Author Topic: SpaceX to begin testing on Reusable Falcon 9 technology this year  (Read 307874 times)

Offline modemeagle

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I'm not sure, but I think that thrusting back along the trajectory will give better results (i.e. a gravity turn in reverse).

I tried this once, and only had about 4 tonnes of fuel to land, which is not enough.  Doing a 180 degree burn leaves ~ 10 tonnes of fuel to land.

Offline RDoc

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Its also multiplied by Cd, and at 30 km my Cd model is nearly zero.  What are you using the calculated Cd?
I'd like to understand how you are calculating that. It's my understanding that Cd mainly varies with Reynold's number, but these are all high turbulent values so if anything the Cd is increasing. I don't believe Cd varies much with altitude and certainly doesn't go to zero, I use 1 as a guess.

Could you provide a reference for Cd going to zero at high altitude?

John Glenn experienced about 9G deceleration at about 36km and 3000 m/s so the Cd would be about 1.2.

If the Cd went to zero at high altitude, LEO could be very low and meteorites wouldn't burn up at over 50 km.

Columbia started to break up at about 70 km and was destroyed at about 60km

30km is only about 100000 ft and drogue chutes are used there.
« Last Edit: 06/09/2012 04:09 PM by RDoc »

Offline MikeAtkinson

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Cd = 2 * F / (rho * v^2 * A)

F = force
rho = mass density of fluid
v = velocity
A = reference area

so F = Cd * rho * v^2 * A / 2

At 30,000m atmospheric density is about 3% of surface.

So to experience the same force as at ground level the stage needs to be going 6x as fast. If terminal velocity is 250m/s (guess) then at 30,000m the stage will experience 1g of acceleration at 1500m/s and 2g at 2100m/s. [numbers rounded]

Offline cordor

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Can it somehow generate enough lift to stay at the same altitude? maybe fire multiple short bursts. Eventually the speed slow down and the pad caught up, by than F9 just need to descent.

Yes, certainly. In that simulation the burn to start the ballistic way back doesn't consume all the propellants, 4.6 mT remain so there's the possibility of slowing down using them.

I'm still thinking about what would be the best way to do reentry because if you start one engine above the atmosphere there might not be enough fuel to reach the surface and do the landing, and if we wait until latter then there's the difficulty of ignition against a supersonic flow.

And you can't wait till the stage is subsonic (1.400 m) either, because then there's not enough time to brake, even at max thrust (one engine).

I see two ways. First, reorient the stage sideways to do reentry so the atmospheric drag slows it down much higher. Second, deploying a supersonic parachute at about 10 km up (mach 3).

Do we have numbers on max-q, velocity and acceleration at max-q of the rocket going up?
...
The question now will be if the engines nozzles would survive 25atm acting on them when falling down?

For the going up max-Q was 30 kPa and max acceleration about 5.5 g.

About max-Q on the way down I made a mistake on my previous post, was not 2,542 kPas but only 295 kPa (3 atm). I just edited the post.

Charliem, why point at 161 degrees instead of 180 to cancel all of the horizontal thrust.  At 161 your adding more to the apogee which will kill the reentry conditions.

You are right about the reentry conditions, but that trajectory was the one that burned less fuel, having also the best range.


Nope, that's not what i mean. No ballistic return, no high speed reentry.  Use angle of attack to generate enough lift at 78,100 m, maybe even steer the flight path toward lower latitude. Stay at that altitude as long as you can, use short bursts of thrust to maintain altitude if you have to, this way you always fly under original max-q . Eventually, you will slow down, then you move to lower altitude, air is getting dense, there will be more drag and more lift, you will slow down faster. Keep doing this until reaching transsonic speed. Then slowly descent downward, the pad is now directly under you.

Offline RDoc

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Cd = 2 * F / (rho * v^2 * A)

F = force
rho = mass density of fluid
v = velocity
A = reference area

so F = Cd * rho * v^2 * A / 2

At 30,000m atmospheric density is about 3% of surface.

