Author Topic: F9v1.1 Stage 1 (and FH booster) recovery trajectory, burns, drag evaluation.  (Read 21519 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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Offline 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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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.

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.

Sorry, I see the return component now. And come to think of it, there may be some advantage to having the entry be a bit more horizontal, in that you have more time in thin atmosphere that can bleed off velocity and reduce the burn 2 duration.

Offline aero

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... come to think of it, there may be some advantage to having the entry be a bit more horizontal, in that you have more time in thin atmosphere that can bleed off velocity and reduce the burn 2 duration.

Maybe. That's another aspect of the optimization problem that I'm glad SpaceX is working and not me.

But come to think of it, my recovery simulator will accept horizontal as well as vertical velocity. I don't know why I went to the trouble of combining the components then calling them all vertical ... not thinking it through I guess. Maybe I'll look at it again because using both components should reduce max q as you suggest.
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Offline AncientU

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Quote
... come to think of it, there may be some advantage to having the entry be a bit more horizontal, in that you have more time in thin atmosphere that can bleed off velocity and reduce the burn 2 duration.

Maybe. That's another aspect of the optimization problem that I'm glad SpaceX is working and not me.

But come to think of it, my recovery simulator will accept horizontal as well as vertical velocity. I don't know why I went to the trouble of combining the components then calling them all vertical ... not thinking it through I guess. Maybe I'll look at it again because using both components should reduce max q as you suggest.
Or eliminate burn 2... Which what they may have already verified on CASSIOPE.
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Offline cambrianera

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aero, thanks for this interesting thread (and nice discussion, thanks to everybody).
Would like to expand slightly on Cd: dante2308 is correct pointing out that Cd changes with Mach number.
Unfortunately small variations of the shape can cause big differences on Cd, therefore it's very difficult to assign a Cd (or worst, a function Cd vs Mn) to a very complicate shape.
Best thing is define some credible values without pretending excessive details.
Obviously the first Cd you used (0.3) was really underestimated, that's why I suggested to change it.

I'm attaching another graph I found on the net,
https://ceprofs.civil.tamu.edu/llowery/woods/Lectures/451/451%20Drag%20Force.PDF (pag 604)
from this one I suggest a Cd of 0.8 for subsonic and 1 for transonic and supersonic.
Once more, really nice job.
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Offline Rocket Surgeon

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Quick Question.: How does Earth's rotation effect the necessary return velocity?

If Earth is towards you at roughly 400 m/s, then that would reduce the neccesary delta-V for the return burn yes? Significantly if your burn is almost 2000 m/s
Of course this number will vary based on your launch sight, and the direction to launch in, but am I right in stating that needs to be taken into consideration?

Offline Jcc

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Quick Question.: How does Earth's rotation effect the necessary return velocity?

If Earth is towards you at roughly 400 m/s, then that would reduce the neccesary delta-V for the return burn yes? Significantly if your burn is almost 2000 m/s
Of course this number will vary based on your launch sight, and the direction to launch in, but am I right in stating that needs to be taken into consideration?

As long as the launch is more or less easterly, not changing latitude much, I don't think it has much impact on stage return. It gets extra velocity going out, so needs to recover the same going back, besides the atmosphere is rotating with the earth, so that 400 m/s would feel like 0 m/s as far as wind resistance goes.

Offline aero

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Quick Question.: How does Earth's rotation effect the necessary return velocity?

If Earth is towards you at roughly 400 m/s, then that would reduce the neccesary delta-V for the return burn yes? Significantly if your burn is almost 2000 m/s
Of course this number will vary based on your launch sight, and the direction to launch in, but am I right in stating that needs to be taken into consideration?

As long as the launch is more or less easterly, not changing latitude much, I don't think it has much impact on stage return. It gets extra velocity going out, so needs to recover the same going back, besides the atmosphere is rotating with the earth, so that 400 m/s would feel like 0 m/s as far as wind resistance goes.

