Supersonic R/C

In article , John Alt wrote: | In article , | snipped-for-privacy@bellsouth.net says... | > Ever heard of terminal velocity: | > | > Sqrt ((2 * Weight)/(Drag Coefficient * Density * Area). | > | > 100,000 feet ain't gonna do it.

You'll never actually reach terminal velocity -- you'll just get closer and closer and closer. At first, you'll accelerate at the full

9.8 m/s^2 (or a tiny bit less because you're getting further from the Earth) ... but as you get faster, your acceleration will decrease, until it's almost zero when you get close to terminal velocity.

Also, the speed of sound decreases as you get up higher, at least to a point. At 40,000 feet, it's really cold, so the speed of sound may be as low as 660 mph or so (it depends on exactly how cold it is.) So Mach 1 there is a good deal slower than Mach 1 at ground level.

In any event, merely giving a formula, and stating `100,000 feet ain't gonna do it' aint gonna do it. If you want to convince anybody of anything, you'll at least need to come up with some reasonable values for those variables and plug them in.

If you want to do the math, figure out what the terminal velocity is at 40,000 feet. Are we talking about a body or a plane? In either case, assume it's pointing straight down (minimizing the area.) I'm not sure what value you'll use for the drag coefficient ...

Neglecting air resistance, an object falling 60,000 feet (from 100k feet to 40k feet, where the speed of sound is the smallest) from rest wil accelerate to 1339 mph -- well above Mach 1. So, given a small enough cross section, or enough weight -- 100,000 feet IS enough. More than enough, actually.

| How quickly the great ones are forgotten. USAF Captain Joe Kittinger | broke the sound barrier 40 years ago in free fall from 102,000 feet. | |

And there you go, trying to confuse the issue with facts! :)

[ about supersonic R/C planes ]

| If the AMA hasen't banned such attempts, they should.

It's not up to the AMA. They have no authority to ban such attempts.

| Lest we not have a hobby after the feds take notice.

I believe that FAA regulations already make such flight illegal, at least except under certain specific conditions (like having official clearance, being in an unpopulated area, etc.)

| You'll never actually reach terminal velocity -- you'll just get | closer and closer and closer. At first, you'll accelerate at the full | 9.8 m/s^2 (or a tiny bit less because you're getting further from the | Earth) ... but as you get faster, your acceleration will decrease, | until it's almost zero when you get close to terminal velocity.

To follow up on my own post, that's not actually true.

Since the terminal velocity will get lower and lower as the pressure gets higher and higher as you fall, you will reach and then exceed terminal velocity, assuming that the terminal velocity of whatever it is you're dropping is less than 1700 mph at sea level and you're dropping from 100,000 ft. (Probably true for just about everything you'd drop.)

let's see if I got this right, You are going to fall faster than terminal velocity which is defined as the top speed of a falling object??

Actually terminal velocity is the point where drag equals the applied force (gravity)

as the drag increases, the speed will decrease, If this was not so, a parachute wouldn't work, and would be just that much more "dead" weight

as you penetrate to a denser atmosphere, the drag will increase. I would be willing to bet that you would be hard pressed to come up with a significant number to represent the difference between actual velocity and terminal velocity at any given altitude or density for an object which had been in freefall for a long enough period of time, say in excess of 120 seconds.

more information available at:

YMMV bob

Theoretically, and practically, there would be no difference between actual velocity and terminal velocity for something that has been falling that long. By the very definition of terminal velocity, they would have to be the same for any given instant. The velocity would be constantly decreasing as the object drops into increasingly denser air.

At least that's how I understand it.

costs.

| You are going to fall faster than terminal velocity which is defined | as the top speed of a falling object??

Yes! :)

(Except that the terminal velocity is not the `top speed', as you mention immediately following.)

| Actually terminal velocity is the point where drag equals the | applied force (gravity)

Yes.

| as the drag increases, the speed will decrease,

You're assuming that the terminal velocity is a constant. It is not. It varies basied on altitude -- which doesn't really matter when you're only falling 200 feet, but from 100,000 feet -- it matters a lot.

