Optimal vehicle movement on planet without air?

[SaturnV]

Let me tell a story first. On my second manned Mun landing explor mission, I got 1225 dV left under a crater. I found if I fly out of it, I can study the highland for extra science. Then I F5, the first try have 784 dV left, not enough to reture home. After some test launch I found I need 970 dV at least to get down to Kerbin atmosphere. I tried many times, finally succeed, return home with research data of two biome of Mun.

My question is, if I want move my vehicle to somewhere near by on a planet without atmosphere, how should I do it to use the least fuel?

In my tests on Mun, I known that I shoul keep my apoapsis as low as possible to minimize gravity lost. I should keep my horizontal speed ALAP too to minimize ‘reentry‘ burn lost. But if my horizontal speed is low, that means more time on the way thus more gravity lost. I didn't do any math but I guess there should be an optimal trajectory.

[Asimov]

Optimal trajectory isn't likely something you can do manually. It would be to take off at a 45 degree angle to the ground. Burn just long enough make the ap exactly 1/2 way to your destination. Then decelerate (using a suicide burn) at a 45 degree angle to the ground, only orienting the lander vertical at the last second when both ground speed and vertical speed are 0.

[Edit: You can do it with mechjeb, I believe. If you want to go that route.]

[Sirine]

To scan Mun 'places' (not Biome), The best way to do it...the old way.

Put a refuel craft orbiting Mun at low orbit. 15~20km.

Send down a manned vehicle, small enough to go down, pick sample, and go back up to orbit dock with fuel craft at Mun orbit. Dock. Refuel, and send down again.

I took 5 places in 1 launch, and still got plenty of fuel left on my craft. Refer to my signature link for detail.

Note: Best engine to go down and go up again to Mun orbit. 1 engine is ALWAYS better than 2 engines of the same type.

*craft (Fuel+Engine) > 6 tons = LV-N,

*craft (Fuel+Engine) < 6 tons = LV-909.

Edited October 20, 2013 by Sirine

[SaturnV]

My design is for one trip, but accidentally I found the second biome is only 5km away. So I try to spare some fuel to get the extra...

[numerobis]

45 degrees is not optimal; you want to be at a pitch as shallow as possible while not falling (nor hitting terrain), so that the remainder all goes into horizontal speed. The required pitch depends on your TWR, and on your horizontal speed -- eventually you reach orbit.

[SaturnV]

Hey guys, since no one gives me a good reply, I did some math eventually.

So it's obviously if you want the least fuel usage you should keep your attitude as low as possible, fight gravity is useless for your movement.

Then we got the mathematic model:

A rocket move horizontally (I drawed a parabola in the figure, however) from A to B, its maximum acceleration is a0, local gravity is g, length AB = S.

Obviously again, you should slice your journey into two same piece, the former part you speed up at angle θ, then you slow down to 0m/s with angle θ to the opposite direction.

We have to keep our vertical speed stick to 0, so we got

a = g / tanθ,

so horizontal acceleration = a*cosθ = g / sinθ = a1

Let's study one part of our journey only, since they are just mirror to each other. In the acceleration part, we move S/2 in time t:

S/2 = 1/2 * a1 * t^2

then t = (S/a1)^(1/2)

Then I assume in the whole move our mass is constant (since it changes not much), the fuel consume rate is proportional to trust, also proportional to acceleration since mass is constant. c is some unimportant constant factor.

Fuel Used = c * a * t

We've found a function, then solve it:

Fuel Used = (sqrt(S * g * c)) / cosθ

That's it:

Fuel Used ∠1 / cosθ (∠is the mark of 'proportional to')

That is to say, the smaller θ is, the bigger cosθ is, the less fuel sued.

The optimal movement is move as horizontal as possible until you reach half way to your target, turn around, deaccelerate with the same angle.

More specifically, fight the gravity and do it fast, but don't conquer it

Since our a0 cannot be infinite, your movement angle θ must bigger than 0 (which require infinite acceleration), then normally optimal movement angle θ is the angle that can barely keeps you from falling

θ = arcsine( g / a0 ), where a0 is your maximum acceleration.

Edited October 21, 2013 by SaturnV

[Nao]

Thats quite interesting question! I would have thought that ships with high TWR would want to burn at 45deg for several seconds at launch and then the same in reverse just before landing, just like artillery shot.

