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Optimal vehicle movement on planet without air?


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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.

Hohmann effect ensures that we should do all of our thrusting at the start (and end) so as much of it is at high speed as possible. The question of when to thrust is therefore already answered. You want to do your burn at the beginning to get going and again at the very end to get stopped.

Gravity turn is really about getting an extra bit of delta-V for when you are trying to reach orbit (which is not what we're talking about). In the majority of moons with no atmosphere the rotation is essentially a minor course correction. Simply make sure you set your heading a few degrees to the east of your target location so that it will arrive there at the same time you do.

And on the original topic; "about 45 degrees" is optimal, but technically it will be slightly more vertical. The higher your T/W ratio, the closer to the 45 degree thrust vector will be most efficient. Essentially it's just adding vectors. You have to point upwards slightly to offset gravity so that when added together the vectors end up at 45 degrees from the horizon.

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And on the original topic; "about 45 degrees" is optimal, but technically it will be slightly more vertical. The higher your T/W ratio, the closer to the 45 degree thrust vector will be most efficient. Essentially it's just adding vectors. You have to point upwards slightly to offset gravity so that when added together the vectors end up at 45 degrees from the horizon.

I may be dumb here, but i'm not seeing how would the vectors go to make optimal attitude more vertical than 45deg. I think it would be the opposite, with more horizontal burn at longer distances due to centrifugal acceleration lowering gravity drag. Also TWR does not seem to change anything at all really, unless it's less than 1,41 so the ship cannot keep 45 deg attitude without crashing.

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I may be dumb here, but i'm not seeing how would the vectors go to make optimal attitude more vertical than 45deg. I think it would be the opposite, with more horizontal burn at longer distances due to centrifugal acceleration lowering gravity drag. Also TWR does not seem to change anything at all really, unless it's less than 1,41 so the ship cannot keep 45 deg attitude without crashing.

It's like how you have to start your burn before a maneuver node, only with vectors. You want your vector in the middle of the burn to be 45 degrees. If you burn at 45 degrees for 10 seconds, your vector at 5 seconds will NOT be 45 degrees. If you burn at (for example) 50 degrees, it could then work.

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Wait wait. Firstly the idea i'm having is that we keep 45deg attitude in relation to the horizon, and after 10 seconds it still will be 45deg. Secondly (for standard orbital operations) i thought burning directly at maneuver node is less efficient than burning prograde. I mean we are in a gravity well, so the path the craft takes is a part of an ellipse. Yes the ship will rotate in relation to fixed reference point in space but the angles will remain the same in relation to gravity vector (and surface).

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Wait wait. Firstly the idea i'm having is that we keep 45deg attitude in relation to the horizon, and after 10 seconds it still will be 45deg. Secondly (for standard orbital operations) i thought burning directly at maneuver node is less efficient than burning prograde. I mean we are in a gravity well, so the path the craft takes is a part of an ellipse. Yes the ship will rotate in relation to fixed reference point in space but the angles will remain the same in relation to gravity vector (and surface).

Not a 45 degree ATTITUDE. A 45 degree VECTOR. Where you're aiming is secondary. You want the direction you are heading (the yellow prograde marker) to be at 45 degrees. I'm not sure, now, if you want it at 45 when you're done burning or midway through, or if it's exactly midway through. However for almost every application, the two should be functionally similar enough for real(digital)-world use.

I'm not sure what you mean by maneuver nodes versus prograde. You do want to burn prograde, usually, to increase an orbit. Sometimes you want to burn a bit to the side to change your orbit in other ways. You do those, though, by setting a maneuver node that will do what you want, and then burning for as long as you need at that maneuver node, in the direction it tells you to go, be it prograde or not. However, this has only SLIGHT similarities to what we're talking here, which is the most efficient liftoff and landing to launch from one place and land at another on the same airless moon.

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So first thing first, Ive already concluded in my previous post (the one with data) that the lowest Dv required to travel a small distance is by burning 45deg attitude no matter the TWR, that also results in horizontal flight at 1,41 TWR, and for many low TWR crafts the trajectory is pretty shallow. I tried some different profiles on the Mun in actual game and the results were similar. If you don't believe me, you can check it in game with some low twr lander.

Second, burning at maneuver node is what actually generates steering losses when TWR is lower than infinity. Imagine a burn with Ion powered probe that takes half the orbit to complete, the node marker would point retrograde at the end of that burn. I can't really explain this in english but being in gravity well means that we cant just add vectors like that, since our velocity vector changes constantly due to gravity.

