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PostPosted: Mon Apr 12, 2010 7:26 am 
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Sky diving from the edge of space and hoping to break the sound barrier on the way down.


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PostPosted: Mon Apr 12, 2010 12:18 pm 
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This man has some serious balls. He's also going to **** kill himself. For science!

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PostPosted: Mon Apr 12, 2010 1:51 pm 
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Isn't this about 20K feet higher than the guy who jumped out of a weather balloon in the late 50s/early 60s? Sounds like a gas.

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PostPosted: Mon Apr 12, 2010 1:55 pm 
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Yeah, but that 4 mile difference is huge.


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PostPosted: Mon Apr 12, 2010 2:10 pm 
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What kind of airplane is going to get him high enough to do that?

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PostPosted: Mon Apr 12, 2010 2:13 pm 
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Diamondeye wrote:
What kind of airplane is going to get him high enough to do that?
From the article:
Quote:
So what will it feel like when Baumgartner opens the hatch of the gondola suspended below the weather balloon that will take him to space, and he steps out into 23 miles of nothingness?


It's going to be a 330 second ride. I would have figured it to take longer.

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PostPosted: Mon Apr 12, 2010 2:21 pm 
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Well, hes hoping to hit 760+ MPH...


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PostPosted: Mon Apr 12, 2010 2:30 pm 
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Ladas wrote:
Well, hes hoping to hit 760+ MPH...

12 2/3 miles a minute, an average of just over 4 miles a min...

That's smoking. I wonder if he'll live? I'd think breaking the sound barrier would be a bit hard on one's limbs...I'd not be surprised if he was ripped apart. Great ride if he makes it though.

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PostPosted: Mon Apr 12, 2010 2:46 pm 
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Acoustic velocity for a given ideal gas is proportional to the square root of temperature:

c = sqrt(k R T)

At 120,000, I'm guess it's quite low.

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PostPosted: Mon Apr 12, 2010 3:27 pm 
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Ladas wrote:
Yeah, but that 4 mile difference is huge.

Oh, definitely. Found a great reference to the guy who did it 50+ years ago: Joseph Kittinger. He made three jumps, it looks like, from 76,000 feet; almost 75,000 feet; and then from 102,800 feet. He hit 614 mph on that jump, and said he "had absolutely no sense of the speed."

Sounds pretty fantastic.

If he makes it, this Baumgartner guy owes him one.

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PostPosted: Mon Apr 12, 2010 5:31 pm 
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Dude has some big brass ones, hope they don't melt.

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PostPosted: Mon Apr 12, 2010 8:27 pm 
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I'm kind of surprised the guy from the 50s even survived the deceleration of the chute opening.

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PostPosted: Tue Apr 13, 2010 7:10 am 
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Well, by the time he got to the elevation to open his chute, he would have been doing considerably less. Isn't the terminal velocity of a human body in an uncontrolled free fall somewhere around 120 MPH?


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PostPosted: Tue Apr 13, 2010 11:07 am 
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In atmosphere density levels levels closer to sea level, yes, about 124 mph/200kph.

Terminal Velocity depends on the drag against the falling body, lower density of the atmosphere at the upper state will let him get to a much higher terminal velocity, his speed will decrease as the atmosphere becomes more dense, creating a whole lot of friction (and heat). By the time he opens his chute, if he is still alive in there, it should only be a bone shaking organ rattling mauling incident.

The romance/thrill of the jump may be worth it to him, but I wouldn't do it unless it was intended to be my terminal act.

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PostPosted: Tue Apr 13, 2010 11:26 am 
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Diamondeye wrote:
I'm kind of surprised the guy from the 50s even survived the deceleration of the chute opening.


He would have been going the same speed as any other skydiver, assuming he pulls the chute fairly low.

From my understanding, at those speeds, even a slight movement of the head could spin you out of control. If the sound barrier breaks, and it shocks him at all, I hope he doesn't move.


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PostPosted: Tue Apr 13, 2010 3:02 pm 
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I thought terminal velocity was the most gravity could accelerate you to in the first place if you hadn't already fallen from such a height that you could accelerate to a higher speed before encountering significant air resistance.

It's actually enough to slow you down from 760+ mph to 120?

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PostPosted: Tue Apr 13, 2010 3:05 pm 
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In fluid dynamics an object is moving at its terminal velocity if its speed is constant due to the restraining force exerted by the air, water or other


Nah, its as fast as you can go due to resistance of air pressure.

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PostPosted: Tue Apr 13, 2010 3:12 pm 
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Diamondeye wrote:
I thought terminal velocity was the most gravity could accelerate you to in the first place if you hadn't already fallen from such a height that you could accelerate to a higher speed before encountering significant air resistance.

It's actually enough to slow you down from 760+ mph to 120?



