Vertical Distance on a Single Tank of Gas

[millsw]

I feel like you\'re trolling me, the best i got was 1330 m/s at 36km. I was throttled way back through most of the atmosphere. I kept my velocity at 100m/s under 15km and then I left the throttle right at the line below the 'E' in the word throttle on the gauge.

Sorry man. Turns out I was using sunday punches fuel efficient LFE

[CarolRawley]

Escape Velocity. Need I say more?

[Rickenbacker]

Escape Velocity. Need I say more?

Yes! How did you do it?

[CarolRawley]

Yes! How did you do it?

Evil results using the 0.11X1 time warp feature.

[Jarnis]

Evil results using the 0.11X1 time warp feature.

I call EXPLOIT! and suggest you should beg for forgiveness...

[CarolRawley]

I call EXPLOIT! and suggest you should beg for forgiveness...

The game already made me beg for forgiveness. It teleported me inside the planet, and I got stuck in a gravity whorl around the core

[Yourself]

From an actual optimal control standpoint this problem is impossible analytically. Numerically is merely difficult. In any case there are some general principles to keep in mind.

If not for the atmosphere, the optimal control would be to just full throttle it all the way. Gravity losses are merely a function of time spent in the gravity well. The longer you spend in higher gravity, the more fuel you waste fighting it.

The real complicating factor here is the addition of the atmosphere, if there\'s any chance for a singular arc in the solution (a singular arc is one where the control, in this case throttle, is not at one of its limits), it\'s introduced by the atmosphere. The atmospheric and drag models will highly influence the optimal solution. Now, I can give you the math, but it won\'t help much. I can tell you that a singular arc exists when:

L2 / m - L3 / Isp = 0

Where L2 and L3 are the Lagrange multiplier costates, m is the total mass of the rocket, and Isp is the specific impulse of the engine. Whoopee. I could give you the differential equations governing the evolution of the costates too, but they\'re too heavily tied up in the exact atmospheric model to be useful.

Now, there\'s a few things I can definitely say about the solution:

It\'ll start at full throttle and it\'ll end at full throttle.

I have a hunch that the optimal control isn\'t bang-bang (meaning you go from full throttle instantly to no throttle), though it could very well be. If the optimal control deviates from full throttle, it\'ll be somewhere in the lower atmosphere.

Now, if someone gave me the drag and atmospheric models I could go ahead and let a computer smack the problem around until it finds a solution. Now, I haven\'t searched the forum too much, but I did find this. However, the atmospheric model has been updated since then (not to mention that the given model is for pressure and I\'m really after density), so I\'m not sure how accurate this is and I\'m still left with the question of the drag model.

[Jarnis]

I do know that for a good result, you have to go full thrust until you reach a velocity where the drag becomes an issue (around 120-150m/s early on) and then manage the throttle until you have cleared the thick parts, pushing it up slowly and then firewalling it to maximum thrust 'at the right altitude'.

The 'right altitude' and the management of throttle after the initial throttle-down is the part which is mostly down to art... and like true artist, I like to keep my secret sauce secret

[Frederf]

From an actual optimal control standpoint this problem is impossible analytically. Numerically is merely difficult. In any case there are some general principles to keep in mind.

If not for the atmosphere, the optimal control would be to just full throttle it all the way. Gravity losses are merely a function of time spent in the gravity well. The longer you spend in higher gravity, the more fuel you waste fighting it.

The real complicating factor here is the addition of the atmosphere, if there\'s any chance for a singular arc in the solution (a singular arc is one where the control, in this case throttle, is not at one of its limits), it\'s introduced by the atmosphere. The atmospheric and drag models will highly influence the optimal solution. Now, I can give you the math, but it won\'t help much. I can tell you that a singular arc exists when:

L2 / m - L3 / Isp = 0

Where L2 and L3 are the Lagrange multiplier costates, m is the total mass of the rocket, and Isp is the specific impulse of the engine. Whoopee. I could give you the differential equations governing the evolution of the costates too, but they\'re too heavily tied up in the exact atmospheric model to be useful.

Now, there\'s a few things I can definitely say about the solution:

It\'ll start at full throttle and it\'ll end at full throttle.

I have a hunch that the optimal control isn\'t bang-bang (meaning you go from full throttle instantly to no throttle), though it could very well be. If the optimal control deviates from full throttle, it\'ll be somewhere in the lower atmosphere.

Now, if someone gave me the drag and atmospheric models I could go ahead and let a computer smack the problem around until it finds a solution. Now, I haven\'t searched the forum too much, but I did find this. However, the atmospheric model has been updated since then (not to mention that the given model is for pressure and I\'m really after density), so I\'m not sure how accurate this is and I\'m still left with the question of the drag model.

I have a fetish for closed-form analytical solutions so color me disappointed. But... I\'m rooting for you! I\'m very interested to see what the optimized throttle profile for this simple system is.

Remember that gravity drag is time spent in the gravity well times the strength of the well at that particular height. Mass is decreasing at a throttle-fraction-specific rate. Don\'t feel bad just throwing in a Cd of 0.3, drag increasing as a function of velocity to the 2.5 power, and just getting a taste for what the answer might be for very assumed aero drag conditions.

