Why are suicide burns better?

[metaphor]

Well yeah, but you're almost never operating in this regime. You're almost always coming down from orbit. Suicide burns are nearly the same efficiency as a constant altitude burn for high TWR vehicles, but for low TWR, constant altitude burns can't be beat. Again, reverse it; when you're ascending from the ground to orbit, you get the best efficiency by getting just high enough to miss the terrain, then burn sideways with just enough upward angle to keep your altitude from dropping. For descent from orbit, most of your burn is going to go toward decelerating, not to braking your descent. A suicide burn necessarily takes longer for deceleration than a constant altitude burn (because you're devoting the max engine vector to deceleration for the constant altitude burn), so it's less efficient because you fight gravity longer. That's what tavert's charts showed on the post linked earlier, and what kosmo-not's video shows. Again, the difference is negligible for high-TWR ships, but is quite significant for low-TWR landers.

I guess it depends what you mean by suicide burn then, and what the initial conditions are. I would say a constant altitude burn is what I described as a suicide burn, since you're burning at the last second at the highest available thrust. You could burn in a different direction, but that's a harder thing to compare since it's a 2D problem instead of a 1D problem.

[xtoro]

I guess it depends what you mean by suicide burn then, and what the initial conditions are. I would say a constant altitude burn is what I described as a suicide burn, since you're burning at the last second at the highest available thrust. You could burn in a different direction, but that's a harder thing to compare since it's a 2D problem instead of a 1D problem.

I think "suicide burn" is considered just falling from orbit and burning retrograde at the last second regardless of your horizontal speed...

[metaphor]

I've had a look around and still haven't found an explanation good enough to satisfy my inquisitive mind! I can't get past the fact that from the same height and starting speed, you convert the same amount of gpe to kinetic energy by the time you reach the surface, so to cancel out that gained ke you have to burn retrograde, and surely it doesn't matter if you do that slowly or in one quick suicide burn close to the surface, it will be the same delta v and therefore the same amount of fuel expended.

What is wrong with my reasoning?

You can't really equate energy that way, since you have to take into account the gravitational potential energy and the kinetic energy of the fuel expended as well, instead of just the ship.

Basically when you burn downwards the gravity field of the planet is cancelling out some of your ship's acceleration, and if you burn slower you lose a higher percentage of your acceleration to the planet's gravity, which means more delta-v used to achieve the same result.

[metaphor]

I think "suicide burn" is considered just falling from orbit and burning retrograde at the last second regardless of your horizontal speed...

So starting with an initial condition of your trajectory already intersecting the surface? In that case you can't really do a constant-altitude landing since you already have a large downward component to your velocity. But still, what you want to do is burn slightly* above retrograde at the last second with the highest thrust available so that you reach both 0 horizontal velocity and zero vertical velocity at 0 altitude. (*How "slightly" depending on your TWR, angle of incidence, velocity, etc.)

[Mr Shifty]

I think "suicide burn" is considered just falling from orbit and burning retrograde at the last second regardless of your horizontal speed...

* Suicide burn is pointing retrograde and burning at maximum thrust so that you reach 0 velocity just when you touch down. For low TWR ships, this could be a long burn, not a 'last second' thing. For instance, orbital velocity on the Mun is about 550 m/s. If you've got a starting Mun-relative TWR of 1.1, it will take 4-5 minutes to slow down from that velocity.

* Constant altitude is doing a shallow de-orbit burn, coasting down to near the surface (but high enough to miss the terrain), then burning with max TWR at whatever angle holds your vertical velocity at 0. Then once you're slow enough that your pitch passes about 45°, do a normal descent and landing.

[The_Rocketeer]

A layman's answer:

You're walking your Alsatian/German Shepherd who is very strong and pulls against the lead/leash as you walk. The dog pulling represents the force of gravity.

Gradual braking:

You hold the lead/leash and try to slow down the dog. By the end of a 20 minute walk you're exhausted, but you're still going slow enough to stay on your feet - that is, unless you get too tired, at which point the dog pulls you off your feet, breaks your grip and disappears across the park.

