Why are suicide burns better?

[airelibre]

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?

Edited September 26, 2014 by airelibre

[TechnicalK3rbal]

Suicide burns are more efficient since you only do one burn, which slows your total velocity, rather than a constant burn, which is "trying" to slow you down and still has to fight the force of the gravity.

Edit: I'm not sure of the actual maths involved, though.

[wasmic]

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.

[dzikakulka]

Okay so let's say you're descending at some speed and some height where 2s suicide burn with TWR=3 at 100m will bring you down safely. Now, what would happen wif you did the same at 1000m instead? Well, you would gain some positive vertical speed, lose it and crash. Now what would happen wif you would prolong that burn, doing it with TWR=1 over 6s? You would crash, retaining whatever speed you had at the beginning of the burn.

All these cases we did burn exact same amount of fuel with same parameters. But it didn't work at all.

We want to minimese the time spent on descent so gravity force will have minimised effect.

[airelibre]

Suicide burns are more efficient since you only do one burn, which slows your total velocity, rather than a constant burn, which is "trying" to slow you down and still has to fight the force of the gravity.

Edit: I'm not sure of the actual maths involved, though.

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.

Hmm, I'm still not sure. I think the maths would actually be useful for me, if someone can provide a simple example.

[airelibre]

Okay so let's say you're descending at some speed and some height where 2s suicide burn with TWR=3 at 100m will bring you down safely. Now, what would happen wif you did the same at 1000m instead? Well, you would gain some positive vertical speed, lose it and crash. Now what would happen wif you would prolong that burn, doing it with TWR=1 over 6s? You would crash, retaining whatever speed you had at the beginning of the burn.

All these cases we did burn exact same amount of fuel with same parameters. But it didn't work at all.

We want to minimese the time spent on descent so gravity force will have minimised effect.

Why would you crash? I need actual numbers to be convinced, otherwise I could just say that in the second case you also would come down safely.

Also, why do you write wif instead of if? I thought it was a mistake the first time but you wrote it twice so I guess it's on purpose.

[capi3101]

Well, the maths are tricky. You have a ship with a certain initial mass (which will change as you burn fuel), a rocket with a certain maximum thrust level (i.e. the amount of force it can provide for braking) and the gravitational force of the body you're landing on (which changes with altitude). The easiest case scenario would assume gravity and your ship's mass are constant; then it'd just be a force-balance and kinematic equation problem using Newton's second law (F=ma) to determine the rate of deceleration (engine thrust minus gravitational pull). Because the mass of your ship and gravity are not constant, you also need to know the rate at which your craft is losing mass and the rate at which gravity is increasing. It becomes a third-order differential equation. Tricky stuff.

Now, if you want to go through it, we can. I'll need some sample parameters for an example case.

[Jovus]

wasmic has the right of it; total acceleration due to a constant force (in this case gravity) is a function of time. Therefore, the longer you spend falling into a gravity well, the more dV (that is, acceleration) you need to counteract it.

Take Earth's gravity at sea level for a very simple example. If I spend 1 second subjected to 10m/s^2 of acceleration, I need 10m/s dV to counteract its effects. If I spend 2 seconds subjected to 10m/s^2 acceleration, I need 20m/s dV to counteract its effects. When you burn too early, you slow yourself downn, meaning you spend more time subjected to the force without the countervailing force of the body's surface.

ETA: It has nothing to do with TWR or the delta mass of your ship at all.

[dzikakulka]

Why would you crash? I need actual numbers to be convinced, otherwise I could just say that in the second case you also would come down safely.

Also, why do you write wif instead of if? I thought it was a mistake the first time but you wrote it twice so I guess it's on purpose.

It was a typo, I don't know how I've did it twice

Okay, second case, TWR=1 so we burn with force to equal to gravity force but in the opposite direction. Result - forces acting on our ship "cancel out". Game follows Newtonian laws so without any force our body keeps moving with constant velocity undisturbed. Basically there is no change of velocity (there would be some due to gravity force if we did not burn). Then it would land safely ONLY if our velocity was considered safe before burn. That's why in practical situation we would crash.

Feel free to ask if you do not undestand something or to point out some stupid things if I did any

[mhoram]

I recommend you to have a look at the following thread of tavert.

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

Actually there are more efficient ways to land on airless bodies than suicide burn.

[dzikakulka]

Actually there are more efficient ways to land on airless bodies than suicide burn.