Taking the Earth's rotation into consideration will be necessary in the real world. I didn't in my evaluation, I did everything relative to the Earth fixed launch pad. The stage goes down range somewhat faster on average than does my lofted return goes up range, so the Landing Pad would gain somewhat on the rocket. A navigation and guidance problem but it might save a little delta-V on burn 1.
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Offline Pete

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So boostback (burn-1) starts at 3.8g, ends at 7.5g
That sounds quite reasonable.

When calculating stage dry mass at the end, how much fuel did you allocate for the landing burn? And how much reserve/remainder at landing? It seems every kg of residual fuel at landing increases the fuel needed at start of boostback by more than 3kg?

hmm, just how close to the bone can we (ahem, i mean spacex) trim the fuel residual without running into trouble...

Offline Pete

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Quick Question.: How does Earth's rotation effect the necessary return velocity?

If Earth is towards you at roughly 400 m/s, then that would reduce the neccesary delta-V for the return burn yes? Significantly if your burn is almost 2000 m/s
Of course this number will vary based on your launch sight, and the direction to launch in, but am I right in stating that needs to be taken into consideration?

The velocities quoted during launch are most definitely relative to the launch site. Thus the effect of earth's rotation are quite negligible during the fraction of an hour the launch and return will require.

Related question:
At what point in a launch do they switch over from launch-site-centric to earth-centric reporting of velocities?
In full correctness, the velocity of the rocket while still strapped in to the pad in Florida should be reported as 1470km/h(408 m/s)
Of course, this would confuse the living daylights out of any non-rocketry person  ;)
« Last Edit: 12/21/2013 06:29 AM by Pete »

Offline pagheca

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very interesting, aero. I'm doing something like that as an interesting exercise but I'm not yet obtained clear results.

One question: did you used the standard atmospheric model to compute drag at various altitudes?
« Last Edit: 12/21/2013 01:25 PM by pagheca »

Offline pagheca

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At what point in a launch do they switch over from launch-site-centric to earth-centric reporting of velocities?
In full correctness, the velocity of the rocket while still strapped in to the pad in Florida should be reported as 1470km/h(408 m/s)
Of course, this would confuse the living daylights out of any non-rocketry person  ;)

I have been thinking about this line too.

I came to the conclusion that for the 1st stage it doesn't really matter because at TO and landing speed must be zero in a RTSL. So, I assumed all the relevant speeds for the 1st stage trajectories posted by SpaceX are launch-site centric.

Of corse if this is just a sea landing, there is a slight difference, due to the different latitude, but quite negligible in the actual situations (Vandenberg, Florida...).

cheers,
« Last Edit: 12/21/2013 01:40 PM by pagheca »

Offline aero

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So boostback (burn-1) starts at 3.8g, ends at 7.5g
That sounds quite reasonable.

When calculating stage dry mass at the end, how much fuel did you allocate for the landing burn? And how much reserve/remainder at landing? It seems every kg of residual fuel at landing increases the fuel needed at start of boostback by more than 3kg?

hmm, just how close to the bone can we (ahem, i mean spacex) trim the fuel residual without running into trouble...

In my evaluation I used a constant acceleration limited 4 g's as that made it easier at the time. BOE calculations instead of running the integration. Using higher accelerations would save some prop. on burn 1 though.

I wonder about that too. I guessed terminal velocity at 200 m/s then used the rocket equation to calculate the mass ratio using dry stage 1 mass as the final mass. Different terminal velocities will require different prop mass for landing of course. I bet terminal velocity is one of the things SpaceX is measuring carefully and will continue to do so with Grasshopper. I speculate the limitation will be the accuracy (precision) they can rely on in measuring remaining prop during engine burn at burn 1 and burn 2. Burn 2 may be more important to deplete excess prop than to slow the descent velocity. That depends very much on real world measurements which they are and will continue to make.
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Offline aero

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very interesting, aero. I'm doing something like that as an interesting exercise but I'm not yet obtained clear results.

One question: did you used the standard atmospheric model to compute drag at various altitudes?

