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#620
by
Alexphysics
on 26 Jun, 2019 21:02
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Some computed values for the STP-2 launch:
Center core cutoff at 11083 km/hr ( = 3079 m/s) at altitude of 123 km and time 3:34. Staging speed is therefore 99 m/s more then Arabsat 6, which was 2980 m/s. Re-entry burn starts at 9:43, so coast was 6:10 or 370 seconds.
Assuming re-entry burn at 60 km, we can solve for vertical component of velocity by solving 123,000 + v*t - 9.8/2*t^2 = 60,000, where t = 370 seconds. We find a vertical velocity at cutoff of 1643 m/s. Given the overall speed of 3079 m/s, horizontal speed is 2604 m/s. Coasting at this speed, in 370 seconds the stage covers 963 km, so all is consistent. (The rest is launch to sep, and entry to landing).
Peak altitude is 261 km at 167.5 seconds after staging. The energy per kg that must be lost by the stage in landing is g*h + 1/2*v^2, and is 5,946,000 joules/kg. For comparison, Arabsat (103 km, 2970 m/s) was 5,449,000 joules/kg. So this landing was 9% more energetic. More significant, most likely, are (a) less fuel left for slowing down, as more was used for boost, and (b) a steeper angle of re-entry from the much more lofted trajectory.
From webcast telemetry I got a vertical velocity at MECO of about 1km/s which corresponds to a tangent velocity of about 2.9km/s and for those values, at the altitude it was at MECO (123km) our friend Newton says the apogee would have been 180km (almost 181km if you round up the number) and not 261km :/
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#621
by
OneSpeed
on 26 Jun, 2019 23:28
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Some computed values for the STP-2 launch:
Center core cutoff at 11083 km/hr ( = 3079 m/s) at altitude of 123 km and time 3:34. Staging speed is therefore 99 m/s more then Arabsat 6, which was 2980 m/s. Re-entry burn starts at 9:43, so coast was 6:10 or 370 seconds.
Assuming re-entry burn at 60 km, we can solve for vertical component of velocity by solving 123,000 + v*t - 9.8/2*t^2 = 60,000, where t = 370 seconds. We find a vertical velocity at cutoff of 1643 m/s. Given the overall speed of 3079 m/s, horizontal speed is 2604 m/s. Coasting at this speed, in 370 seconds the stage covers 963 km, so all is consistent. (The rest is launch to sep, and entry to landing).
Peak altitude is 261 km at 167.5 seconds after staging. The energy per kg that must be lost by the stage in landing is g*h + 1/2*v^2, and is 5,946,000 joules/kg. For comparison, Arabsat (103 km, 2970 m/s) was 5,449,000 joules/kg. So this landing was 9% more energetic. More significant, most likely, are (a) less fuel left for slowing down, as more was used for boost, and (b) a steeper angle of re-entry from the much more lofted trajectory.
From webcast telemetry I got a vertical velocity at MECO of about 1km/s which corresponds to a tangent velocity of about 2.9km/s and for those values, at the altitude it was at MECO (123km) our friend Newton says the apogee would have been 180km (almost 181km if you round up the number) and not 261km :/
Are you including the sub-orbital nature of the trajectory? Altitude = v²/2g only applies in the idealized case where horizontal velocity is an inconsequential fraction of orbital velocity.
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#622
by
edkyle99
on 26 Jun, 2019 23:36
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Given how close it got to the "X" on deck before aborting I have to wonder when the TVC failed and how.
I keep reading statements like this - that the stage "aborted" at the last second. Is there a basis for such assertions?
Wouldn't the stage behave in similar fashion if TVC failed in the final moments (which is the failure mode described by Elon)?
- Ed Kyle
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#623
by
Coastal Ron
on 26 Jun, 2019 23:57
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Given how close it got to the "X" on deck before aborting I have to wonder when the TVC failed and how.
I keep reading statements like this - that the stage "aborted" at the last second. Is there a basis for such assertions?
Yes, Elon Musk on Twitter:
Everyday Astronaut: And did the computer know that and know to divert?!?!
Elon Musk: Most likely. It is programmed to do so.
