Efficient landing procedure?

[Snark]

Perhaps I wasn't clear. AFTER I've placed my PE as low as possible over the landing site. And I AM burning retrograde, and I do start at full throttle. But I quickly have to back off the throttle or I'll end up hovering (or rising) and that MUST be inefficient.

Are you saying I should build a ship with a lower TWR and run all the way down at full throttle? Why would that be better?

or are you saying I should free-fall more and do shorter burns?

Ding ding ding

That's exactly it. An efficient suicide burn means that you run at zero throttle as you fall towards the surface. And fall. And fall. (And your kerbals are looking nervously at the altimeter and wondering, okay, he's gonna start the retro-burn soon, right? Like, pretty soon? For example, right now would be good. Aaaaany time. Gosh, look at that thing coming up towards us very fast, so very very fast, that would be the ground, when are you going to TURN ON THE FREAKING ENGINES ALREADY.)

And then at the last possible instant (and the higher your TWR, the better, so "last" is later), you slam on the throttle full power, which brings you to a halt right when you happen to reach ground level. And then cut thrust the moment you touch down at zero velocity. (And your passengers all high-five each other and reach for a fresh pair of underwear.)

You don't "end up hovering" because the point at which you reach zero velocity (i.e. "hover") is designed to be precisely at ground level, a.k.a. "landed".

The key, of course, is knowing exactly when the "last possible instant" is. The later you start your thrust, the more efficient your landing... but if you start too late, you end up lithobraking. That's why they call it a "suicide burn."

and again, Why would that be better?

I'm trying to understand why.

The reason why this is more efficient than going at half-throttle all the way down is that by falling fast and waiting until the last moment to slow down, you reach the surface sooner. When you spend a long time descending on half-throttle, essentially you're hovering already (at least, it's a "partial" hover). You're wasting fuel keeping yourself airborne (well, okay, it's a vacuum, but you know what I mean) longer than necessary. You're piling up gravity losses, since your engine is wasting a lot of fuel fighting gravity while it's trying to slow the ship down.

Edited November 6, 2015 by Snark

[Brainlord Mesomorph]

Still not understanding the why. Granted, hovering is a waste of fuel. And coming in very slowly is a similar waste to fuel.

But if a 15 second burn at full throttle will stop my ship. Wouldn’t a 30 burn at half throttle, or a one minute burn to at ¼ throttle be exactly the same? Except perhaps without the changing my shorts part?

My goal is always zero velocity just at zero altitude. How is doing it at panic more efficient?

EDIT:Wait, scratch that, the longer I’m in the air the longer gravity is working against me? Gravity is building up by the second, so it’s like Judo, I just go with it.

Edited November 6, 2015 by Brainlord Mesomorph

[Starhawk]

Burning any sooner than necessary increases the amount of time before you contact the surface. But this increases the time that the gravity acts on the craft and, therefore, increases the amount of fuel that will be spent only on fighting gravity. So burning any sooner than you need to is inefficient.

Of course you need some inefficiency to land safely.

Happy landings!

EDIT: That's it. You've got it!

Edited November 6, 2015 by Starhawk

[KerikBalm]

I saw it mentioned in another thread that landing is super inefficient and can consume upwards of 2-3x the dV given on a dV map.

Only if you are super inefficient... I find the dV maps to be a good guide.

So I've heard two (three if you count atmospheric methods but let's assume no atmo for now) methods here so far that are interesting:

Method A: Chase the retrograde marker all the way down (suicide burn), under thrust the whole way, looking to attain 0m/sec just as your gear hits the dirt.

Method B: De-orbit from gonzo-low Periapsis first, then over-burn on your way to the ground.

I've been using Method C: When you're near where you want to land, burn laterally to cancel out your horizontal velocity and then regulate your descent based on available TWR. This sounds rather like it splits the difference between these two, is it more or less efficient than either, do we think?

Its worse than both... but safer if you have a lot of fuel.

