Most efficient vertical flight speed?

Yes.

The results are in books called "Flight sheet guides" and "Malewicki Charts."

Jerry

Hint: this is a classic problem problem known as the Goddard problem.

What did you find surprising?

Alan

I'm not asking to find a ideal weight for a given engine... What I'm talking about is doing it in reverse. Say you're given a rocket with parameters Cd and D. Now you're given "x" amount of fuel (with an isp of 'you pick')... if you could control the thrust curve, what's the highest you can make it go?

For example, say if I use a K1100 with a rocket with a diameter of

It's not as applicable to solids as it is to hybrids (or liquids). It's not a really big difference until you start looking at optimizing curves for spaceshots, where fuel is the majority of your weight and atmospheric density changes significantly.

Dave

Jerry Irv> >

For typical model/HPR conditions, optimum thrust/time for altitude seems to be a bit on the long duration side, compared to what you really want for a clean hard liftoff - but a "kicker" motor generates high velocity at low altitude, which coasts off rapidly... some motors seem to hit a "sweet spot" (in "typical" consumer rocketry airframes) - Cesaroni Pro38's, Econojet G38FJ, and Apogee 18mm D10 mini composites come to mind: they come off the pad clean and hard, and then keep going and going like the energizer bunny...

-dave w

"dave.harper" wrote:

That the thrust curve that maximizes altitude seeks to satisfy: V(t)=Terminal Velocity (C,A,rho(t))

i.e. the most efficient vertical velocity is the same as terminal velocity at a given altitude for a given rocket.

Dave

The results are in books called "Flight sheet guides"

What is your test case?

Assuming you don't care about practical considerations, and are only doing

In practice it doesn't quite work the same way, since you have to handle perturbations.

Brett

Close enough. The optimal thrust for most MR and HPR is bang-sigular arc-coast. In practical terms it is max trust (ideally an impulse) to get you up to speed, followed by approximately T=2W until burn out, followed by coast to apogee.

Now then, assuming a case where your optimal trust time exceeds 15 seconds, how does the optimal thrust change when a 15 second burn time limit is imposed?

Alan

Yeah, T=2W gives you terminal velocity, only in the positive direction. As you start getting into the higher atmosphere, that ratio starts to increase, and in a vaccuum it'd be infinite.

Given the same amount of fuel as the thrust curve that exceeded 15 seconds? I imagine it would be boost, sustain at terminal velocity (i.e. T=2W), then at 14.9 seconds, an impulse of all remaining fuel.

Dave

My test case? You mean this?

optimum

Dave

--------- --------- --------- --------- ----- --------- --------- 8.0 4641.3 1665.6 1319.9 1.2 0.9 10.3 9.0 4896.3 1559.3 1287.5 1.1 0.9 11.1 10.0 5130.8 1460.8 1256.0 1.1 1.0 11.8 11.0 5326.7 1368.3 1223.2 1.1 1.0 12.4 12.0 5495.7 1279.8 1191.4 1.1 1.0 13.0 13.0 5644.4 1196.4 1154.4 1.0 1.1 13.5 14.0 5735.9 1119.0 1108.6 1.0 1.1 13.9 15.0 5764.0 1048.3 1057.6 0.9 1.1 14.2 16.0 5697.3 984.3 1000.1 0.9 1.1 14.4 17.0 5603.7 926.8 947.2 0.8 1.1 14.5 18.0 5490.8 874.9 898.5 0.8 1.0 14.6 19.0 5362.0 828.0 853.8 0.8 1.0 14.7 20.0 5221.0 785.3 812.6 0.7 1.0 14.6 TOTAL IMPULSE POUND-SECONDS = 575 TOTAL IMPULSE NEWTON-SECONDS = 2557.6 AVERAGE THRUST NEWTONS = 1278.8 AVERAGE THRUST POUNDS = 287.5 SPECIFIC IMPULSE (LB-SEC/LB) = 203.496094 TIME (SEC)= 0 THRUST (LBS)=287.5 TIME (SEC)= 2 THRUST (LBS)=287.5

--------- --------- --------- --------- ----- --------- --------- 8.0 4637.5 1561.1 1450.7 1.3 0.9 10.4 9.0 4888.6 1449.6 1399.2 1.2 0.9 11.2 10.0 5110.0 1348.9 1349.2 1.2 1.0 11.9 11.0 5292.5 1253.7 1297.4 1.1 1.0 12.5 12.0 5465.6 1166.6 1246.6 1.1 1.0 13.1 13.0 5595.8 1086.7 1197.5 1.1 1.1 13.6 14.0 5714.5 1013.6 1147.1 1.0 1.1 14.0 15.0 5743.8 947.0 1087.8 1.0 1.1 14.3 16.0 5714.8 887.5 1029.2 0.9 1.1 14.5 17.0 5617.5 834.0 972.3 0.9 1.1 14.7 18.0 5503.5 786.0 920.3 0.8 1.0 14.7 19.0 5373.3 742.7 872.8 0.8 1.0 14.8 20.0 5230.5 703.4 829.3 0.7 1.0 14.8 TOTAL IMPULSE POUND-SECONDS = 574.8 TOTAL IMPULSE NEWTON-SECONDS = 2556.7104 AVERAGE THRUST NEWTONS = 1278.3552 AVERAGE THRUST POUNDS = 287.4 SPECIFIC IMPULSE (LB-SEC/LB) = 203.425312 TIME (SEC)= 0 THRUST (LBS)=189.8 TIME (SEC)= 2 THRUST (LBS)=385