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#624
by
Vettedrmr
on 27 Jun, 2019 00:25
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Well, IMO, "most likely" doesn't translate into "that's what it did." It *might* have, but think about it: how can you tell from an external video source if the pitch over was a commanded response or just a result of the failure itself?
If S1 *did* abort, then I'd re-examine the criteria for an abort. It appeared that the vertical velocity was right on target for an attempted landing, so if it did fail on landing the damage to ASDS would be no worse than any other failed landing attempt.
HUGE GUESSING ALERT: I'm not saying this is what happened, it's just my guess. And I'd say I've got a really small chance of getting this right:
I'm guessing that the damage to the TVC was in the actuator hydraulic. If there was a leak on the return line of the hydraulics loop, the supply could have been depleted over time, finally running out right at the end.
I don't know how much authority the RCS has during the terminal phase compared to TVC, but if it's large then that shoots a pretty big hole in my theory.
Have a good one,
Mike
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#625
by
ulm_atms
on 27 Jun, 2019 00:27
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Doesn't the TVC use RP-1 for the hydraulic fluid? I wonder if the hydraulic pump pulled air and stalled the TVC...or reentry fried a feed line to the pump for the air pull to happen. They were already so close on fuel to begin with for this. Maybe Elon will give more info in the future.
I do see them drawing a 1000km line however between reusable/expendable for ASDS landings. Since heat breached the engine bay (from Elon), they would have to redesign some of the protection down there...with all their money into Starship...I don't see it being a high priority at this point. What was the longest successful ASDS? I think this was the first ASDS over 1000km
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#626
by
ulm_atms
on 27 Jun, 2019 00:32
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I don't know how much authority the RCS has during the terminal phase compared to TVC, but if it's large then that shoots a pretty big hole in my theory.
The TVC is like 98% of the control authority in the last few seconds. Air speed is too slow for the fins to do too much and the GN2 can't do much at ground level.
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#627
by
Vettedrmr
on 27 Jun, 2019 00:35
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I think they use normal hydraulic fluid to drive the actuators. But they're driven by turbo-pumps, so if the fuel system burped then it could affect the system. BUT, I think we would have seen some significant engine instability at the same time, which I didn't see until the stage was almost horizontal.
Have a good one,
Mike
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#628
by
Herb Schaltegger
on 27 Jun, 2019 01:09
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Just to toss out support for Elon's tweet (like that should even be necessary but here we are) ...
Hydraulic TVC systems can have multiple failure modes. Elon's initial tweet indicating that the system failed does not specify which of the numerous failure mode or modes actually occurred, nor does his tweet say the system was completely non-functional. It is quite possible that the system failed in one axis, or that hydraulic pressure was falling and thus unlikely to be able to control safely to landing, thus leading the guidance software to command a divert maneuver.
In short, we ought to respect his comment and realize we don't know nearly as much about the detailed hardware designs or the control software as he does.
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#629
by
Vettedrmr
on 27 Jun, 2019 01:25
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I have no issue with what Elon said in his tweet: "Most likely." He didn't say "yes it did", he said that was a likely scenario.
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#630
by
Okie_Steve
on 27 Jun, 2019 02:28
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The early Merlins did use RP-1 for gimbal as I recall, but no longer.
I should have said "divert" instead of "abort" in my original post. As I understand it, much like RTL landings aim off shore until fairly close and the retargets to the LZ so the debris path in case of failure, ASDS landings aim for a near miss to avoid lawn darts on failure and retarget to the "X" when fairly close. Since we could see exhaust painting said "X" from less than one Falcon length* up I think that qualifies as close. Only then did it divert. While it is hard to measure nearly empty tanks they clearly thought there was enough propellant to go for the landing. And given the length of the divert burn I think they were correct in that thought.
Elon said the root cause was reenty heat damage, but when did the TVC actually pack it in? Did it fail during the landing burn or was it only discovered during the landing burn? Or, possibly, was it known ahead of time and they tried to compensate as long as possible anyway but ultimately had to punt and thus diverted. Which now that I think about it might be good practice toward landing on Mars where divert is not an option, along the lines of "Any landing you can walk away from ..."