With bigger distances the speed will be closer to orbital velocity so probably lower launch angle would be better. But for a low TWR craft doing what Asimov mentioned, could actually be the right thing to do!

This needs more science!

[Stochasty]

I'm reasonably certain that constant altitude is the right approach here. This is just like a partial takeoff followed by a partial landing, and for both of those cases you want to minimize altitude changes during the process.

EDIT: On second thought, I take it back. I can easily come up with a counter-example. Consider a craft with infinite TWR. Such a craft, if it wishes to hold constant altitude, will instantly put itself into orbit, traverse the needed distance, and then deorbit and land. However, that means delta-v expenditure equal to twice orbital velocity - no matter how long or how short the trip. So, for an infinite TWR craft, ballistic is the better approach.

Presumably, for less-than-infinite TWR craft, there will be a tradeoff between the extra gravity losses incurred by altitude changes and the extra delta-v spent to maintain an artificially low trajectory. So, as Nao said - an interesting problem.

EDIT 2: There's a second possibility for infinite TWR craft holding constant altitude. Accelerate horizontally to some velocity below orbital, point up and burn slowly to hover during the transit, then decelerate to a stop at the target.

Optimum should be either this or the ballistic case; now I need to set up the problem for real and do some math.

Edited October 21, 2013 by Stochasty

[numerobis]

Good point.

If you have a short trip and high TWR, then you can analyze it like Galileo. Then a 45 degree impulse will be optimal.

If you have a very long trip and low TWR, you go into orbit. Then a constant-altitude trajectory will be optimal.

In the middle: ????

[Nao]

After some thoughts on the infinite TWR craft, i wonder if the right path to take would be a part of minimal energy orbit (assuming point mass at the center) that that intersects the surface at start and finish.

In the extreme of going to the other side of the moon/planet, it would just be half of the circular orbit at ground level. And on the other end (short distance) probably it would be very close to ballistic patch (with 45deg angles at start/end) with the Pe of said orbit would be very close to the point mass in center.

edit2: after messing around in KSP orbit mechanic it looks like the orbit approach can work but not for "minimal energy" but rather minimal speed at ground intersects.

Using kerbin as base (not counting atmosphere) 20km distance has minimal launch velocity of 439m/s at 45deg. And 628km distance has minimal launch velocity of around 1980m/s at 28deg. Also 1256km needs around 2338m/s at only 14deg.

Edited October 21, 2013 by Nao
made it more clear

[Stochasty]

For the constant altitude trajectory:

dV = 2v + (d/v)*(g-v^2/r)

Where dV is the total delta-V expended, v is the horizontal velocity of the flight, d is the distance across the surface, r is the radius of the planetoid, and g is local surface gravity. This has a minima at either sqrt(dg/(2-d/r)) (for non-orbital flights) or 2sqrt(gr) (for orbits), with the crossover at d = 1.6r between whether it's more efficient to orbit or not orbit.

For the ballistic trajectory: Nao, you are correct that we want to find the minimum energy orbit that crosses the surface at the two points; the total delta-V is then just twice the velocity of the orbit at the surface. Still need to work that out for comparison purposes.

Note that this analysis assumes a non-rotating planetoid. Factoring in a rotating body makes things quite a bit more complicated and I'm not yet willing to hazard a guess as to the nature of the correct solution in that case.

EDIT:

To give an example of the problems involved in rotation, imagine trying to make a transfer from the equator to a pole of a spherical planetoid rotating such that the equator is moving at exactly orbital speed. The right way to travel in that case might very well be a bi-elliptic transfer.

Edited October 21, 2013 by Stochasty

[Nao]

Yeah, rotation of the body is a can of worms to include in the maths, fortunately for most bodies its quite low compared to surface orbital velocity.

But i did a simple test on Mun at 1,8 start (mun) TWR. First flight: constant altitude, using 755m/s Dv gave distance of 37km, then using constant attitude of 45deg result was 57km. And the first flight even had a little hoop as i needed to clear a hill so it was launched at 30deg attitude.

You must construct additional Pylo... Science!

edit: yep did a bunch more test in 200m/s Dv range and 1,8 TWR ... longest range was with constant 45deg burn then small coast and another 45deg landing burn (~4km of distance). Other tries included: constant altitude flight (at constant throttle), staying close to velocity vector (quite bad) and a constant altitude burn with a mid section of upward burn (similar to constant alt burn).

Edited October 21, 2013 by Nao

[Nao]

More science complete!