--------- --------- --------- --------- ----- --------- --------- 8.0 8760.2 6614.4 676.9 0.6 1.7 9.8 9.0 8603.4 6294.1 660.9 0.6 1.6 10.3 10.0 8409.1 5977.9 644.1 0.6 1.6 10.7 11.0 8183.0 5667.5 626.4 0.6 1.5 11.0 12.0 7929.3 5364.6 607.5 0.5 1.5 11.3 13.0 7652.5 5070.5 587.5 0.5 1.4 11.4 14.0 7355.6 4786.5 566.5 0.5 1.4 11.5 15.0 7042.6 4513.4 544.6 0.5 1.3 11.5 16.0 6716.8 4251.8 522.1 0.5 1.3 11.4 17.0 6381.8 4001.8 499.2 0.4 1.2 11.3 18.0 6040.9 3763.7 476.1 0.4 1.1 11.2 19.0 5697.5 3537.3 452.9 0.4 1.1 11.0 20.0 5354.9 3322.3 429.8 0.4 1.0 10.7 TOTAL IMPULSE POUND-SECONDS = 575.85 TOTAL IMPULSE NEWTON-SECONDS = 2561.3808 AVERAGE THRUST NEWTONS = 213.4484 AVERAGE THRUST POUNDS = 47.9875 SPECIFIC IMPULSE (LB-SEC/LB) = 203.796914 TIME (SEC)= 0 THRUST (LBS)=100 TIME (SEC)= 1 THRUST (LBS)=100 TIME (SEC)= 1.1 THRUST (LBS)=43 TIME (SEC)= 12 THRUST (LBS)=43

Jerry

---------

First, are you accounting for reduction in mass due to propellant burn? Just checking to make sure we're using the same methods.

Anyway, using this one as an example, I get 7966 ft max altitude, 5816 ft at burnout, and a velocity of 608 ft/s at burnout, so we're relatively close (within about 5%).

Using an optimized thrust profile with the same amount of propellant, same mass, Cd, dia, isp, etc... I get a max alt of 9487 ft, burnout alt of 8565 ft, and a burnout velocity of 292 ft/s. That's almost a 20% increase in altitude using the same amount of fuel (and using mine as the reference). That's what I mean by optimizing the thrust.

Dave

Well, yes T=-2W would not get you much positive altitude. "Terminal velocity" is a poor description in this case. I usually refer to the optimal cruise (or singular arc) speed to distinguish it from the initial optimal acceleration phase, and the final coasting phase.

That is true, but even large HPR flights seldom exceed 20K Ft. T=2W is quite practical for typical MR, HPR flights. There are two interesting consequences of thrusting to stay on the optimal singular arc. On the one hand, as mass burns off, optimal thrust tends to decrease, while on the other hand as density decreases with altitude, trust tends to increase to accelerate the rocket and maintain "optimal drag' (or even "terminal velocity"). For some things like MR altitude with an AT E6 sustainer, constant thrust can be just about right.

Yes, and assume constant ISP for all thrusting, just to keep it simple. The 15 seconds is cumulative for all thrust phases, but impulse thrusting is OK. Actually I'd like to suggest an example of a

OK. I had a more impractical solution in mind. I should probably just solve this problem, but it is more fun to pose it as an open problem, especially when you said that you had a simulation program set up to optimize thrust.

Alan

I'd have to add some coding to make it a simulation program to "optimize thrust" given specific limitations... the simulation I use right now only limits the max thrust of the engine.

Dave

Yes using the thrust curve as a proxy.

But you do not say what that thrust profile is.

I do.

Also my run is a practical buildable motor.

Your run if ever disclosed could probably be approximated by a buttkicker booster motor cluster (0.4s) and a central endburner with a very low probabibility of vertical flight.

I find your maximum thrust limitation odd, and interesting.

A related problem is to assume a high thrust motor that is not throttleable, but that can be turned on and off at will. E.g. an 80N motor that would accelerate the model to speed and then be turned on and off in some duty cycle to maintain the desired speed.

Alan

BTW, which Dave Harper are you? Is you dad's name Roy? Where do you live now?

COAST

TIME

propellant,

Well, I'll post the full profile when I get home tonight, if that'll make it more believable.

Considering I haven't stated what my thrust profile is (as you mention above), this is a big assumption, isn't it?

Actually, it is pretty similar (in profile), not nessesarily in a that configuration... Much like your "Stepped Thrust Motor" profile, only with a lower, longer sustained flight (which is what the optimization simulation calculates). And it hits 200mph in under a second, so I don't know why you say it'd probably fail to go vertical...?

Dave

We'll look forward to it.

Nope. A unlateral statement.

The low thrust combined with gravity turn and greatly accelerated by wind perturbation.

Tech Jerry

program

If I didn't limit thrust, the rocket would experience over 1000g's (10,000+ lbs of thrust) during the first loop iteration. The optimizing routine would try to get the rocket to it's most efficient vertical speed in 1 step. Secondly, if it did accelerate to it's most efficient velocity in one step, it would leave drag out of the picture during that first step, since drag is determined by the initial velocity at the begining of the loop. That's why I put the limitation in.

That's certainly possible... similar to pulse detonation engines...

Sorry, but no, my dad's name is not Roy. And I'm in Georgia, but Birmingham originally.

Dave