Every second of "regulating your velocity on the way down" (i'll take that to mean a constant velocity descent) costs you 1.63 m/s to land on Mun,

So if you cancel your horizontal velocity at 500 m over the Mun, and descend at 10 m/s... it will take 50 seconds to touch down, and it will cost you 81.5 m/s lost to gravity drag.

Next is the oberth effect/the fact that burns at higher hvelocity are more efficient... If you take your orbital velocity at 100km vs 1km... its not so different

However, if you "stop and drop" from 100 km... how fast do you think you'll be falling when you impact?

x= 1/2 at^2

1/2 at^2= 100,000

0.815 t^2= 100,000

t^2= 122,700m

t= 350 seconds

a= 1.63 m/s/s velocity = a*t -> 570 m/s... nearly orbital velocity again... (can anyone confirm? it seems a bit much to have it be orbital velocity.. I guess gravity isn't 1.63 m/s/s at 100km, but its close enough?) you've just about doubled the dV requirement.

If you "regulate" your descent from 100km... your gravity losses will be extreme

Then there's the trigonometry.

Its much more efficient to make an angled burn, then 1 horizontal burn, and 1 vertical burn.

ie... I need to cancel out 400 m/s horizontal velocity and 300 m/s vertical velocity... if I do a 400m/s and then a 300m/s burn, I spend 700... when I could just do a 500m/s burn at the right angle between them.

The most efficient way to land is to set your AP and PE right above the terrain, and burn horizontal... calculating about how long it will take to reach 0 surface velocity.

Take your altitude, divide it my that time... that's the descent rate you want to go for... angle your engine slightly downward while still burning mostly horizontally as needed so that you don't exceed that descent rate by much.

If the ground is coming up too fast, angle your thrust down more to slow your descent rate.

*yes, as you start angling your thrust, it will take a bit longer to actually come to a stop, but it will take a while just to reach your desired descent rate, and your TWR willl increase during the burn, so its a good enough approximation.

With an infinit TWR, the difference between this and a suicide burn disappears... its just set a PE that touches the ground, and cancel velocity at PE.

A suicide burn seems to be more precise as to where you will land... you pretty much point the trajectory line at just past where you want to land, and then burn at the last moment.

Suicide burns are pretty efficient... and as you come in from a lower and lower orbit, the difference between them and a constant descent rate approach gets pretty small

[qoonpooka]

One thing I want to have cleared up before I go check the "Answered" box. As an aside, thank you guys for this great discussion it's been very helpful!

Given This:

Free-fall more and do shorter burns is exactly what I'm saying. The more time you spend burning with any vertical component, the more fuel is being wasted.

I'm oversimplifying this, but: All the fuel burned horizontally is efficient. All of the fuel burned vertically is inefficient. You want to minimize the latter as much as possible.

One way to think of it is this. When you burn horizontally, you change your velocity relative to the surface. However, you can stay in one place burning vertically and burn all your fuel just hovering and not get anywhere. Time spent burning horizontally fights only inertia. Time spent burning vertically fights gravity. And gravity just eats delta-v.

Why should this be true?

Then there's the trigonometry.

Its much more efficient to make an angled burn, then 1 horizontal burn, and 1 vertical burn.

ie... I need to cancel out 400 m/s horizontal velocity and 300 m/s vertical velocity... if I do a 400m/s and then a 300m/s burn, I spend 700... when I could just do a 500m/s burn at the right angle between them.

I understand why it should be true when driving on the ground, where all motion is subject to friction losses and whatever else. But in vacuum, it seems to me like an object in orbit has both the horizontal velocity and vertical acceleration to contend with. But, assuming I have enough TWR, why should it matter when I cancel out each component? KerikBalm's example of dropping from 100km is extreme but useful here: I get that the longer I spend in gravity's grasp the more velocity I'll need to counter but when I lower my orbit it looks like all I do is trade some of that vertical velocity for a higher horizontal component as a result of my higher orbital speed.

Does the Oberth effect not help make up the difference? My higher velocity when I begin my landing burn should make that burn more efficient as I begin and by then end it should match up?

Why does 400 + 300 = 500 in KerikBalm's example?