Math was done under assumption of a flat surface (no centripetal force and no initial surface motion). While this is pretty big generalization, horizontal speeds of 155m/s reduces gravitational pull by ~7,5% so i think trends shown by the data are meaningful.

In real flights Dv numbers at higher max speeds will be somewhat lower.

Constant altitude burn				
Mun g		1,61	m/s2		
Distance	4	km

Hacc - horizontal acceleration
H Vmax - maximum speed, at the half distance point.

angle	TWR	Hacc	H Vmax	Dv m/s
5	11,5	18,4	271,3	544,7
15	3,9	6,0	155,0	321,0
25	2,4	3,5	117,5	259,3
35	1,7	2,3	95,9	234,2
45	1,4	1,6	80,2	227,0
55	1,2	1,1	67,2	234,2
65	1,1	0,8	54,8	259,3
75	1,0	0,4	41,5	321,0
85	1,0	0,1	23,7	544,7


Constant angle burn with coast				
Mun g		1,61	m/s2		
Distance	4	km		
angle		45	deg

Hacc - horizontal acceleration
Co-ang - starting coast angle
Vmax - speed at the beginning of coast phase

TWR	Hacc	Co-ang	Vmax	Dv m/s
11,5	13,1	41,2	76,0	161,7
3,9	4,4	32,4	71,2	170,1
2,4	2,7	21,9	70,6	185,4
1,7	2,0	10,7	74,0	205,7
1,4	1,6	0,0	80,2	227,0

Ballistic calculation (TWR = infinity), 45 deg angle.
Dv needed = 160,5 m/s

As always I'm not 100% sure about those since i tend to make mistakes, but the results are quite realistic.

What surprised me actually was that if we insist on flying at constant altitude then accelerating constantly and turning around at half the distance, is worse than just burning at 45deg with lower throttle. Because of that I think the constant altitude flight could be made more efficient by first accelerating at full throttle and then coasting for a little while just burning upwards. (Although earlier tests indicated otherwise).

So not counting orbital mechanics, just burning at 45deg full throttle, coasting and burning opposite direction at 45deg is the most efficient way to travel short distances with limited TWR.

Edited October 22, 2013 by Nao

[Superfluous J]

More science complete!

...

So not counting orbital mechanics, just burning at 45deg full throttle, coasting and burning opposite direction at 45deg is the most efficient way to travel short distances with limited TWR.

Haha you just proved with numbers what anybody who's played any number of artillery games (Scorched Earth, Worms, Gorilla.bas) knows by feel. 45 degrees gets you the furthest distance per thrust used when there's no air resistance.

[numerobis]

Galileo first proved it in the 17th century, and the fact was first noted (in the west, anyway) in the 16th century.

I bet what we're interested in was worked out for the V2 rocket and other IRBMs.

[Stochasty]

Nao, I think you are making a mistake with the constant altitude burn. You don't want to keep accelerating until the halfway point. You want to accelerate only up until some optimal speed (which, for infinite thrust, I gave in my last post) and then point vertical and hover until it's time to decelerate.

The problem is that you are spending more delta-v here on acceleration and deceleration than is necessary. Once you get going fast enough, it's more efficient to simply counteract gravity while you travel than it is to keep accelerating horizontally. This is trivially obvious for infinite TWR, since continual acceleration would have you orbiting for arbitrarily short distances.

[numerobis]

I highly doubt you ever want to point straight up. You're burning to get horizontal acceleration, and to get vertical acceleration. There's a total amount of each you want. You should sum them up and burn along that vector.

[Stochasty]

For non-infinite TWR, this is probably true. There should be some optimum thrust/angle function which minimizes delta-v. For infinite TWR, I suspect that an impulse acceleration to optimum horizontal velocity, followed by hovering until the destination and then an impulse deceleration is in fact optimal (it's one of the edge cases, the other being holding a constant angle of attack the way Nao is doing).

EDIT: Hmmm... I just worked through the delta-v for accelerate + hover + decelerate (taking r to infinity to give a flat surface) for the numbers Nao is using above and got 227 m/s delta-v; exactly equal (up to my rounding errors) to the 45 degree case. That's... very odd.

EDIT 2: So, since I'm getting the same numbers for infinite TWR that Nao did for the 45 degree case, this would appear to suggest that for high TWR craft (above 1.41) ballistic is better.

