Anyone suggesting to use drag to settle the propellant might want to take out an envelope and do a quick estimate of the associated heating...
In October 2020 up-thread, the best guess was that the necessary settling acceleration is on the order of 1e-4 m/s2.
Also, keeping the depot pointing toward the sun to minimize illuminated surface area and therefore boil-off is probably important. Either way, using drag for the acceleration doesn't really make sense, because the ship would have to be brought up to altitude again using propulsion. Besides, boil-off will create ullage gas that will have to be vented anyway; they might as well use it for settling.
I agree, logically VLEO is the best place for refilling, at least for the lowest "rung" of any refilling ladder. However I'm not convinced about using drag for propellant settling.
You may find this graph handy: https://space.stackexchange.com/questions/18223/where-can-i-find-data-for-atmospheric-density-vs-altitudeAt hypersonic velocities you can assume a Cd of 2.2, so for a 9 m diameter 100 tonne vehicle you achieve 1 mm/s2 at just over 100 km altitude, and for a 2,000 tonne vehicle it's almost exactly 100 km.
differential drag on the 350t refill ship with the 2300t almost full StarshipV3, as their drag acceleration will be different, the heavier one at 0.1mm/s2 and the lighter one at .66mm/s2. The difference being .56mm/2.F=ma, so the differential force on the lighter one will be 200N. I would assume a mating connector can handle 200N.
Quote from: InterestedEngineer on 12/07/2024 12:47 amdifferential drag on the 350t refill ship with the 2300t almost full StarshipV3, as their drag acceleration will be different, the heavier one at 0.1mm/s2 and the lighter one at .66mm/s2. The difference being .56mm/2.F=ma, so the differential force on the lighter one will be 200N. I would assume a mating connector can handle 200N.I had assumed that, short of a total redesign of the plumbing, the settling acceleration should be axial. If so, then that differential drag creates a torque ≈2kNm. (assuming 1m of docking hardware, or else to 1sf)
I'm curious whether this VLEO co-incidentally (depending on solar weather) is close to the altitude corresponding to 16 complete orbits each node (<24h per sidereal rotation and precession), and therefore daily launch opportunities with zero phasing. (Not at all, it seems that's ≈208km for 28.6⁰ inclination)
Quote from: eriblo on 12/06/2024 10:23 pmAnyone suggesting to use drag to settle the propellant might want to take out an envelope and do a quick estimate of the associated heating...Sure, why not. The deltaV is 7m/sec over a period of 2 hours.KE of say 500t combined Starships + fuel at 7,814 m/sec is 15,264,649,000,000 joules (3.6kt of TNT)KE of say 500t of combined Starships + fuel at (7814 - 7) mm/sec is 15,237,312,250,000 joules (3.6kt of TNT)That's a net change of -27,336,750,000 over 7200 seconds, or 3.8e6 joules/second (3.8MW)Like re-entry heating, 90% of the heat is dissipated in the passing particles, not on the heat shield tiles themselves, so that leaves us with 380kW of heat flux for the whole Starship.Assuming the tiles have the same heat capacity as the Space shuttle (628 J/kg-K), and there are 10,000 kg of tiles on a Starship, that's a heating rate of 380kJ/sec / 10,000kg / 628 J/kg-K = 0.06degK/sec.Now the surface area of the heat shield tiles on Starship is roughly 400m2, which gives us a heat flux of 380kW/400 = 950 watts/m2We now have to calculate the stefan-boltzmann equilibrium of the tiles that gets us to a heat flux of -950W/m2 and thus equilibrium.Let's say they start at 100K, How much do they have to heat up for them to emit 950w/m2?The answer is they must be heated to 360K. That's going to be hardly any heat at all coming through the tiles to the tank, they are below the boiling point of water and their thermal conductivity is very low. (there's also an insulation blanket and back up heat shield behind them).It will take 260K / .06K/sec = 4333 seconds to engage the full heat capacity of the tiles. This suggests that they will never reach the 360K equilibrium, but the calculation for the actual final temperature is a 4th order polynomial and I really don't feel like solving one of those right now.TL;DR - no problemo on heating from the drag.