I used two different density models splined at about 10,500 meters altitude. Then I ran the models and compared them by eye to a standard atmosphere table. I used Excel to do this. Models and table are here:

http://en.wikipedia.org/wiki/Density_of_air#Altitude Low altitude spline.

I can't find the URL for the source of the high altitude spline, but the model is: rho = rho0*exp(-h/H).
roh0= 1.752 kg/m^3   and H=6.7 km
   
http://www.engineeringtoolbox.com/standard-atmosphere-d_604.html Used the second table, SI units.

Good luck and let us know how it comes out.
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Offline go4mars

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I'm attaching another graph I found on the net,
https://ceprofs.civil.tamu.edu/llowery/woods/Lectures/451/451%20Drag%20Force.PDF (pag 604)
By strange coincidence that looks quite a bit like a Van Krevelin diagram. 
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Offline Pete

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In my evaluation I used a constant acceleration limited 4 g's as that made it easier at the time. BOE calculations instead of running the integration. Using higher accelerations would save some prop. on burn 1 though.


It would save a lot of prop.
Starting with same mass, Burn-1 would be about 7.5 seconds shorter, thus consuming more than 5 tons less fuel.

But starting with less fuel needed, allows higher acceleration, slightly shorter total downrange, etc...

I figure 28 tons fuel needed total, 37 seconds burn-1 (vs your 51 sec). Same speed, fuel, angle etc.. for entry after that.

The rest of your calc is so complete, do yourself the favor of considering the decreasing rocket mass during the initial burn. It makes a *lot* of difference.

Offline aero

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In my evaluation I used a constant acceleration limited 4 g's as that made it easier at the time. BOE calculations instead of running the integration. Using higher accelerations would save some prop. on burn 1 though.


It would save a lot of prop.
Starting with same mass, Burn-1 would be about 7.5 seconds shorter, thus consuming more than 5 tons less fuel.

But starting with less fuel needed, allows higher acceleration, slightly shorter total downrange, etc...

I figure 28 tons fuel needed total, 37 seconds burn-1 (vs your 51 sec). Same speed, fuel, angle etc.. for entry after that.

The rest of your calc is so complete, do yourself the favor of considering the decreasing rocket mass during the initial burn. It makes a *lot* of difference.

As a practical matter you may be right. Also as a practical matter there may be acceleration limits on the stage one hardware. If so, I don't know what they are. But mostly, I did this and posted it as food for thought here on the forum. You have thought about it and now point out to everyone a potential way to save prop. My objective is satisfied on this point.

I'm surprised no one has commented about my instant deceleration upon staging. I suspect there will need to be some delay just to allow Stage 2 to clear the area and maybe a longer delay while Stage 1 re-orients to point up range. Thoughts anyone? Any delay at this point in recovery will cost fuel as the stage is coasting down range at relatively high velocity and the time to return to point 2 is limited. The further the stage coasts the more prop needed for the higher velocity return though it might be on the order of 10's of meters/second.
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Offline macpacheco

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Take this comment with a grain of salt, I'm no aerospace or even a mechanical engineer.
Wouldn't the Cd for the stage be even higher than a flat object ?
The shape of the engines are like 9 hard shaped parachutes. They're essentially creating higher turbulence to the airflow
Those calculations are very interesting to help us understand the actual speeds the landing legs might be extended.
Can you tell us vertical speed at 15km, 10km and 5km altitude ?
Thanks
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Offline aero

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Take this comment with a grain of salt, I'm no aerospace or even a mechanical engineer.
Wouldn't the Cd for the stage be even higher than a flat object ?
The shape of the engines are like 9 hard shaped parachutes. They're essentially creating higher turbulence to the airflow
Those calculations are very interesting to help us understand the actual speeds the landing legs might be extended.
Can you tell us vertical speed at 15km, 10km and 5km altitude ?
Thanks
I don't know what the Cd would be in reality. Do you know anyone with a wind tunnel and a model?  :) Or a 3-D printer to print up the model of the base, that would be quick.