I wonder if the actual math is more like 400 + 300 = 700m/sec suicide burn, but stop and drop is more like 400 + 300 = 900?

[Brainlord Mesomorph]

Ok, I think I'm following this. (god, I am learning sooo much physics playing this game.)

Then would this be a method:

(start in very low orbit)

1 Do a deorbit burn that gives me an impact site about 500 m past my landing target.

2 Plot a burn over the landing site to stop.

3 Look at the burn time, add a 25% safety margin

4 that many seconds before the mode, throttle up

5 change shorts.

??

[Starhawk]

But, assuming I have enough TWR, why should it matter when I cancel out each component?

...

Why does 400 + 300 = 500 in KerikBalm's example?

This is just geometry. Force and velocity add just like distance on the ground. Two orthogonal velocity changes can be added with vector addition to compute the final resulting velocity change. It is always cheaper to make a single burn in the resulting direction than it is to make two orthogonal burns. It's just Pythagoras.

Ok, I think I'm following this. (god, I am learning sooo much physics playing this game.)

Then would this be a method:

(start in very low orbit)

1 Do a deorbit burn that gives me an impact site about 500 m past my landing target.

2 Plot a burn over the landing site to stop.

3 Look at the burn time, add a 25% safety margin

4 that many seconds before the mode, throttle up

5 change shorts.

??

Well, mostly. The first burn doesn't set an impact site. It only drops your periapsis very low (maybe 2 or 3 km depending on conditions). And you don't start the final burn until very low in your orbit (almost at periapsis). But you're still in orbit at that point. As you burn your trajectory will change so that eventually it intersects the ground. As you keep burning you eventually reach the point where you have no horizontal velocity to get rid of and all you need to do to land is manage the remaining vertical velocity and the change in that while you traverse the distance to the ground.

So the final burn is actually the deorbit burn.

Happy landings!

ETA: As far as the shorts go, it gets less scary once you practice it enough.

Edited November 6, 2015 by Starhawk
clarity

[Empiro]

I understand why it should be true when driving on the ground, where all motion is subject to friction losses and whatever else. But in vacuum, it seems to me like an object in orbit has both the horizontal velocity and vertical acceleration to contend with. But, assuming I have enough TWR, why should it matter when I cancel out each component? KerikBalm's example of dropping from 100km is extreme but useful here: I get that the longer I spend in gravity's grasp the more velocity I'll need to counter but when I lower my orbit it looks like all I do is trade some of that vertical velocity for a higher horizontal component as a result of my higher orbital speed.

Does the Oberth effect not help make up the difference? My higher velocity when I begin my landing burn should make that burn more efficient as I begin and by then end it should match up?

Why does 400 + 300 = 500 in KerikBalm's example?

I wonder if the actual math is more like 400 + 300 = 700m/sec suicide burn, but stop and drop is more like 400 + 300 = 900?

There's a couple of different factors at work here. First, 400^2 + 300^2 = 500^2, which is trigonometry. It's no different than if you take 400 steps forward, and then turned right and took 300 steps, you could have just walked diagonally 500 steps. Friction has nothing to do with it -- you simply traveled less distance when you took the diagonal route. Likewise, you simply spend less delta-V if you burn diagonally than if you burn horizontally then vertically.

The Oberth Effect comes into play too. The easiest way to take advantage of the Oberth Effect is simply that the faster you're moving, and you burn directly prograde/retrograde, the more efficient your burn is (in terms of kinetic energy changed). When are you moving fastest when you're attempting to land? It's when you're super close to the surface. No matter what velocity you were going when you were high up, you'll be going even faster once you're close to the surface because of gravity. Therefore, it's best to try to spend most of your fuel there.

You might think that you're just trading off vertical velocity for a higher horizontal velocity, but that trade-off is not 1:1.

Edited November 6, 2015 by Empiro

[Brainlord Mesomorph]

Well, mostly. The first burn doesn't set an impact site. It only drops your periapsis very low (maybe 2 or 3 km depending on conditions). And you don't start the final burn until very low in your orbit (almost at periapsis). But you're still in orbit at that point. As you burn your trajectory will change so that eventually it intersects the ground. As you keep burning you eventually reach the point where you have no horizontal velocity to get rid of and all you need to do to land is manage the remaining vertical velocity and the change in that while you traverse the distance to the ground.