Any one have a good estimate how close to the deck the feet got before the divert?
I am still curious about the relative expense of ASDS lawn dart repairs vs the value of a recovered booster in (apparently) borderline situations like this.
*New unit of measure
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#631
by
Vettedrmr
on 27 Jun, 2019 02:49
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Elon said the root cause was reenty heat damage, but when did the TVC actually pack it in? Did it fail during the landing burn or was it only discovered during the landing burn? Or, possibly, was it known ahead of time and they tried to compensate as long as possible anyway but ultimately had to punt and thus diverted.
Well, the landings are totally autonomous, so the booster would have to determine when to divert (maybe that's what you meant). ASSUMING my theory of hydraulic fluid depletion is correct (btw, this is just a fun conversation until we learn more), then they probably saw the fluid level dropping on telemetry, but obviously couldn't do anything about it.
<engage Daffy Duck voice> "of coase, I cud be wong."
Have a good one,
Mike
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#632
by
rubicondsrv
on 27 Jun, 2019 02:57
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I am still curious about the relative expense of ASDS lawn dart repairs vs the value of a recovered booster in (apparently) borderline situations like this.
*New unit of measure
Steel is cheap even when tens of tons are being replaced. the propulsion system is more costly than steel plate, but likely still fairly inexpensive. IMO the biggest concern would be schedule impacts due to repairs.
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#633
by
LouScheffer
on 27 Jun, 2019 03:13
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Some computed values for the STP-2 launch:
Center core cutoff at 11083 km/hr ( = 3079 m/s) at altitude of 123 km and time 3:34. Staging speed is therefore 99 m/s more then Arabsat 6, which was 2980 m/s. Re-entry burn starts at 9:43, so coast was 6:10 or 370 seconds.
Assuming re-entry burn at 60 km, we can solve for vertical component of velocity by solving 123,000 + v*t - 9.8/2*t^2 = 60,000, where t = 370 seconds. We find a vertical velocity at cutoff of 1643 m/s. Given the overall speed of 3079 m/s, horizontal speed is 2604 m/s. Coasting at this speed, in 370 seconds the stage covers 963 km, so all is consistent. (The rest is launch to sep, and entry to landing).
Peak altitude is 261 km at 167.5 seconds after staging. The energy per kg that must be lost by the stage in landing is g*h + 1/2*v^2, and is 5,946,000 joules/kg. For comparison, Arabsat (103 km, 2970 m/s) was 5,449,000 joules/kg. So this landing was 9% more energetic. More significant, most likely, are (a) less fuel left for slowing down, as more was used for boost, and (b) a steeper angle of re-entry from the much more lofted trajectory.
From webcast telemetry I got a vertical velocity at MECO of about 1km/s which corresponds to a tangent velocity of about 2.9km/s and for those values, at the altitude it was at MECO (123km) our friend Newton says the apogee would have been 180km (almost 181km if you round up the number) and not 261km :/
I agree that the altitude number was changing more slowly than expected, but computing velocity by differentiating a delayed, filtered, and quantized number is susceptible to lots of potential errors.
If the altitude at cutoff was 123 km, and the vertical velocity 1000 m/s, then we can calculate the altitude 370 seconds later at entry burn. It is h0 +v*t - 1/2*g*t^2, or 123000 + 1000*370 - 9.8/2*370^2 = -177.8 km. It's pretty clear that the entry burn was in the atmosphere, not the earth's mantle, so the vertical velocity must be quite a bit higher. Since we know for sure the entry burn was above sea level, we know vertical velocity must be at least (-123000 + 0.5*9.8*370^2)/370 = 1480 m/s. (But see the sub-orbital correction below.) A more realistic assumption of a 60 km entry burn gives the numbers above.
Are you including the sub-orbital nature of the trajectory? Altitude = v²/2g only applies in the idealized case where horizontal velocity is an inconsequential fraction of orbital velocity.
This is a reasonable correction. Since horizontal v is about 2600 m/s, and the Earth's radius about 6356 km, then the acceleration induced is v^2/r =2600^2/6456000 = 1.05 m/s^2. So instead of 9.8 m/s^2, we should use 8.75 m/s^2. We then solve for the vertical v where 123000+v*370 - 1/2*a*370^2 = 60000, and get a corrected vertical velocity of 1448 m/s.
