You, my very lucky friend, live in the not-to-distant future where people can easily withstand any number of G-forces. You have been strapped to a rocket capable of moving at the speed of sound. It goes off, you’re on top of it, “riding the wave”. You don’t burst into flames or any crap like that because… you’re very lucky. All in all, you survive. Suddenly, you break off of the rocket. A few questions pop into you’re head, and someone here will answer them for you…
How quickly will you decelerate from SS (Speed of Sound) down to TV (Terminal Velocity)
What is S.S. (For easy reference please)
What is T.V.
4a) What is the heat, in joules, that you may encounter during decelleration?
4b) And the g-forces?
I know I worded it akwardly. I have an underdeveloped lobe. Or something. I just want to know the straight dope, can anyone help me? Also: This is just a plea that anyone answering this question provides valid proof
In everyday conversation, Terminal velocity usually seems to refers to the point at which the forces related to wind resistance reach a balance with the accelerating force of gravity upon a human falling through air. On the face of it, this doesn’t seem very applicable to your rocket question, or is the rocket travelling downwards at the speed of sound when the passenger breaks loose?
Well, I imagine that there is air friction in the atmosphere no matter what you do, because eventually you will slow down no matter what unless you are in a vacuum.
So basically, if one is screaming through the air horizontally, the slowdown would be rather fast just due to friction relating to the atmosphere, then would slow down to a more desirable err… slowdown thingy.
However, I would like to know what would happen in both instances. I’m not sure how to phrase it if we are talking about horizontally, but I’m interested in both sides of the arguement.
It’s essentially impossible to put a number to terminal velocity without actual experiments with the actual object that will be doing the falling. In the case of a human being, it will vary based on the clothing worn, the build of the person, whether they’re in the fetal position or spread-eagled, and a hundred other variables.
The speed of sound varies based a lot of weather phenomena as well (temperature, humidity, etc.) but I suppose one could state a value under ‘standard’ conditions.
In other words, I don’t think anyone could just calculate an answer for you. Either you need to specify some ideal values (state T.V. and S.S., use a smooth sphere instead of a person, something like that) that can be worked on with aerodynamics equations, or you’ll just have to do the experiment a few times with a stopwatch handy.
I bet a ballistics expert would have some numbers to look at for how this happens with bullets.
It’ll depend in large part on what altitude you’re at when all this happens.
Terminal velocity for a person falling in a relaxed arch, with normal clothing, at lower altitudes (say less than 13,500 feet) is about 110-120mph.
The higher you go, the thinner the atmosphere and thus the higher your terminal velocity. Captain Joe Kittinger’s balloon jump back in 1960 showed a terminal velocity at 90,000 feet of over 600mph. That’s a person in freefall
If your rocket is moving 600mph at 90,000 feet and you let go, well, it won’t be like hitting a brick wall or anything - that’s how fast you’d fall anyhow. You’ll start to decellerate as you start to descend into thicker air but again, judging by Kittinger’s jump, you’d be fine.
On the other hand if you were going 600mph at 10,000 feet then you’d suddenly have to start slowing down to about 120mph. Unfortunately I don’t know how long that would take but perhaps the results of early supersonic ejections (where pilots went from 0.0mph (relative to the air) to over 600mph (relative to the air) in a fraction of a second would provide some guidelines. Those were generally people in good physical condition, with flight suit and helmet. IIRC nobody burst into flame from air friction but there were a lot of nasty injuries. I read one reference to the first supersonic ejection, they calculated that due to G forces the victim’s eyeballs weighed approximately 2 pounds apiece, ouch.
Hm, let’s assume a human eye weighs about an ounce, then 2 pounds represents about 32G. Definite ouch but survivable if it wasn’t for extended periods of time.
To figure out exact energies involved, I think you’d want a rough drag coefficient for the person. This could be found by assuming a stable arch position, rough estimate of the surface area of the person and then noodling out Cd given a known terminal velocity (say 120mph at a mile or two up) and air density at that altitude.
Once you’ve got Cd and the area of the person, look up air density at the altitude in question. Given all those numbers you can figure out the decelleration that will be experienced by said person hitting air of that density at whatever speed your rocket is moving.
Sadly I don’t have those numbers in front of me - I’m an idea man [grin] - but get thee to Google my son.
Ok, forget half the stuff I said. Let’s make this generic.
You are moving at a constant speed that is greater then you’re terminal velocity. at least 4X that velocity. Suddenly, whatever is keeping you at said speed is gone. You begin to approach terminal velocity from that speed going back towards it. How fast does it happen?
Sorry to be picky but it really depends on the density of the atmosphere that you’re moving through. IIR my college physics and fluid dynamics correctly, drag force is proportional to air density. Air density changes a lot based on altitude so you’ve got to know how high you are before you can figure out force and thus decelleration.
Unless you just want a bunch of equations that you’ll have to plug all the numbers into?
I don’t think that is the case. I think that the speed of sound varies, solely, as the square root of absolute temperature in a gas. I could be wrong, but I don’t think I am on this one.
Ok, sorry, I’m just asking this for a person in general, how fast is the decelleration after losing the speed-up.
I’m sorry if I made it sound like I wanted specifics, all I want to know is if the person in question returns to terminal velocity very quickly or gradually. Let us think they are on sea level if that helps, an average 5’9’’ male, 120 pounds.
I’m pretty sure I know what you’re asking, but I think you might not have a clear picture of what happens. First of all, there is no such thing as a horizontal terminal velocity; or, rather, horizontal terminal velocity is zero. “Terminal velocity” is the speed at which all forces on a body cancel, resulting in zero acceleration. That makes sense if you’re thinking about vertical motion, because there will be a speed where gravitational forces exactly balance aerodynamic drag forces, giving some net positive terminal velocity. For horizontal motion, though the only force is drag, and nothing balances it.
Second of all, even for purely vertical motion, you never (technically) really “reach” terminal velocity. You’ll approach it asymptotically, and pretty soon you’ll get so close as makes no practical difference, but you don’t quite reach it.
So, that said, let me rephrase your question. How about: “If you jumped out of an airplane, which was travelling near the speed of sound, what would your speed and acceleration be over time, ignoring compressible (i.e., speed of sound) effects?”
So, to answer that, I have a quick-n-dirty ballistics spreadsheet ginned up. For those of you who wish to repeat these calcs, I assume: air density at sea level (0.0023 slugs/ft[sup]3[/sup]), a 120 lb man with 5 sq. ft. area and 0.7 drag coefficient, and a starting horizontal velocity of 1000 ft/sec and 5000 ft in the air. Some of the numbers are SWAGs (and area will change a lot if you’re spread-eagled vs. nose-diving), but note that this combination gives terminal velocity of ~173 ft/sec, which is within a reasonable range. Note also that I’m not including air density changes with altitude or compressible effects. Selected results (x = horizontal, y = vertical):
So note that the total velocity is about double terminal (vertical) velocity after 2 seconds, and pretty near it at about 5 seconds. What surprises me a little is that the total velocity drops below the terminal (173 ft.sec) for a few seconds, and works its way back upwards from below. Interesting. But not unreasonable.
The problem is that a human is an irregular shape and would likely be tumbling if not broken up pretty badly when dumped off at supersonic speed. If the subject didn’t tumble you’d need to know know his coefficient of drag. That combined with his mass could be used to calculate deceleration.