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Second, burning at maneuver node is what actually generates steering losses when TWR is lower than infinity. Imagine a burn with Ion powered probe that takes half the orbit to complete, the node marker would point retrograde at the end of that burn. I can't really explain this in english but being in gravity well means that we cant just add vectors like that, since our velocity vector changes constantly due to gravity.

EDIT: Don't just point at the (blue) node marker; if you're trying to be efficient you need to pay attention to your direction of travel (yellow) and the horizon. I thought you were referring to burning at a maneuver node. Only after my post did I realize you meant burning at the blue marker on the nav ball.

Steering losses are caused by burning in a direction that is different than the one you are travelling.

If you are travelling along a perfect equatorial orbit at 866 m/s and you burn at a right angle straight north or south for 500 m/s of delta v your new velocity is only 1000m/s. That is a steering loss. If you had burned in the direction you were travelling you would have been up to 1366m/s.

If you were pointed in the exact opposite direction of travel your speed would be reduced to 366m/s and you would have suffered even greater steering losses.

To be clear, that is where steering loss comes from. Doesn't matter when/where you burn. What matters is that you are not pointed in the direction your ship is already moving.

Gravity losses are caused by not burning toward the horizon. The portion of thrust spent "fighting" gravity is essentially always wasted.

If you keep your burn always pointed in the direction of your velocity on a long burn you do not suffer steering losses (though you have to constantly keep adjusting to stay pointed in the direction of the yellow circle. BUT what happens is that your elevation will be increasing or decreasing and you will be thrusting above/below the horizon causing gravity losses.

This is why after getting out of the thick atmosphere an ideal ascent usually involves keeping your apoapsis "just" ahead of your vessel. At apoapsis your ship is travelling directly at the horizon. This means you can burn in the direction of travel AND at the horizon and suffer minimal steering and minimal gravity losses.

The way the vectors add, you get minimal loss during ascent when thrusting at a point half way between your direction and the horizon... and the closer your direction of travel is to the horizon the more efficient... and if you had infinite T/W you would just coast up to apoapsis and do your perfectly efficient impulse to reach orbit. (but most people need a fair bit of time to build up orbit speed...)

But the benefit of the impulse at apoapsis is offset by the hohmann effect of getting your burns done lower in the gravity well... (since you are constantly slowing down if you coast up to apoapsis...) but I digress...

Edited by Alistone
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Gravity losses are caused by not burning toward the horizon. The portion of thrust spent "fighting" gravity is essentially always wasted.

<snip> BUT what happens is that your elevation will be increasing or decreasing and you will be thrusting above/below the horizon causing gravity losses.

Gravity losses don't depend (directly) on thrust direction or magnitude, only on velocity.

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Gravity losses don't depend (directly) on thrust direction or magnitude, only on velocity.

I enjoy many of you're posts; but could you clarify what you mean by "only on velocity".

Are you trying to reinforce the importance of hohmann effect during "liftoff".

Or are you focusing on the way in which the longer your flight takes (meaning the slower your velocity across the surface) the greater your losses due to gravity.

I'm pretty sure it all corresponds exactly with having to burn in a direction other than the horizon; since they go together it becomes a bit of the chicken and the egg to say which causes which.

I'd probably understand it better if I used mechjeb instead of manualy piloting...

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@Alistone yup thats what i meant :)

Gravity losses are caused by not burning toward the horizon. The portion of thrust spent "fighting" gravity is essentially always wasted.
For me gravity and steering losses are pretty similar and both base around the idea of thrust vector that isn't in line with current velocity vector.

From what i understand right now If we are not at Ap or Pe when burning prograde/retrograde we won't be burning toward horizon, and both steering and and gravity losses will be zero. The lack of efficiency of such burn would come from Oberth effect and not from gravity losses.

edit: im actually not sure what is Hohmann effect, i know about Hohmann transfer and Oberth effect... can you link a wiki article or something about Hohmann effect, cant find it on google.

Edited by Nao
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I enjoy many of you're posts; but could you clarify what you mean by "only on velocity".

Are you trying to reinforce the importance of hohmann effect during "liftoff".

Or are you focusing on the way in which the longer your flight takes (meaning the slower your velocity across the surface) the greater your losses due to gravity.

I'm pretty sure it all corresponds exactly with having to burn in a direction other than the horizon; since they go together it becomes a bit of the chicken and the egg to say which causes which.