Can you please post an example? AFAIK suicide burn is basically closest practical approach at having "ideal impulse" that would get us to safe speed on ->zero altitude over ->zero seconds which would minimise time spent on descent to lowest amount possible (without burning in any other direction). Crew splat and stucture shatter with infinite G not taken into account

Edited September 26, 2014 by dzikakulka

[KerikBalm]

Lets suppose you just hover at 10 meters, for 10 minutes... how much fuel would that take.

"But thats different!" you'll say "I'm talking about a descent"

-Ok, Lets say you're now at 10 meters, descending at .01 m/s... It will take you 1,000 seconds to reach the surface - approximately 1600 m/s of dV on the Mun.

"but I'm talking about deceleration! not the final descent speed"

-Ok, but you acknowledge a controlled descent at a slower speed will use more fuel than a controlled descent at a higher speed

Lets say you have to decelerate 1,000 m/s in the suicide burn. If you do this in 10 seconds, or in 100 seconds, what is the difference?

Lets assume a 10 m/s^2 constant deceleration rate (as you get lighter, unless you throttle back, this won't be true, but the math is a lot simpler if it is).

Lets sum up in 10 second increments (not doing an integral to take it to zero, just to make it simpler)

T=10 seconds of the slow burn, goinging 900 m/s now- you essentially suffer the losses of descending at 900 m/s instead of 1,000 for 10 seconds

+

losses of 10 seconds of an 800 m/s descent

+

losses of 10 seconds of an 700 m/s descent

+

...

+

losses of 10 seconds of an 200 m/s descent

+

losses of 10 seconds of an 100 m/s descent

Not coincidentally, I believe these losses will add up to the time spend burning * the gravity of the body * sin of the angle of the burn (which will likely change throughout the burn, especially with a slower burn)

[mhoram]

The link in my previous post contains a description of this method and there is also a video available demonstrating this landing method.

The idea is basically to keep the altitude until the horizontal velocity is zeroed and afterwards to descend to the landing.

[airelibre]

Thanks all, all the answers are starting to "click" in my head, I just think I need more time for it to sink in, and maybe some experiments in KSP to see it happen before my eyes.

[Kryxal]

Incidentally, you can put it into energy terms in the same fashion as the Oberth Effect ... you are at a certain energy state at the beginning of your descent, and a lower energy state on landing, the faster you're moving when you make your burn, the greater the change in energy for the same change in delta-v.

[Tw1]

Can you please post an example?

Maybe lithobraking? Save a few hundred m/s by making use of that tough, bouncy lower stage.

[MarvinKitFox]

, 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?

Simply take the difference to an extreme, and compare.

Make your slow burn take one year to get to the surface.

Do you still think it takes the same fuel?

[xtoro]

Try doing a suicide burn on Tylo and then tell me it's effective...

Look at the video on post number #2 of the thread that Mhoram posted http://forum.kerbalspaceprogram.com/threads/39812-Landing-and-Takeoff-Delta-V-vs-TWR-and-specific-impulse?p=510523&viewfull=1#post510523

This is how I land except on bodies with low gravity...

Edited September 26, 2014 by xtoro

[dzikakulka]

@xtoro

Well executed suicide burn should be more effective, Jovus explained that in the most clear way I think in this thread, #8 post.

That thread is informative and interesting but few things: it specifically mentiones that it is efficient with low TWR, when suicide burn is NOT possible. Second thing, it implies landing from 0 altitude circular orbit. To do a suicide burn like we discussed in the thread, you set yourself on (ideally zero horizontal velocity) impact trajectory first, then we talk about landing.

[metaphor]

I thought we were assuming that the initial speed is constant and straight up/down, and the spacecraft does not have any sideways velocity (so you're only moving up/down). In that case a suicide burn (burning at the last second at full thrust so that your velocity reaches zero exactly as your position reaches zero) is always the most efficient.

For example, let's say you start out stationary 1 km up from a body that has a gravitational acceleration of 1 m/s^2. If you don't do anything, it would take you 44.7 seconds to hit the surface, and you would be moving at 44.7 m/s when you hit it. Let's say your spacecraft can accelerate at 4 m/s^2 at full throttle. Then while it is burning, its effective acceleration with respect to the ground is 3 m/s^2 upwards (since the acceleration due to gravity is 1 m/s^2). So starting at 250 m up, when its velocity is 38.7 m/s, it would burn full throttle for 12.9 seconds, ending up on the ground at 0 m/s speed. The total delta-v used by the engines is 12.9 s * 4 m/s^2 = 51.6 m/s.