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#634
by
OneSpeed
on 27 Jun, 2019 03:43
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Since horizontal v is about 2600 m/s, and the Earth's radius about 6356 km, then the acceleration induced is v^2/r =2600^2/6456000 = 1.05 m/s^2. So instead of 9.8 m/s^2, we should use 8.75 m/s^2. We then solve for the vertical v where 123000+v*370 - 1/2*a*370^2 = 60000, and get a corrected vertical velocity of 1448 m/s.
So, updating:
Flight path angle sin-1(1448/3079) = 28°
Horizontal component = cos(28°) x 3079 = 2,717m/s
Downrange distance covered = 370s x 2,717m/s = 1,005kms
Plus a bit more due to the increased induced acceleration.
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#635
by
Okie_Steve
on 27 Jun, 2019 04:43
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I am still curious about the relative expense of ASDS lawn dart repairs vs the value of a recovered booster in (apparently) borderline situations like this.
*New unit of measure
Steel is cheap even when tens of tons are being replaced. the propulsion system is more costly than steel plate, but likely still fairly inexpensive. IMO the biggest concern would be schedule impacts due to repairs.
Very good point, I had not thought about that
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#636
by
CorvusCorax
on 27 Jun, 2019 05:25
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When would the stage abort a landing and how would it know when to do so?
I don't know how SpaceX implements it, but if I were to design it, instead if trying to cope with many different possible damage models and their expected symptoms, I'd go by a simple criteria similar to the autonomous LAS:
Constantly check whether the vehicles position, velocity and attitude is within a safe corridor to make a safe landing. Try the best to stay within this corridor with whatever actuators are available.
The stage would have done exactly that, and while still going fast enough for grid fins having sufficient control authority over a malfunctionibg TVC, it was seemingly spot on, definitely within the corridor.
The corridor can be calculated dynamically. It would simply be the space from which in an intact, functioning stage has sufficient control authority to make a succesful landing. If you ever go outside of that, you know you can't possibly make it anymore.
The corridor becomes narrower and narrower towards touchdown. At the same time any issue with TVC would become worse due to decreasing grid fin authority.
We did not see the stage in that phase, but my guess is the attitude rate and/or horizontal speed became too high, it left the corridor, knew it wouldnt make it and throttled up for the emergency fallback (safe ocisly)
That would look pretty much the same regardless of the exact type of failure or its cause.
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#637
by
Alexphysics
on 27 Jun, 2019 06:02
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Some computed values for the STP-2 launch:
Center core cutoff at 11083 km/hr ( = 3079 m/s) at altitude of 123 km and time 3:34. Staging speed is therefore 99 m/s more then Arabsat 6, which was 2980 m/s. Re-entry burn starts at 9:43, so coast was 6:10 or 370 seconds.
Assuming re-entry burn at 60 km, we can solve for vertical component of velocity by solving 123,000 + v*t - 9.8/2*t^2 = 60,000, where t = 370 seconds. We find a vertical velocity at cutoff of 1643 m/s. Given the overall speed of 3079 m/s, horizontal speed is 2604 m/s. Coasting at this speed, in 370 seconds the stage covers 963 km, so all is consistent. (The rest is launch to sep, and entry to landing).
Peak altitude is 261 km at 167.5 seconds after staging. The energy per kg that must be lost by the stage in landing is g*h + 1/2*v^2, and is 5,946,000 joules/kg. For comparison, Arabsat (103 km, 2970 m/s) was 5,449,000 joules/kg. So this landing was 9% more energetic. More significant, most likely, are (a) less fuel left for slowing down, as more was used for boost, and (b) a steeper angle of re-entry from the much more lofted trajectory.
From webcast telemetry I got a vertical velocity at MECO of about 1km/s which corresponds to a tangent velocity of about 2.9km/s and for those values, at the altitude it was at MECO (123km) our friend Newton says the apogee would have been 180km (almost 181km if you round up the number) and not 261km :/
I agree that the altitude number was changing more slowly than expected, but computing velocity by differentiating a delayed, filtered, and quantized number is susceptible to lots of potential errors.