Quote from: InterestedEngineer on 12/07/2024 12:18 amQuote from: eriblo on 12/06/2024 10:23 pmAnyone suggesting to use drag to settle the propellant might want to take out an envelope and do a quick estimate of the associated heating...Sure, why not. The deltaV is 7m/sec over a period of 2 hours.KE of say 500t combined Starships + fuel at 7,814 m/sec is 15,264,649,000,000 joules (3.6kt of TNT)KE of say 500t of combined Starships + fuel at (7814 - 7) mm/sec is 15,237,312,250,000 joules (3.6kt of TNT)That's a net change of -27,336,750,000 over 7200 seconds, or 3.8e6 joules/second (3.8MW)Like re-entry heating, 90% of the heat is dissipated in the passing particles, not on the heat shield tiles themselves, so that leaves us with 380kW of heat flux for the whole Starship.Assuming the tiles have the same heat capacity as the Space shuttle (628 J/kg-K), and there are 10,000 kg of tiles on a Starship, that's a heating rate of 380kJ/sec / 10,000kg / 628 J/kg-K = 0.06degK/sec.Now the surface area of the heat shield tiles on Starship is roughly 400m2, which gives us a heat flux of 380kW/400 = 950 watts/m2We now have to calculate the stefan-boltzmann equilibrium of the tiles that gets us to a heat flux of -950W/m2 and thus equilibrium.Let's say they start at 100K, How much do they have to heat up for them to emit 950w/m2?The answer is they must be heated to 360K. That's going to be hardly any heat at all coming through the tiles to the tank, they are below the boiling point of water and their thermal conductivity is very low. (there's also an insulation blanket and back up heat shield behind them).It will take 260K / .06K/sec = 4333 seconds to engage the full heat capacity of the tiles. This suggests that they will never reach the 360K equilibrium, but the calculation for the actual final temperature is a 4th order polynomial and I really don't feel like solving one of those right now.TL;DR - no problemo on heating from the drag.I think those are some optimistic assumptions. For aerobraking it's typically 90% of the energy that goes into the vehicle skin rather than into the gas. For Starship the ballistic coefficient is high enough that you might get a decent shock but it will still be a higher fraction than on reentry (and if you have shocks you have hot spots).
There is also the problem of relying on the tanker heat shield to take all the heating. That orientation requires extra internal piping and is likely to be unstable with partially filled tanks. If you go tail first there is a lot less area and likely less insulation.
Any part of the depot tankage exposed to the flow will be hampered by having low emissivity and will get significantly hotter.
The biggest problem with using atmospheric drag is of course that you don't gain anything because you'll lose more propellant compensating than you save.
Quote from: eriblo on 12/07/2024 06:29 pmQuote from: InterestedEngineer on 12/07/2024 12:18 amQuote from: eriblo on 12/06/2024 10:23 pmAnyone suggesting to use drag to settle the propellant might want to take out an envelope and do a quick estimate of the associated heating...Sure, why not. The deltaV is 7m/sec over a period of 2 hours.KE of say 500t combined Starships + fuel at 7,814 m/sec is 15,264,649,000,000 joules (3.6kt of TNT)KE of say 500t of combined Starships + fuel at (7814 - 7) mm/sec is 15,237,312,250,000 joules (3.6kt of TNT)That's a net change of -27,336,750,000 over 7200 seconds, or 3.8e6 joules/second (3.8MW)Like re-entry heating, 90% of the heat is dissipated in the passing particles, not on the heat shield tiles themselves, so that leaves us with 380kW of heat flux for the whole Starship.Assuming the tiles have the same heat capacity as the Space shuttle (628 J/kg-K), and there are 10,000 kg of tiles on a Starship, that's a heating rate of 380kJ/sec / 10,000kg / 628 J/kg-K = 0.06degK/sec.Now the surface area of the heat shield tiles on Starship is roughly 400m2, which gives us a heat flux of 380kW/400 = 950 watts/m2We now have to calculate the stefan-boltzmann equilibrium of the tiles that gets us to a heat flux of -950W/m2 and thus equilibrium.Let's say they start at 100K, How much do they have to heat up for them to emit 950w/m2?The answer is they must be heated to 360K. That's going to be hardly any heat at all coming through the tiles to the tank, they are below the boiling point of water and their thermal conductivity is very low. (there's also an insulation blanket and back up heat shield behind them).It will take 260K / .06K/sec = 4333 seconds to engage the full heat capacity of the tiles. This suggests that they will never reach the 360K equilibrium, but the calculation for the actual final temperature is a 4th order polynomial and I really don't feel like solving one of those right now.TL;DR - no problemo on heating from the drag.I think those are some optimistic assumptions. For aerobraking it's typically 90% of the energy that goes into the vehicle skin rather than into the gas. For Starship the ballistic coefficient is high enough that you might get a decent shock but it will still be a higher fraction than on reentry (and if you have shocks you have hot spots).Your criticism is correct, further research shows there's no bow shock above what I think the broadcast commenters call the "entry interface", or about 100km or so, so 90% of the deceleration energy is going to go into the heat shield above 100km. It switches to 10% below the entry interface as the bow shock forms.You know what they say about assumptions...Since it appears we should be surfing the thermosphere in the 120-160km range to get the desired acceleration of 0.1mm/s2, I shall redo the calculations.Let's stick to the 2 hour refuel, which is probably pessimistic (27kg/sec seems really slow). We'll go axial, nose first, so the bare steel is protected. We don't have to worry about non-impinging surfaces heating, there's no bow shock.That's a deltaV of 7200s * 0.0001m/s2 = 0.72m/sec. Thanks for that order magnitude of margin!KE of say 500t combined Starships + fuel at 7,814 m/sec is 15,264,649,000,000 joules (3.6kt of TNT)KE of say 500t of combined Starships + fuel at (7814 - .721) mm/sec is 15,261,832,182,960 joules (3.6kt of TNT)That's a net change of -2,816,817,040 joules over 7200 seconds, or 391 kj/second (391kW).90% of that goes to heat shield, 10% to the passing particles, leaving us with a net heat flux of 350kW.I note if we are not refueling axial but rather heat-shield-facing-forward, the result is nearly identical to above, about 360K equilibrium temperature. Thanks for that order of magnitude smaller settling acceleration.Let's stick with axial tail-to-tail, since that seems to be the most likely without significant plumbing changes. The nose has surface area of 64m2, so the heat flux is 5.5kw/m2The equilibrium temperature is now 560K. This is well within the range of the tiles. furthermore the heat is in the nose cone, with vacuum separation from the header tanks, and tiles have a layer of insulation underneath them. The tiles will do their job, and the tanks won't heat up at all.TL;DR - heat build up is still not a problem when accounting for the non-bowshock behavior of surfing the thermosphere, because the requirements dropped by an order of magnitude, saving my bacon (which alas will be undercooked)
You know, even if you can refuel using the atmosphere to settle fuel, you'll have to pay it all back when you do a burn to get back to a stable orbit. Using thrusters for settling, on the other hand, will give you a small boost towards your destination.This is to say nothing of exciting problems such as aerodynamic occlusion resulting in one vehicle seeing more drag than the other. And of course, since the vehicles will be different masses due to the weight of the fuel, they'd be slowed at different rates even if the drag was identical.
Quote from: Twark_Main on 12/06/2024 11:43 pmI agree, logically VLEO is the best place for refilling, at least for the lowest "rung" of any refilling ladder. However I'm not convinced about using drag for propellant settling.why won't drag work for propellant settling?
Quote from: InterestedEngineer on 12/07/2024 12:25 amQuote from: Twark_Main on 12/06/2024 11:43 pmI agree, logically VLEO is the best place for refilling, at least for the lowest "rung" of any refilling ladder. However I'm not convinced about using drag for propellant settling.why won't drag work for propellant settling?I didn't say it won't work. I'm just unconvinced that it's the best solution.By flying so low you're incurring a lot of unnecessary drag whenever you're not currently in the process of refilling. However in all your calculations you're only calculating the loss during that 2 hour duration ("paying back 0.72m/s2 of deltaV"). You actually need to use the total duration of flight, integrated over the density of the atmosphere.
As for Oberth, you can still get all those same benefits without using drag to settle your propellant. You can simply wait for your orbit to decay down to 120 km (or whatever) and then perform the TMI burn. If you counter that the wait is too long, then that's actually good news because it means you can lower your original parking orbit.
Quote from: Narnianknight on 12/06/2024 08:44 pmIn October 2020 up-thread, the best guess was that the necessary settling acceleration is on the order of 1e-4 m/s2. Cool, thanks for the numbers! That means much less deltaV will be needed, so any heating from drag will be negligible, and the refuel can take place at a much higher altitude than 100km.QuoteAlso, keeping the depot pointing toward the sun to minimize illuminated surface area and therefore boil-off is probably important. Either way, using drag for the acceleration doesn't really make sense, because the ship would have to be brought up to altitude again using propulsion. Besides, boil-off will create ullage gas that will have to be vented anyway; they might as well use it for settling.I'm not quite sure how you conclude that it doesn't make sense, you just made the problem 10 times easier than my calculations above show. thanks for the 10x margin... but why won't it work?As far as pointing heat shield at the sun, if we are going to belly to belly one heat shield is pointed in the direction of travel and the other heat shield is pointed the opposite, so the sun will never hit the large unshielded surface area, so good news there.Note that the heat flux for the sun was similar to what I calculated for the drag - you probably won't hit the heat shield thermal capacity in two orbits.