Hence the "(directly)"... instantaneous gravity losses at any given moment don't depend on whether you're thrusting at that moment or in which direction, they depend on the direction and magnitude of your velocity vector at that moment - and on how far you are from the body. If you have zero horizontal velocity, then gravity will accelerate you downward at mu/r^2. If you have zero vertical velocity and horizontal velocity equal to orbital speed, then the net effect of gravity on your velocity is just to change its direction, but not your speed, and we say gravity losses are zero.

There is as you said a chicken-and-egg problem that your velocity changes depending on the thrust forces you apply. But when you're considering gravity losses, they don't stop when you turn off the throttle on your engines.

Anyway, regarding the question at hand, I seem to remember somebody did the math on point-to-point suborbital hops assuming infinite TWR and neglecting rotation. You take the family of orbits that intersect the surface of the body at the start and end points, and minimize over speed at the surface intersection. For small distances you get an approximately parabolic answer with 45 degree angle, but I don't know off the top of my head what the solutions look like as you start traveling further away. The thread may have been lost back in April, not sure.

And when you start considering surface rotation, then the destination intersection position depends on the hang-time, and it matters whether you're traveling east or west, or non-equatorial, generally messy. Messier still with finite thrust. Probably solvable with the Pontryagin principle, the thrust should be in the direction of the costates and I believe the switching function will be bang-bang in the absence of an atmosphere.

Edited by tavert
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Hence the "(directly)"... gravity losses at any given moment don't depend on whether you're thrusting at that moment or in which direction, they depend on the direction and magnitude of your velocity vector at that moment - and on how far you are from the body. If you have zero horizontal velocity, then gravity will accelerate you downward at mu/r^2. If you have zero vertical velocity and horizontal velocity equal to orbital speed, then the net effect of gravity on your velocity is just to change its direction, but not your speed, and we say gravity losses are zero.

There is as you said a chicken-and-egg problem that your velocity changes depending on the thrust forces you apply. But when you're considering gravity losses, they don't stop when you turn off the throttle on your engines.

This suggests that an elliptical orbit where gravity is always slowing you down or speeding you up is always suffering from gravity losses and gains...

I never though about an elliptical orbit as constantly having gravity losses and gains... but I suppose it would neatly account for orbital "sling-shotting" as positive gravity gains.

Thanks for the explanation.

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Hmm in orbital motion we have transfers from potential energy to kinetic and vice versa, the word losses does not fit the above description. For me gravity loss, would result in irreversible loss of energy, like for example ship hovering over ground, expends chemical energy without gaining both potential nor kinetic one.

Also at the bottom of this page i've found a following definition (...) For a finite thrusting period perpendicular to the position vector of the spacecraft no gravity losses occur. (...) I assume this includes moment when the craft's velocity vector is not at right angle to gravity vector.

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Think of it as

(ending speed) - (starting speed) = (delta-v applied) - (drag losses) - (steering losses) - (gravity losses)

The gravity losses aren't really permanent, in an elliptical orbit the kinetic/potential energy shifts back and forth so speed is higher at some points, lower at others.

Gravity losses there depend on flight path angle gamma, which is angle to velocity vector, not thrust angle alpha. Thrust perpendicular to position vector would be alpha = 0...

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Sorry to nitpick, but Im trying to understand this clearly. So if gravity losses aren't permanent (i understand why) then what are the looses during vertical ascent ? I mean is there other term for that? Ive never read about gravity drag as a part of energy exchange, it was always about loss of Dv during finite TWR thrust.

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Well there's going to be a certain design-dependent minimum amount of gravity loss that is required to get from stationary on the surface to the lowest-altitude stable circular orbit. If you consider the minimum-altitude stable circular orbit your reference point, you have to climb out of the gravity well (or the lack-of-horizontal-speed well...) to reach that point. From then on, around that reference, you can go into elliptical orbits and things will fluctuate up and down.

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The gravity losses aren't really permanent, in an elliptical orbit the kinetic/potential energy shifts back and forth so speed is higher at some points, lower at others.

I don't like this way of thinking about things. The problem is that rockets are constant thrust devices - rather than constant power - so you can have losses from gravity drag that do nothing to the energy budget of the system. The trivial example here is hovering; the rocket is thrusting continuously, and thus expending delta-v, but doing no work. In this sense, hovering is mathematically similar to radial/anti-radial or normal/anti-normal burns, so you can consider this type of gravity loss to be a form of steering loss.

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This suggests that an elliptical orbit where gravity is always slowing you down or speeding you up is always suffering from gravity losses and gains...

I never though about an elliptical orbit as constantly having gravity losses and gains... but I suppose it would neatly account for orbital "sling-shotting" as positive gravity gains.