THAT's what I originally described. One long burn to the ground and "manage the remaining vertical velocity"

Snark says what I need to do is one short burn right before impact. I'm saying PLOTTING A NODE for that burn tells you what your speed WILL be, Tell you how long a burn it is and gives you countdown for it. Taking all the guesswork out of it.

I just tried it, it worked, took less fuel, and was very exciting,

(flying horizontally (toward a mountain) at 550 m/s about 1 km up, with "Well, it says I'll stop in 70 seconds..." aaaaand go!

EDIT: Just did it again, (definitely more fun!) only problem so far is that I'm have trouble landing on target because of planetary rotation between plotting the node and getting there.

Edited November 6, 2015 by Brainlord Mesomorph

[Snark]

Ok, I think I'm following this. (god, I am learning sooo much physics playing this game.)

Then would this be a method:

(start in very low orbit)

1 Do a deorbit burn that gives me an impact site about 500 m past my landing target.

2 Plot a burn over the landing site to stop.

3 Look at the burn time, add a 25% safety margin

4 that many seconds before the mode, throttle up

5 change shorts.

??

Step 1 is correct, just try to do the deorbit burn early so that you're coming in shallow, almost horizontally, at your projected impact site.

Step 3 is actually backwards, you want to reduce it. If the burn meter is telling you that it will be a 30 second burn, you want to wait until it's like 20 or 25 seconds to maneuver until you start your burn. Shallower angle or lower TWR means longer safety margin. The time you need will always be greater than 1/2 of the estimated burn, so don't wait that long. .

[mabarry3]

I'm not grasping why using the Oberth effect for landing (doing the deorbit burn at periapsis), or even a Hohmann transfer to a lower "orbit" is more efficient way to land than doing the burn at apoapsis until your periapsis is in the correct spot for you to do a suicide burn or whatnot. Even if the burn itself is more efficient at periapsis, you have to lower the apoapsis a greater amount (sometimes significantly) than the periapsis would have to be lowered in order to put yourself on a suborbital trajectory. Why is the burn done at periapsis? The only reason I can think of is then you're already bleeding off velocity, but even then it feels like the lower the altitude reduction required to be on a suborbital trajectory, the better; by definition, that would make the burn at apoapsis.

What am I missing?

[qoonpooka]

This is just geometry. Force and velocity add just like distance on the ground. Two orthogonal velocity changes can be added with vector addition to compute the final resulting velocity change. It is always cheaper to make a single burn in the resulting direction than it is to make two orthogonal burns. It's just Pythagoras.

Geometry involves static shapes. If I was expending fuel for distance traveled, instead of for velocity, this would make sense. But all the while I'm changing my velocity, and gravity is working to do the same. So if I burn diagonally, my thrust is less effective at canceling my horizontal motion, AND less effective at canceling my vertical motion. To me this seems like it should work like cornering in racing: At any given moment I can only have access to 100% of my traction. Some of that I spend turning, some of that I spend braking, but the moment I ask my tires for more than 100%, I crash.

Some of my thrust I spend horizontally, some I spend vertically, but when I thrust diagonally against my retrograde marker, there is nothing in geometry that says that I should somehow be able to apply more than 100% of my thrust - just that I should be able to travel a shorter total distance (that's the hypotenuse). But burning in one direction to cancel out my horizontal, and then burning the vertical off involves traveling more total distance, and involves taking more time, but you're coasting for part of that, and falling while you burn out your horizontal. That extra velocity is velocity that /would have/ been canceled out by the diagonal thrust, I'm just waiting until the horizontal element is gone before I set about burning that off.

There's a couple of different factors at work here. First, 400^2 + 300^2 = 500^2, which is trigonometry. It's no different than if you take 400 steps forward, and then turned right and took 300 steps, you could have just walked diagonally 500 steps. Friction has nothing to do with it -- you simply traveled less distance when you took the diagonal route. Likewise, you simply spend less delta-V if you burn diagonally than if you burn horizontally then vertically.

