People are sometimes mystified at why skydivers get up to a
Terminal Velocity (usually of around 120 mph) and then
do not accelerate to a higher velocity. After all, gravity
is still working!|
This presentation will show what happens, and even how to calculate such things!
When a skydiver first jumps out of an airplane, he/she initially accelerates at g, the acceleration due to gravity, just as we might expect. However, there is actually another force in play, the air resistance effect. This air resistance creates a Force which is actually upward, because it has to oppose the motion, and in this case, the direction of that motion and acceleration is downward.
After a little while, the skydiver stops accelerating, and from then on falls at relatively constant velocity. This occurs when the velocity has gotten high enough that the air resistance effect exactly cancels out the effect of the free-fall due to gravity.
The SPEED at which that happens can be changed! If a person is positioned to be relatively flat, there is enough frontal area for that air resistance force to act on, that there is a Terminal Velocity of around 120 mph (for a human). But if that same person changes to a vertical position, where there is very little frontal area for the air resistance to work against, then the person can fall at MUCH higher speed, apparently over 200 mph and maybe much higher than that!
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Technical StuffIn mathematical terms, we might say that the downward velocity at any instant is described by v(t), where t is the time. We can simplify things by saying that v(0) was 0, that is, the starting downward velocity was zero, just before jumping. v(t) will therefore increase downward, due to gravity, and we could consider that the positive axis direction. This does not say WHAT that function v(t) is, just that the downward velocity is somehow dependent on time. However, we DO specify that v(t) must be greater to or equal to zero, for a real situation where the person could not ever wind up having an upward velocity, at any moment. We can similarly say that there is a v'(t) which represents the RATE of the velocity changing, the apostrophe symbol indicating the mathematical DIFFERENTIAL of the quantity v(t). Differential simply refers to the quantity v(t) changing, and that the DIFFERENCE over some period of time is the RATE of the velocity changing. In a second, or a millionth of a second however much change of velocity occurred is therefore the differential during that interval of time. We have another term for this, and we call it acceleration. That just means the change of velocity, in some specific interval of time.
We are describing in mathematical terms what Physics usually calls Newton's Laws of Motion. IF a force is applied to a mass (weight), then it accelerates, per Newton's F = m * a. In this case, the force is that of gravity, and so F is m * g, or g = a, which is why we call g the acceleration due to gravity. The other term we have been considering, the air resistance, can be considered as am Aerodynamic Drag or Frontal Drag like on a vehicle. In the case of vehicles, we know that the Aerodynamic Drag is very close to being proportional to the air velocity (of impact) times the total amount of air impacted each second (which also happens to have as a factor the speed).
If we say that the air resistance force is PROPORTIONAL to the (relative) velocity (at that instant), we can then say that:
v'(t) = g - k * v(t)
where k is some constant number.
This is also sometimes written with the time dependence assumed:
v' = g - k * v
In either form, this is called a Differential Equation.
If we had no air resistance, this would say that v'(t) = g, the acceleration due to gravity. This is just describing the long accepted "acceleration due to gravity" as being g, which is around 32 ft/sec/sec or 9.8 meters/sec/sec.
The second term in our Differential Equation simply says that the (opposite) effect of the air resistance is due to a constant factor k times whatever the air velocity happens to be at that instant.
We can now solve this equation, which means to mathematically Integrate the equation, which essentially means to add up the effects of the accelerations for every fraction of a second to see what the total result would be.
In Calculus, there are many formulas for helping to solve various forms of Differential Equations. There are also a number of rules which apply for certain types of Differential Equations. We will not get into the Calculus here, but we will mention that this particular Differential Equation is of a type which is called Separable, and it is therefore very easy to solve.
SOLVING any Differential Equation means "mathematically Integrating" the equation. This really means adding up the effect of all those seconds or millionths of seconds mentioned above. For a simple example, which is involved below, the Integral of "t" (the time) is just the total of adding up all those individual millionths if a second, which then totals the whole time interval being considered, in other words, t. When you actually calculate an Integral, you specify a starting value and an ending value, in this case the time of starting and the time of ending. Therefore, the result of that Integral of the time is simply the TOTAL of what happens in the WHOLE interval of time specified.
(You can find examples of how this Integration process works in either our presentation on Radiometric Time Dating (or Carbon-14) or the one on an Introduction to Calculus.)
Our solution is:
- 1 / k * ln(g - k * v) = t + C1
g - k * v = C * e-k * t
Where e is the base of the logarithm system.
We know that v = 0 when t = 0, so we know that the constant C is equal to g.
We can solve this for v:
v = (g / k) * (1 - e-k * t
We therefore know what the velocity is at any time!
We could do another mathematical process of finding the LIMIT as t approaches infinity. The final term gets smaller and smaller as t gets larger, and so the right hand of the equation becomes ONE times (g / k). Therefore, the: Limiting Velocity or Terminal Velocity:
v = g / k.
For an average size and weight human, flat with arms spread, k is about 0.18 (second). In the English system, g = 32 f/s/s, and so the limiting value of v is around 178 feet/second or 121 mph.
A bodybuilder would likely have less frontal area, and therefore for him, k would be a little lower, maybe 0.16 (second). This would result in his terminal velocity being a little higher at 136 mph.
Also, if a person holds his arms down along his sides, that similarly reduces the frontal area, which again reduces the coefficient k. In aerodynamics, such as with vehicle design and aircraft design, the factor k is the combination of two factors, one which is the frontal area presented to the wind and the other is called the Aerodynamic Drag Coefficient.
We are not quite done! We can now know the VELOCITY at each instant, by the equation we derived above. We know that the velocity is related to the velocity as v(t) = x'(t). Our solution equation above is therefore:
x' = (g / k) * (1 - e-k * t)
We now do the Integration again and get the solution:
x = (g / k) * (t + 1 / k * e-k * t) + C2
We had initially set x = 0 when t = 0, so we have:
C2 = - g / k2
Therefore, we have:
x = (g / k) * t + (g / k2) * (e-k * t - 1)
THIS is the complete Equation of Motion which applies.
We could consider this same problem with a different assumption, that the air resistance (or drag) is proportional to the SQUARE of the velocity and not just the velocity itself.
v' (or a, the acceleration) = g - k2 * v2
This is a little more complicated Differential Equation to solve, but still fairly easy. The solution, as the equation of motion is:
x = g / r2 * ln((er * t + e-r * t)/2)
Where the constant quantity r2 is defined as g * k2
A different value for the constant k (or r) applies with this Equation of Motion. Again, there are variations in that constant for different people and different orientations.
We might also note that:
x'' = g * e-k * t
and that the LIMIT of this equation at t approaches infinity simply says that the acceleration which occurs (x'') always goes to zero, for any object falling through any fluid in a gravitational field, showing us that there is a Terminal Velocity for every object in every fluid. It may be faster achieved for a feather than for a lead ball, but that is only because the constant k is different for those two objects in air. As was demonstrated on the Moon, where there is no air, a feather falls just as quickly as any other object.
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C Johnson, Theoretical Physicist, Physics Degree from Univ of Chicago