How much can the falling speed of starship approximately be reduced by the bellyflop fall with fins? Are we talking about numbers around 5% or 50% here?
Wouldn't it make sense to put holes where the fins are attached to the body and blow out the air at the end of the fin through channels to "increase" the fin size with air?
Let's look at the ballistic coefficients of a bunch of reentry vehicles. This is the mass divided by the cross section area. The drag coefficients of all these blunt bodies will drop from very roughly 1.0 while hypersonic to 0.5 near terminal velocity, except for the Pershing II MARV which being a sharp cone is smaller.
| Dry Mass | Wet mass | Length | Diameter | Xsec area | Ballistic coefficient | |
|---|---|---|---|---|---|---|
| Falcon 9 S1 end-on | 24 mt | 49 mt | 3.7 m | 10.2 m^2 | 4803 kg/m^2 | |
| Pershing II MARV | 0.64 mt | 0.5 m | 0.2 m^2 | 3260 kg/m^2 | ||
| Starship end-on | 120 mt | 150 mt | 50 m | 9 m | 63.6 m^2 | 2358 kg/m^2 |
| Dragon 2 | 9.5 mt | 12.5 mt | 4.0 m | 12.6 m^2 | 992 kg/m^2 | |
| Apollo CM | 5.5 mt | 3.9 m | 11.9 m^2 | 465 kg/m^2 | ||
| Shuttle | 78 mt | 37.2 m | 23.8 m | 250 m^2 | 312 kg/m^2 | |
| Starship sideways | 120 mt | 150 mt | 50 m | 9 m | 520 m^2 | 288 kg/m^2 |
| Shuttle SRB | 91 mt | 590 mt | 45.5 m | 3.7 m | 603 m^2 | 151 kg/m^2 |
By descending sideways, Starship decreases the ballistic coefficient by 8.2x, which should decrease the terminal velocity of the vehicle by 2.8x with the fins fully outstretched.
The vehicle will not actually get down to terminal velocity.
The decrease in velocity when the landing burn starts will decrease the amount of propellant needed during that burn, approximately linearly.
On Falcon 9, every 7 kg of propellant used for the landing burn reduces the potential payload into low Earth orbit by 1 kg. The landing propellant is approximately equal to the stage dry mass.
On Starship, it's 1 for 1, so they are far more sensitive to landing propellant usage. The landing propellant is 20% of the stage dry mass. I'm not sure what accounts for the propellant savings beyond lower terminal velocity, but the most obvious candidate would be greater deceleration which reduces gravity losses.
Can't speak to Starship. With the shuttle, the belly of the craft is the primary drag energy dissipation device. In the initial stages the essential goal is to keep the force of drag matched to the centre of mass so that the ship doesn't tumble.
(I think this is why missing a tile or so is a big problem. A single tile will have the air fill the space and following air will essentially flow around it. (think dragging a cup of milk through a pool of water very evenly) Lose either a bunch of tiles in a row, or tiles on the leading edge then there is no cushion effect and now you have turbulence which makes drag; irregular asymmetry, which which distorts the drag, and the potential for burn through at the unprotected part of the ship.)
Initial reentry attitude corrections are done with the steering rockets. Craft has to slow down and get into thicker air before the wing's control surfaces work.
Remember that a lot of this is happening in VERY thin air and very high speed. I'm not sure how you would go about taking in air that's coming in at Mach 15 or so, then blowing it out through slots fast enough to make a difference. Seems like a lot of equipment and huge engineering effort for a fairly small gain. Just make the fins larger.
Slowing down faster increases the acceleration by definition. Ship has to be stronger to afford the additional stress. But you may ask, "It has to be strong for launch!" True, but coming in on its belly it has to be strong enough for stress in a different direction.
(The Atlas rocket, designed for 10+ G's vertically, couldn't support it's own tanks lying on it's side. They had to keep the tanks pressurized with inert gas to keep them from collapsing.)