And that's the part I don't understand. Why is that? Why are 'speed I am traveling' and 'distance I am traveling' necessarily the same? Why can't I take more time, travel more distance, but because there's no drag, spend the same amount of fuel overall? It's not like part of a triangle is being constantly accelerated away from the rest of the triangle so saying things like "It's trigonometry" isn't actually helpful to me in this instance. I learned trig on a chalkboard, not in a blender or tilt-o-whirl... at least not within my frame of reference.

[RocketPropelledGiraffe]

Geometry involves static shapes. If I was expending fuel for distance traveled, instead of for velocity, this would make sense. But all the while I'm changing my velocity, and gravity is working to do the same. So if I burn diagonally, my thrust is less effective at canceling my horizontal motion, AND less effective at canceling my vertical motion. To me this seems like it should work like cornering in racing: At any given moment I can only have access to 100% of my traction. Some of that I spend turning, some of that I spend braking, but the moment I ask my tires for more than 100%, I crash.

Geometry works vor any vectors, it has the same laws for distance vectors as well as velocity. This also goes for acceleration, which starts to play a role in landing due to the body's gravity, obviously.

And that's the part I don't understand. Why is that? Why are 'speed I am traveling' and 'distance I am traveling' necessarily the same? Why can't I take more time, travel more distance, but because there's no drag, spend the same amount of fuel overall? It's not like part of a triangle is being constantly accelerated away from the rest of the triangle so saying things like "It's trigonometry" isn't actually helpful to me in this instance. I learned trig on a chalkboard, not in a blender or tilt-o-whirl... at least not within my frame of reference.

The landing in space is a complicated process with which we don't have day-to-day experience, and our intuition often does not give good results in such cases.

Evene if there is no drag, there still is gravitational pull. You want to remember that you have to take all three things into account: distance traveled, velocity, and acceleration (due to gravity). In this sense "speed I am traveling" and "distance I am traveling" are not the same, but they are linked- you want to go down and slower, all the while you are accelarated towards the surface, which is not helping - so try to spend as little time being accelerated toward the surface as possible:

Keep in mind, that you can hover in a gravitational field, not traveling any distance at all, but still spending a lot of delta-v vertically to fight the accelaration, which does not help at all for landing. You can never do such "useless" burns horizontally, all delta-v applied in this direction helps with your landing effort.

[Snark]

I'm not grasping why using the Oberth effect for landing (doing the deorbit burn at periapsis), or even a Hohmann transfer to a lower "orbit" is more efficient way to land than doing the burn at apoapsis until your periapsis is in the correct spot for you to do a suicide burn or whatnot. Even if the burn itself is more efficient at periapsis, you have to lower the apoapsis a greater amount (sometimes significantly) than the periapsis would have to be lowered in order to put yourself on a suborbital trajectory. Why is the burn done at periapsis? The only reason I can think of is then you're already bleeding off velocity, but even then it feels like the lower the altitude reduction required to be on a suborbital trajectory, the better; by definition, that would make the burn at apoapsis.

What am I missing?

The real killer for Oberth effect is how and where you circularize. The fact is, you already spent quite a bit of dV to get into that circular orbit in the first place. So consider these two alternatives:

Method A: You approach the Mun on a trajectory from Kerbin that will have your Pe out at a few hundred km altitude from the Mun. You do burn #1 to circularize in that high orbit. Then you do burn #2 to drop your Pe down to the surface. Then you do suicide burn, #3, to land.

Method B: You approach the Mun on a trajectory from Kerbin that will have your Pe really close down low, almost grazing the surface. You do burn #1 to circularize in that low orbit. Then you do burn #2 to drop your Pe down to the surface. Then you do suicide burn, #3, to land.

B1 > A1. But B2 < A2, and B3 < A3. And it turns out that the extra you spend on burn #1 is outweighed by what you save on burns #2 and #3. That's where you get the benefit from Oberth.

