Mun Landing Descent

[Blaarkies]

8 hours ago, Should-not-be-tagged-person said:
  " No, you can't reverse your velocity without expending energy, that's called magic. "

Low Mun orbit, full fuel tank, burn full-throttle retrograde and lose fuel(and speed) until we come to a suicide burn.
On Mun surface, empty fuel tank, burn full-throttle following gravity turn prograde with engine that produces fuel until we come to a perfect circularised low Mun orbit.

Reverse either of those videos and please tell us what the difference is. I cannot see any difference in the amount of energy difference between the start and end, so how could one be more/less efficient than the other?

Edited May 3, 2017 by Blaarkies
removed tag as requested

[Spricigo]

3 hours ago, Alshain said:

The Oberth maneuver is what I have been discussing the whole time though.  Read the first thing you quoted from me.

Nonetheless there is a misconception from your part. The oberth effect happens even without a oberth maneuver. As I said early they are related but not quite the same. And while @kryxal may be incorrect invoking Oberth*, the vessel is indeed falling a gravity well to perform the later burn at higher orbital speed .

The Oberth effect its the fact a given change in velocity result in a increased change in knetic energy if the starting speed is also increased. As explained here , here and here).

It can be better visualised with this graph (kinetic energy vs velocity)  or this image

 

 

3 hours ago, Alshain said:

No, you can't reverse your velocity without expending energy, that's called magic.

Its not magic, nor even changing the the movement itself. Im talking about a mathematical 'trick' that let me describe someting as if happening in the reverse order of events called Time Reversal Transformation. Its using -t instead of t in the equations of motion and ending with a correct mathematical description because of some symmetries. (in the case the fact that a landing looks exactly like a liftoff in reverse). That is also the situation Kryxal describe, a liftoff with an engine generating fuel (negative consumption)

But , since I indeed said something incorrect, let me rectify myself: if you do a time reversal transformation some variable don't change (even variables) while others change signal/direction (odd variables). Velocity its a odd variable, if you do a time reversal transformation you end with -V instead of V.

 

 

*Notice the error there is not that Oberth Effect dont plays a role. The error its the force, or rather the orientation, you are considering.

Divide your thurst T in two components: Ty in the direction of movement (horizontal) and Tz perpendicular to the movment (vertical). Tz will only chage the direction of your movimentl not change the intensity(no change in kinetic energy), wich means that Tz have oberth effect = 0   **

However you are using Tz to counteract you weigth W so it also dont cause any change*** in the intensity of the velocity, only change in direction. W also have oberth effect = 0  **

 

** Ok oberth effect its not a quantity. What I mena by "oberth effect = 0" is "it dont produces a change in velocity that causes a higher increase in kinetic energy compared with a object in rest"

***an aproximation... if your're not Niel Armstrong.

[Spricigo]

26 minutes ago, Alshain said:

Guys, believe what you want to believe.  I just don't care anymore.  Please stop tagging me in this thread.

Well you put arguments and examples and we replied with counter-arguments and counter-examples. Unfortunatelly we didnt reach a common understanding.

But asking us to ignore all the discussion after it happened becuase you dont care...Sorry, but I care. Not if I'm right or wrong, but if that is a opportunity of learning something and I'm losing it.

[Reusables]

19 hours ago, Starman4308 said:

I'm pretty sure the theory boils down to kinetic energy and the Oberth effect, that by allowing negative vertical velocity to build up, you're going to need to deal with it later by spending excess delta-V. That said, for high TWR and small bodies, I'm pretty sure constant-altitude and perfect suicide burns come pretty close to each other.

Sorry, but I can't get this part. Won't you get cosine loss from misalignment of thrust and velocity, resulting in less energy loss? Moreover, as gravity turn let vertical velocity build up, it will have more speed overall resulting in more energy loss. What part of this is wrong?

Also, I can't find the dv of gravity turn landing from the article you referred to. Would you tell me where I can get it?

[Starman4308]

4 hours ago, Abastro said:

Sorry, but I can't get this part. Won't you get cosine loss from misalignment of thrust and velocity, resulting in less energy loss? Moreover, as gravity turn let vertical velocity build up, it will have more speed overall resulting in more energy loss. What part of this is wrong?

Also, I can't find the dv of gravity turn landing from the article you referred to. Would you tell me where I can get it?

Do note that I'm not the one who came up with the theory, but my guess:

With infinite TWR, constant-altitude and suicide-burn descents are identical: a Hohmann transfer down to the desired altitude, followed by an instantaneous cancellation of horizontal velocity. That is the ideal, that you cancel out excess gravitational potential energy with Hohmann transfers preceding descent, and the "descent" is just about killing horizontal velocity.

