Author Topic: F9v1.1 Stage 1 (and FH booster) recovery trajectory, burns, drag evaluation.  (Read 17647 times)

Offline aero

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My estimate of fuel required for recovery of the F 9 v1.1 first stage is 36.7 tonnes. This is based on my evaluation of a "Loft back and drop" recovery trajectory. Here follows my logic and calculations for your critique.

I started with the Cassiope launch data, staging at 100.2 km altitude and 7400 km/hr = 2056 m/s velocity. The complete set of data given on the Livestream video suggests that at staging, Vx ~= Vz ~= 1450 m/s and that the booster was about 132 km down range at staging.

I imposed a 4 g acceleration limit on the boost back as it only uses 3 engines giving thrust to only boost 56,700 kg at 4 gs acceleration. This choice of a 4 g limit provided a very strange but fortuitous result.

I calculated the loft time for the stage. Upward velocity of 1450 m/s decelerated by gravity slows to zero and returns to the same 100.2 km altitude in 295.72 seconds. Estimating that burn 2 will occur at 50 km altitude I calculated an additional 31.19 seconds for the booster to plummet to 50 km altitude where it will reach 1,755.89 m/s downward velocity. (This condition is referred to as Point 2.) This loft and drop allows about 327 seconds for the booster to return to the vicinity of the landing pad, at 50 km altitude.

I ignored the coasting time needed to reorient the stage. (How much time should it take to reorient the stage?). (Staging is point 1 in the following.) I boosted horizontally up range at 4 gs acceleration, and calculated that the horizontal velocity was killed in 37 seconds over a distance of 26.8 km. The stage is now 158.3 km down range. Traveling up range to point 2 in the remaining 290 seconds requires an average horizontal velocity of 546 m/s from burn 1. That is, burn 1 needs 1,995.96 m/s delta V.

Using 56,700 kg as m-1, mass at point 1, then 1,997 m/s delta V and the rocket equation gives m-2 = 29,469 kg. The booster is oriented and plummeting downward at 1,756 m/s at 50 km altitude at point 2. Now it is time to skip ahead to point 3 and the landing burn.

I assume a terminal velocity of 200 m/s and a dry mass of 20,000 kg. The rocket equation gives the mass ratio of about 1.07 so the mass, m-3 is 21,355. Therefore the mass ratio for burn 2 is (m-2/m-3) is 29,469 kg/21,355 kg. Using this in the rocket equation gives the burn 2 delta V = 982 m/s. The magnitude of the horizontal and vertical velocities at point 2 = 1839 m/s so after burn 2 the velocity is 856.70 m/s generally downward in whatever direction it needs to be in order to arrive at the landing pad.

I put these numbers, 50 km and -856.7 m/s into my recovery model of my trajectory integration program. The atmospheric density values in the model included are close to the standard atmosphere below 100 km altitude. I used a reference area A = 10.5209 m^ and Cd = 0.3.

The model calculated that at 32,400 m altitude, drag overcame gravity and the booster started to slow down. Max q came at 17,230 m altitude imparting a deceleration of 33.11 m/s^2 (3.38 gs). Finally, at one meter altitude, velocity was -108.3 m/s with acceleration =0.61 m/s^2, that is, the booster was still slowing slightly. I did not try to use the recovery model to do burn-3 for this model run.

My conclusion is that this that the vehicle should survive this trajectory and that stage 1 propellant reserved for recovery needs is about 10%.
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Offline Lars_J

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Can you graph the trajectory?

Offline aero

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Not conveniently as I relied on the rocket equation and existing velocities at staging. Boost back thrust was horizontal, maximum altitude due to the vertical velocity at staging, about 207 km, velocity at point 2, is 546 m/s up range and 1,755.89 m/s downward.

A graph would look something like the trajectory of a ball if you held it by your thigh then threw it underhanded high over your head and behind you. The upward height would be about 2/3 of the distance behind you and 4 times the length of your arm.
« Last Edit: 12/19/2013 07:52 PM by aero »
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Offline hrissan

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Great work, but can you make calculation for less benign trajectory? I may be wrong, but 6g or even 10g during Max Q down may be survivable, because the forces trying to destroy the rocket are acceleration times mass, and the mass of almost empty rocket is small in comparison to the almost full rocket on the Max Q up.

Could you please instead of the velocity of -856.7 m/s down @ 50km altitude you used, try different velocities and see what velocities result in 6g and 10g during Max Q down? Are those velocities much larger?

If they are, then the second (braking) burn would require less fuel, the rocket will weight less at first burn, and you would probably be able to increase the acceleration of first burn using the same 3 Merlins, saving some fuel again.