The general tendency seems to be that the Cd goes down as speed gets very low. I am using Cd = 0.8 throughout the descent and get these numbers but I don't really believe them due to the unknown variation in Cd at low speed.

15 km alt., Vertical velocity = -88.7 m/s
10 km alt., Vertical velocity = -58.6 m/s
05 km alt., Vertical velocity = -43.6 m/s

I plan to code the Cd as a function of Mach number curves into my model soon. The curves seem to cover this low speed range as well as high speeds, so that will be helpful. Right now I'm back to looking at the Raptor 9 rocket so maybe later.
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Offline aero

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Ok! I finished programming temperature(altitude), speed of sound(temperature), Mach number(speed of sound) and coefficient of drag(Mach #) into my simulation, one Mach curve for moving toward the pointy end, and another curve for going backwards. It made quite a difference in terminal velocity so I'm more confident in the numbers now. The drag model for recovery is the one for a sphere. Numbers now for terminal velocities.

Altitude = 15 km, Vz= 88.7 m/s (using Cd = 0.8)  Vz = 124.5 m/s using spherical drag model
Altitude = 10 km, Vz= 58.6 m/s (using Cd = 0.8)  Vz =   89.3 m/s using spherical drag model
Altitude =   5 km, Vz= 43.6 m/s (using Cd = 0.8)  Vz =   70.0 m/s using spherical drag model

70 m/s is Mach 0.2 and still seems a little slow but that empty stage doesn't weigh much for its size ...
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Offline macpacheco

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Ok! I finished programming temperature(altitude), speed of sound(temperature), Mach number(speed of sound) and coefficient of drag(Mach #) into my simulation, one Mach curve for moving toward the pointy end, and another curve for going backwards. It made quite a difference in terminal velocity so I'm more confident in the numbers now. The drag model for recovery is the one for a sphere. Numbers now for terminal velocities.

Altitude = 15 km, Vz= 88.7 m/s (using Cd = 0.8)  Vz = 124.5 m/s using spherical drag model
Altitude = 10 km, Vz= 58.6 m/s (using Cd = 0.8)  Vz =   89.3 m/s using spherical drag model
Altitude =   5 km, Vz= 43.6 m/s (using Cd = 0.8)  Vz =   70.0 m/s using spherical drag model

70 m/s is Mach 0.2 and still seems a little slow but that empty stage doesn't weigh much for its size ...

Edit... See more recent corrections by aero, numbers were underestimated by 120%, or about 1/5th the kinetic energy, which is a big deal. After the corrections my comments loose some sense.

Anyhow, the original post:

Exactly, those speeds are surprisingly low for a large free falling object of slender shape without wings or chutes.

Typical terminal velocity for a skydiver in a basic arched position (at low altitude), doing nothing special to speed up or slow down his fall is 180-200 Km/h, 43.6m/s = 157 Km/h, 88.7 m/s = 320Km/h.

It's been 15 years that I don't jump, but the adrenaline of a jump allows me to recall all sensations with extreme detail.

I've been to 300Km/h in a forward, head down or standing up dive, and still the forces are far from anything that the muscles of a young man can't handle.

For me if this is correct, settles it, the legs will be extended just soon enough the stage can transition from terminal velocity (legs retracted) to terminal velocity (legs open) plus some time to spare. No thermal issues at this point. The skydiving experience also suggests it shouldn't take more than 2 minutes for the legs to effect 99% of it's slow down (maybe even a single minute).

This would also tell me that even at the very thin atmosphere 15km ~ FL450 (air density less than 1/4 of sea level), the the vast majority of the heating from re-entry is gone, very chilly air even at the hottest day of the year in the cape, or any other launch site candidates.

The only issue will be control authority for yaw (to prevent spinning). On that topic, I really can't help.
Control authority on the lateral plane shouldn't be an issue, the length of the rocket should give it some minimum neutral static stability, the more sideways the stage is vs the relative wind, drag should go up, trying to straighten the stage, should be larger than the tail of the rocket trying to increase the movement. Even with the legs open, the length of the stage should do wonders.