Quote from: Twark_Main on 12/07/2024 10:25 pmQuote from: InterestedEngineer on 12/07/2024 12:25 amQuote from: Twark_Main on 12/06/2024 11:43 pmI agree, logically VLEO is the best place for refilling, at least for the lowest "rung" of any refilling ladder. However I'm not convinced about using drag for propellant settling.why won't drag work for propellant settling?I didn't say it won't work. I'm just unconvinced that it's the best solution.By flying so low you're incurring a lot of unnecessary drag whenever you're not currently in the process of refilling. However in all your calculations you're only calculating the loss during that 2 hour duration ("paying back 0.72m/s2 of deltaV"). You actually need to use the total duration of flight, integrated over the density of the atmosphere.I'm pretty sure I multiplied out the drag by calculating a drop of 1km per two orbits, then guessing at one refill every 6 orbits, so 60 orbits to fill Starship V3 to 2000t of prop and a total of 30km drop. Did I do the math wrong?
Is 60 orbits, or over 4 days too optimistic?
for station keeping one could burp an engine for a second and throw away a half a ton of propellant to get a few m/s to raise the orbit, it's still cheaper to do that at double the Isp of cold gas thrusters.
Quote from: Twark_Main on 12/07/2024 10:25 pmAs for Oberth, you can still get all those same benefits without using drag to settle your propellant. You can simply wait for your orbit to decay down to 120 km (or whatever) and then perform the TMI burn. If you counter that the wait is too long, then that's actually good news because it means you can lower your original parking orbit. Once you've paid the expense to get the fuel up over 200km, there's no Oberth benefit.
Make sure you use a C3 calculation, I got fooled when I didn't.
If you want to refill below 200km, you are back in in the regime of "you have atmospheric drag, you might as well use it".
The greatest benefit turns out to be the not having the lift the fuel from the surface of the earth to over 200km, but to an average of 135km instead.
The second benefit is not wasting tons of prop on cold gas thrusters.
Quote from: InterestedEngineer on 12/08/2024 02:03 amI'm pretty sure I multiplied out the drag by calculating a drop of 1km per two orbits, then guessing at one refill every 6 orbits, so 60 orbits to fill Starship V3 to 2000t of prop and a total of 30km drop. Did I do the math wrong?Yes, you only multiplied by 7200 s (2 hours), not 6 hours.
I'm pretty sure I multiplied out the drag by calculating a drop of 1km per two orbits, then guessing at one refill every 6 orbits, so 60 orbits to fill Starship V3 to 2000t of prop and a total of 30km drop. Did I do the math wrong?
For sending thousands of Starships to Mars per launch window, yes it's hilariously optimistic.
Quote from: Twark_Main on 12/08/2024 02:36 amQuote from: InterestedEngineer on 12/08/2024 02:03 amI'm pretty sure I multiplied out the drag by calculating a drop of 1km per two orbits, then guessing at one refill every 6 orbits, so 60 orbits to fill Starship V3 to 2000t of prop and a total of 30km drop. Did I do the math wrong?Yes, you only multiplied by 7200 s (2 hours), not 6 hours.No, I converted a constant drag of 0.1mm/sec to altitude drop per orbit, which is about 500m/orbit. Sorry if that was confusing.
That's a deltaV of 7200s * 0.0001m/s2 = 0.72m/sec.
We are paying back 0.72m/s2 of deltaV, rounding error
Quote from: Twark_Main on 12/08/2024 02:36 amQuote from: InterestedEngineer on 12/08/2024 02:03 amIs 60 orbits, or over 4 days too optimistic?For sending thousands of Starships to Mars per launch window, yes it's hilariously optimistic.Only passenger ships need a full fuel drop. Cargo ships need half of that, they take the slow route, so it's only 1/10 ships that need the full load.
Quote from: InterestedEngineer on 12/08/2024 02:03 amIs 60 orbits, or over 4 days too optimistic?For sending thousands of Starships to Mars per launch window, yes it's hilariously optimistic.
You think it'll take well over 4 days to fill up one passenger ship's tanks? Passengers aren't going to wait around that long anyways, the passengers will hate it.
If it's 100 ships that's 1000 refuelings over 4 days or 250/day, which will keep 250 starships busy at refueling, assuming they have a 24 hour cycle, and about 32 boosters and launch mount pairs if cycle time is 3 hours. That doesn't sound completely insane by Elon standards.
With a launch window of 3 weeks (21 days), that gives 5 iterations of this, and because the fuel for cargo is a half load taking the slow route to Mars, there's your 10x. 32 booster/launch mount pairs, 250 fuelers, and add 10% for inevitable snafus.
Shorter is also much better for boil off reasons. I don't think we want things sticking around in VLEO for very long due to that alone, so an orbital drop of 8km/day seems reasonable.