Faster than terminal velocity

I had so much fun creating graphs for the Red Bull Stratos Space Jump calculation, that I figured I should make some more.

Can you fall faster than terminal velocity? That is the question.

Air Resistance

Air resistance is a force exerted on an object as it moves through some stuff - air in this case. The magnitude is usually modeled as:


  • Rho is the density of the stuff the object is moving through
  • A is the cross sectional area of the object
  • C is the drag coefficient of the object - this depends on the shape (a cone would be different than a flat disk)
  • v is the magnitude of the velocity of the object

The direction of this air resistance force is in the opposite direction as the velocity.

Terminal Velocity

Here is a diagram of a sky diver that just jumped out of a stationary balloon.


Here there is the gravitational force (weight) and a small air resistance force. The air resistance is small because the jumper just started falling and is not moving too fast. The net force is in the downward direction. Since this is in the same direction as the velocity, the speed increases.

In a little bit more time, the diagram would look like this:


Since the jumper is going faster, there is a greater air resistance force. This means that the net force is still down, but much smaller. Maybe I should remind you of Newton's Second law:


Since the net force is smaller, the acceleration is smaller and the jumper does not speed up as much. Essentially, the jumper will reach a speed where the air resistance is the same magnitude as the gravitational force (weight). At this time, the net force will be zero (vector) and the acceleration will be zero (vector). The velocity will not change. It will not speed up, it will be terminated - terminal velocity.

So, here is an expression for terminal velocity (the magnitude).


Great. So, essentially the terminal velocity just depends on stuff about the object - mass, C A. But! What if the gravitational force is not constant? What if the density of air is not constant? In this case, the terminal velocity will change also.

Back to the Space Jump

If you jump out of a balloon at 120,000 feet above the ground, some things are different. Mostly, the density of air is really low so the jumper can really get going fast. When falling to a lower altitude, the density would then increase.

I will go ahead and modify my python calculation. Here is a plot of speed and terminal velocity (magnitude) vs. time. I am plotting the terminal velocity for the altitude the jumper is at that instant.


I am not showing the speeds from time zero seconds. This is because when the jumper starts, the terminal velocity is HUGE. At at about 46 seconds, the jumper is going at terminal velocity, however as the height gets lower, the terminal velocity is also getting smaller. So right after this, the jumper is going faster than terminal velocity.

What about the acceleration?

One more plot, I promise. Here is a plot of the acceleration of the jumper as a function of time.


When the jumper starts - the acceleration is essentially -9.8 m/s2. After the jumper goes faster than terminal velocity, the air resistance force is greater than the weight so that the acceleration is in the positive direction. The greatest positive acceleration is somewhere around + 8 m/s2. This is important because this is the acceleration the jumper will "feel". The gravitational force pulls the same (per unit mass) on all parts of the body, so you don't really feel that. Just imagine what it feels like in free-fall with no air resistance, you are weightless just like in orbit. Ok - I lied. Here is one more plot. This is a plot of the air resistance force divided by mass in units of "g's". So, if the air resistance is equal to your weight, you would experience 1 g.


The shape looks the same because the gravitational force is essentially constant. Here though, you can see his max "g-force" will be less than 2 g's.

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Great. So, essentially the terminal velocity just depends on stuff about the object - mass, C A.

No. The drag coefficient C is also a property of the medium. I mean, think about it, will a ball bearing have a higher terminal velocity in air or in glycerol?


Glycerol would have a higher density - also, the model of resistance that says the force is proportional to v^2 probably doesn't work in that kind of material.

A diver who wanted to get faster than terminal velocity could first orient themselves to develop a significant horizontal velocity, and then give themselves a negative angle of attack, so that lift forced them downward.

Sorry, but you're way off on this one.
Assuming the jumper does not change his orientation or body shape, it is impossible to exceed the terminal velocity the way you show, because of the definition of terminal velocity. "Terminal velocity" is the speed you get to when all the forces balance out in steady state (i.e. the force due to gravity equals the force due to drag). As the jumper gets to higher and higher velocities, the aerodynamic drag force increases until it balances the force due to gravity and the jumper no longer accelerates. Even if they changed their configuration, they'd only see a momentary exceedance of the terminal velocity of the new configuration because the drag in the new configuration would be higher than the force due to gravity and declerate them to the new terminal velocity.