The formula given for terminal velocity

Sqrt ((2 * Weight)/(Drag Coefficient * Density * Area)

includes a density figure. That's the density of the air. Since it's under a sqrt(), dividing the density by four will double the terminal velocity.

Some examples --

At 10k feet, the pressure is approximately 0.66 atm, so the terminal velocity would be 23% higher than at sea level.

At 50k feet, the pressure is approximately 0.13 atm, so the terminal velocity would be 180% higher than at sea level.

At 100k feet, the pressure is approximately 0.016 atm, so the terminal velocity would be 684% higher than that at sea level.

| If this was not so, a parachute wouldn't work, and would be just | that much more "dead" weight

Fortunately, the pressure at sea level is usually approximately a full atmosphere, so your terminal velocity will be as expected. But if you were skydiving and landing at the peak of Mount Everest (26,000 feet?), you'd land a lot harder than expected (about 70% faster than you would at sea level.)

| as you penetrate to a denser atmosphere, the drag will increase.

Yes. And the terminal velocity will drop. But what if you're already going faster than the current terminal velocity? You'll slow down, but will never *quite* reach terminal velocity, for two reasons -- 1) the terminval velocity will keep getting smaller as you fall, and 2) you never quite reach terminal velocity anyways -- you just get closer and closer.

| I would be willing to bet that you would be hard pressed to come up | with a significant number to represent the difference between actual | velocity and terminal velocity at any given altitude or density for | an object which had been in freefall for a long enough period of | time, say in excess of 120 seconds.

The terminal velocity formula has nothing to do with time falling or current velocity -- the only variables are weight, drag, and the force of gravity.

To actually calculate the speed at each point as an item was dropped from say 100,000 feet would require knowing the drag coefficient and weight for your item, and require a lot of computation. It's certainly doable, but I'm not going to do it here. But certainly, if somebody were to jump out of a balloon at 100k feet, and open his parachute immediately, he would spend most of his trip down at *above* the current terminal velocity.

Besides, this has gotten way off topic :)

| more information available at: |

And I used to estimate the pressure (and therefore the change in the terminal velocity) at a given altitude. (This page was meant for the GURPS game system, but it's theory seems sound. If it's estimate is wrong, then my examples will be wrong, but the general idea won't be affected.)

Ob R/C:

is a new web site that covers pretty much what the name says. If you're in or near Austin, find out where we're flying and join us! :)

There's a nice java applet on this page that lets you get the predicted temperature at different altitudes:

It says to expect -69 F at 40,000 feet.

This page has a calculator for Mach 1 at 40,000 feet:

It suggests that Mach 1 would be 660 mph.

Marty

actually, YOU are ASSuming that I assumed something I did not.

"term>You're assuming that the terminal velocity is a constant. It is not.

A human has broken the sound barrier in free fall. Jumped from over 100,000 feet and exceeded Mach 1 before entering the denser atsmophere.

costs.

Doug McLaren opined

I wrote a program to do free fall jumps by human beings. It took into account both gravity and air density. For a jump from 30,000 meters the statistics are

-- the fall from 30000 meters took 256 secs max v = -253.368234 m/sec at 23758.89 meters max a = 3.05881450 m/sec**2 at 19001.98 meters

It's a lot more than lots of power, autopilots and strong structures. Supersonic flight involves a different set of aerodynamic rules, since the air has no time to respond to an approaching body, and the response of the airplane to control inputs is not so simple. The Concorde was an example of the terrific expense of supersonic flight. I don't think it would be feasible for anyone except someone with an awful lot of money.

Dan

I think we're in complete agreement: a hard problem, but technically soluble (since it has been solved many times by many aircraft), given enough time and money.

I'm not going to put any of my time and money into this project. I'm just answering the narrowly focused question that was asked at the beginning of this thread: could it be done. My amateur guess is "yes."

Marty