Quote from: OTV Booster on 01/14/2023 07:16 pmI'm assuming that an 'ideal' mature transfer scenario allows 6 hours (21,600 s) for transfer. The depots orbit can be matched up every 12 hours, assuming a single launch site. Figure three hours for approach and docking and the same to undock and clear.I think the single site has two launch opportunities a day, but they are not 12 hours apart. There are also two landing opportunities a day. This assumes the Depot orbit crosses the latitude of the launch site. Am I missing something? If those two launch opportunities are far enough apart, then you still get an average of 12 hours for the full transfer operation. Each tanker will need some time (maybe 24 hours?) to sync with Depot's location in its orbit, and will need maybe the same amount of time after completion of fueling to move to the correct spot along the orbit to start its deorbit. With all of this, I guess each tanker mission is about three days, so you need six tankers to maintain a 12-hour cadence. I'm pretty sure you also need separate launch and catch towers.

I'm assuming that an 'ideal' mature transfer scenario allows 6 hours (21,600 s) for transfer. The depots orbit can be matched up every 12 hours, assuming a single launch site. Figure three hours for approach and docking and the same to undock and clear.

Surely you mean sun shielding, not insulation, since the transfer is occurring in a vacuum?

I'm assuming that an 'ideal' mature transfer scenario allows 6 hours (21,600 s) for transfer. The depots orbit can be matched up every 12 hours, assuming a single launch site. Figure three hours for approach and docking and the same to undock and clear.I'm also assuming that only the depot will have active cooling and that PV and maybe radiators will have to tuck away during transfer ops. These assumptions point to wanting a loading campaign done either as fast as possible (a tanker every 12 hours) or very relaxed with a chance to literally chill between loads. There is no aspect of Elon that is relaxed so a fast cadence looks like a good bet. Fast also means less wear and tear on the panel deployment mechanism. There's a lot of assumptions here but you might think about tucking some of them into your noodling.

My key takeaway/question is whether propellant loss due to milli-G acceleration is the only reason to do prop transfers as fast as the system allows. Everything else (I think) pushes you towards taking your time - you probably wouldn't launch a tanker more than once a day (at least at first) so a 6+ hour prop transfer isn't going to be the limiting factor, and lower transfer speed means less sloshing/swirling in the tank.

Quote from: mikelepage on 01/15/2023 03:11 amMy key takeaway/question is whether propellant loss due to milli-G acceleration is the only reason to do prop transfers as fast as the system allows. Everything else (I think) pushes you towards taking your time - you probably wouldn't launch a tanker more than once a day (at least at first) so a 6+ hour prop transfer isn't going to be the limiting factor, and lower transfer speed means less sloshing/swirling in the tank.For the parameters you have set here, the transfer mass flow is only 6.9kg/s. Also, the longer you take at a particular acceleration, the more delta-v you wind up spending. If you take 21,600s at 5mm/s², you're expending more than 1600m/s of delta-v. Your depot/tanker will wind up in a much higher orbit, and both vehicles will have to spend a bunch of delta-v to get back to the depot's parking orbit.

If we need nearly continuous low acceleration and we think this will waste too much propellant, perhaps we need to look a high efficiency thrusters. Specifically, ion thrusters of some sort. Something like the Gateway PPE. This would be part of the Depot, not the tankers, so the cost of lifting it to orbit is only incurred once. Ideally, we would use some magical thruster that uses CH_{4} or O_{2} instead of Xenon.Question: how much thrust is needed to maintain 0.5 mm/s^{2} for a max-loaded Depot/tanker combo?

I think the maneuver during propellant transfer is a big circle, so the net change in position is zero. This introduces a very small lateral acceleration which should not affect the transfer.

Quote from: DanClemmensen on 01/15/2023 03:39 amQuestion: how much thrust is needed to maintain 0.5 mm/s^{2} for a max-loaded Depot/tanker combo?F = ma, baby. m = 95t + 1450t + 120t + 150t = 1,815,000kg. a = 0.005m/s². Thrust = 9075N.Electric propulsion usually generates 60mN/kW. So you'd need 151MW of power to drive the coupled system. Nope.

Question: how much thrust is needed to maintain 0.5 mm/s^{2} for a max-loaded Depot/tanker combo?