So to experience the same force as at ground level the stage needs to be going 6x as fast. If terminal velocity is 250m/s (guess) then at 30,000m the stage will experience 1g of acceleration at 1500m/s and 2g at 2100m/s. [numbers rounded]
For the stage going end on, SL terminal velocity is about 220 m/s, sideways it's about 60 m/s. The density at 30km is about 1.4% so end on to get 1G is about 1900m/s and sideways about 500 m/s.

That's assuming a constant Cd of 1 which is likely low.

Offline modemeagle

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Its also multiplied by Cd, and at 30 km my Cd model is nearly zero.  What are you using the calculated Cd?
I'd like to understand how you are calculating that. It's my understanding that Cd mainly varies with Reynold's number, but these are all high turbulent values so if anything the Cd is increasing. I don't believe Cd varies much with altitude and certainly doesn't go to zero, I use 1 as a guess.

Could you provide a reference for Cd going to zero at high altitude?

John Glenn experienced about 9G deceleration at about 36km and 3000 m/s so the Cd would be about 1.2.

If the Cd went to zero at high altitude, LEO could be very low and meteorites wouldn't burn up at over 50 km.

Columbia started to break up at about 70 km and was destroyed at about 60km

30km is only about 100000 ft and drogue chutes are used there.

I calculate CD based on this document.  It was the best I could find when I was writing my simulator.

Offline charliem

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[I calculate CD based on this document.  It was the best I could find when I was writing my simulator.

Thanks a lot Modemeagle. Really interesting. Much better than trying to guess Cd.

Nope, that's not what i mean. No ballistic return, no high speed reentry.  Use angle of attack to generate enough lift at 78,100 m, maybe even steer the flight path toward lower latitude. Stay at that altitude as long as you can, use short bursts of thrust to maintain altitude if you have to, this way you always fly under original max-q . Eventually, you will slow down, then you move to lower altitude, air is getting dense, there will be more drag and more lift, you will slow down faster. Keep doing this until reaching transsonic speed. Then slowly descent downward, the pad is now directly under you.

Sorry Cordor, I did not understand you the first time.

Just a couple of things. A cylinder has very little lift, to be able to counter gravity and maintain altitude at 78 km, the stage would have to have an enormous horizontal velocity. In fact the lift is so low that that's not possible even much lower into the atmosphere, and the friction would be so high that most probably it would melt (remember the [mostly] titanium skin of SR-71? At mach 3 and 25 km high parts of it needed active cooling or they would melt).

The alternative, using thrust to keep altitude, looks more promising although I suspect there's not enough fuel but I'll try it in my simulator and see what comes out.

Offline modemeagle

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[I calculate CD based on this document.  It was the best I could find when I was writing my simulator.

Thanks a lot Modemeagle. Really interesting. Much better than trying to guess Cd.

Nope, that's not what i mean. No ballistic return, no high speed reentry.  Use angle of attack to generate enough lift at 78,100 m, maybe even steer the flight path toward lower latitude. Stay at that altitude as long as you can, use short bursts of thrust to maintain altitude if you have to, this way you always fly under original max-q . Eventually, you will slow down, then you move to lower altitude, air is getting dense, there will be more drag and more lift, you will slow down faster. Keep doing this until reaching transsonic speed. Then slowly descent downward, the pad is now directly under you.

Sorry Cordor, I did not understand you the first time.

Just a couple of things. A cylinder has very little lift, to be able to counter gravity and maintain altitude at 78 km, the stage would have to have an enormous horizontal velocity. In fact the lift is so low that that's not possible even much lower into the atmosphere, and the friction would be so high that most probably it would melt (remember the [mostly] titanium skin of SR-71? At mach 3 and 25 km high parts of it needed active cooling or they would melt).

The alternative, using thrust to keep altitude, looks more promising although I suspect there's not enough fuel but I'll try it in my simulator and see what comes out.


