# Why not increase contact surface when reentering the atmosphere?

If a craft were to increase the surface area where contact is made with air during reentry, I imagine the heat quantity per area unit would decrease, making the use of (heavy) heat shields less of a problem. It would probably come at the expense of greater g-forces, is it the limiting factor ? If not, where am I wrong ?

• If you increase surface area and this increases the g-force, an increased heat load is the result. So heat load per area would not neccessarily decrease. – Uwe Mar 26 '19 at 21:13
• – leftaroundabout Mar 28 '19 at 0:08

I've done a lot of work on this subject with researchers and engineers at JPL, NASA Langley, and NASA Ames. There are some interesting things that come out of high-fidelity CFM (Computational Fluid Mechanics) modeling of entries or re-entries, and also from flight experience. This FAA tutorial segment is a good general reference for the principles involved.

There is no single entry circumstance (atmosphere-relative entry velocity, entry flight path angle, atmospheric temperature profile, atmospheric composition, etc.) that tells unambiguously what the entry conditions (heat flux profile, maximum heat flux, integrated heat load, inertial force profile ["g-load" profile], stagnation pressure profile, etc.) will be. It's a multivariate problem.

One of the important entry circumstances is the vehicle's ballistic coefficient, the vehicle's mass divided by its effective area, i.e. the actual frontal area times the drag coefficient. This online paper by Dinesh Prabhu (one of my Ames colleagues) shows a type of chart they use a lot, this one specific to an Earth entry and heat shield geometry. The axes are two entry circumstances: entry flight path angle and ballistic coefficient $$\beta$$. The three sets of curves you can read off that chart are entry conditions: peak heating rate, integrated heat load, and peak deceleration rate in g's.

One interesting result is that for an exponential atmosphere without huge deviations from the typical temperature profile, the deceleration profile is only mildly sensitive to the ballistic coefficient! "How can that be??" you say. As @Uwe pointed out, a larger surface area for roughly the same mass (and hence lower ballistic coefficient) will produce a larger drag force, thus a larger deceleration rate.

If the velocities and air densities are the same.

But they're not. The vehicle with the lower ballistic coefficient starts seeing non-trivial deceleration at lower atmospheric densities, and thus higher altitudes. By the time it gets down to the altitude where the higher-$$\beta$$ vehicle is seeing non-trivial deceleration, the lower-$$\beta$$ vehicle is now traveling slower.

The net result is that for the lower-$$\beta$$ vehicle the time profile of deceleration looks very much like that of the higher-$$\beta$$ vehicle, just earlier and at higher altitudes. This means that the lower-$$\beta$$ vehicle's deceleration is, on average, in less dense atmosphere, so the heating rate and integrated heat load per unit area do indeed decrease.

This fact has led to the "ballute" concept for atmospheric entry and aerocapture (example papers here, here, and here). The fundamental concept is that you use a balloon-like structure to create a huge surface area, so the major part of deceleration occurs at very high altitudes and at very low atmospheric densities, so heating rates and heat loads are much lower. If they get low enough, flexible materials can handle them, so the deceleration system can be enclosed in a relatively small container until needed, and its total mass is less than that of a system that must handle more challenging heating environments.

NASA and collaborators are still looking for a material that is light enough, flexible enough, and sufficiently resistant to heating that you can build such a ballute that actually saves mass. For a while they were considering polybenzoxazole but found there were problems in manufacturing thin, flexible, uniform sheets. The search continues.

There is a fixed amount of energy which has to be dissipated. You can, to some extent, choose how fast this is done -- more air resistance (either by getting into thicker air or having a bigger surface) dissipates it faster, with higher g forces. Less air resistance dissipates it slower, but you do have to make sure to get rid of it all before you hit the ground. Choosing a slower (or at least not faster) dissipation with a bigger area (eg a pointy vehicle with wings that spreads the load out along the sides and wings) will mean you can use a lighter heat shield per unit area, but, you need more of it, so it might be heavier overall. Finally, aerodyanmic stability across a huge range of speeds may limit the shapes you can use.

• I suppose you don't HAVE to get rid of all the energy before you hit the ground, but the people inside would really appreciate it if you did. – corsiKa Mar 27 '19 at 18:20
• Afaik the possibility of hitting the ground with a significant speed (coming from the orbital kinetical energy and not from the low athmosphere freefall) is practically impossible. 0.001 bar athmosphere could kill the Columbia, I don't think that anything man-made could exist what could reach the ground with living humans therein. I think most of the deceleration should happen in $\approx$ 40-60 km altitude. – peterh - Reinstate Monica Mar 27 '19 at 18:50
• @corsiKa well technically you don't have to get rid of ALL of the energy, considering some energy is transferred to the earth or ocean on touchdown in anything other than a retropropulsive hoverslam landing. Still, the energy left is several orders of magnitude smaller than before re-entry. – Sdarb Mar 27 '19 at 21:40
• If you've got a wheeled gliding vehicle, you could tolerate quite a bit of residual horizontal speed (up to several hundred meters per second), the exact amount depending on the maximum certificated tyre speed and the heat tolerance of the tyres and brake assemblies. – Sean Mar 27 '19 at 22:33
• @Sean: So basically the Space Shuttle. Still, the problem is that shedding 99% of the speed means shedding 99.99% of the kinetic energy. And those are realistic figures - starting at Mach 25 and landing at Mach 0.25. – MSalters Mar 28 '19 at 11:07