Electric propulsion usually generates 60mN/kW. So you'd need 151MW of power to drive the coupled system. Nope.

Quote from: TheRadicalModerate on 01/15/2023 04:43 amQuote from: DanClemmensen on 01/15/2023 03:39 amQuestion: how much thrust is needed to maintain 0.5 mm/s^{2} for a max-loaded Depot/tanker combo?F = ma, baby. m = 95t + 1450t + 120t + 150t = 1,815,000kg. a = 0.005m/s². Thrust = 9075N.Electric propulsion usually generates 60mN/kW. So you'd need 151MW of power to drive the coupled system. Nope.Dan asked about half a millimeter per second squared, while you calculated for five. So you would only need 15 megawatt."Only"...So your conclusion is correct anyway; that is still a "Nope", probably even a "No way in h***".Heck, even an empty Starship at 120 tonne would need 60 N thrust for that acceleration, which would require 1 MW electric power, which is probably about an order of magnitude more than you can expect from solar panels carried on a Starship...

Quote from: OTV Booster on 01/14/2023 07:16 pmI'm assuming that an 'ideal' mature transfer scenario allows 6 hours (21,600 s) for transfer. The depots orbit can be matched up every 12 hours, assuming a single launch site. Figure three hours for approach and docking and the same to undock and clear.I'm also assuming that only the depot will have active cooling and that PV and maybe radiators will have to tuck away during transfer ops. These assumptions point to wanting a loading campaign done either as fast as possible (a tanker every 12 hours) or very relaxed with a chance to literally chill between loads. There is no aspect of Elon that is relaxed so a fast cadence looks like a good bet. Fast also means less wear and tear on the panel deployment mechanism. There's a lot of assumptions here but you might think about tucking some of them into your noodling.Thanks, and agreed that it's a huge trade space. I've also been assuming that only the depot will have active cooling, and will in fact have a number of optimisations for keeping prop cool that it doesn't make sense to install on every Starship (OT: whether that would lead to sending a single depot with each fleet of Starships that makes the Earth-Mars transit is an interesting question to me: the ability to keep one ship uncrewed and cold the whole trip seems like a good energy saving measure, then also, you have the capability to top up the rest of the fleet's header tanks with fresh cryo prop in the days before landing).Just attaching a screenshot of what happens if you plug 21,600s into TheRadicalModerate's spreadsheet. It now requires 15 tanker trips to fill a depot, and thats with 300s hot gas thrusters going the whole time to create 5mm/s^{2}. I somehow doubt that losing nearly 40% of the propellant launched is really acceptable. With 70s ullage thrust, the transfers lose so much prop that the process never completes. If you go to 1mm/s^{2} and 300s thrust you get to 5.7% losses over 12 tanker loads which is (maybe) acceptable.My key takeaway/question is whether propellant loss due to milli-G acceleration is the only reason to do prop transfers as fast as the system allows. Everything else (I think) pushes you towards taking your time - you probably wouldn't launch a tanker more than once a day (at least at first) so a 6+ hour prop transfer isn't going to be the limiting factor, and lower transfer speed means less sloshing/swirling in the tank.

On geysering and slosh reduction: The early Jupiter missiles used a system of floating bodies inside the tank covering the propellant/ullage gas interface to reduce propellant slosh (after observation that transporting water tanks with lumber floating in them significantly reduced slosh) prior to the introduction of anti-slosh and anti-vortex baffles to the tank walls. I've previously suggested this floating body system may be repurposed as an insulator layer for reduction of ullage gas heating, but it could also serve to reduce the effects of geysering from propellant pumping, by absorbing stream and droplet energy in trying to move the more massive (in discrete terms) floats rather than allowing droplets to continue unimpeded.