Again, I'm not an aerospace engineer, no fluid dynamics formal education, but my gut feel is the last few minutes before landing might be the easiest of the times for the stage (before the slam landing burn), and extending the legs to get maximum drag slow down might be totally doable and useful given the cumulative effects of any fuel savings on the final burn have on usable launch payload.
« Last Edit: 12/23/2013 11:53 PM by macpacheco »
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Offline cambrianera

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@aero, please check your calcs, because with a simple formula ( http://en.wikipedia.org/wiki/Terminal_velocity )
sea level terminal velocity for Falcon 9 should be 200 m/s.
(mass 20000 kg, air density 1.2 kg/m3, area 10.5 m2, Cd 0.8 ).
Oh to be young again. . .

Offline MP99

Altitude = 15 km, Vz= 88.7 m/s (using Cd = 0.8)  Vz = 124.5 m/s using spherical drag model
Altitude = 10 km, Vz= 58.6 m/s (using Cd = 0.8)  Vz =   89.3 m/s using spherical drag model
Altitude =   5 km, Vz= 43.6 m/s (using Cd = 0.8)  Vz =   70.0 m/s using spherical drag model

You need to tick the "Don't use smileys." box!

cheers, Martin

Offline MP99

I'm linking this mostly because of the fab Calvin & Hobbes strip, but it may also be of interest:-

https://sugarshotsolidworks.wordpress.com/2013/12/22/numeric-analysis-of-nose-cone-heating-first-steps/

cheers, Martin

Offline aero

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@aero, please check your calcs, because with a simple formula ( http://en.wikipedia.org/wiki/Terminal_velocity )
sea level terminal velocity for Falcon 9 should be 200 m/s.
(mass 20000 kg, air density 1.2 kg/m3, area 10.5 m2, Cd 0.8 ).

What would I check? Oh, I see one thing. Been fooling with my Raptor model so had the wrong ref. area in the recovery portion. (I thought my Launch and recovery segments were independent) With the F 9 v1.1 ref. area = 10.5209 m^2 I now have:

15 km----------236.9 m/s   Mach 0.8
10 km----------156.4 m/s   Mach 0.53
  5 km----------121.5 m/s   Mach 0.38
SL ---------------98.96 m/s Mach 0.29

My drag Coefficient  drops from ~ .62 to ~ .35 over that range of Mach numbers.

This is much better. I did have a thought although I'm not addressing it in my simulations. Aren't the propellant tanks filled with helium during recovery? That should give them about 400 kg of buoyancy if so.

As for checking my calculations, what else could be wrong? If the Mach # is correct and it is, see:

http://www.grc.nasa.gov/WWW/k-12/airplane/mach.html

Mach numbers check, so the Cd look-up is correct, checked against the graphs for Cd for a sphere up thread, dynamic pressure and drag calculations are very hard to mess up and that's it. I did verify my simulation against the advertised F 9 v1.1 payload to orbit and it does reach orbit with 13150 kg of payload, just barely but it makes it, so I'm confident of my integrator. That means to me that my latest numbers are probably as close to reality as I'll get.

I should try to add legs, which will be easy to do if we can decide at what altitude they are deployed. First thing would be to change the area at a given altitude, then if we can figure it out, change the Cd model to represent the stage with the legs extended.
« Last Edit: 12/23/2013 05:55 PM by aero »
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Offline guckyfan

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You could try to do the evaluation backwards, assuming Elon Musks statement is correct. He said the final burn will consume less than 5% of the empty stage weight. So calculate the burn for 1t of fuel and see what terminal speed can be killed with that burn. Then you can calculate the drag required to achieve that terminal speed. Should be interesting.


Offline Pete

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 Aren't the propellant tanks filled with helium during recovery? That should give them about 400 kg of buoyancy if so.


Their content is irrelevant, as the first stage is a volume-constrained rigid object.
Its buoyancy would be exactly that of the air mass displaced, which is about 440m3. At sea level, that is about 525kg. At anything over 10km altitude, a very very good approximation of zero.