Granted, none of this quite works out when you start dealing with things like "terrain", but the theory has to make some simplifications.

With finite TWR, you're always going to need to expend some extra delta-V perpendicular to the direction of travel to avoid smacking into the ground, although both do a good job of burying a lot of that via the magic of trigonometry. With a constant-altitude descent, this is pretty easy to see: as you start to pitch up to maintain altitude, your gravity losses are basically cosine losses from not burning directly retrograde. With a suicide burn, it's slightly subtler, as while you're always burning retrograde, effectively a local optimum* for eliminating unwanted kinetic energy, some of that unwanted kinetic energy is a side effect of earlier in the burn, as you were eliminating horizontal velocity without taking care of the vertical acceleration it would induce.

*Which is why I suspect an idealized suicide burn boils down to being a greedy algorithm.

Effectively, while constant-altitude is not eliminating your horizontal velocity with perfect efficiency, you avoid causing unwanted vertical velocity that you need to deal with later, and via the magic of math that presumably works*, it can be shown that the cosine losses of maintaining 0 vertical velocity is preferable to suicide burns, where you must start from a higher altitude, and generate some vertical velocity that you don't deal with until later.

*Unfortunately, the linked post's proof is no longer available. I know I've seen at least one other forum topic about it: wish I'd kept the link to it. I think I vaguely understand the concepts, but otherwise I'd need to do that math myself.

[Spricigo]

4 hours ago, Abastro said:

Sorry, but I can't get this part. Won't you get cosine loss from misalignment of thrust and velocity, 

Yes.  You let the cosine loss occur while focusing in reducing gravity loss. While in a gravity turn you let the gravity loss occur while focusing In reducing cosine loss. 

Something to notice: in a constant descent trajectory the resulting force (thrust - weight)  is aligned with velocity. You use 'just enough' of your thrust to counteract weight and any extra to reduce horizontal speed. Another point of constant descent ftrajectory is that while you velocity is high enough (orbital)  you are  constantly falling and missing the ground(orbital freefall) .  Ideally you are in a trajectory with a periapsis barely above the ground and all your thrust will be used entirely to kill your orbital velocity at the periapsis.

 Requires instantaneous deltaV expenditure,  thus impossible in practice. What you do is come as close as possible to this,  dealing also with the 'details'  like terrain and piloting errors. 

[Reusables]

36 minutes ago, Starman4308 said:

effectively a local optimum* for eliminating unwanted kinetic energy, some of that unwanted kinetic energy is a side effect of earlier in the burn, as you were eliminating horizontal velocity without taking care of the vertical acceleration it would induce.

Would you clarify this part? Specifically, what's the 'unwanted kinetic energy'? Do you mean the excess energy converted from the potential energy(as gravity turn needs losing altitude)?

Also, is it local optimum? I'm looking forward to seeing the graph, please let me know where is it or how to find it...

 

So it's much more complex than Oberth effect if I understand it right.

28 minutes ago, Spricigo said:

Ideally you are in a trajectory with a periapsis barely above the ground and all your thrust will be used entirely to kill your orbital velocity at the periapsis.

I don't get what you mean by 'periapsis'. With its original meaning, shouldn't it be underground?

Edited May 3, 2017 by Reusables
small correction

[Starman4308]

1 minute ago, Abastro said:

Would you clarify this part? Specifically, what's the 'unwanted kinetic energy'? Do you mean the excess energy converted from the potential energy(as gravity turn needs losing altitude)?

Kinetic energy that you have from going down. Down is the enemy, because up is the enemy.

The ideal is a Hohmann transfer to 0 altitude circular orbit followed by instantaneous cancellation of surface velocity. Any thrust applied upwards is waste. Any downwards velocity you let accumulate requires upwards thrust. In a suicide burn, you're letting some downwards velocity accumulate early in the burn that you have to cancel later.

I suppose you could think of it as this: in a suicide burn, you don't start pitching over until relatively late in the process, plus you have to start with a periapsis above the landing altitude, incurring gravity loss, whereas a constant-altitude descent, you start pitching over almost immediately, but you wind up not having to pitch as far up because you've never let downwards velocity accumulate.

10 minutes ago, Abastro said:

Also, is it local optimum? I'm looking forward to seeing the graph, please let me know where is it or how to find it...

If you're burning pure retrograde, that is inherently the instantaneous local optimum for eliminating energy. It kind of builds suicide burns into being a greedy algorithm, because while there is no better way to shed your kinetic energy at that precise moment, you can cause yourself larger problems down the line.

[Reusables]

3 minutes ago, Starman4308 said:

Kinetic energy that you have from going down.