If he performs the jump correctly, yes. If not, he'll burn up or something crazy.


Even though he will be going insanely fast, the distance of the jump and gradual increase of air resistance will gradually slow him down and by the time he's ready to open his chute he'll be able to, unless he sneezes.

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PostPosted: Tue Apr 13, 2010 4:06 pm 
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Diamondeye wrote:
I thought terminal velocity was the most gravity could accelerate you to in the first place if you hadn't already fallen from such a height that you could accelerate to a higher speed before encountering significant air resistance.

It's actually enough to slow you down from 760+ mph to 120?


Yes, but the net deceleration will be a function of the difference in independent deceleration produced by the air pressure and the independent acceleration resulting from gravity. How significant the difference is will determine how fast he decelerates to terminal velocity. This may or may not occur in time, I don't know.


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PostPosted: Tue Apr 13, 2010 8:44 pm 
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There are two drag forces exerted when moving in a fluid: the stagnation pressure and "skin friction" which occurs where the boundary layer exists.

The force due to gravity is proportional the product of the masses being attracted together and inversely proportional to the square of the distance of separation.

A very basic relationship might be:

F = +F(gravity) - F(drag)

F = G m(earth)*m(diver+suit) / r^2 - 1/2 * rho * v^2 * C(coeff of drag) * A

However, finding C requires empircal experimentation for a given shape. A is a function of the diver's orientation toward the direction of flow (i.e. how headfirst or bellow side down he dives) as is C.

This isn't very useful because rho is a function of r and F is a function of v.


F = G m(earth)*m(diver+suit) / r^2 - 1/2 * rho(r) * v^2 * C(coeff of drag) * A

= m(diver + suit)*a

a=dv/dt

dv/dt*m(diver+suit) = G m(earth)*m(diver+suit) / r^2 - 1/2 * rho(r) * v^2 * C(coeff of drag) * A

So we have to solve a second order non homogeneous ordinary, linear differential equation. We could solve it with a laplace transform since we know the initial conditions (x(0) = -r if we consider direction of travel forward and 0 altitude is r = 0 and v(0) = 0)

But we sould at least be able to determine v (velocity) as a function of r (altitude). We then have to simultaneously solve it with an equation which describes the acoustic velocity at a given altitude, since acoustic velocity in some medium is a function of temperature. So we solve (speed of sound) = sqrt(k R T(r)) where T is a function of r.

The problem is, the aerodynamic perspective is much too simplistic. Drag is more complicated in a medium where compressibility becomes a concern. Unfortunately for us, air behaves in compressible at low speed since pertubations caused by an object moving through it are communicated outward much more quickly than moving through the medium. As we approach acoustic velocity, these pertubations being to "stack" causing incompressible behavior.

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PostPosted: Tue Apr 13, 2010 9:04 pm 
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I'd just like to point out how fast this post went from totally awesome to somewhat geeky =P


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PostPosted: Tue Apr 13, 2010 9:32 pm 
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I'm sure we could come up with some sort of relationship for that, too.

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PostPosted: Tue Apr 13, 2010 9:35 pm 
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I just want to know if he's gunna break the sound barrier, and if so, will he have his arms ripped off?

Inquiring minds want to know!

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PostPosted: Tue Apr 13, 2010 10:20 pm 
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technically if his arms rip off than the chances of his head and legs ripping off are just as high... think human cannon ball.


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PostPosted: Tue Apr 13, 2010 10:48 pm 
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I'm going to have to nerd rain this party a bit:

The percent of sonic velocity at altitude(2) to that at sea level(1) can be defined as:

C2/C1

Given we assume an ideal gas (a good assumption, even for somewhat humid air which consists of 70% nitrogen, 29% oxygen and trace other gases and particulates), C(n)=sqrt(R(n)T(n)K)

So C2/C1 = sqrt(R(2)T(2)K) / sqrt (R(2)T(2)K)

Since K's are equal and R(1)=R(2) (we're assuming air is basically isoproprietary at all altitudes, a fairly good assumption), we are left with:

C2/C1 = sqrt(T2)/sqrt(T1).

From here: http://www.engineeringtoolbox.com/air-a ... d_461.html

We can see air @ sea level is 518 degrees R and @ only 16k ft is 461 R.

Therefore the sonic velocity at 16a is sqrt(461)/sqrt(518) or about 94% of the sea level velocity.

I actually just found this chart, here it shows how dramatically acoustic velocity decreases at altitude:

Image

1.466666 ft/sec = 1 mph

On top of this, air resistance decreases as a function of air density which also decreases at altitude. It's why high mach numbers are achievable at higher elevations easier.

Not to discredit what this guy is doing, it's still ... **** sick. I'd like a shot at it.

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