If the altitude at cutoff was 123 km, and the vertical velocity 1000 m/s, then we can calculate the altitude 370 seconds later at entry burn. It is h0 +v*t - 1/2*g*t^2, or 123000 + 1000*370 - 9.8/2*370^2 = -177.8 km. It's pretty clear that the entry burn was in the atmosphere, not the earth's mantle, so the vertical velocity must be quite a bit higher. Since we know for sure the entry burn was above sea level, we know vertical velocity must be at least (-123000 + 0.5*9.8*370^2)/370 = 1480 m/s. (But see the sub-orbital correction below.) A more realistic assumption of a 60 km entry burn gives the numbers above.
Are you including the sub-orbital nature of the trajectory? Altitude = v²/2g only applies in the idealized case where horizontal velocity is an inconsequential fraction of orbital velocity.
This is a reasonable correction. Since horizontal v is about 2600 m/s, and the Earth's radius about 6356 km, then the acceleration induced is v^2/r =2600^2/6456000 = 1.05 m/s^2. So instead of 9.8 m/s^2, we should use 8.75 m/s^2. We then solve for the vertical v where 123000+v*370 - 1/2*a*370^2 = 60000, and get a corrected vertical velocity of 1448 m/s.
I didn't use a parabolic equation for that mainly because the Earth is not flat at 1200km of distance. If you use the conservation of energy equation you just need the radial component of the velocity at MECO and the altitude to get the apogee altitude. It might be off by maybe 10 or even 20km mainly because I used the mean earth radius because I don't know the actual radius of the earth at those positions but one can do an estimate with the maximum radius of the earth and one at the minimum radius and do an average of that. And for one speed: I just did a quick interpolation frame by frame of the altitude. I just took 2 seconds of data which is short enough to consider vertical speed to still be sort of constant (I know, not entirely constant, but in such short amount of time one could even improve that and take an average value but within such a short amount of time the average would end up being very close to the actual value you get the other way).
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#638
by
OneSpeed
on 27 Jun, 2019 07:12
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I just did a quick interpolation frame by frame of the altitude. I just took 2 seconds of data which is short enough to consider vertical speed to still be sort of constant (I know, not entirely constant, but in such short amount of time one could even improve that and take an average value but within such a short amount of time the average would end up being very close to the actual value you get the other way).
Assuming 2,717m/s is the horizontal component at 123kms, the induced acceleration is v²/r = 2717²/6456000 = 1.14m/s upwards. So, the total acceleration is 9.8 - 1.14 = 8.66m/s. The vertical velocity is 1,448m/s, so from apogee = v²/2g, the core stage apogee would be at 123,000 + 1,448²/(2*8.66) = 244,057m or 244kms.
I will try to complete a simulation of STP-2 by the weekend, it might be interesting to compare results then.
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#639
by
Alexphysics
on 27 Jun, 2019 09:03
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I just did a quick interpolation frame by frame of the altitude. I just took 2 seconds of data which is short enough to consider vertical speed to still be sort of constant (I know, not entirely constant, but in such short amount of time one could even improve that and take an average value but within such a short amount of time the average would end up being very close to the actual value you get the other way).
Assuming 2,717m/s is the horizontal component at 123kms, the induced acceleration is v²/r = 2717²/6456000 = 1.14m/s upwards. So, the total acceleration is 9.8 - 1.14 = 8.66m/s. The vertical velocity is 1,448m/s, so from apogee = v²/2g, the core stage apogee would be at 123,000 + 1,448²/(2*8.66) = 244,057m or 244kms.
I will try to complete a simulation of STP-2 by the weekend, it might be interesting to compare results then.
Yeah that's the problem. I got about 1.04km/s of vertical speed and about 2.89km/s of horizontal speed at MECO and not those other numbers. At apogee the velocity is just the tangent component of the velocity which is constant over the flight at 2.89km/s. Using that as the final velocity you can figure out that the apogee can be calculated as:
rf = (1/ri - (vi^2-vf^2)/2GM)^-1
And then you substract the radius of the earth from rf and you get the apogee altitude.