You mean kinetic energy element from downward part of the velocity? Sorry, but this is really confusing..

Wouldn't it be better compensated when actually moving downward? Also I think potential energy, horizontal kinetic energy and vertical kinetic energy will exchange some energy with each other, which confuse me even more. Would you provide more explanation on this?

 

5 minutes ago, Starman4308 said:

If you're burning pure retrograde, that is inherently the instantaneous local optimum for eliminating energy. It kind of builds suicide burns into being a greedy algorithm, because while there is no better way to shed your kinetic energy at that precise moment, you can cause yourself larger problems down the line.

So it's not actual graph.. I was looking forward to one

[Spricigo]

46 minutes ago, Abastro said:

I don't get what you mean by 'periapsis'. With its original meaning, shouldn't it be underground?

It's that meaning,  the lowest point in a orbit.  

No it shouldn't be underground, that is too low for a constant descent.  The idea it's to never have enough vertical momentum for your trajectory intercept the ground (Except rigth before touching down). 

Assume you have perfect timing/piloting and instantaneous change of velocity.  You set your periapsis to 1m and exactly at the periapsis reduce velocity to zero. Now you are falling,  but it's so close to ground that you only reach 4-5m/s at touching down. 

I'm practice we don't have instantaneous change of velocity or perfect piloting so we can't be that good. We do as close as possible to this. 

[Reusables]

1 minute ago, Spricigo said:

It's that meaning,  the lowest point in a orbit.  

No it shouldn't be underground, that is too low for a constant descent.  The idea it's to never have enough vertical momentum for your trajectory intercept the ground (Except rigth before touching down). 

Assume you have perfect timing/piloting and instantaneous change of velocity.  You set your periapsis to 1m and exactly at the periapsis reduce velocity to zero. Now you are falling,  but it's so close to ground that you only reach 4-5m/s at touching down. 

I'm practice we don't have instantaneous change of velocity or perfect piloting so we can't be that good. We do as close as possible to this. 

You mean, when you have infinite TWR? Otherwise it'll be highest point of the orbit during the burn.

[Spricigo]

12 minutes ago, Abastro said:

You mean, when you have infinite TWR? Otherwise it'll be highest point of the orbit during the burn.

Assuming infinite TWR it's just to simplify the maths. A theoretical constructo. 

In practice you start with a periapsis above the ground (but not much)  and start the burn before periapsis.  As you burn your periapsis is lowering slowly becoming 0 at touch down (ideally) 

Alternatively you star with periapsis slightly below ground and start you burn and start the burn before hitting the ground.  As you burn periapsis is raising slowly and become 0 at touch down. 

The idea there its to prevent abruptly changing your periapsis while decelerating. 

 

Notice that don't exclude a suicide burn (performing the burn as close,  in time, as possible to the touch down.) The alternative landing method* is the gravity turn,  which also can be performed with or without a suicide burn. 

*other alternative methods are what we do when failing to follow a gravity turn/constant descent.  

[Reusables]

6 minutes ago, Spricigo said:

Assuming infinite TWR it's just to simplify the maths. A theoretical constructo. 

In practice you start with a periapsis above the ground (but not much)  and start the burn before periapsis.  As you burn your periapsis is lowering slowly becoming 0 at touch down (ideally) 

Alternatively you star with periapsis slightly below ground and start you burn and start the burn before hitting the ground.  As you burn periapsis is raising slowly and become 0 at touch down. 

The idea there its to prevent abruptly changing your periapsis while decelerating. 

 

Notice that don't exclude a suicide burn (performing the burn as close,  in time, as possible to the touch down.) The alternative landing method* is the gravity turn,  which also can be performed with or without a suicide burn. 

*other alternative methods are what we do when failing to follow a gravity turn/constant descent.  

Ah, now I got it. So it's somewhat similar to low-TWR hohmann transfer.

In the case, I heard that pro/retrograde hold burn costs less dv than pure manuever burn, and I think constant altitude burn is closer to the pure manuever burn. This doesn't apply in this case, right? Please explain how it goes.

 

Besides, is it possible that there is more efficient profile somewhere in between gravity turn and constant descent?

[Spricigo]

33 minutes ago, Abastro said:

Ah, now I got it. So it's somewhat similar to low-TWR hohmann transfer.

Not similar to a Hohmann transfer.  It Is a Hohmann transfer. 

What happens it's that with the 2nd burn we combine the burn to rendezvous (match velocities)  with the ground. 

Need to go now. But coming back later for the other points. 