All this will affect is the weight of the stage, not its mass. So the only real difference is a reduction of final terminal velocity of about 1.3%. *way* under the error bars of our guesstimates for mass, drag, and even thrust.

Offline Pete

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You could try to do the evaluation backwards, assuming Elon Musks statement is correct. He said the final burn will consume less than 5% of the empty stage weight. So calculate the burn for 1t of fuel and see what terminal speed can be killed with that burn. Then you can calculate the drag required to achieve that terminal speed. Should be interesting.

Matches our figures, as a first approximation.
5% of empty stage mass is about 1000kg.
At sea level, full throttle, the single merlin 1d impart 3.1g to empty stage + 1 ton.
It would burn through that 1 ton of fuel in 4.25 seconds, delivering total delta of about 129m/s.

We are assuming terminal velocity without legs as about 100m/s, and terminal with legs of about 70 m/s.
Stopping 70m/s under these conditions would require only 2.3 seconds of full thrust, leaving a fuel reserve of only 2 seconds.. not enough? Maybe we need to leave a bit more reserve fuel, to give us some leeway for inexact fuel sensors, fuel line burps, etc.
This is totally out of my expertise... When you have a device that pumps 235 per second, and you dare not let it run dry, how empty can you run your tankage before the inevitable airbubble, unexpected pressure due to empty plumbing, etc? Just how much juice is available to the turbopump, when the tank itself is exactly empty? Even worse, its not just one fluid you need to worry about, but two completely separate sources that *must* be in the correct ratio. A hiccup in either would shred the pump, thus RUD the engine.

How much propellant will be left when the engine starts complaining about it?
  and how much more do we need to keep in reserve, due to various inaccuracies?
(I'm sure this would have been discussed previously, I just could not locate the thread)

Offline guckyfan

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You could try to do the evaluation backwards, assuming Elon Musks statement is correct. He said the final burn will consume less than 5% of the empty stage weight. So calculate the burn for 1t of fuel and see what terminal speed can be killed with that burn. Then you can calculate the drag required to achieve that terminal speed. Should be interesting.

Matches our figures, as a first approximation.
5% of empty stage mass is about 1000kg.
At sea level, full throttle, the single merlin 1d impart 3.1g to empty stage + 1 ton.
It would burn through that 1 ton of fuel in 4.25 seconds, delivering total delta of about 129m/s.

Thanks, good comparison.

Offline MP99


 Aren't the propellant tanks filled with helium during recovery? That should give them about 400 kg of buoyancy if so.


Their content is irrelevant, as the first stage is a volume-constrained rigid object.
Its buoyancy would be exactly that of the air mass displaced, which is about 440m3. At sea level, that is about 525kg. At anything over 10km altitude, a very very good approximation of zero.

Yes, that is the buoyancy. If the stage contained air at ambient pressure, then the internal gas would weigh the same as the displaced gas, and there would be no net buoyancy.

However, if the stage's O2 tank contains O2 at 5 atmospheres, then 5x the weight of that volume of O2 at sea level would be added to the tank's dry mass (and this would be more than 5x the weight of that pressure of air, given the population of molecules involved).



Think of this another way. In vacuum, the mass/weight of the stage constitutes the hardware, any remaining prop, and any pressurisation gas.

As the stage goes lower in the atmosphere it experiences a certain amount of buoyancy, until at ground level it experiences lift equating to 1x atmosphere over the volume of the tank. But that lift may be less or greater than the mass of gas in that tank, depending on which gases are involved, and their pressure.

cheers, Martin

Offline Aerospace Dilettante

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Thanks to the OP, even though I don't understand all of it.  While read through this thread, I remembered something Elon said at the National Press Club:

Quote
(Also worth noting,) you dont need wings to steer aerodynamically, you just need some lift over drag numbers and lift vector.


Does anyone think that the boost-back burn could be reduced and stage could be sort of glided part of the way back to the launch site.  Don't know what the L/D is for a ginormous cylinder is.

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