Aerobraking and reentry

There’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 – 1/ro ) ) + vo ( 1 – k hs exp( ( RE – ro ) / hs ) ) / ( 1 – k hs exp( RE – 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 cross-sectional 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 re-entry, particularly if they start approaching a hundred-fold.

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 non-constant k-factor 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.

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 – 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.

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.

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 “gâ€

[quote="rpspeck"] Note for “gâ€

I'd imagine that any vehicle designed for greater-than-one-gee 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 eyeballs-in or eyeballs-up would be very good (as the other two orientations are painful at best and inurious or deadly at worst).

OK I took the trouble to enter the exact ballute size needed to keep the G load at or below 3 for every second of the reentry. Starting with drag 1000 times the that of the default value, I allow the G load to build to 3 and then start reducing the drag just enough each second to keep the G load right at 3. I had to reduce drag to 0.2245 before the vehicle slowed down to 0.81 km/s at 14 km altitude. After that I had to INCREASE the drag to keep the vehicle slowing at 3 Gs. At a speed of 0.07 km/s and an altitude of 4 km I continued increasing drag, but slowly enough to allow the G load to drop below 3. Finally, at an altitude of 3 km and with the drag back up to 1000, the speed was 3 meters per second and the G load was 1. Perhaps this portion of the flight could be considered dropping the ballute and deploying a regular parachute. Here is a graph. Each X tic is 1 minute. The vertical scale varies depending on which curve you are looking at.

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.

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)

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.