Offline cambrianera

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Nice job, very interesting!
Found the calculations consistent with estimates done by others (with different logic, obviously).
Only one thing to discuss with you: I think that Cd 0.3 for a blunt object is too low, a better value should be between 0.5-0.8.
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Offline aero

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Nice job, very interesting!
Found the calculations consistent with estimates done by others (with different logic, obviously).
Only one thing to discuss with you: I think that Cd 0.3 for a blunt object is too low, a better value should be between 0.5-0.8.

I wondered about the Cd. Yes, I can run it again with higher values of Cd.

I also wondered about deploying the landing legs early. To model the deployed legs in supersonic flow (at the end of burn-2) what would the drag coefficient be? the ref. area? How would the shock interaction from the 4 legs feet interact with the engines/octaweb? Wouldn't those shock waves cause very high pressure about the octaweb hence deceleration even at high altitudes? I don't know any numbers to use but I can change the reference area and drag coefficient as often and at altitudes of my choice

Quote
Great work, but can you make calculation for less benign trajectory? I may be wrong, but 6g or even 10g during Max Q down may be survivable, because the forces trying to destroy the rocket are acceleration times mass, and the mass of almost empty rocket is small in comparison to the almost full rocket on the Max Q up.

Could you please instead of the velocity of -856.7 m/s down @ 50km altitude you used, try different velocities and see what velocities result in 6g and 10g during Max Q down? Are those velocities much larger?

If they are, then the second (braking) burn would require less fuel, the rocket will weight less at first burn, and you would probably be able to increase the acceleration of first burn using the same 3 Merlins, saving some fuel again.

Yes, I can increase the velocity (shorter burn-2) and search for 6 to 10 g's maximum deceleration from aero drag. It will take a few hours as I would like to get a better handle on drag coefficients and leg deployment, too. But maybe I'll just up the velocity first as that is straight forward to do.

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Offline cambrianera

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Supersonic behaviour is a complicate thing, but frankly speaking I don't think the actual legs are strong enough to stand a supersonic deployment; my suggestion is avoid that modeling, I guess it's useless.
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Offline rockinghorse

My estimate of fuel required for recovery of the F 9 v1.1 first stage is 36.7 tonnes.

How much does this reduce payload to LEO?

Offline aero

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@hrissan

I increased velocity after burn 2, first to 1000 m/s, then just did not do burn 2. Here are the results:

50 km alt, -1000 m/s velocity - Max q gives 4.1 g's deceleration at -734 m/s and 17,526 m altitude.
50 km alt, -1,838.81 m/s Vel. - Max q gives 10.76 g's deceleration at -1,197 m/s and 18,486 m altitude.
Edit add,   -1300  m/s velocity- Max q gives 6.02 g's deceleration at -887.8 m/s and ~ 18,000 altitude.

So it looks to me like we need a burn at point 2, how much is subject to debate until we learn more about the hardware. It also says that the parachute recovery could have worked with the F 9 v1.0 but we know it didn't. We don't know why it didn't work though, it could have been insufficient control authority of the RCS.

@rockinghourse
Quote
How much does this reduce payload to LEO?

Over the expendable version. I don't know but I guess you could use the rocket equation and play with the staging mass and payload mass until it calculated the same total payload delta V as the expendable rocket stages give.

« Last Edit: 12/19/2013 10:35 PM by aero »
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Offline cambrianera

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My estimate of fuel required for recovery of the F 9 v1.1 first stage is 36.7 tonnes.

How much does this reduce payload to LEO?

Approximately you lose 1 kg payload every 7-10 kg increase in dry mass of first stage (additional fuel is equivalent to dry mass increase).
This lead to 3.7 - 5 t loss (again, that's approx).
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Offline aero

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@cambrianera

I reset parameters back to 50 km altitude and -856.7 m/s then ran with Cd = .5 and Cd = .8. As the drag coefficient goes up, the altitude of max q goes up, velocity of max q goes down as does max q. Here are the numbers.

Cd = 0.5, Max q=3.21 g's at 20,600 m altitude and 649.8 m/s velocity,
Cd = 0.8, Max q=3.06 g's at 23,680 m altitude and 634.4 m/s velocity
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Offline dante2308

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@cambrianera

I reset parameters back to 50 km altitude and -856.7 m/s then ran with Cd = .5 and Cd = .8. As the drag coefficient goes up, the altitude of max q goes up, velocity of max q goes down as does max q. Here are the numbers.

Cd = 0.5, Max q=3.21 g's at 20,600 m altitude and 649.8 m/s velocity,
Cd = 0.8, Max q=3.06 g's at 23,680 m altitude and 634.4 m/s velocity

I just wanted to point out that the drag coefficient is a function of Mach number.

