conductivity Watts/meter/Kelvin A = Area meter^2  A = Area meter^2 Z = Distance meter  Z = Distance meter^2 J = Current flux Amperes/meter^2  Jtherm= Thermal flux Watts/meter^2 = I / A  = Ittherm / A = C * E  = Ctherm * Etherm V = Voltage Volts  Temp = Temperature difference Kelvin = E Z  = Etherm Z = I R  = Itherm Rtherm R = Resistance Volts/Ampere = Ohms  Rtherm= Thermal resistance Kelvins/Watt = Z / (A C)  = Z / (A Ct) H = Current heating Watts/meter^3  = E J  P = Current heating power Watts  = E J Z A  = V I 
L = Length of wire meters A = Cross section of wire meters^2 _______________________________________________________________________________________________________  Electric quantities  Thermal quantities  Q = Charge Coulomb  Etherm= Thermal energy Joule I = Current Amperes  Itherm= Thermal current Watts E = Electric field Volts/meter  Etherm= Thermal field Kelvins/meter C = Electric conductivity Amperes/Volt/meter  Ctherm= Thermal conductivity Watts/meter/Kelvin A = Area meter^2  A = Area meter^2 Z = Distance meter  Z = Distance meter^2 J = Current flux Amperes/meter^2  Jtherm= Thermal flux Watts/meter^2 = I / A  = Ittherm / A = C * E  = Ctherm * Etherm V = Voltage Volts  Temp = Temperature difference Kelvin = E Z  = Etherm Z = I R  = Itherm Rtherm R = Resistance Volts/Ampere = Ohms  Rtherm= Thermal resistance Kelvins/Watt = Z / (A C)  = Z / (A Ct) H = Current heating Watts/meter^3  = E J  P = Current heating power Watts  = E J Z A  = V I 
Continuum quantity Macroscopic quantity E <> V C <> R = L / (A C) J = C E <> I = V / R H = E J <> P = V I
Critical Critical Type temperature field (Kelvin) (Teslas) MagnesiumBoron2 39 55 2 MRI machines Niobium3Germanium 23.2 37 2 Field for thin films. Not widely used MagnesiumBoron2C 34 36 Doped with 5% carbon Niobium3Tin 18.3 30 2 Highperformance magnets. Brittle Vanadium3Gallium 14.2 19 2 NiobiumTitanium 10 15 2 Cheaper than Niobium3Tin. Ductile Niobium3Aluminum Technetium 11.2 2 Niobium 9.26 .82 2 Vanadium 5.03 1 2 Tantalum 4.48 .09 1 Lead 7.19 .08 1 Lanthanum 6.3 1 Mercury 4.15 .04 1 Tungsten 4 1 Not BCS Tin 3.72 .03 1 Indium 3.4 .028 Rhenium 2.4 .03 1 Thallium 2.4 .018 Thallium 2.39 .02 1 Aluminum 1.2 .01 1 Gallium 1.1 Gadolinium 1.1 Protactinium 1.4 Thorium 1.4 Thallium 2.4 Molybdenum .92 Zinc .85 .0054 Osmium .7 Zirconium .55 Cadmium .52 .0028 Ruthenium .5 Titanium .4 .0056 Iridium .1 Lutetium .1 Hafnium .1 Uranium .2 Beryllium .026 Tungsten .015 HgBa2Ca2Cu3O8 134 2 HgBa2Ca Cu2O6 128 2 YBa2Cu3O7 92 2 C60Cs2Rb 33 2 C60Rb 28 2 2 C60K3 19.8 .013 2 C6Ca 11.5 .95 2 Not BCS Diamond:B 11.4 4 2 Diamond doped with boron In2O3 3.3 3 2The critical fields for NiobiumTitanium, Niobium3Tin, and Vanadium3Gallium are for 4.2 Kelvin.
All superconductors are described by the BCS theory unless stated otherwise.
Boiling point (Kelvin) Water 273 Ammonia 248 Freon R12 243 Freon R22 231 Propane 230 Acetylene 189 Ethane 185 Xenon 165.1 Krypton 119.7 Oxygen 90.2 Argon 87.3 Nitrogen 77.4 Threshold for cheap superconductivity Neon 27.1 Hydrogen 20.3 Cheap MRI machines Helium4 4.23 Highperformance magnets Helium3 3.19The record for Niobium3Tin is 2643 Amps/mm^2 at 12 T and 4.2 K.
Titan has a temperature of 94 Kelvin, allowing for superconducting equipment. The temperature of Mars is too high at 210 Kelvin.
The maximum current density decreases with temperature and magentic field.
Maximum current density in kAmps/mm^{2} for 4.2 Kelvin (liquid helium):
Teslas 16 12 8 4 2 Niobium3Tin 1.05 3 Niobium3Aluminum .6 1.7 NiobiumTitanium  1.0 2.4 3 MagnesiumBoron2C .06 .6 2.5 4 MagnesiumBoron2 .007 .1 1.5 3Maximum current density in Amps/mm^{2} for 20 Kelvin (liquid hydrogen):
Teslas 4 2 MagnesiumBoron2C .4 1.5 MagnesiumBoron2 .12 1.5
1898 Dewar liquefies hydrogen (20 Kelvin) using regenerative cooling and his invention, the vacuum flask, which is now known as a "Dewar". 1908 Helium liquified by Onnes. His device reached a temperature of 1.5 K 1911 Superconductivity discovered by Onnes. Mercury was the first superconductor found 1935 Type 2 superconductivity discovered by Shubnikov 1953 Vanadium3Silicon found to be superconducting, the first example of a superconducting alloy with a 3:1 chemical ratio. More were soon found 1954 Niobium3Tin superconductivity discovered 1955 Yntema builds the first superconducting magnet using niobium wire, reaching a field of .7 T at 4.2 K 1961 Niobium3Tin found to be able to support a high current density and magnetic field (Berlincourt & Hake). This was the first material capable of producing a highfield superconducting magnet and paved the way for MRIs. 1962 NiobiumTitanium found to be able to support a high current density and magnetic field. (Berlincourt & Hake) 1965 Superconducting material found that could support a large current density (1000 Amps/mm^2 at 8.8 Tesla) (Kunzler, Buehler, Hsu, and Wernick) 1986 Superconductor with a high critical temperature discovered in a ceramic (35 K) (Lanthanum Barium Copper Oxide) (Bednorz & Muller). More ceramics are soon found to be superconducting at even higher temperatures. 1987 Nobel prize awarded to Bednorz & Muller, one year after the discovery of hightemperature superconductivity. Nobel prizes are rarely this fast.