Edited October 22, 2013 by Stochasty

[numerobis]

Say you want to accelerate to X m/s horizontal, and you need to coast for t seconds. Then you need to expend Y = t*g m/s to coast. If you burn it all in one impulse, it costs you sqrt(X^2 + Y^2) m/s. If you burn it as an impulse of X m/s then you hover, you burn X + Y m/s. By the triangle inequality, hovering loses.

If you have bounded TWR, then you may need to apply some extra hovering at the start of the burn. I'm not sure how to fully compute that yet.

Edited October 22, 2013 by numerobis

[Stochasty]

Say you want to accelerate to X m/s horizontal, and you need to coast for t seconds. Then you need to expend Y = t*g m/s to coast. If you burn it all in one impulse, it costs you sqrt(X^2 + Y^2) m/s. If you burn it as an impulse of X m/s then you hover, you burn X + Y m/s. By the triangle inequality, hovering loses.

Ah. Yes. I beilieve this is sufficient proof that ballistic wins for infinite TWR. You have to fight gravity losses either way, but by the triangle inequality it's cheaper if you do everything all at once, rather than incurring steering losses by hovering. I suspect that travel time for the two optimum trajectories (ballistic and constant altitude) is actually identical.

[Nao]

For the instant horizontal burn and coast on power, it feels like we are making big steering losses with burning twice at 90deg angle to each other. I would be surprised if it was equal or better to the ballistic path.

Science continues.

[Alistone]

For the instant horizontal burn and coast on power, it feels like we are making big steering losses with burning twice at 90deg angle to each other. I would be surprised if it was equal or better to the ballistic path.

Science continues.

Burning at 90 deg angle to each other is MUCH more efficient than burning at 180 deg to each other with a direct horizontal arc. (You meant 180 degree in your post... since you were talking about horizontal burns, but the balistic 45 degree burns end up being 90 degrees different from eachother like ^)

It's a balance of gravity loss (from hovering) and steering loss (from thrusting and then turning to thrust in a different or even opposite direction).

If you try to go straight horizontal you end up with very high steering loss by having to turn completely around and burn at nearly 180 degrees of your initial impulse. Then you suffer gravity losses for however long you have to "hover" to avoid crashing into the surface.

So in an "ideal" trajectory you balance the total horizontal thrust to speed up and then to slow down with your total vertical thrust to "hover" for the duration of the maneuver.

If you try to go nearly straight up you never get anywhere and all your fuel is wasted due to gravity losses. Also, hohmann effect tells us it's better to do it as an "impulse" instead of wasting fuel "hovering".

So in the end you realize that the most efficient method will be an impulse at about 45 degrees that splits losses equally between steering losses (horizontal thrust) and gravity losses (vertical thrust).

Regarding time; it is fastes to go nearly horizontal and thrust prograde for the first half and retrograde for the last half. But it is slowest to go nearly straight up and then to slowly drift over to the new location. The optimal trajectory is "Ballistic" and while slower than the direct horizontal it is far more efficient.

Edited October 22, 2013 by Alistone

[allmhuran]

Yep yep. Low vs high TWR won't change the optimum trajectory (symmetrical ballistic launch and landing at highest possible acceleration on both ends). A flatter trajectory is only advantageous if you're willing to spend more fuel to get there faster. If this is the case, then TWR obviously matters

[Nao]

The steering and gravity looses balance is a really nice way to show why 45 deg works. Thanks . That'll help making next set of equations that include orbital motion.

edit: to post above. I would say that trajectory (path that craft takes) is what is changing with TWR, what is constant is attitude in relation to horizon. And I think that's what caused (at least for me) some confusion.

Edited October 22, 2013 by Nao

[SaturnV]

Hey guys, nice discussion here! The problem is more difficult than I thought!

For a infinite TWR rocket, it can accelerate in no time, looks like a bullet with initial speed, then of course the best trajectory is 45 degree free fall. But the bullet have no thrust power, that's a big difference, since a rocket can decide when to thrust and when not to. Another possible optimal trajectory is a combination of gravity turn ascend and ballistic. I can tell if a 45 degree is optimal for a rocket since it tradeoff more gravity loss for longer flight time. As for gravity turn, it can only be solve with numeric integration.

So, since I'm not good at math, I'll try to do some simulation.

More complicated, if we consider the orbit thing... if you're moving to other side of a planet, I think goto there with a ellipse orbit is much more fuel saving than moving slow on ground. But this looks rather complicated, I will not include it in my next simulation.