Offline Kabloona

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What dynamic pressure are you getting at max Q (in psi or psf)? Since you've already done the calcs.
« Last Edit: 12/19/2013 11:12 PM by Kabloona »

Offline aero

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@cambrianera

I reset parameters back to 50 km altitude and -856.7 m/s then ran with Cd = .5 and Cd = .8. As the drag coefficient goes up, the altitude of max q goes up, velocity of max q goes down as does max q. Here are the numbers.

Cd = 0.5, Max q=3.21 g's at 20,600 m altitude and 649.8 m/s velocity,
Cd = 0.8, Max q=3.06 g's at 23,680 m altitude and 634.4 m/s velocity

I just wanted to point out that the drag coefficient is a function of Mach number.

Yes but can you tell be the function?
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Online eriblo

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Just to add what we know they did on the CASSIOPE flight:

Relight at close to 63.5 km and 1893 m/s, burning for at least 22 s.

From your numbers it would seem that heating is a larger concern than g-forces, going somewhat against the "pancaking/bellyflopping on the atmosphere" qotes. Apparently a good role of thumb is that stagnation temperature (in K) is about the same as reentry speed (in m/s). Checking with NASAs online calculator gives just over 1800 K for the numbers above, although I know nothing about the actual heat flow or how much of the stage will see those temperatures.
« Last Edit: 12/19/2013 11:19 PM by eriblo »

Offline dante2308

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@cambrianera

I reset parameters back to 50 km altitude and -856.7 m/s then ran with Cd = .5 and Cd = .8. As the drag coefficient goes up, the altitude of max q goes up, velocity of max q goes down as does max q. Here are the numbers.

Cd = 0.5, Max q=3.21 g's at 20,600 m altitude and 649.8 m/s velocity,
Cd = 0.8, Max q=3.06 g's at 23,680 m altitude and 634.4 m/s velocity

I just wanted to point out that the drag coefficient is a function of Mach number.

Yes but can you tell be the function?

Here is a little bit of a guide.

Offline starsilk

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you could also cancel out some of the vertical velocity during burn 1 - which would result in lower velocity at entry.

interesting optimization problem to see if using more fuel for burn 1 can save you any later on (assuming a burn 2 is necessary for safe reentry).

Offline aero

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What dynamic pressure are you getting at max Q (in psi or psf)? Since you've already done the calcs.

Well, density is in kg/m^3 and velocity is m/s so dynamic pressure would be in kg m^1 s^2 or Newtons/m^2 or Pascal. Yes, I've done the calcs but the numbers get overwritten for each change.

For the most recent one - Cd = 0.8, Max q=3.06 g's at 23,680 m altitude and 634.4 m/s velocity, I have dynamic pressure as 10,279.41 Pa which converts to 1.490 psi, or 214.689 psf.

Edit add: and density was 5.10890E-02 kg/m^3
« Last Edit: 12/19/2013 11:48 PM by aero »
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Offline Jcc

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If I understand the OP correctly, this is not return to launch site, it is landing in the water.

If you are trying to return, you have a very definite target to hit, and need to coordinate the reversal of down range velocity, and the amount of return velocity, and the residual vertical velocity that is needed to keep the stage out of the atmosphere long enough to make a RTLS trajectory, then the amont of burn #2 needed depends on total velocity (although vertical component may be dominant), so you might be able kill some vertical during burn#1, but it might not be wise to do so if you need the loft time to return.

Offline aero

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If I understand the OP correctly, this is not return to launch site, it is landing in the water.

If you are trying to return, you have a very definite target to hit, and need to coordinate the reversal of down range velocity, and the amount of return velocity, and the residual vertical velocity that is needed to keep the stage out of the atmosphere long enough to make a RTLS trajectory, then the amont of burn #2 needed depends on total velocity (although vertical component may be dominant), so you might be able kill some vertical during burn#1, but it might not be wise to do so if you need the loft time to return.

It was for a return to the launch site, I just used Cassiope launch data because it is easily available in enough detail to estimate the orientation of the vehicle at staging.

As for the burn to kill vertical velocity I pretty much agree with you. Killing vertical velocity would shorten the time to return to the vicinity of the landing pad hence require a higher velocity, higher delta-V for the boost back. Of course there might be some benefit derived from the cosine of the thrust. Point the thrust vector at a slight downward angle and get almost full thrust up track while getting a significant component of downward thrust. That is an optimization problem that I'm not set up to solve but it still shortens the loft time. And it could work the other way. Point the thrust vector at a slight upward angle and increase the loft time and reduce the thrust needed for boost back at the expense of more thrust from burn 2.

I would be happier if I had a drag profile for a blunt cylinder across the velocity regime to better identify the needed delta-V from burn 2. dante2308's drag profiles are indicative but I'd feel much more comfortable with a blunt cylinder. I guess I should check Google now that this point has risen to the surface.
« Last Edit: 12/20/2013 12:09 AM by aero »
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