Thanks for the explanation.

The better way of thinking about this (or rather the proper way) is in terms of energy. Energy in a (closed) system is always conserved, this is a universal rule. So your rocket in an elliptical orbit has two types of energy at all times, Kinetic energy and Gravitational Potential energy. At the high end of the elliptical orbit the craft has a lot of gravitational energy and therefor cannot have much kinetic. Conversely at the low points the craft has the opposite, high kinetic and low gravitational. (The formulas for these being 0.5mV^2 and mgh if you're interested)

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Sorry to nitpick, but Im trying to understand this clearly. So if gravity losses aren't permanent (i understand why) then what are the looses during vertical ascent ? I mean is there other term for that? Ive never read about gravity drag as a part of energy exchange, it was always about loss of Dv during finite TWR thrust.

Air resistance/drag. Without it; if you take of horizontally all of the "gravity loss" could be regained by falling back down.

For vertical thrusting gravity acts as a "drag" because it directly counteracts the force being applied by your engine and that portion cannot be regained (hence hovering loss), but even when "coasting" upward vertically there is a "gravity loss of velocity" which you regain when coasting back down.

The better way of thinking about this (or rather the proper way) is in terms of energy. Energy in a (closed) system is always conserved, this is a universal rule. So your rocket in an elliptical orbit has two types of energy at all times, Kinetic energy and Gravitational Potential energy. At the high end of the elliptical orbit the craft has a lot of gravitational energy and therefor cannot have much kinetic. Conversely at the low points the craft has the opposite, high kinetic and low gravitational. (The formulas for these being 0.5mV^2 and mgh if you're interested)

And I know all about PE vs KE and conservation of energy; but in my extremely limited use of mechjeb I didn't realize that it was calling gravity losses due to elevation change "losses" because it implied to me "wasted" energy. I see now it really is just explaining where you spent (or perhaps acquired if you do a slingshot...) your delta-V; not that the energy was wasted.

Edited by Alistone
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This is a simple simulation with rather low accurate, just lazy to precisely implement every factor.

1jhji.png

This is the profile of three method movement. Quick conclusion: constant altitude is an EXTREME BAD idea, never try that; The ascending part is somehow lesser significant compare to your fire angle, make sure it's 45 degree.

Let me explain this image then.

The simulation assume a vehicle weighs 15t, 5t of it is fuel. Engine is Mainsail (1500kN thrust, 544kg/s fuel flow at max thrust).

The constant altitude version, the vehicle launch at 65 degree for a gravity turn end with 45 fire angle

The 45 degree version, the rocket change its angle continuously to keep the velocity points to 45 degree until stop burning.

The gravity turn version's launch is similar to constant altitude version, launch at 85 degree to make sure the velocity angle is 45 degree when stopping burning.

Detailed record here:https://docs.google.com/spreadsheet/ccc?key=0Ahme0iIpl_3ZdE9LMkw0QzQ1cFpqNDh4aVA1WlRaRVE&usp=sharing

Columns:

Time...............MET

Distance...........Fly distance until land (or crash :/)

Altitude...........As is

Mass...............Current mass

Thrust Accel X.....X-component of acceleration from thrust

Thrust Accel Y.....Y-component of acceleration from thrust

Net Accel Y........Y-component of acceleration from thrust minus gravity acceleration

Velocity X.........X-component of velocity

Velocity Y.........Y-component of velocity

DeltaV.............Current delta V left

Note that this simulation is simulating physical at 10000Hz, but record sample rate is much lower (10Hz for constant altitude version, 1Hz for others)

Source code here:http://pastebin.com/cbb0RFQc

(Only core logic, wrapper and I/O things is not included

As the gravity turn version is slightly further than thrust at constant 45 degree, I think the optimal movement is launch at angel X which will make your fire angle be 45 degree when you use half of your delta V up and stop burning. Coast until your reach the optimal suicide burning point (Burn to retrograde and your speed decreased to 0 just when you reach the ground, you can see how I did this in the source code, however it's an approximate implement)

Edited by SaturnV
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Thanks for the chart SaturnV. Could you try one more profile? With the rocket keeping 45deg attitude continuously through the burn. I think it will go even further, or at least the same as the gravity turn one.

Also since we are doing over 100m/s2 acceleration this is almost like an artillery shot (5s burn time), it would be more interesting to do much longer burn. Doing a simple test with your numbers, with engine throttled down to 20% thrust, range with keeping velocity vector roughly 45deg is 18,82km (attitude of 65->62deg) and with keeping constant attitude of 45deg is 21,52km

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