Let's put some math on it. Take the Mun. Its SoI is about 2.4e6 m, and its GM is 6.5e10 m3/s2. Let's say you have a ship that's approaching the Mun, and has a velocity of 1000 m/s a the time that it enters the Mun's SoI. To take the two extremes of Oberth, consider two cases: one with a high-periapsis trajectory, one with a low-periapsis trajectory.

High-periapsis trajectory is where you're basically just grazing the SoI. So you need to stop nearly dead in order to capture to the Mun and drop your Pe down to the surface. So that's a retro-burn of about 1000 m/s out at the SoI boundary, then when you get down to the surface you're falling at close to the Mun's escape velocity of 800 m/s, which you kill with a suicide burn right at the surface. So that's a total dV of 1800 m/s.

Now let's consider the low-periapsis trajectory, where you enter the SoI aimed right at the Mun on a path that grazes the surface. So you just fall all the way and do a massive suicide burn right at the surface to land. If you're going 1000 m/s almost straight down when you enter the SoI, you're just going to keep accelerating all the way down and you'll be going faster when you get to the surface, right? Well, how much faster? Doing the math, it turns out to be a bit under 1300 m/s.

So by doing your burn low rather than high, you save 500 m/s.

Edited November 7, 2015 by Snark

[Empiro]

And that's the part I don't understand. Why is that? Why are 'speed I am traveling' and 'distance I am traveling' necessarily the same? Why can't I take more time, travel more distance, but because there's no drag, spend the same amount of fuel overall? It's not like part of a triangle is being constantly accelerated away from the rest of the triangle so saying things like "It's trigonometry" isn't actually helpful to me in this instance. I learned trig on a chalkboard, not in a blender or tilt-o-whirl... at least not within my frame of reference.

Your goal is to reduce your velocity to 0, which is most efficient when you directly oppose your direction of motion. Lets take another example:

If you're going 10 m/s forward, and you want to stop, you'd spend 10 m/s backward to stop, correct? You wouldn't spend 10 m/s backwards and to your left to try to stop -- it wouldn't result in you stopping. It's a similar concept here -- if you're travelling 300 m/s horizontally, and 400 m/s vertically, by burning just horizontally (or just vertically), you're not directly opposing your direction of motion, so you end up spending more delta-V. If you're having trouble visualizing this, try drawing it on a sheet of paper, and rotating the paper so that the velocity vector points "forward".

In addition, as some have pointed out here, you have to contend with things like gravity drag and the reduction in the Oberth effect, making things much less efficient.

Edited November 7, 2015 by Empiro

[G'th]

GoSlash27's "Reverse Gravity Turn" is the most efficient non-suicidal method. I do it a bit differently but the concept is the same.

http://forum.kerbalspaceprogram.com/threads/104638-Reverse-gravity-turn-landing-technique-for-airless-bodies

This. I also like to build some extra dV into my landers to bring the starting orbit down in altitude. Makes it a lot easier to land because its easier to hit a specific target and you have a bit less speed to kill off.

[qoonpooka]

Your goal is to reduce your velocity to 0, which is most efficient when you directly oppose your direction of motion. Lets take another example:

If you're going 10 m/s forward, and you want to stop, you'd spend 10 m/s backward to stop, correct? You wouldn't spend 10 m/s backwards and to your left to try to stop -- it wouldn't result in you stopping. It's a similar concept here -- if you're travelling 300 m/s horizontally, and 400 m/s vertically, by burning just horizontally (or just vertically), you're not directly opposing your direction of motion, so you end up spending more delta-V. If you're having trouble visualizing this, try drawing it on a sheet of paper, and rotating the paper so that the velocity vector points "forward".

In addition, as some have pointed out here, you have to contend with things like gravity drag and the reduction in the Oberth effect, making things much less efficient.

Thank you for this. I think this is going to be one of those things I just don't understand and have to memorize, rote. Every time someone has tried to explain this, it sounds to me like they're contradicting themselves or other things taken as read.

I'ma tag this 'Answered' and trust that the discussion is useful for folks.