Cd is not a constant, but would change based on conditions.  I looked everywhere for any data on drag and Cd calculation and finally came across this.  Since calculating wave drag requires integrals I will probably not try that in Excel as I don't know of any way to do those in excel.

Offline charliem

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Cd is not a constant, but would change based on conditions.  I looked everywhere for any data on drag and Cd calculation and finally came across this.  Since calculating wave drag requires integrals I will probably not try that in Excel as I don't know of any way to do those in excel.

Yes, Cd is a tough parameter. In my case tried to guess it using some tables I found with results from wind tunnel tests of other rockets. Not exactly the best way but could not find any better ... until know. Thanks again.

Changing subjects. I  did the simulation that you suggested for the boost-back of the F9 first stage.

As I suspected thrusting heading 180 deg results in burning more propellants, but it has some advantageous side effects that might make that unimportant.

I changed a little the conditions for the simulation. Allowed acceleration to reach 8 g instead of the previous 6, and started with 4 engines as you indicated.

The initial parameters (at MECO) where the same that for the previous round:

Speed relative to the pad: 2,600 m/s
Angle of ascent: 23 deg
Altitude: 78,100 m
Distance from the pad: 111,000 m
Propellant unburned: 40.9 mT

Thrusting at 161 deg the amount of propellants remaining after the burn was 5.8 mT, while thrusting at 180 deg was 3.5 mT, but a flatter trajectory allows for a much gentler reentry. Max-speed, max-Q, and max-g are significantly lower:

thrust heading ________ 161 deg _______________ 180 deg _____
max-speed _____ 2020 m/s (33 km high) ___ 1720 m/s (34 km high)
max-Q ___________ 302 kPa (12 km) _________ 172 kPa (13 km)
max-g ______________ 132 m/s2 ______________ 86 m/s2
subsonic at altitude______800 m ________________ 3400 m
prop for landing ____ 5.8 mT (955 m/s) ______ 3.5 mT (620 m/s)

All in all I'd say your trajectory is better. It ends with less fuel but the stresses are much lower, and the stage is subsonic at an altitude that allows it to do the final brake using only thrust, while mine would require some type of assist (I thought of a supersonic parachute, but with your alternative that's not needed).

P.E. Numerical integration with Excel is easy. The Simpson algorithm is quick and simple, and fairly accurate for most applications. I have a couple of examples if you want them.

Offline modemeagle

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Cd is not a constant, but would change based on conditions.  I looked everywhere for any data on drag and Cd calculation and finally came across this.  Since calculating wave drag requires integrals I will probably not try that in Excel as I don't know of any way to do those in excel.

Yes, Cd is a tough parameter. In my case tried to guess it using some tables I found with results from wind tunnel tests of other rockets. Not exactly the best way but could not find any better ... until know. Thanks again.

Changing subjects. I  did the simulation that you suggested for the boost-back of the F9 first stage.

As I suspected thrusting heading 180 deg results in burning more propellants, but it has some advantageous side effects that might make that unimportant.

I changed a little the conditions for the simulation. Allowed acceleration to reach 8 g instead of the previous 6, and started with 4 engines as you indicated.

The initial parameters (at MECO) where the same that for the previous round:

Speed relative to the pad: 2,600 m/s
Angle of ascent: 23 deg
Altitude: 78,100 m
Distance from the pad: 111,000 m
Propellant unburned: 40.9 mT

Thrusting at 161 deg the amount of propellants remaining after the burn was 5.8 mT, while thrusting at 180 deg was 3.5 mT, but a flatter trajectory allows for a much gentler reentry. Max-speed, max-Q, and max-g are significantly lower:

thrust heading ________ 161 deg _______________ 180 deg _____
max-speed _____ 2020 m/s (33 km high) ___ 1720 m/s (34 km high)
max-Q ___________ 302 kPa (12 km) _________ 172 kPa (13 km)
max-g ______________ 132 m/s2 ______________ 86 m/s2
subsonic at altitude______800 m ________________ 3400 m
prop for landing ____ 5.8 mT (955 m/s) ______ 3.5 mT (620 m/s)

All in all I'd say your trajectory is better. It ends with less fuel but the stresses are much lower, and the stage is subsonic at an altitude that allows it to do the final brake using only thrust, while mine would require some type of assist (I thought of a supersonic parachute, but with your alternative that's not needed).

