That's what I did, I took the equilibrium and "goal seeked" it. You can make a table with the two going in opposite directions. You can write a Python script for that. modern LLMs can write a python script (but I always verify the result).
I had already assumed the heat shield faced earth shine - using the boiled methalox for that kind of positioning.
Note you just point the nose at the sun ALL THE TIME and I'm pretty sure that's the optimal orientation, and then you rotate around the axis to keep the heat shield pointed at the earth. which is at its worst when 90 degrees from the sun (at the terminator). When you are on the opposite side of the sun the nose is facing the earth, when you are on the side of the sun the tail is facing the earth. You angle it just enough to hide the bare steel section, no more. That's why I went with an average of 200W/m^2.
As far as TUFI, I used the Shuttle tiles, which was 0.05 W/m-K at room temperature - is TUFI really that much worse? Maybe you used the high temperature thermal conductivity?
Quote from: InterestedEngineer on 10/29/2025 11:24 pmI had already assumed the heat shield faced earth shine - using the boiled methalox for that kind of positioning.Do we have an order-of-magnitude of how much propellant this would require?
Quote from: TheRadicalModerate on 10/29/2025 07:45 pmQuote from: InterestedEngineer on 10/28/2025 03:50 pmAssume we face the sun, so the sunshine never hits the tanks, so we only have to deal with earthshine.Note that this requires some amount of attitude control to oppose tidal forces. It's small but probably not trivial.Well, the boiled methalox has to go *somewhere*, attitude control seems like a good place for it.
Quote from: InterestedEngineer on 10/28/2025 03:50 pmAssume we face the sun, so the sunshine never hits the tanks, so we only have to deal with earthshine.Note that this requires some amount of attitude control to oppose tidal forces. It's small but probably not trivial.
Assume we face the sun, so the sunshine never hits the tanks, so we only have to deal with earthshine.
Quote from: InterestedEngineer on 10/28/2025 03:50 pmAssume we face the sun, so the sunshine never hits the tanks, so we only have to deal with earthshine.Note that this requires some amount of attitude control to oppose tidal forces. It's small but probably not trivial.QuoteNow the tank area facing the earth is about 270m2 facing 100W of earthshine. assuming none of this is radiated back out (it will be), that's a rate of heat absorption of 27kW.Just taking the Google AI answer at face value, IR re-radiation from Earth is 150-350W/m², average 230W/m². Average Earth albedo is about 0.3,¹ which would translate to ~400W/m². So your average luminance from Earth is 630W/m².QuoteOTOH, stefan-boltzmann for the entire starship at the boiling point of methane (130K) is -10kW/m2.The tiles themselves will have a heat gradient on them and stay at 170K while conducting little heat to the inside, so half the ship is really at -20kW/m2. So that's 34kW/m2 net just to have the entire starship at boiling point of methane (an equilibrium).I'm on shaky ground here, but I think you're doing this wrong. Yes, the skin itself will be at 90K-130K, but you can't really calculate the emittance without knowing the thermal resistance of the insulation (SOFI, MLI, who knows?), which should give the gradient, and eventually the outer surface temperature. The net result should give a better result than what you're estimating, i.e., the outside of the insulation should be hotter than you estimated.________¹I'm unclear if that number is for the whole Earth, or just the illuminated side. If the former, then my back-of-napkin should be about right. If the latter, divide by roughly 2.
Now the tank area facing the earth is about 270m2 facing 100W of earthshine. assuming none of this is radiated back out (it will be), that's a rate of heat absorption of 27kW.
OTOH, stefan-boltzmann for the entire starship at the boiling point of methane (130K) is -10kW/m2.The tiles themselves will have a heat gradient on them and stay at 170K while conducting little heat to the inside, so half the ship is really at -20kW/m2. So that's 34kW/m2 net just to have the entire starship at boiling point of methane (an equilibrium).
Quote from: InterestedEngineer on 10/29/2025 11:14 pmQuote from: TheRadicalModerate on 10/29/2025 07:45 pmQuote from: InterestedEngineer on 10/28/2025 03:50 pmAssume we face the sun, so the sunshine never hits the tanks, so we only have to deal with earthshine.Note that this requires some amount of attitude control to oppose tidal forces. It's small but probably not trivial.Well, the boiled methalox has to go *somewhere*, attitude control seems like a good place for it.A better place for it to go is into a cryocooler.
Quote from: Twark_Main on 10/29/2025 11:39 pmQuote from: InterestedEngineer on 10/29/2025 11:24 pmI had already assumed the heat shield faced earth shine - using the boiled methalox for that kind of positioning.Do we have an order-of-magnitude of how much propellant this would require?well the boil rate is about 0.2kg/sec, at an Isp of 70 that's a thrust of 137N.That should be plenty. I would assume most of it will be wasted on offsetting outgassing.
Quote from: TheRadicalModerate on 10/30/2025 03:52 amA better place for it to go is into a cryocooler.That's a shoo in for the depot, but for a tanker or an end user ship?
A better place for it to go is into a cryocooler.