[Brainlord Mesomorph]

I thought of another way of saying this may help anyone who still not getting it.

They measure gravity in meters per second per second (m/s/s).

So for every second you’re off the ground (EDIT: AND NOT IN ORBIT) you're adding that many m/s to the total amount of work you’re doing to land.

Hovering is a complete waste of fuel. (Agreed?) So, coming in at half of free fall is half of that complete waste of fuel. Even taking 25% off of free fall, is 25% of a complete waste of fuel. So you need to free fall absolutely as much as possible, and run your engines absolutely as little as possible.

How much fuel can this save?

Well, so far, two different vehicles that I have that I thought had enough fuel to land, but not make orbit again (without refueling), turns out they have enough fuel to land and take off, as long as I don’t waste it on landing.

Sci fi movies are missing a bet, by not accurately depicting a minimum delta V landing. That is, a very dramatic suicide burn!

So, I sure learned something in this thread. Thanks everyone

Edited November 7, 2015 by Brainlord Mesomorph

[KerikBalm]

Not really.... being in a stable orbit indefnitely does't cause required dV to go arbitrarily high.

Its every second you spend fighting gravity... thrusting down instead of horizontally.

The rest is related to oberth and hohman.

A "stop and drop" with infinite TWR will waste less than a stop and drop with a 1.5:1 TWR.

capturing into an orbit with a PE right off the ground takes less dV than capturing to a 50km orbit, and then lowering it

Think about it.. your craft's acceleration is measured in m/s/s - so is gravity.

Assumy tour craft can accelerate at 2 m/s

Fight gravity on mun (ie, thrust straight down, to go straight up), and the net acceleration is 2-1.63 = 0.37 m/s/s ... 0.37/2*100% = 18.5%... 81.5% of your acceleration is lost to gravity.

Do it again with a 100m/s/s acceleration, and most of your thrust actually changes the velocity of your craft.

A suicide burn is basically the burn you do when your AP is significantly higher than your PE.

In the absolutely ideal case of a perfectly spherical body, not mountains or bumps or anything, to land you would set your PE right above the ground, and then with your infinite TWR, come to a stop right at PE.

With a limited but still much higher than gravity TWR, you could basically reverse perapsis kick until you are in circular orbit, 10 meters off the ground with basically no loss in efficiency.

At this point, decreasing velocity by half would have you drop into the ground, so as you slow down, you need to angle your thrust down as well.

At a 2:1 TWR, this angle wouldn't exceed 30 degrees before horizontal velocity is cancelled out.

A suicide burn would have you dropping, likely following retrograde (maybe thrusting down a bit more which can turn the terminal part of the landing to a case much like the above case) so in this case, the angle of your engine thrust gets farther and farther from horizontal, ending vertical if you do the reverse gravity turn profile

A suicide burn can be done with the PE just touching the ground, or well under the ground... its more efficient the shallower you come in.. and you can't come in shallower than parallel to the ground as in the constant altitude approach... but that doesn't work so well if the ground is not flat.

In both cases, your goal is to be right above the ground right at the end of your burn.

In the constant altitude approach... you are always right above the ground

As a suicide burn approach becomes shallower and shallower, and the TWR goes up, the distinction between the two goes away.

[Brainlord Mesomorph]

Not really.... being in a stable orbit indefnitely does't cause required dV to go arbitrarily high.

I didn't mean in orbit. Fixed.

I don't think anything in KSP (or reality) is a perfect sphere. (edit: Jool doesn't count, we're taking about landing)

And I don't think everyone here is using the same definition of "suicide burn." A horizontal burn with lots of TWR isn't suicide.

A Pe well underground and not enough TWR to change that and hoping you get the timing right to avoid crashing... that's my definition of "suicide burn."

Edited November 7, 2015 by Brainlord Mesomorph

[qoonpooka]

Not really.... being in a stable orbit indefnitely does't cause required dV to go arbitrarily high.

Its every second you spend fighting gravity... thrusting down instead of horizontally.

The rest is related to oberth and hohman.

