For RTLS, they can try optimizing boostback burn initiation delay after MECO. Thus far it was quite long, some 30-40 seconds. During this time, S1 distance from LS nearly doubled!

Quote from: gospacex on 03/06/2016 10:38 pmFor RTLS, they can try optimizing boostback burn initiation delay after MECO. Thus far it was quite long, some 30-40 seconds. During this time, S1 distance from LS nearly doubled!Does it really matter? If it's mostly above the atmosphere, then isn't it just delta-V that matters?

Has anyone figured out the T/W ratio of the RTLS F9FT landing through video analysis?

Quote from: Robotbeat on 03/06/2016 11:09 pmHas anyone figured out the T/W ratio of the RTLS F9FT landing through video analysis?Sorry, haven't been paying attention to this section for a few days. But to answer your question, yes. Hrissan did a pretty good analysis of the landing and it is what I used as the basis for all my tinkering in the OP.

A simple model (constant mass, constant thrust, instant start, no air drag, * ) says the landing burn would take 20 seconds (=150/7.5) and start at 1500 m altitude (0.5*7.5*20^2). Gravity loss would be 20 sec *~10 m/sec^2 = 200 m/sec which is more than the terminal velocity. Total velocity killed would be 350 m/secIf this landing used three engines at the same thrust, T:W~ 5.25 and a~42.5 m/sec^2.The simple model says that this would take only ~3.5 sec (=150/42.5) and start at ~265 m altitude. (Talk about BPL!)

Quote from: Comga on 03/10/2016 05:02 amA simple model (constant mass, constant thrust, instant start, no air drag, * ) says the landing burn would take 20 seconds (=150/7.5) and start at 1500 m altitude (0.5*7.5*20^2). Gravity loss would be 20 sec *~10 m/sec^2 = 200 m/sec which is more than the terminal velocity. Total velocity killed would be 350 m/secIf this landing used three engines at the same thrust, T:W~ 5.25 and a~42.5 m/sec^2.The simple model says that this would take only ~3.5 sec (=150/42.5) and start at ~265 m altitude. (Talk about BPL!)I had no idea a 3 engine burn would save that much. Thanks.

if the engines don't light up properly, you can be hitting the barge less than two seconds later.

Quote from: Robotbeat on 03/09/2016 04:01 amLet's say it's a = 50m/s^2 (with respect to the surface, not freefall), and the stage has to be within v=2m/s of zero in order to land safely. How accurate do you have to be within the z-direction?a = v^2/(2*d) becomes: 2*d*a = v^2 becomes d = v^2/(2*a) = (2m/s)^2/(2*50m/s^2) = 4 centimeters (!)You have to be within 4 centimeters in the z-direction in order to stay within your landing velocity constraint when you're hoverslamming with 3 engines. If something doesn't throttle up fast enough or you start too early or late, you're toast. This isn't impossible, but it's DANG challenging.I think this analysis is too pessimistic. It's OK for the ends of the legs to hit the ground faster, provided the body of the rocket reaches 0 vertical speed before the legs run out of travel (or the engine bell hits the ground, whichever comes first). Assuming the legs can absorb one meter of bend before breaking, then you need the lower vertex of the parabola to be between the deck and a point one meter below. Is this practical? With 3 engines, 30 tonnes mass, your acceleration varies from 3.8G at 70% throttle to 5.5G at 100%. Assume you plan your burn for 4.5Gs so you have leeway in both directions. If you are falling at 250 m/s (about what you'd guess from the one engine landings) you'd want the engine to start at 82% throttle at 5.5 seconds before impact, at a height of 694 meters. You get about a 1/2 second of slop since as long as you start before 568 m you can still stop at full thrust.Once (if) your engines start you are in good shape. On this time scale the radar altimeter and calculations should be instantaneous, so you immediately know the desired acceleration to place the vertex 50 cm below the landing pad (or whatever your target). You don't know the exact mass of the stage, nor the actual thrust for a commanded amount, but measuring the achieved acceleration tells you the proportionality constant. Now you start adjusting the commanded thrust to get the acceleration right.At 1 second before landing at 4.5 Gs , you are 22.5 meters up. A 1% acceleration error will move the vertex +- 22 cm. That's about all you can afford, since it's already half your error budget. So you need to have the acceleration right to the 1% level by 1 second to go. You get 4.5 seconds of correction to do this. If the initial error is 20% (say 10% for throttle and 10% for mass) the you need to reduce the error by a factor of 20. Assuming a linear system, this level of correction requires 3 time constants (e^3 = 20) so if your time constant for throttle response is 1.5 seconds or less, it should be possible. Given that the engine can get to (nearly) steady state during either a static fire or the short time before liftoff, such a time constant seems possible.Now this analysis assumes you are coming straight down with the rocket vertical, no attempt to steer horizontally, no errors in the radar altimeter or IMU, etc. But even given these errors, it seems possible to make this work.