For an end user ship, if not needed elsewhere, leave the depot attached for cryo services. This could be some combination of circulating propellant for cooling and providing shade until its time for the end user to light off and go.
During a depot filling campaign the most reasonable goal for the tanker is to git 'er in and git 'er transferred as fast as possible. My takeaway from the settling discussion of the last week or two is the tanker approach has to be slow and gentle. Twisting about to minimize thermal input seems to be at variance with this.
Related would be tank pressure, specifically in the tankers. Has Elon shared any numbers from later tank tests to destruction? More mass towards higher tank pressure would give a higher boiloff temp but enough to be worthwhile? The hassle of special handling in the sausage factory has to be part of the equation.
For solar panel calcs total insolation at earth distance is 1.3kW in orbit and 1.0 at the surface. This may include near IR - or maybe not. Reflected light would be for half the orbit while reemitted IR would be for the entire orbit. The 230W/m^2 average sounds like a useful BoE place holder for the entire orbit but cloud cover, specular reflections and high vs low beta orbits would add so much variability that every situation would be a situation.
Quote from: OTV Booster on 10/30/2025 03:40 pmFor solar panel calcs total insolation at earth distance is 1.3kW in orbit and 1.0 at the surface. This may include near IR - or maybe not. Reflected light would be for half the orbit while reemitted IR would be for the entire orbit. The 230W/m^2 average sounds like a useful BoE place holder for the entire orbit but cloud cover, specular reflections and high vs low beta orbits would add so much variability that every situation would be a situation.I think you're mixing apples and oranges here. There are two separate issues:1) How much irradiance (flux) is the Starship receiving, in W/m²? Three contributors to this:a) Solar flux (1362W/m² average). Received when not in eclipse.b) The outgoing longwave radiation (OLR) emitted by the Earth. Received always in LEO. This is the ~230W/m². Received all the time in LEO.c) Albedo. Mostly shortwave, and measured as a coefficient = incident/reflected radiation = ~0.3. My question is whether that's an average across all phases of Earth illumination, or whether it's normalized to the full Earth.2) Emittance from Starship. That's strictly a function of temperature at the exterior of the Starship, which is a function of the gradient, which is a function of tank temperature and the thermal conductivity (or resistance, take your pick).Heat absorbed by the Starship is (irradiance - emittance)*A.
There's also the idea that the tank is what matters. The cross section of steel between the nose cone and the tanks is very small and 304L stainless is a poor heat conductor as well, so that's why you can round off the solar flux by pointing the nose at the sun.
The new HLS info published by SpaceX and the updated "moon" section contain the following two attached pictures.Two important things to note:1) The depot has the sockets and the tanker has the struts, which is kinda surprising to me. I can't figure out why this would be useful. Maybe the sockets weigh more?
2) The tanker and the depot are offset from one another, so the tanker will be farther forward than the depot when they're docked. I assume that implies that there really are two QDs on the depot, one for launch and one, farther forward, for refueling.
Note that there's no sign of a QD in the artwork.
Quote from: TheRadicalModerate on 10/30/2025 08:53 pmNote that there's no sign of a QD in the artwork.Refilling through the HSR struts and docking struts confirmed.
Is this the first time there's been an official render of a depot?
I think it's just a trick of the camera angle. If you trace out the ellipse the separation plane on both vehicles, it looks like they're just next to each-other. Just a couple grain silos on flat ground.
The Depot Starship has 4 extra panels that are not visible on the HLS Starship, are these fold out radiators?
Starship V3 vehicles come equipped with docking ports and can be configured to act as tanker vehicles with the addition of docking probes. Starship also has a connection point where propellants are loaded onto the vehicle in preparation for launch that has been updated to enable on-orbit propellant transfer.
I had already assumed the heat shield faced earth shine - using the boiled methalox for that kind of positioning. Note you just point the nose at the sun ALL THE TIME and I'm pretty sure that's the optimal orientation, and then you rotate around the axis to keep the heat shield pointed at the earth. which is at its worst when 90 degrees from the sun (at the terminator). When you are on the opposite side of the sun the nose is facing the earth, when you are on the side of the sun the tail is facing the earth. You angle it just enough to hide the bare steel section, no more. That's why I went with an average of 200W/m^2.
Quote from: InterestedEngineer on 10/29/2025 11:24 pmI had already assumed the heat shield faced earth shine - using the boiled methalox for that kind of positioning. Note you just point the nose at the sun ALL THE TIME and I'm pretty sure that's the optimal orientation, and then you rotate around the axis to keep the heat shield pointed at the earth. which is at its worst when 90 degrees from the sun (at the terminator). When you are on the opposite side of the sun the nose is facing the earth, when you are on the side of the sun the tail is facing the earth. You angle it just enough to hide the bare steel section, no more. That's why I went with an average of 200W/m^2.This is not as easy as you think. Either a massive gyro system (mass penalty), or near constant small thrusters (fuel cost). It would be interesting to do the actual calculations, but shouldn't be assumed to be cheap.