totm nov 2023 SpaceX Discussion Thread

[tater]

14 minutes ago, sevenperforce said:

The math only BARELY works for this, I think....

Depends a lot on what the actual mass is.

They could expend the SH, or land it at sea someplace.

Assuming a roughly F9 like profile that would gain 20-40% payload to LEO (propellant in this case).

[tater]

Wow, assume LSS is 90t (you used 80 or 85, right @sevenperforce?)

Looks like an expended SH might be able to get it to LEO with ~5km/s dv remaining. Means they could send a SS of some sort as "Gateway" in 1 launch, and it grossly exceeds actual Gateway in everything.

 

Or a nominal 120t vehicle (does that include landing props?).

1. SH RTLS gets SS to LEO with 150t cargo (props). ~3022 m/s remaining

2. SH ASDS (assume that's a thing) gets SS to LEO with ~180t cargo (props). ~3415 m/s remaining (assumption that ASDS would be a ~20% payload improvement from RTLS)

3. SH expended gets SS to LEO with ~210t cargo (props). ~3770 m/s remaining (assumes expended is like F9 a 40% improvement over RTLS)

That's awful close to being able to do a Gateway RT with no refilling at all. If nominal SS was 100t, #3 can easily do that. SS can do it if the nominal mass is maybe 105t. Earth surface to Gateway round trip. No refilling, but expended booster.

 

Edited May 25, 2021 by tater

[SOXBLOX]

So, for this mission, Starship will reenter itself, with the crew on board? And also, does it have enough ∆v to insert into LEO after returning from the Moon? 

[StrandedonEarth]

16 minutes ago, sevenperforce said:

The math only BARELY works for this, I think....

 

Well, I suppose if nothing else it means that they shouldn't have to completely refill it. Just a splash in the tanks  (Half full? Quarter?) and away you go. 

[tater]

Just now, SOXBLOX said:

So, for this mission, Starship will reenter itself, with the crew on board? And also, does it have enough ∆v to insert into LEO after returning from the Moon? 

That looks like crew from surface to surface.

No, it doesn't have the dv to propulsively enter LEO. Presuming it's a real SS (not LSS), it could aerobrake, though it might take a few passes, then it needs just a little to circularize.

 

 

[Deddly]

On 5/23/2021 at 6:58 PM, RealKerbal3x said:

Technically it would still need SL Raptors for control during ascent as RVacs don't have gimbal. (I don't think RCS would provide enough authority)

I believe the plan is to use differential thrust for control authority using the vacuum engines. 

[tater]

 

[Beccab]

2 minutes ago, tater said:

 

It also seems the final number of sections for the tower will need are 7, not 8, so 5 more to go (plus presumably the starship mating stuff itself)

[tater]

1 minute ago, Beccab said:

It also seems the final number of sections for the tower will need are 7, not 8, so 5 more to go (plus presumably the starship mating stuff itself)

Will be interesting to watch. Once they get this first one done, I think they will know the drill and it goes faster.

[RCgothic]

2 hours ago, sevenperforce said:

The math only BARELY works for this, I think....

 

What's the absolute minimum DV required for a lunar free return trajectory insertion? I know we generally use 3200m/s for TLI, but it varies a bit.

What's the absolute minimum?

[tater]

47 minutes ago, RCgothic said:

What's the absolute minimum DV required for a lunar free return trajectory insertion? I know we generally use 3200m/s for TLI, but it varies a bit.

What's the absolute minimum?

Might be able to shave 100 m/s off that?

[RCgothic]

2 hours ago, tater said:

Might be able to shave 100 m/s off that?

I did the math for the Hoffman transfer to lunar altitude. Although the altitude required for free return is difficult to calculate, the difference in DV required to lunar apogee or lunar perigee is only 10m/s, 3125m/s minimum from a 200km reference orbit to lunar minimum of 363230

So they aren't gaining the extra margin from timing.

[tater]

 

2 hours ago, RCgothic said:

I did the math for the Hoffman transfer to lunar altitude. Although the altitude required for free return is difficult to calculate, the difference in DV required to lunar apogee or lunar perigee is only 10m/s, 3125m/s minimum from a 200km reference orbit to lunar minimum of 363230

So they aren't gaining the extra margin from timing.

Yeah, so 75 m/s lower than the oft-quoted 3200 m/s. Some margin for midcourse corrections, etc.

[RCgothic]

A free return probably requires a little more altitude and therefore DV than an exact match to lunar altitude, but probably only on the order of a couple 10s of m/s.

So that's not where they're getting the margin from. Either they expend Superheavy for extra props to LEO (nominal time of MECO didn't seem to agree), Starship plus crew cabin outfitting weighs less than 120t, or Raptor has found some extra ISP.

[tater]

6 hours ago, RCgothic said:

So that's not where they're getting the margin from. Either they expend Superheavy for extra props to LEO (nominal time of MECO didn't seem to agree), Starship plus crew cabin outfitting weighs less than 120t, or Raptor has found some extra ISP.

I didn't even look at the times, lol.

You are right, #dearmoon and the SN20 test flight both have 171 seconds as sep.

That could be the nominal time, and dearmoon is not accurate. Else with fewer people maybe the tank is stretched?

[sevenperforce]

7 hours ago, RCgothic said:

A free return probably requires a little more altitude and therefore DV than an exact match to lunar altitude, but probably only on the order of a couple 10s of m/s.

So that's not where they're getting the margin from. Either they expend Superheavy for extra props to LEO (nominal time of MECO didn't seem to agree), Starship plus crew cabin outfitting weighs less than 120t, or Raptor has found some extra ISP.