Let's say it's a = 50m/s^2 (with respect to the surface, not freefall), and the stage has to be within v=2m/s of zero in order to land safely. How accurate do you have to be within the z-direction?a = v^2/(2*d) becomes: 2*d*a = v^2 becomes d = v^2/(2*a) = (2m/s)^2/(2*50m/s^2) = 4 centimeters (!)You have to be within 4 centimeters in the z-direction in order to stay within your landing velocity constraint when you're hoverslamming with 3 engines. If something doesn't throttle up fast enough or you start too early or late, you're toast. This isn't impossible, but it's DANG challenging.

What would be the issues with a 2 engine landing burn?

Simplistic model in table form Engines T/W Accel t h g-loss ratio fuel (m/s^2) (sec) (m) (m/s) 1 1.75 7.5 20 1500 200 2.33 100% 2 3.5 25.0 6 450 60 1.40 60% 3 5.3 42.5 3.5 265 35 1.24 53% 4 7.0 60.0 2.5 188 25 1.17 50% 5 8.8 77.5 1.9 145 19 1.13 48% 6 10.5 95.0 1.6 118 16 1.11 47% 7 12.3 112.5 1.3 100 13 1.09 47% 8 14.0 130.0 1.2 87 12 1.08 46% 9 15.8 147.5 1.0 76 10 1.07 46% Conclusion: This shows quickly diminishing returns. Most of the fuel savings happens in the first step.It seems obvious why the SpaceX went from 1 to 3 engines. Three engines are rigged to restart for boostback and entry burns.But there is 85% of the fuel savings, with almost twice the time for adjustments, if it uses 2 engines instead of 3. (A symmetric configuration would be the outer 2 of the 3 restarting engines.)What would be the issues with a 2 engine landing burn?

Quote from: Comga on 03/14/2016 02:39 pmSimplistic model in table form Engines T/W Accel t h g-loss ratio fuel (m/s^2) (sec) (m) (m/s) 1 1.75 7.5 20 1500 200 2.33 100% 2 3.5 25.0 6 450 60 1.40 60% 3 5.3 42.5 3.5 265 35 1.24 53% 4 7.0 60.0 2.5 188 25 1.17 50% 5 8.8 77.5 1.9 145 19 1.13 48% 6 10.5 95.0 1.6 118 16 1.11 47% 7 12.3 112.5 1.3 100 13 1.09 47% 8 14.0 130.0 1.2 87 12 1.08 46% 9 15.8 147.5 1.0 76 10 1.07 46% Conclusion: This shows quickly diminishing returns. Most of the fuel savings happens in the first step.It seems obvious why the SpaceX went from 1 to 3 engines. Three engines are rigged to restart for boostback and entry burns.But there is 85% of the fuel savings, with almost twice the time for adjustments, if it uses 2 engines instead of 3. (A symmetric configuration would be the outer 2 of the 3 restarting engines.)What would be the issues with a 2 engine landing burn?Another solution for 3 engine landing is to throttle lower. I think I originally calculated that the 1.75 T/W regime of the OG2 landing was with a throttle setting around 90% (though it might have been as high as 95% with slightly different assumptions). We know for a fact that the center engine can throttle at least to 80% of the final thrust setting (based on the first landing burn regime of about 4 m/s^2 deceleration). If all three engines can throttle to the same extent (and there is some consideration that they cannot, or maybe just not as precisely), then you should be able to get a 3 engine landing burn with 31 m/s^2 deceleration. This would net you at least another second of burn time without having to sacrifice much in terms of control authority or fuel.Complicating this is the fact that with a dry mass of around 30000kg, coming in for a landing with 4-6k kg less fuel (assuming saved fuel is burned before meco and assuming a fuel reserve/ballast on OG2 that would also be mostly burned prior to meco) affects the landing T/W in non trivial terms. At the first landing burn of OG2 I assumed 2500kg of fuel reserve plus 5500kg of fuel used in the burns. A 38000kg stage with a 54kgf (about 72% of 75kgf full thrust) will nicely create a 1.4 T/W, but if you pull 3000kg of fuel off the stage, then your 90% throttled engine will produce about 1.93 T/W, not 1.75 which is going to speed up all of your landing assumptions.Anyway, just some food for thought. Personally, I think they could start all three engines at absolute minimum thrust (no throttle down margin), and if they accidentally lit the engines too high up and need to throttle down further just cut either the center engine (after zeroing enough lateral error to let the outer engines gimble range take over) or the outer two if needed. This would allow for the greatest throttle up margin (which apparently is what doomed this last landing) while giving as much time as possible to resolve ignition transients and compute a landing solution. IMHO