Suicide burn:

The dog pulls the lead/leash out of your hands as soon as you close the front door. 2 minutes later, you see the dog running back up the road chasing a cat. You jump out and grab the trailing lead/leash and with a show of brute force stop the dog. It was very exciting and took a bit of effort, but you stopped the dog and you're not completely worn out.

Make sense?

[Jovus]

Just to throw a wrench in the works a bit: If we're talking about the most efficient method (in this simple one-body gravity system we have) then suicide burns are also the best method because you further maximize the gravity turn effect.

And to address the earlier discussion, yes, a 45 degree line is a better tangent approximation to a landing ellipse, which is why it conserves fuel over 2 perpendicular burns. But you know what's an even better approximation? An ellipse. Any landing burn, suicide or otherwise, is going to use much less dV if you point it retrograde at all times.

[dzikakulka]

@Jovus

Now I'm curious about maths on this and when would killing horizontal velocity + suicide burn on vertical fall would be more efficient than continuous retrograde burn with fixed throttle adjusted to finish at landing. I'll probably duct tape some kOS scripts and try in practice cause it's pretty messed up.

And well, obviously getting your landing site precise would be hard with retro burn but lets just disregard that.

[John FX]

I land like this.

It`s really peaceful...

Also quite efficient if you have a low TWR on an airless body. Your craft is supported through most of the burn by your orbital velocity giving you fuel savings.

Of course on bodies with air I land like this...

[bandi94]

The simples answer is :

- if you start burning at 10km the gravity will take effect on you until to 0km.

- Your speed without burn will not ecxed 600m/s

So if you slow down to 100m/s at 5km you still have 5km where the gravity will speed you up again up to 600m/s , but if you do a suicid burn at 1k you will slow down to <100m/s until you hit the ground this way you will fight with gravity only 1 time , and not 4-5 times from 10km to 0 km.

[John FX]

The simples answer is :

- if you start burning at 10km the gravity will take effect on you until to 0km.

- Your speed without burn will not ecxed 600m/s

So if you slow down to 100m/s at 5km you still have 5km where the gravity will speed you up again up to 600m/s , but if you do a suicid burn at 1k you will slow down to <100m/s until you hit the ground this way you will fight with gravity only 1 time , and not 4-5 times from 10km to 0 km.

This suggests that even with high TWR craft the most efficient descent is to set your Pe as low as possible to reduce the fuel needed to get to the `drop point` for the suicide burn then apply just enough downward thrust (by adjusting the attitude of your craft) to counter the rising effect from gravity as your orbital speed decreases. If done perfectly then there are literally just a few meters to drop after scrubbing the orbital speed.

[StrandedonEarth]

Think about it this way: if you burn slowly, your trip down will take a longer time to complete. If it takes a longer time to complete, gravity has a longer time to exert its force on you. You'll want to fall for as short a time as possible, and you'll achieve that by doing a suicide burn.

I think this summed it up best. You want the math to back it up, you say:

Change in velocity is governed by dV=at (acceleration times time). In this case acceleration is provided by (the local) gravity (g), so dV=gt. Slowing down sooner means your descent takes longer, so gravity has longer to act, as already stated. So if t (descent time) is, for instance, doubled, then the dV accrued from gravity also doubles, meaning your ship needs to expend twice the dV to counteract gravitational acceleration.

I was also unsure why a suicide burn would be better until I read wasmic's post, at which point I had to smack my forehead and think "Of course, duh! that makes total sense"

[lincourtl]

I think this summed it up best. You want the math to back it up, you say:

Change in velocity is governed by dV=at (acceleration times time). In this case acceleration is provided by (the local) gravity (g), so dV=gt. Slowing down sooner means your descent takes longer, so gravity has longer to act, as already stated. So if t (descent time) is, for instance, doubled, then the dV accrued from gravity also doubles, meaning your ship needs to expend twice the dV to counteract gravitational acceleration.

I was also unsure why a suicide burn would be better until I read wasmic's post, at which point I had to smack my forehead and think "Of course, duh! that makes total sense"

That's true if your spacecraft is falling straight down, but coming from orbit you also have a horizontal velocity component which isn't accounted for in that analysis. If you think of it as a right triangle, the two legs represent your horizontal and vertical velocity components, and the sum of the two legs is always greater than the hypotenuse, except for special cases.

