[Space, Spaceships] Armor needed for Aerobraking/Re-entry

[DaltonS]

I searched for the answer to these questions, but the results I got were more confusing than enlightening. How much armor dDR is required for a ship capable of atmospheric re-entry/aerobraking? Can it all be up front, or does there need to be minimum armor on center and rear hull sections? I'm assuming a streamlined hull here; an unstreamlined one would need more. (How much?)

Dalton “is there a Spaceships FAQ?” Spence

[RogerBW]

Originally Posted by DaltonS

I searched for the answer to these questions, but the results I got were more confusing than enlightening. How much armor dDR is required for a ship capable of atmospheric re-entry/aerobraking? Can it all be up front, or does there need to be minimum armor on center and rear hull sections? I'm assuming a streamlined hull here; an unstreamlined one would need more. (How much?)

Dalton “is there a Spaceships FAQ?” Spence

My rule of thumb, from back in the 3e days, is DR 100 (i.e. dDR 10) on either the front or the back.

To be fair, the longer you're prepared to take over re-entry or the more dV you're willing to spend, the less heat shielding you need. (An extreme form of this is Vertical Landing, see the box on Spaceships p. 40; that basically brakes to a stop at orbital altitude, then descends slowly.) If you don't mind taking many hours, and you have enough manoeuvre capacity to stay under control, you can decelerate over multiple passes getting a little lower and a little slower each time, and use very little heat shielding.

[Anthony]

Realistically, being re-entry capable is a specialized system that is only incidentally related to armor; no material armor is capable of soaking the total heat involved in a re-entry on an earth-sized planet, so you need specialized construction to dissipate as much heat intot he atmosphere (instead of you) as possible.

[DaltonS]

Originally Posted by RogerBW

My rule of thumb, from back in the 3e days, is DR 100 (i.e. dDR 10) on either the front or the back.

That's a bit much for a small ship with medium advanced (TL9) armor.

Originally Posted by RogerBW

To be fair, the longer you're prepared to take over re-entry or the more dV you're willing to spend, the less heat shielding you need. (An extreme form of this is Vertical Landing, see the box on Spaceships p. 40; that basically brakes to a stop at orbital altitude, then descends slowly.) If you don't mind taking many hours, and you have enough manoeuvre capacity to stay under control, you can decelerate over multiple passes getting a little lower and a little slower each time, and use very little heat shielding.

Ah, but aerobraking is all about using atmospheric friction to provide the ΔV to transition from escape to orbital velocity. There should be an equation to calculate how long it takes. Maybe N=k*ΔV/A where N is the number of passes required, ΔV is velocity that needs to be shed to attain orbit, A is the ship's armor in dDR and k is a constant. Not sure how to convert N into time units. (Atmosphere density is probably a factor too.)

Dalton “because 'how long it takes' can be critical” Spence

[weby]

By spaceships RAW one of:
-One streamlined armor module in front or middle regardless of the DR value.
-Soft landing system.

[Fred Brackin]

Originally Posted by DaltonS

I searched for the answer to these questions, but the results I got were more confusing than enlightening. How much armor dDR is required for a ship capable of atmospheric re-entry/aerobraking? ]

Look at the Life Pod/Drop Pods. I think they have DR1oo but 80 of that is ablative with DR 20 (dDR2 in Spaceships) remaining after ablation.

There was an old Ve2 rule requiring DR 20 for hypersonic flight.

[thrash]

Originally Posted by DaltonS

There should be an equation to calculate how long it takes.

In reality, "orbital velocity" and "escape velocity" are situationally dependent -- the formulas in the books are generalizations. For any particular case, the only basic requirement is to change from an open, hyperbolic orbit (e > 1) to a closed, elliptical orbit (e < 1). That, in turn, depends on how close to parabolic (e = 1) your original approach orbit is. (All this relative to the body you intend to orbit.) The math is messy, since it's a variation on Kepler's problem with deceleration based on atmospheric density. You would normally solve it through numeric methods, rather than analytically.

Once you are in closed orbit, you may be able to use subsequent passes through the atmosphere to lower your apoapsis still further and circularize your orbit. But that's a different maneuver.

[Anthony]

Originally Posted by thrash

The math is messy, since it's a variation on Kepler's problem with deceleration based on atmospheric density.

You can approximate close enough as no interesting difference without advanced math. Your velocity at closest approach starts out as sqrt ( approach velocity ^2 + escape velocity ^ 2), and must drop below escape velocity for capture to occur.

[thrash]

That tells you the delta-V required. It doesn't solve the problem of how long you are in the atmosphere, what rate of deceleration you experience, and thus how much heat damage you can expect. The deceleration is a function of atmospheric density, which is (in turn) a function of altitude, which varies non-linearly as you approach and decelerate.

[Anthony]

Originally Posted by thrash

That doesn't solve the problem of how long you are in the atmosphere, what rate of deceleration you experience, and thus how much heat damage you can expect.

In some ways that's even easier. You need to dissipate energy equal to your approach energy, and you have the amount of time you are in atmosphere to do so. This is enough energy that, if absorbed, it will reliably kill you, so you need to figure out how fast you can dissipate heat -- which is a tricky problem but not for reasons of orbital mechanics.