Page 6 of 6 
[ 85 posts ] 
Aerobraking and reentry
Author  Message 

Spaceflight Trainee
Joined: Mon Aug 29, 2005 6:37 pm
Posts: 23 Location: Lake Charles, LA 
Thereâ€™s an error in my previous post. The general form of the differential equation is correct, but the final solution I gave for velocity is wrong. I made two mistakes in the integration: failure to account for initial conditions and omitting a square root. The equation for velocity should be:
v = ( sqrt( 2 G M ( 1/r â€“ 1/ro ) ) + vo ( 1 â€“ k hs exp( ( RE â€“ ro ) / hs ) ) / ( 1 â€“ k hs exp( RE â€“ r / hs ) ) where RE is the radius of the earth and the ro and vo are the starting values for r and v, respectively. My apologies. Regarding the earlier mystery of why we found the accelerations so high, the constant k used in my formula for velocity can be expanded to: k = Rho cd / sigma where sigma is the surface density of the vehicle (mass per unit crosssectional area). Peter's spreadsheet appears to assume a value of 1 T/m^2. The space shuttle probably reaches about 0.3 T/m^2, and there are designs like the German Mars Society's Archimedes space probe that are shooting for less than 0.03 T/m^2. Variations in sigma could significantly affect the acceleration curve during reentry, particularly if they start approaching a hundredfold. _________________ “The next generation of engineers, astronauts and scientist are not going to appear out of thin air. They need to be inspired and educated, and the best way to do that is to get them involved.”  John M. Powell 
Back to top 

Spaceflight Trainee
Joined: Mon Aug 29, 2005 6:37 pm
Posts: 23 Location: Lake Charles, LA 
nihiladrem wrote: C M Edwards wrote: and the time for the descent is then just the integral of dt = dr / v (which I didnâ€™t bother to doâ€¦ ). Ha! Good luck to anyone who tries to get an integral for that Considering that neither of us has yet come up with a velocity equation that matches the other's, I'm inclined to just let that gauntlet lie. I still claim it can be done for the simple case we're discussing, but a more accurate model would have a nonconstant kfactor to account for hypersonic effects, and I am NOT grinding that one out just for an internet forum! Numerical integration is definitely the method of choice here. _________________ “The next generation of engineers, astronauts and scientist are not going to appear out of thin air. They need to be inspired and educated, and the best way to do that is to get them involved.”  John M. Powell 
Back to top 

Space Station Member
Joined: Tue Dec 07, 2004 6:50 am
Posts: 265 Location: UK 
C M Edwards wrote: Considering that neither of us has yet come up with a velocity equation that matches the other's What I'm saying is that neither of the previous equations for v works, nor even this new one. Even after fixing the second bracket to ( (RE â€“ r) / hs ) the differences between it and the true solution are very noticeable. It's particularly obvious when for any starting condition where vo is zero, there is no drag anywhere. If it even is possible to integrate this to give some kind of formula made of standard functions(which isn't something I would assume) my guess is that it would be a complicated enterprise. 
Back to top 

Moon Mission Member
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361 Location: Austin, Texas 
It looks like everyone ran out of steam on this, just like I did. I guess I'll keep doing these simulations the same quick and simple way and just use smaller time increments when more accuracy is needed.

Back to top 

Space Station Member
Joined: Thu Jan 27, 2005 12:34 am
Posts: 450 
campbelp2002 wrote: Well, the reducing ballute didn't work out after all. First I set up the best reentry I could with a 100x ballute. Actually, your results are quite promising! The problem with your 10% steps is that they were FIXED MAGNITUDE, not fixed ratio. The last step reduced the 100X BALUTE to 10X, then dropped it to 1X. If you use 0.9 ratio reductions (46 of them to reduce the 100X BALUTE to 1X) then the results will look very good. This is a reasonable approximation to steady deflation of the BALUTE, which only needs to have its diameter reduced by 10X to achieve the 100X area reduction. A smaller number of steps could be used with similar results, but the steady reduction can be easily achieved in practice. Note for â€œgâ€ 
Back to top 

Moon Mission Member
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361 Location: Austin, Texas 
[quote="rpspeck"] Note for â€œgâ€

Back to top 

Moon Mission Member
Joined: Tue Feb 10, 2004 2:56 am
Posts: 1104 Location: Georgia Tech, Atlanta, GA 
I'd imagine that any vehicle designed for greaterthanonegee loads would have acceleration seats (quite possibly made out of that stuff that they use to make beds with  it should be reasonably cheap). Crash couches are doubtful, although somehow maneuvering the seats so the acceleration was taken eyeballsin or eyeballsup would be very good (as the other two orientations are painful at best and inurious or deadly at worst).
_________________ American Institute of Aeronautics and Astronautics Daniel Guggenheim School of Aerospace Engineering In Memoriam... Apollo I  Soyuz I  Soyuz XI  STS51L  STS107 
Back to top 

Moon Mission Member
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361 Location: Austin, Texas 
rpspeck wrote: Actually, your results are quite promising! The problem with your 10% steps is that they were FIXED MAGNITUDE, not fixed ratio. The bottom line is that the area of the ballute has to start out at least 1000/0.2245 = 4,454 times that of the reentering capsule to keep peak deceleration at 3 Gs. If it starts out any smaller, the capsule simply does not slow down enough before the ballute is fully deflated. Assuming a circular ballute, the diameter when fully inflated would have to be almost 67 times the diameter of the capsule. 
Back to top 

Moderator
Joined: Thu Jun 03, 2004 11:23 am
Posts: 3745 Location: Hamburg, Germany 
Just because the thought came to my mind (may have been talked about already: What about using a parachute deployd behind the reentering vehicle?
The air is that thin at the beginning of reentry that this might be applied then. Short before regions are reached where the air is too dense the parachute would have to be taken onboard again. What about it? Dipl.Volkswirt (bdvb) Augustin (Political Economist) 
Back to top 

Moon Mission Member
Joined: Mon Nov 01, 2004 6:15 pm
Posts: 1233 Location: London, England 
What about using a regular parachute with slits in it that lengthen/shorten to adjust the amount of drag. In simple terms I was thinking of something like a number of vents attached to elastic and as the force against the chute increased the elastic would stretch and open vents which would reduce drag.
If memory serves hot air balloons have a zipped slit that lets them be deflated quickly. _________________ A journey of a thousand miles begins with a single step. 
Back to top 

Page 6 of 6 
[ 85 posts ] 
Who is online
Users browsing this forum: No registered users and 39 guests 