[Kryxal]

9 hours ago, Blaarkies said:

8 hours ago, Should-not-be-tagged-person said:
  " No, you can't reverse your velocity without expending energy, that's called magic. "

Low Mun orbit, full fuel tank, burn full-throttle retrograde and lose fuel(and speed) until we come to a suicide burn.
On Mun surface, empty fuel tank, burn full-throttle following gravity turn prograde with engine that produces fuel until we come to a perfect circularised low Mun orbit.

Reverse either of those videos and please tell us what the difference is. I cannot see any difference in the amount of energy difference between the start and end, so how could one be more/less efficient than the other?

 

Actually, I think it came out as constant-altitude run up to orbital velocity was more efficient, a gravity turn mostly is used on bodies with an atmosphere because pointing off-prograde does BAD things to your craft if you're not careful.  Does anybody have solid data on that?

Edited May 3, 2017 by Kryxal
weird doubled text

[Blaarkies]

52 minutes ago, Kryxal said:

Actually, I think it came out as constant-altitude run up to orbital velocity was more efficient, a gravity turn mostly is used on bodies with an atmosphere because pointing off-prograde does BAD things to your craft if you're not careful.  Does anybody have solid data on that?

Yeah sure, i was just explaining or giving analogies to the "reverse time" trick before that post, but i would rather do the constant-altitude burn for atmospherless bodies, even for pinpoint landings.

Is it beter? I don't see the equations at the moment to prove it, but i think a thought experiment can shed some light:
Start on the Surface of Mun, ready for launch...
Gravity turns start pointing straight up and slowly point horizontal, following the movement trajectory. So they spend some time "hovering" upwards (and lose dv on gravity drag) before they start gaining horizontal speed to "escape" gravity drag.
Constant-altitude burns start by immediately pointing as horizontal as possible to "escape" gravity drag as quickly as possible, only burning vertical enough to counteract the current amount of gravity drag.

That doesn't make it sound like Constant-altitude burns are much better then, until you think about combining the x and y components of the burn. See, gravity turns do a lot of y burning at the start, then transitions through to end in lots of x burning. That in itself has some "2 separate maneuver nodes" burning incorporated in it(not much though due to the time spent doing 45degree burning). Contrary to that, the Constant-altitude burn tries to do all this in a single burn, which is combining those "2 separate  maneuvers"

So i am confident that Constant-altitude burns are better(if skill and mission specs allow), but i cannot prove it...that "gravity drag vs cosine losses while it is transitioning with a variable speed from y to x..." thing is killing me 

[Spricigo]

1 hour ago, Abastro said:

In the case, I heard that pro/retrograde hold burn costs less dv than pure manuever burn, and I think constant altitude burn is closer to the pure manuever burn. This doesn't apply in this case, right? Please explain how it goes.

Maneuver nodes assume instantaneous change in velocity. With that in mind at each burns of a hohmann transfer the maneuver node is exactly in the same direction of prograde/retrograde. However in real maneuvers change in velocity its not instantaneous and you will lose efficiency anyway, if you are pointing in direction of maneuver node there is a difference in direction between velocity and burn, if you are holding pro/retrograde there is change in direction of your burn. For what I know both cases are equivalent in regard of efficiency, if something holding pro/retrograde its more prone to inaccuracies due to piloting/SAS.

Its the same situation for a constant descent trajectory, for the part of our thrust we are not using to counteract gravity.

 

6 hours ago, Abastro said:

Besides, is it possible that there is more efficient profile somewhere in between gravity turn and constant descent?

Not enough data to know (from my part at least). But seems plausible a combination of both that end with a better overall efficiency (eg we start following a constant descent trajectory and finish a gravity turn)

2 minutes ago, Kryxal said:

Actually, I think it came out as constant-altitude run up to orbital velocity was more efficient, a gravity turn mostly is used on bodies with an atmosphere because pointing off-prograde does BAD things to your craft if you're not careful.  Does anybody have solid data on that?

More to the point if there is a atmosphere you want to reduce drag what you archive reducing the cross-sectional area perpendicular to the airflow. In a gravity turn you are already keeping the thurst pointed in the (opposite) direction of the airflow so you get two benefits(the other being minimal cosines loses) with the same trajectory.

Without delving in maths or experimenting we don't know which one is better for a airless body . To me it seems that if the gravity losses are a major concern constant altitude is better, if steering losses are a major concern gravity turn is better. But there is several catches:

1.It may be the case that  we are just trading one kind of loss for another in similar amounts. It helps nothing if to reduce gravity loss in 200m/s we lose the same200m/s due steering.

2.We may be doing a efficient maneuver in a inefficient way, (eg not burning at full throttle, letting the craft wobble)

3.Other variables may be a greater concern than either gravity or cosine loss (eg low control authority).

4.Its possible that the ideal trajectory is part constant descent, part gravity turn.