P.E. Numerical integration with Excel is easy. The Simpson algorithm is quick and simple, and fairly accurate for most applications. I have a couple of examples if you want them.

I noticed today the document I refer to has non integral equations for wave drag calculations.  I will try to integrate  ;) the math into my Cd calculation.  My simulator was really written for launch and not reentry/recovery hence I only used friction drag in my simulation.  I will see if the addition will change the Cd at altitude.


Offline modemeagle

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Here is my final recovery simulation. 

I used a base .75 Cd.
Boost back is with 4 engines 100% thrust
Min throttle on center engine was 35% thrust (below published minimum for M1D)

Edit:  Edited graphs for accuracy.
« Last Edit: 06/10/2012 05:24 PM by modemeagle »

Offline RDoc

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I calculate CD based on this document.  It was the best I could find when I was writing my simulator.
Everything I've read has said that until the mean free path of the medium is large compared to the size of the object, Cd is pretty much constant. As the velocity goes transonic it greatly increases, then decreases again but to a higher level at supersonic speeds.

Beyond that, I'm not sure how to make a convincing argument, although if you are interested you might want to do web searches linking Cd and density or altitude. You won't find many hits, it's generally considered a constant.

Anyway, these are my comments:

First, the paper never discusses Pressure Drag which is what Q*A is about. I think what he's talking about is additional factors added to the subsonic Pressure Drag, but it's hard to say since the paper doesn't define its terms. Base drag by the way is the drag due to the tail of the object creating a vacuum at high speed.

Second, that paper is from the Jackson Model Rocket Club (jmrc) which isn't really a particularly impressive source.

Third, the NASA papers never mention anything about Cd varying with altitude, and in fact just take it as a constant. Personally, I consider NASA a better source.

Fourth, demonstrably Cd isn't anywhere near zero at high altitude.

If you are making a model that varies Cd, test it by using the Mercury capsule data at least. The capsule had a frontal area of about 3m^2, mass of 1200km, and at 36km and 3000 m/s generated 9G. That is seriously inconsistent with a Cd approaching zero.

http://www.google.com/url?sa=t&rct=j&q=atmospheric%20reentry&source=web&cd=2&sqi=2&ved=0CFoQFjAB&url=http%3A%2F%2Fexoaviation.webs.com%2Fpdf_files%2FAtmospheric%2520Re-Entry.pdf&ei=tR7UT4DgHMTW6gHy6aS8Aw&usg=AFQjCNEtKyc-oXS5vEOK6FdMC8lYqe22OA&cad=rja

If a model is wildly off from measured reality, it's the model not the reality that needs correction.

Offline RDoc

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Here is my final recovery simulation. 

I used a base .75 Cd.
Boost back is with 4 engines 100% thrust
Min throttle on center engine was 35% thrust (below published minimum for M1D)
I'm sorry to keep disagreeing with you, but this chart is pretty clearly not correct.

It still shows no dynamic pressure above about 50km and a sudden peak at 20km. That cannot be correct.

What was it that broke up the Columbia well above that altitude and caused the Mercury capsule to decelerate at 2G at 60km and over 8 at 36km?

Also, is the G chart in absolute numbers? It shows positive G values both increasing and decreasing velocity.

Offline modemeagle

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Here is my final recovery simulation. 

I used a base .75 Cd.
Boost back is with 4 engines 100% thrust
Min throttle on center engine was 35% thrust (below published minimum for M1D)
I'm sorry to keep disagreeing with you, but this chart is pretty clearly not correct.

