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.

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.

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?

Quote from: aero on 12/19/2013 06:47 pmMy 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?

@cambrianeraI 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

Quote from: aero on 12/19/2013 10:59 pm@cambrianeraI 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 velocityI just wanted to point out that the drag coefficient is a function of Mach number.

Quote from: dante2308 on 12/19/2013 11:05 pmQuote from: aero on 12/19/2013 10:59 pm@cambrianeraI 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 velocityI just wanted to point out that the drag coefficient is a function of Mach number.Yes but can you tell be the function?

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

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.

Quote from: Jcc on 12/19/2013 11:46 pmIf 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.

... 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.

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.

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/sOf 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?

Quote from: Rocket Surgeon on 12/20/2013 11:06 pmQuick 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/sOf 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.

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

So boostback (burn-1) starts at 3.8g, ends at 7.5gThat 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...

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'm attaching another graph I found on the net, https://ceprofs.civil.tamu.edu/llowery/woods/Lectures/451/451%20Drag%20Force.PDF (pag 604)

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.

Quote from: aero on 12/21/2013 03:42 pmIn 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.

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 airflowThose 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

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. Vz = 124.5 m/s using spherical drag modelAltitude = 10 km, Vz= 58.6 m/s (using Cd = 0. Vz = 89.3 m/s using spherical drag modelAltitude = 5 km, Vz= 43.6 m/s (using Cd = 0. Vz = 70.0 m/s using spherical drag model70 m/s is Mach 0.2 and still seems a little slow but that empty stage doesn't weigh much for its size ...

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

@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 ).

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

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.

Quote from: guckyfan on 12/23/2013 06:54 pmYou 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.

Quote from: aero on 12/23/2013 05:19 pm 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.

“(Also worth noting,) you don’t need wings to steer aerodynamically, you just need some lift over drag numbers and lift vector.”