Part of your problem is that you used an invalid assumption in your drag calculations. The coefficient of drag is not constant, but rather is a function of Mach number and will increase as Mach increases to about 1.3 then will start going down again. This is due to the fact that the definition of the coefficient of drag is:
Cd = F_drag / (1/2 * rho * A * v^2)

The drag coefficient can't be computed, it has to be measured as a function of rho and velocity, so you'd have to look it up in some book to find out what the drag coefficient of a human is at different Mach numbers, etc (more precisely, you'd probably have to approximate it using some similar shape).

Once you start talking about Mach numbers above 0.3, you can no longer assume air is an incompressible fluid and have to start taking compressibility effects into account (like the above mentioned change in drag coefficient). I don't remember enough about compressible flow to be able to correct your math for you right now, maybe when I have more free time.

No--see above definition of "terminal velocity". No matter what a skydiver did, at some point their velocity would eventually max out and that velocity is the terminal velocity for that configuration (whether they're pointed straight down, laid out flat, curled up in a ball, or trying to develop "negative" lift like you mentioned, which all have different terminal velocities). When they go from a lower drag configuration to a higher drag configuration (e.g. going from pointed straight down to laid out flat), they will momentarily be going faster than the terminal velocity for the new configuration, but will quickly slow down to the new terminal velocity. The same is true for developing "negative lift", which is just a force that would add to the force due to gravity. The jumper would see a higher terminal velocity but would not exceed the terminal velocity for that specific configuration.


I agree that the model I used was most likely not valid. I pointed that out in my other space jump post. However, it doesn't seem like it is a crazy-wacked out model either. My calculations for the 1960 jump are fairly close to the actual data.

Also, you can look at it this way. If the air resistance force is ever greater than the weight, he will exceed the terminal speed. This also means that he will slow down. He has to slow down at some point during this fall.

True, but he wouldn't blow through the terminal velocity the way your graph shows (exceeding it by ~75 m/s at max), instead he would slowly approach it and would go beyond it by a fairly small amount and remain slightly above it because the the air is constantly getting getting thicker. He would technically remain slightly above it all the way to the ground, but once he got below about 10,000 feet, the amount by which he exceeded the terminal velocity would be so small as to be almost immeasurable.

It is an interesting problem, unfortunately the complexities of compressible flow and high speed aerodynamics makes it pretty difficult. I looked at your other post, but haven't given it a lot of thought yet. One of these days, when I have enough free time to do the refresher studying I'd need from my college aero classes, maybe I'll try to tweak your calculations.


I think there is a way - let me try to figure that out.

At first glance, it seems to me that d(v-v_term.)/dt should always be negative, which it isn't right after the graphs cross here. What do you think?


I assume you are talking about the dv/dt for felix? Right when the two lines cross, his dv/dt is zero (about). After that dv/dt is negative (but it really should be positive since I plotted the magnitude of the y-velocity so that it would look better).

So - after that point, Felix is slowing down since he is moving faster than the terminal velocity. This means the air resistance is greater than his weight and the net force is in the opposite direction to his motion.


You sparked an idea. I figured if there was acceleration data for the 1960 jump then I could "experimentally" get a sense of the air resistance model. There was a rumor that the 1960 issue of National Geographic had a (or some) graphs in it.

After wasting some time trying to find this article online, I went to the library. There is a graph in that article, but it is just Joe's breathing rate and heart beat and stuff. Oh well.

It is great to know that it is possible to gain speed even after attaining a maximum terminal speed

By vinod arora (not verified) on 24 Feb 2010 #permalink

Actually, I'm looking at the time derivative of the difference between his v and the terminal v. That derivative being negative means that his speed (when v>v_term.) should be decreasing relative to v_term (approaching v_term. from above). I changed my mind about it, though, I think - that's what you get from first glances!


I get it now - I just didn't understand your notation. Maybe I should just plot the derivative of the velocity differences - I will when I get a chance.