It still shows no dynamic pressure above about 50km and a sudden peak at 20km. That cannot be correct.

What was it that broke up the Columbia well above that altitude and caused the Mercury capsule to decelerate at 2G at 60km and over 8 at 36km?

Also, is the G chart in absolute numbers? It shows positive G values both increasing and decreasing velocity.

Substitute G for acceleration and that is really what that graph is for recovery.  I just forgot to relabel the chart and change the units from G to m/s.  Multiply the value by 9.80665.

As for dynamic pressure.  This stage is not entering at orbital velocity but much lower.  If I input 5900 m/s at 61km I get 4,806 pa for dynamic pressure.

Offline modemeagle

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I calculate CD based on this document.  It was the best I could find when I was writing my simulator.
Everything I've read has said that until the mean free path of the medium is large compared to the size of the object, Cd is pretty much constant. As the velocity goes transonic it greatly increases, then decreases again but to a higher level at supersonic speeds.

Beyond that, I'm not sure how to make a convincing argument, although if you are interested you might want to do web searches linking Cd and density or altitude. You won't find many hits, it's generally considered a constant.

Anyway, these are my comments:

First, the paper never discusses Pressure Drag which is what Q*A is about. I think what he's talking about is additional factors added to the subsonic Pressure Drag, but it's hard to say since the paper doesn't define its terms. Base drag by the way is the drag due to the tail of the object creating a vacuum at high speed.

Second, that paper is from the Jackson Model Rocket Club (jmrc) which isn't really a particularly impressive source.

Third, the NASA papers never mention anything about Cd varying with altitude, and in fact just take it as a constant. Personally, I consider NASA a better source.

Fourth, demonstrably Cd isn't anywhere near zero at high altitude.

If you are making a model that varies Cd, test it by using the Mercury capsule data at least. The capsule had a frontal area of about 3m^2, mass of 1200km, and at 36km and 3000 m/s generated 9G. That is seriously inconsistent with a Cd approaching zero.

http://www.google.com/url?sa=t&rct=j&q=atmospheric%20reentry&source=web&cd=2&sqi=2&ved=0CFoQFjAB&url=http%3A%2F%2Fexoaviation.webs.com%2Fpdf_files%2FAtmospheric%2520Re-Entry.pdf&ei=tR7UT4DgHMTW6gHy6aS8Aw&usg=AFQjCNEtKyc-oXS5vEOK6FdMC8lYqe22OA&cad=rja

If a model is wildly off from measured reality, it's the model not the reality that needs correction.

I went through that PDF and did not find a single chart that had dynamic pressure at 200kft.  Can you point out the relevant data.  Also, if you look at the graphs for mercury friendship 7 notice the velocity at 200kft and then see how that does not match the velocity on my graphs.  That mission was orbital and the F9 SI return to site is suborbital.  Please advise with your findings.  If you have the time can you make me a chart for dynamic pressure for that mission for comparison.

Offline charliem

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Here is my final recovery simulation. 

I used a base .75 Cd.
Boost back is with 4 engines 100% thrust
Min throttle on center engine was 35% thrust (below published minimum for M1D)
I'm sorry to keep disagreeing with you, but this chart is pretty clearly not correct.

It still shows no dynamic pressure above about 50km and a sudden peak at 20km. That cannot be correct.

What was it that broke up the Columbia well above that altitude and caused the Mercury capsule to decelerate at 2G at 60km and over 8 at 36km?

I see your objection is based in the striking difference between the numbers for F9 we are talking, and Mercury and Shuttle behaviors during reentry.

There's no contradiction. It's a question of weight, area, and velocity.

For Mercury you mentioned an speed of 3 km/s, area 3 m2, and mass 1.2 mT. That means a ballistic coefficient ~400 kg/m2 (if Cd ~1).

For a F9 first stage reentry is much slower, half of those 3 km/s. Moreover area is ~11 m2 and mass ~20 mT, so the ballistic coefficient is huge in comparison, ~1,800 kg/m2 (also supposing Cd near 1).