[Kryxal]

That's nit-picking under the circumstances, and doesn't counter the main point of the example, that it costs MORE delta-v from a longer drop.

[lincourtl]

That's nit-picking under the circumstances, and doesn't counter the main point of the example, that it costs MORE delta-v from a longer drop.

Then why do you think the Apollo missions used a long braking maneuver (mhoram's video is a good replica) to minimize fuel use?

On edit: Apollo 11 Lunar Descent profile from NASA Technical Memorandum TM X-58040 Apollo Lunar Descent and Ascent Trajectories; March 1970.

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Edited September 28, 2014 by lincourtl

[Mr Shifty]

That's nit-picking under the circumstances, and doesn't counter the main point of the example, that it costs MORE delta-v from a longer drop.

It's not nitpicking. Argh. A suicide burn -- that is a burn where you point retrograde for the whole burn -- is not the most efficient drop precisely because it's not reducing your horizontal velocity as fast as a burn that focuses on reducing horizontal velocity as much as possible. This has been empirically and mathematically shown many times.

[metaphor]

I think there's two different things being discussed here. How to most efficiently get down from a circular orbit (which is a retroburn so that Pe is at 0 followed by constant-altitude max-thrust burn at Pe), or how to most efficiently land once you're already on a trajectory intersecting the surface (which is a suicide burn, burning at max thrust and slightly above retrograde so that both your horizontal and vertical velocity reach 0 when your height reaches 0).

[Kryxal]

I think you're right, two different things are getting mixed up. I'm perfectly happy doing the Apollo-style landing if I get the chance (except my TWR is always serious overkill for it) but StrandedOnEarth's statement is accurate at least in terms of greater/less than, if not actual values. Applying your thrust as late as possible conserves delta-v, but the Apollo landers were constructed with a low TWR, so for them that IS pretty much as late as they can thrust.

It's actually analogous to the discussion about launches, I think (and can be modeled as such if you hack the engine so it produces fuel): straight up and straight over vs 45 degrees and circularize vs lowest possible takeoff and get up to speed. The last is most efficient, but the second isn't so bad and a lot less complicated (more so on the landings than takeoffs).

[Dieter_G]

I've had a look around and still haven't found an explanation good enough to satisfy my inquisitive mind! I can't get past the fact that from the same height and starting speed, you convert the same amount of gpe to kinetic energy by the time you reach the surface, so to cancel out that gained ke you have to burn retrograde, and surely it doesn't matter if you do that slowly or in one quick suicide burn close to the surface, it will be the same delta v and therefore the same amount of fuel expended.

What is wrong with my reasoning?

I must admit, I didn't read through all of this thread, so maybe someone already explained it...

The easiest way of thinking about it (for me) is as follows :

Your engine generates a certain ammount of thrust. When burning retrograde your effective thrust (that is going to slow down your ship) however is :

EngineThrust - GravitationalForce. So "gravity" is always subtracted from your engines thrust. Only what remains will go into slowing down your ship. So it will be a good idea to get as much of that remaining thrust as possible.

In other words : the thrust needed to counter gravity is wasted. E.g. hovering 10m above the surface for 1 minute yields zero physical work, but needs a lot of fuel anyway.

So, by suicide burning, one shortens the time where your engine must fight gravity by as much as possible, thats why it saves fuel.

Your consideration :

so to cancel out that gained ke you have to burn retrograde, and surely it doesn't matter if you do that slowly or in one quick suicide burn...

is also correct, if applied to the "effective thrust" (after gravity is subtracted).

[EdFred]

I always defined a suicide burn as "waiting until the last possible moment to burn at maximum thrust, so your velocity reduces 0 just as you touch down." Because if you don't do it correctly, you die. Hence the term suicide burn. I've been using that term since before a lot of you were born. It is not (necessarily) a continuous retrograde burn starting from apoapsis. Maybe you are in a 100% vertical descent, maybe you aren't.

Edited September 29, 2014 by EdFred

[lincourtl]

So, by suicide burning, one shortens the time where your engine must fight gravity by as much as possible, thats why it saves fuel.