Okay, so I think I may have figured it out.

The seminal work on free-return trajectories was written by Arthur Schwaninger in 1963 for Marshall Space Flight Center. Here's a screengrab from Technical Note D-1833:

While not noted in the image, these calculations assume a LEO parking orbit of 185 km, where the orbital velocity is 7,798 m/s. 

Paradoxically, a closer pass to the moon (which you'd think requires a lower apogee) actually requires more energy. Using the moon's gravity to boost you farther without actually coming as close to the moon, which means a higher periselenum, takes less energy. And we know that Elon said #dearMoon will go "farther from Earth than humans have ever gone" which suggests a higher periselenum.

For a co-rotation injection with a very high periselenum, the TLI can be as low as 3,052 m/s (or 3,060 m/s for the circumlunar path depicted in the #dearMoon graphic). They might be able to shave a little bit more off of that by using a lower parking orbit.

[tater]

10 minutes ago, sevenperforce said:

For a co-rotation injection with a very high periselenum, the TLI can be as low as 3,052 m/s (or 3,060 m/s for the circumlunar path depicted in the #dearMoon graphic). They might be able to shave a little bit more off of that by using a lower parking orbit.

So with some margin ~3100 m/s? We have to remember that there will always be some residuals in the tanks, as well, which counts as "dv" that might go unused (that or we add that residual mass to the dry mass plus landing props).

Seems pretty doable with an expended booster, but the #dearmoon implies some sort of soft touchdown for the booster since they use the same stage sep time as the first orbital test. The orbital test has a boostback, but NOT RTLS. I would imagine as a full test of SH, they want to send SS to orbit using the props SH would nominally use, so it likely touches down with residuals for that little bit of unused boostback with the at sea landing (ASDS or not).

Above I guestimated 3022 m/s with RTLS though, and that was a 120t SS that gets 150t of residuals to LEO. If it can get 160t residuals, it's got almost 3200 m/s. Basically, it needs to buy 10t via mass reduction from 120t, or additional payload. Maybe they can uses some 3mm steel?

[Geonovast]

4 minutes ago, tater said:

The orbital test has a boostback, but NOT RTLS.

Isn't that more of a safety/testing thing and not as a limitation of the launch profile?  Just because it's not doing RTLS doesn't mean it can't, right?  It could just be doing a short boostback so they can test landing without destroying... everything.

[RealKerbal3x]

3 minutes ago, Geonovast said:

Isn't that more of a safety/testing thing and not as a limitation of the launch profile?  Just because it's not doing RTLS doesn't mean it can't, right?  It could just be doing a short boostback so they can test landing without destroying... everything.

Given that the eventual plan is to have it be caught by the launch tower, it's definitely capable of full RTLS. They're only landing it in the sea here because, as you said, in the event of a landing failure, they don't want it to destroy the launch infrastructure they spent a year building. (Also, the tower catching arms probably won't be ready in time for the first orbital launch)

[tater]

9 minutes ago, Geonovast said:

Isn't that more of a safety/testing thing and not as a limitation of the launch profile?  Just because it's not doing RTLS doesn't mean it can't, right?  It could just be doing a short boostback so they can test landing without destroying... everything.

Yeah, I would assume they are doing a partial boostback for safety. Really is the best test to fully load the thing. It could RTLS, but won't for safety.

Looks like the free return is very, very close to doable for a 120t SS. And certainly possible with even a small mass reduction, or better than 150t of residuals.

[SpaceFace545]

26 minutes ago, RealKerbal3x said:

(Also, the tower catching arms probably won't be ready in time for the first orbital launch)

I really think this is just an empty musk tweet and probably won't happen. Like seriously how would any of this work. It seems like a massive rube Goldberg machine just to save a few tons that landing legs would cost.

[tater]

11 minutes ago, SpaceFace545 said:

I really think this is just an empty musk tweet and probably won't happen. Like seriously how would any of this work. It seems like a massive rube Goldberg machine just to save a few tons that landing legs would cost.

The original plan was literally to land on the launch mount. Catching it is less precise that that, and they manage to land F9 in the same space required (SH actually has a smaller footprint than F9), but the F9 core almost has to land in the inner circle or it's going for a swim).

Also, unlike F9, SH might well be able to hover. It can throttle to 880kN, which is pretty much a hover for SH assuming it masses ~200t at landing.

[RealKerbal3x]

22 minutes ago, SpaceFace545 said:

I really think this is just an empty musk tweet and probably won't happen. Like seriously how would any of this work. It seems like a massive rube Goldberg machine just to save a few tons that landing legs would cost.

They're really doing it. A job posting for Boca Chica a while back was asking for engineers to work on the 'Super Heavy launch/catch tower'. It's not just Elon rambling on twitter, they are working on it.

It does have quite a few merits over just landing it, as well. Catching the booster means that they can move all of the shock absorber hardware out of the vehicle (where it's mass-limited) into the launch tower (where it can be as massive as they want). It also allows the booster to be moved directly back to the launch mount after landing, which is great for rapid reusability.

Now, the whole 'catch Starship without a landing burn' thing was definitely just Elon thinking out loud. But catching Super Heavy is a genuine project.

Edited May 25, 2021 by RealKerbal3x

[sh1pman]

14 minutes ago, RealKerbal3x said:

Super Heavy launch/catch tower

Should call it "The Trampoline".

[CatastrophicFailure]

Just now, sh1pman said:

Should call it "The Trampoline".

Operate the shock absorber mechanism in reverse at the moment of liftoff and pick up a tiny performance noose too.