With a ballistic coefficient like that deceleration only gets high much lower into the atmosphere.

Here's a reference to an excel file showing it, by a source you might find more trustworthy.

Using it you find that for a F9R reentry like the one we are talking about dynamic pressure only starts to grow enough to count well under 40,000 m.

Executing that chart with our initial conditions it says that deceleration starts while crossing the 34000 m level, and max-Q is at 12300 m (deceleration 7.8 g), which is near the altitudes we were discussing.

EDIT: grammar
« Last Edit: 06/10/2012 03:28 PM by charliem »

Offline RDoc

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Here's a reference to an excel file showing it, by a source you might find more trustworthy.

Using it you find that for a F9R reentry like the one we are talking about dynamic pressure only starts to grow enough to count well under 40,000 m.

Executing that chart with our initial conditions it says that deceleration starts while crossing the 34000 m level, and max-Q is at 12300 m (deceleration 7.8 g), which is near the altitudes we were discussing.

Thanks for that reference, it's quite interesting to play with. One thing to keep in mind is that that's calculating a lifting reentry which requires keeping the vehicle under control and at the correct angle of attack.

The discussion modemeagle and I are having is WRT the stage going out of control after the retro burn and reentering essentially side on with no attitude control. It would almost certainly start to tumble, but we're assuming the same kinds of forces as a straight sideways flight.

In modemeagle's simulation, he's adjusting the Cd for altitude such that above about 30km the Cd is essentially zero. That in combination with low dynamic pressure results in essentially zero deceleration.

It is my understanding that for the stratosphere and lower, Cd is pretty much constant except for transonic and supersonic effects both of which increase Cd. That makes me think the stage will crush and start breaking up at high altitude, between 40 and 50 km.

If you set up your spreadsheet for a sidewise entry, Area = 1521, entry angle 45, Cd=1, Cl=.5, it will hit 1G at about 40km and rapidly go up to 6G at 30km.

That's probably pretty conservative since a better scenario is likely the stage hitting at 90 with no lift in which case it goes over 1G at 42km and over 5G at 33km.

WRT Cd, if you calculate the Cd at each altitude, it stays very close to the input value, although generally a bit higher. I imagine that's due to lift and compression effects.

Offline RDoc

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2. Engine forward configuration would not go above 1G until impact with the water (at over 1,900 m/s)..
If you use the spreadsheet charliem referenced, an engine first reentry would exceed 1G at about 26km, max out at about 9.5G at 11km and hit the ground at about 300m/s.

Edit - I reran this using an entry angle of 70 degrees to match your simulation, Cd of 1.2 (supersonic), and Cl of .1.
« Last Edit: 06/12/2012 02:46 PM by RDoc »

Offline malu5531

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In the reusability video from SpaceX, we see the landing gear deploy right before landing.

Now, what if they where to be deployed (or "half-deployed") at apogee through supersonic phase, acting as some type of ballute, while keeping attitude of stage nose-first? It might help with the deceleration and mitigate high Gs?

Modemeagle; How would that influence your simulation?

I attach image to illustrate possible landing gear dual function as ballute.

(Edit; Ballute might be the wrong word, since it's not a balloon / parachute combo, but for lack of better word. :))
« Last Edit: 06/10/2012 10:17 PM by malu5531 »

Offline Joel

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As per suggestion, I will forward a thought from a different thread (http://forum.nasaspaceflight.com/index.php?topic=29021.0).

For the moment, F9 does not have engine-out capability in the very beginning of the flight. But, if the first stage carries extra propellant for a propulsive landing, would it be possible to use this propellant to give engine-out capability throughout the flight?

That is, if an engine needs to be shut down early in the flight, could the rocket switch to an expendable mode and still make orbit? Thus sacrificing the launch vehicle to save the payload.
"Success is stumbling from failure to failure with no loss of enthusiasm." - Winston Churchill

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