OK, let's start over. Assume we need 5 mm/s^{2} accelleration, and we want to use centrepital force. We mount the tanker atop the depot and then sling them from a cable with tanker toward the center of rotation, and start spinning. We need another mass at the far end of the cable. let's say that otehr mass is 1/10 the mass of our tanker+depot+propellant. For a cable that is 1100 meters, the tanker+depot CoM will be 100 meters from the center of rotation. The formula is accel = (rotational rate squared) * radiusI get a rotational rate of 0.42 RPM. Is this correct?Tension in the cable in Newtons is mass the complex times 0.005 m/s^{2}, which is not a lot.Practical considerations: How to connect the cable to the depot so as to still allow the tanker to dock atop it? I think this will require a wide set of spreaders. CoM must stay "below" the attachment point even for a full tanker on an empty depot unless the tanket is stabilized to the cable using the tanker's chopstick catchpoints. How to spin up after docking the tanker and spin down before undocking? I think thrusters on the depot. How to keep the cable from tangling when not spinning? I think make it slightly rigid, possibly using a pressurized pipe. What is the right length and the right mass for the counterweight? I think 100 meters is too short for CoM to center of spin since the gradient is too high (tanker too light, depot too heavy). A longer cable means a lower rotation rate. How much energy is needed to spin up and spin down? ( too much math for me right now).

While the concept looks promising in theory, tethers are hard. Your inflatable tube tether is ingenious and has the seeds of a deployment system embedded in it. Deployment is one of the really hard parts.

It occurred to me that, if you engineer your pumping power so that it achieves the needed flow rate but causes the liquid to exit the outlet pipe at exactly ambient pressure, then there's no geysering possible.You can figure out that pump power via the Poiseuille equation,¹ which tells you the pressure drop from the pump outlet to the ambient fill pipe outlet, and then you can derive the pump power² from the pressure drop. So it requires a requires a variable-power pump to compensate for the ambient hydrostatic head at the outlet, but otherwise the prop will ooze out of the outlet at just the right rate with basically no kinetic energy.______________¹Poiseuille equation:Δp = 8μLQ/(πR⁴)

It occurred to me that, if you engineer your pumping power so that it achieves the needed flow rate but causes the liquid to exit the outlet pipe at exactly ambient pressure, then there's no geysering possible.You can figure out that pump power via the Poiseuille equation,¹ which tells you the pressure drop from the pump outlet to the ambient fill pipe outlet, and then you can derive the pump power² from the pressure drop. So it requires a requires a variable-power pump to compensate for the ambient hydrostatic head at the outlet, but otherwise the prop will ooze out of the outlet at just the right rate with basically no kinetic energy.______________¹Poiseuille equation:Δp = 8μLQ/(πR⁴)Where:Δp is the pressure drop through the pipeμ is the dynamic viscosity (5.0E-5 Pa-s for LOX, 1.1E-5 Pa-s for LCH4)L is the pipe lengthQ is the volumetric flow rate (in m³/s, not kg/s)R is the pipe radius²The pump power is:P = (Δp + p_{outletHead})Q/ηwhere P is the power (W)η is the pump efficiency.

Quote from: TheRadicalModerate on 01/16/2023 05:00 amIt occurred to me that, if you engineer your pumping power so that it achieves the needed flow rate but causes the liquid to exit the outlet pipe at exactly ambient pressure, then there's no geysering possible.You can figure out that pump power via the Poiseuille equation,¹ which tells you the pressure drop from the pump outlet to the ambient fill pipe outlet, and then you can derive the pump power² from the pressure drop. So it requires a requires a variable-power pump to compensate for the ambient hydrostatic head at the outlet, but otherwise the prop will ooze out of the outlet at just the right rate with basically no kinetic energy.______________¹Poiseuille equation:Δp = 8μLQ/(πR⁴)Interesting, I didn't realise this (/R^{4}!!). Key takeaway being that fat pipes are your friend, since one assumes SpaceX were already going to pick the most efficient pumps possible.Thought I should show you the current progress of my noodling (not the final product - the revisions of these animations will form part of my next youtube video). I should say that this began as a revision of my spin-G Starship work, but the configuration that I came up with seemed to be applicable to the propellant transfer problem.This particular spin config - with the offset nose docking and dorsal-ventral rotation - was one that I hadn't considered until recently, but I'm liking it more and more (solar panels/radiators not yet shown, bi-directional ports assumed). It means you don't need to move the header tank at all. Even if you really did want to put the docking port on the nose, you'd have to figure out how to protect the port with the heat shield during EDL.The offset port at the nose also avoids intermediate axis issues if you rotate in the dorsal-ventral plane. Obviously the pipes on the depot variant are pretty long, but they are all on the leeward side of the vehicle. This animation depicts 4 pipes on the arm (2x Methane and LOX, liquid and ullage) each with a diameter of 20cm.