Pushing the limits of the hoverslam landing

[gospacex]

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!

[Robotbeat]

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

[cmcqueen]

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!

Does it really matter? If it's mostly above the atmosphere, then isn't it just delta-V that matters?

[sewebster]

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!

Does it really matter? If it's mostly above the atmosphere, then isn't it just delta-V that matters?

After cancelling out the horizontal velocity, it needs to add some back in the opposite direction to get back to the launch site... more if it is further away, right?

[PreferToLurk]

Has 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.   

[Kabloona]

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!

The boostback burn has to be delayed that long because that's how long it takes for the relatively weak GN2 thrusters to reorient the stage and settle propellant for the burn.

[Comga]

Has 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.   

I agree that Hrissan's post is terrific work and a great place to start.

He finds terminal velocity is 150 m/sec
Baseline deceleration is 7.5 m/sec^2  or ~3/4g, which means the Thrust to Weight ration is ~1.75. (ignoring drag)

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/sec

If 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!)

Gravity loss would be only 3.5 sec *10 m/sec^2 = 35 m/sec, or about a seventh of that for a single engine landing.

Total velocity killed would be ~185 m/sec, which is 53% or just over half, of the single engine landing, so it would need little more than half the fuel.  Hence the motivation to try.

As for the thread title, "pushing the limits" imagine the absurd case of using all nine engines.  T:W~16.  a~150 m/s^2. The landing burn takes ~1 second, and starts around 75 meters!  Gravity loss is 10 m/sec, so total deceleration is 160 m/sec.  This is 86% of the three engine case and 43% of the one engine case.  That's only an additional 7% savings,  definitely diminished returns.

* I was going to add "a spherical cow in a vacuum" but that's sort of an in joke among people who took, or worse yet majored in, physics in college.   It implies gross over-simplification.

edit: corrected error in 9 engine calculation.  Now shows even less benefit

[guckyfan]

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/sec

If 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.

[the_other_Doug]

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/sec

If 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.

Yes, thanks!  I was trying to get a concept of when the landing burn would begin if it was a 3-engine burn, and I was estimating between 500 and 750 meters in altitude.  A landing burn starting at 265 meters altitude means that, if the engines don't light up properly,  you can be hitting the barge less than two seconds later.

On SES-9, we obviously had an ignition attempt (the bright light coming down off-center in the barge video), but if the engines never came up to thrust, we would have had an impact within just a few seconds of ignition.  If the hole in the barge is any indication, the engines couldn't have come up to thrust and left the stage with enough kinetic energy to punch that hole...

[Okie_Steve]

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

Which is about the time from ignition to release at launch after they verify engine thrust etc.
I'm sure they don't hold longer than necessary. Insane suicide burn indeed.

[LouScheffer]

Here's an analysis of the accuracy needed, from the SES-9 discussion thread:

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.

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.

The potential fuel savings are similar:   A single engine landing burn might have 25 sec at full thrust followed by 5 sec at 70%, so 28.5 engine-sec.  For three engines, 5.5 sec x 82% x 3 = 13.5 engine-sec.  So 15 engine-sec savings means 1.66 more seconds of first stage burn.  At about 5G, that's 80 m/sec more to the payload.  Well worth it if you can make it work.

[Comga]

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?                                          

[The_Ronin]

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

The outer engines do not have the same gimble range as the center.  Make make corrections difficult.

[PreferToLurk]

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?

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

[CraigLieb]

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?

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

It seems with only a 3 second burn, thrust variability during start-up and shut-down could play a much larger role in reduced reliability of the landing scenario.   

[drzerg]

gimbal range wich is limited due to other engines is irrelevant because they could gimbal outwards just before shutdown

[Comga]

The_Ronin:  Two engines with limited gimbal range looks like it is still more force than one with full range
PerferToLurk: This model is woefully insufficient to use for such subtleties, but that misses the point.  There is little reason to use more than two engines.  The is not much more to gain in terms of fuel savings.
CraigLeib: Precisely, even at 3+1 seconds
drzerg:  That doesn't work because it counts on cosine losses, and for the small gimbal angles Cos~1.

[meekGee]

I like the idea of a planned nominal one engine burn, but have all three engines spun up, so that if the center engine fails, you can still switch to the other two in time to land.

On more demanding missions, go for a planned 2 engine burn, with three engines as a fall back...

[the_other_Doug]

My only concern is that, with the gimbal limitations of the outer engines, you just don't have the required control authority to actually hit the target unless you use the center engine.  We're talking about the end of the descent, when the grid fins are losing their control effectiveness and the control is shifting more and more to the engine gimbals, until right at the end it is entirely controlled by engine gimballing.

As a caveat, though, only SpaceX knows the control authority allowed by using two opposing outer engines for a hoverslam.  Without this knowledge, anything we say about it is nothing more than gross speculation.

[Kabloona]

On the subject of the hoverslam, there has been some debate in other threads about whether the stage reached terminal velocity before the landing burn started, and what that velocity might be.

For reference, two NSF members did earlier independent video analyses of the OG2 landing and derived "terminal" velocities before landing burn of 150 m/sec and 180 m/sec on that flight.

In the following SES-9 simulation, landing burn starts at a velocity of about 170 m/sec, and a relative acceleration of 1.18 g, so the stage is decelerating relatively slowly, therefore almost at terminal velocity (which would be 1.00 g relative acceleration).

Also, the entry burn starts at 65 km altitude at about 2,300 m/sec and decelerates to about 1,300 m/sec. From that point on, aero drag takes over and decelerates the stage continuously as it falls towards the ASDS.

The simulation was done by someone (Zach) who says (in the comments section) that he compared the output of his simulation against telemetry data scraped from the webcast in order tweak his aero model. I don't know how accurate the landing portion of the trajectory is, but it seems credible and consistent with what we saw on OG2.

Update: In response to my question about how he modeled the trajectory after entry burn, Zach replied with this comment:

"There is no public information or telemetry on the first stage after MECO. The only hints we get from the webcast are the callouts for when the burns start and sometimes when the stage goes transonic. My simulation currently doesn't take lift into account so I would say the trajectory is only accurate until it starts getting deep into the atmosphere."

"The only thing we know for sure is that the landing burn used three engines this time. My simulation shows that burning 3 engines on 60% thrust will generate 4.2G. I'm assuming the dry mass of the stage is ~25,000kg and the engines generate 756kN of thrust at sea level (known value). If the burn lasted 4-5 seconds then the kinematic equations will show it had to be moving around 170 m/s."