A "stop and drop" with infinite TWR will waste less than a stop and drop with a 1.5:1 TWR.

capturing into an orbit with a PE right off the ground takes less dV than capturing to a 50km orbit, and then lowering it

Think about it.. your craft's acceleration is measured in m/s/s - so is gravity.

Assumy tour craft can accelerate at 2 m/s

Fight gravity on mun (ie, thrust straight down, to go straight up), and the net acceleration is 2-1.63 = 0.37 m/s/s ... 0.37/2*100% = 18.5%... 81.5% of your acceleration is lost to gravity.

Do it again with a 100m/s/s acceleration, and most of your thrust actually changes the velocity of your craft.

A suicide burn is basically the burn you do when your AP is significantly higher than your PE.

In the absolutely ideal case of a perfectly spherical body, not mountains or bumps or anything, to land you would set your PE right above the ground, and then with your infinite TWR, come to a stop right at PE.

With a limited but still much higher than gravity TWR, you could basically reverse perapsis kick until you are in circular orbit, 10 meters off the ground with basically no loss in efficiency.

At this point, decreasing velocity by half would have you drop into the ground, so as you slow down, you need to angle your thrust down as well.

At a 2:1 TWR, this angle wouldn't exceed 30 degrees before horizontal velocity is cancelled out.

A suicide burn would have you dropping, likely following retrograde (maybe thrusting down a bit more which can turn the terminal part of the landing to a case much like the above case) so in this case, the angle of your engine thrust gets farther and farther from horizontal, ending vertical if you do the reverse gravity turn profile

A suicide burn can be done with the PE just touching the ground, or well under the ground... its more efficient the shallower you come in.. and you can't come in shallower than parallel to the ground as in the constant altitude approach... but that doesn't work so well if the ground is not flat.

In both cases, your goal is to be right above the ground right at the end of your burn.

In the constant altitude approach... you are always right above the ground

As a suicide burn approach becomes shallower and shallower, and the TWR goes up, the distinction between the two goes away.

does this mean that (to a point, obviously) higher TWR means you waste less fuel?

[KerikBalm]

yes, absolutely...

But there is a point where the added mass of the engines makes it not worth it.

[Snark]

does this mean that (to a point, obviously) higher TWR means you waste less fuel?

For landing on a vacuum world, yes. The higher your TWR, the less fuel you waste in gravity losses on landing.

The balancing factor (the one that favors lower TWR) is engine mass: too much engine = wasted dead weight.

In practice, it's usually not a problem, because except for Tylo, most vacuum worlds have a much lower gravity than Kerbin, which means you can get a very high local TWR with quite moderately sized engines. If you can manage a craft with 10 m/s2 acceleration (pretty easy for a small lander), that's already a TWR of 6 on the Mun, for example.

There's also very much a diminishing benefit. You're going to take a certain total dV hit for doing the suicide burn to land even if you have infinite TWR: that's your base cost, there's no way to reduce it, and you're just trying to reduce the extra "tax" that gets put on for gravity losses. Once your local TWR gets above, say, 5 or so, the gravity losses are quite small if you're doing a good suicide burn; boosting your TWR from 5 up to 10, for example, would give you only a very small gain, while requiring double the engines.

So in practice, keep the TWR fairly high and you're good to go.

[davidpsummers]

I'm not convinced that the Oberth effect will be a major effect here. For any low orbit, the difference in orbital speed won't be that high. A suicide burn has the most efficient Oberth effect for vertical velocity. Yes, if you angle down, you can combine killing orbital velocity and speed from falling in one burn, but that is part of the difference of going down the hypotenuse of the triangle rathter than the legs. (I find the increased accuracy in landing worth the cost).

[numerobis]

Where is that video from Kosmo-not? It was very instructive as to what the ideal landing really implies: a low-TWR spacecraft (because engines are heavy), then, as termed above, a reverse gravity burn -- at full throttle.

It's scary when topography starts to matter: all is well, all is well, oh .... that's a hill!

tavert had a formal proof that it was better than any other recommended approach, but I don't think we ever got a formal proof it was optimal.

The video is here:

Edited November 12, 2015 by numerobis