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.