So how do you explain tavert's results (and math -- see the derivation PDF linked at end of his OP)?

http://forum.kerbalspaceprogram.com/threads/39812-Landing-and-Takeoff-Delta-V-vs-TWR-and-specific-impulse

That's not rhetorical. I'm actually asking, because it seems to me the constant altitude descent actually minimizes fighting gravity. In fact, it uses the gravitational curve to aid in the descent; i.e. a true gravity turn.

[Dieter_G]

@lincourtl

What I said is simplified a lot. I didn't account for different trajetories. I'm just comparing two cases, both with e.g. a pure vertikal trajectory, where you do a "suicide burn" in one case and a [slow, gradual burn] in the other.

Of course, theoretically the most ideal trajectory for landing should be the same than the [ideal trajectory for ascending into an orbit], just reverse.

The [ideal trajectory for ascending into an orbit] involves gradually turning your thrust-vector from [pure vertical] to [pure horizontal]. At the end of this process the centrifugal forces compensate gravity, so you don't have to "fight" it anymore. So, basically, at the beginning of such a trajectory you are affected by 100% of gravity. Then, by building up horizontal velocity, gravity starts to become gradually offset by the centrifugal forces. Thus you only need to "fight" like 100% ... 90% ... 80% ...... 10% ... 0% of gravity over time. I hope you know what I mean...

Thats what makes such a trajectory the most economical one. But that is a different topic, and of course you are right. As I said, I just talked about a pure vertical path, suicide burn vs. nromal burn.

Also, when I said that "gravity is subtracted from your engines thrust", thats somewhat misleading. In fact, it doesn't matter if your engine is activated or shut down. You need to shorten the time where gravity acts upon your ship (without accelerating towards the planet/moon of course) : Fly through the gravitational field as fast a nature provides (not artificially accelerating toward the plantet/moon!), then do a hard brake at the very, very end.

if you break to early (so your ships stops e.g. 200m above the surface and you have to do an additional burn shortly after) you'll also need more fuel, because for every second you are in the gravitational field (e.g.) 10m/s of vertical velocity will be added to your "bill"...

Damn... could we switch over to german language for a second ?

Its really hard to explain it in a clean and proper manner.

[xtoro]

To those who are convinced that a suicide burn is absolutely the most efficient way to land; Perhaps you should apply to work at NASA, because apparently you found the best and most efficient way to land, and none of the thousands of geniuses who work at NASA have ever figured it out. I guess the people at NASA really don't know what they're doing and you should call them right now! Their phone number is (202) 358-0001. Tell them you've figured it out and that they're wrong! Oh, and tell them that all the math that proves it wrong, is also wrong, because, you know, maths, pffft.....

[magnemoe]

To those who are convinced that a suicide burn is absolutely the most efficient way to land; Perhaps you should apply to work at NASA, because apparently you found the best and most efficient way to land, and none of the thousands of geniuses who work at NASA have ever figured it out. I guess the people at NASA really don't know what they're doing and you should call them right now! Their phone number is (202) 358-0001. Tell them you've figured it out and that they're wrong! Oh, and tell them that all the math that proves it wrong, is also wrong, because, you know, maths, pffft.....

Suicide burns has some downsides, the obvious one is safety, you have very small margins and very limited abort options.

You will also require an more powerful engine who would be heavier and more bulky.

In atmosphere however a suicide burn is the best option as your speed is limited to terminal speed who goes down as you decent, the manned dragon pod will use something like an suicide burn however not at full trust to have an margin for error.

[xtoro]

Suicide burns has some downsides, the obvious one is safety, you have very small margins and very limited abort options.

You will also require an more powerful engine who would be heavier and more bulky.

In atmosphere however a suicide burn is the best option as your speed is limited to terminal speed who goes down as you decent, the manned dragon pod will use something like an suicide burn however not at full trust to have an margin for error.

Yeah but in an atmosphere doesn't really count here because the atmosphere slows you down to anywhere from 5-20% your original sub-orbital trajectory speed, so you're not using ANY Delta-V for that. And also, all the way to the ground, you have the atmosphere constantly slowing you down on top of the thrust from your engines. Also, in atmo, chutes are used, so retro burns aren't really an issue here.