[nivvydaskrl]

I don\'t have a proof supporting this statement, but my gut feeling is that the optimal speed at any given point will be the point where decelerating forces from gravity equal decelerating forces from drag (i.e: terminal velocity). At this point we minimize the amount of time spent fighting gravity whilst ensuring that we don\'t push the velocity so high that the energy savings gained by minimizing time fighting gravity is countered by the exponential increase in drag forces.

I set up a spreadsheet and attempted to follow the computed figures at given altitudes for this and while I think my computations were off, I got ~80,000m on the first try.

If I feel ambitious, I\'ll try to write up a formal proof of my hunch and see if it holds water.

Edit: Hmm. The drag model is questionable, is the main issue. I may have to compare to a real life example, even if it\'s not usable due to the specifics of the drag model not being known.

[illectro]

Yep I pretty much headed up at about 50-30% thrust until the drag became negligible and then boosted at max up to 86,800m

[illectro]

With just 1 Tank and one Engine (no decoupler or chute) I\'ve just managed to get to 96,528m

[Alasseo]

Just managed an altitude of 1275346m, using 1 2m LFT and a Medium Bertha LFE, plus a 2m-3m shroud decoupler as a launching section. Admittedly, they aren\'t stock parts, but I figured the lack of streamlining would leave it fairly unstable and tumble in atmosphere.

Strangely, it barely tumbled at all until around the 60km mark, and tumbled even less on re-entry, although every kerman aboard died on impact.

I probably should have added a \'chute, come to think of it...

[Rickenbacker]

What a difference thrust management (and leaving off the chute and decoupler - sorry Jeb!) makes. I managed the throttle to stay around 200m/s until I left the rightmost atmosphere band, then full thrust until the engine quit - 94,007m!

[Jarnis]

What a difference thrust management (and leaving off the chute and decoupler - sorry Jeb!) makes. I managed the throttle to stay around 200m/s until I left the rightmost atmosphere band, then full thrust until the engine quit - 94,007m!

Indeed. It can also make huge differences with larger rockets. It is just pointless to burn perfectly good fuel against atmospheric drag...

Basic 'save fuel' flight plan;

- Full thrust off the pad until you hit somewhere around 150m/s. With larger rockets this can actually take a while. No problem.

- Throttle down until you no longer accelerate. Again, with larger rockets this can be a fairly small amount.

- As the rocket gets lighter, it again starts to accelerate slowly. Manage it so it stays below 200m/s (I usually keep it under 170m/s) until you are out of the rightmost atmospheric band, then floor it.

I\'m sure someone could work out with math the exact 'perfect' throttle at each altitude, but this seems to work as a good rule-of-thumb.

[Yourself]

I don\'t have a proof supporting this statement, but my gut feeling is that the optimal speed at any given point will be the point where decelerating forces from gravity equal decelerating forces from drag (i.e: terminal velocity). At this point we minimize the amount of time spent fighting gravity whilst ensuring that we don\'t push the velocity so high that the energy savings gained by minimizing time fighting gravity is countered by the exponential increase in drag forces.

I set up a spreadsheet and attempted to follow the computed figures at given altitudes for this and while I think my computations were off, I got ~80,000m on the first try.

If I feel ambitious, I\'ll try to write up a formal proof of my hunch and see if it holds water.

Edit: Hmm. The drag model is questionable, is the main issue. I may have to compare to a real life example, even if it\'s not usable due to the specifics of the drag model not being known.

I don\'t have a proof supporting this statement, but my gut feeling is that the optimal speed at any given point will be the point where decelerating forces from gravity equal decelerating forces from drag (i.e: terminal velocity. At this point we minimize the amount of time spent fighting gravity whilst ensuring that we don\'t push the velocity so high that the energy savings gained by minimizing time fighting gravity is countered by the exponential increase in drag forces.)

I think this line of reasoning is questionable. What\'s so special about making the magnitude of the drag force and the gravitational forces equal? It doesn\'t follow that this minimizes the amount of time fighting gravity. I\'ve run some tests and this strategy does offer improvement, but I\'ve noticed that I can get an improvement on that (albeit a small one) if I keep the aerodynamic forces about 3% lower than the gravitational forces. I\'m unconvinced that this is optimal.

In any case, here are the results:

http://imgur.com/a/ZiZNP

I think my model is reasonably close to the actual one (which is what the first few images are about, I was just tweaking model parameters to get things close).

I haven\'t actually attempted any optimization so far, just testing out models.

[Zeroignite]

Interesting. According to that simulation, altitudes over 100km are definitely possible. Still not really clear if that\'s the optimum flight path (though it well might be), but it definitely implies that the current records can be beaten.

[nivvydaskrl]

I think this line of reasoning is questionable. What\'s so special about making the magnitude of the drag force and the gravitational forces equal?

Absolutely nothing is special about it, other than a hunch.

I think what needs to happen is that we need to integrate the negative acceleration curves for gravity and aerodynamic forces -- with respect to a certain ratio between them -- so we can identify total impulse loss in each case over the whole of the powered flight, then test various values to find the ratio which provides the minimum total impulse loss. The model can then be run for that ratio to show us the actual desired figures and peak performance possible.