Now let's say we use the same spacecraft but instead of burning at full throttle we burn at half throttle, or 2 m/s^2. Then while it is burning, its effective acceleration with respect to the ground is 1 m/s^2 upwards. So it would have to start at 500 m up, at a velocity of 31.6 m/s, and burn at half throttle for 31.6 seconds, ending up on the ground at 0 m/s speed. The total delta-v used by the engines in this case is 31.6 s * 2 m/s^2 = 63.2 m/s.

If the spacecraft has infinitely high TWR, then it can stop right when it hits the surface, with a delta-v of 44.7 m/s. If the spacecraft burns at 1/4 throttle or less, that means it has a TWR of less than 1 on this body and it will not be able to slow down. So the delta-v used has a minimum of 44.7 m/s (with infinitely high TWR), and goes up to infinity as you go down in throttle to 1 m/s^2 of acceleration.

Equations used here: kinematic equations

You can also think of a landing as being the same as an ascent going backwards in time. It's less efficient going lower than full throttle during an ascent from an airless body.

[capi3101]

Equations used here: kinematic equations

The kinematic equations can't be directly applied, though; as I said in my previous post, the ship's mass is not constant (which by Newton's Second Law has the implication that the braking acceleration will increase with respect to time) and the gravitational acceleration is also not constant (by definition, g = GM/R^2, where GM is the body's gravitational parameter and R is the distance to the center of mass - including altitude; since R decreases as you descend and it's inversely proprtional to g, g will increase with respect to time). In both of your example cases, the end effect of these factors would be that you wouldn't be on the ground when you hit zero m/s; you'd be close, but not quite there (the rate at which the acceleration due to gravity increases should be less than the rate at which the deceleration due to thrust increases, though I'd have to run a proof mathematically to be sure).

[xtoro]

@xtoro

Well executed suicide burn should be more effective, Jovus explained that in the most clear way I think in this thread, #8 post.

That thread is informative and interesting but few things: it specifically mentiones that it is efficient with low TWR, when suicide burn is NOT possible. Second thing, it implies landing from 0 altitude circular orbit. To do a suicide burn like we discussed in the thread, you set yourself on (ideally zero horizontal velocity) impact trajectory first, then we talk about landing.

Yes, I agree. If we're talking about the landing and ONLY the landing, then I suppose a suicide burn is best.

However, when thought of as a whole, when considering the design of the craft, then it's not effective. When higher TWR is required, building up the ship just so that you're able to do a suicide burn is ineffective in respect to craft design when you can keep that lower TWR craft, and adjust your landing. I mentioned earlier the Tylo landing.... I can land a much smaller lighter ship by flying it the way it's shown in the video I linked earlier. To do a suicide burn down to Tylo, I need a larger craft with more TWR and staging just to land.

With my craft, it's 2-stage, the descent stage is 2700 Delta-V with a TWR of 1.4. After landing, I still have 190 Delta-V left in the descent stage for ascent plus 1730 in my ascent stage. For suicide burns, I had much larger craft which I had to cart around with my mothership, but it was an ineffective craft for the mission.

Don't forget, wherever you're landing your ship, you have to first get it there and spend some fuel.

Edited September 26, 2014 by xtoro

[dzikakulka]

You're right, I'm just used to pick landing spots and then pinpoint them which pretty much requires negating all horizontal movement. Which is indeed a waste - that's partially why way you've shown is better, you get 100m/s horiontal and vertical when burning 144m/s 45 degrees between them, my way would take 200m/s to do the same. So every thing has it's applications

[Mr Shifty]

I thought we were assuming that the initial speed is constant and straight up/down, and the spacecraft does not have any sideways velocity (so you're only moving up/down). In that case a suicide burn (burning at the last second at full thrust so that your velocity reaches zero exactly as your position reaches zero) is always the most efficient.

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.

[metaphor]

The kinematic equations can't be directly applied, though; as I said in my previous post, the ship's mass is not constant (which by Newton's Second Law has the implication that the braking acceleration will increase with respect to time) and the gravitational acceleration is also not constant (by definition, g = GM/R^2, where GM is the body's gravitational parameter and R is the distance to the center of mass - including altitude; since R decreases as you descend and it's inversely proprtional to g, g will increase with respect to time). In both of your example cases, the end effect of these factors would be that you wouldn't be on the ground when you hit zero m/s; you'd be close, but not quite there (the rate at which the acceleration due to gravity increases should be less than the rate at which the deceleration due to thrust increases, though I'd have to run a proof mathematically to be sure).

Yes, I was assuming the ship was in a constant gravitational mass and didn't lose a significant mass, to make it simpler to get the point across. If you're landing on the Mun with an Isp of 350s, the changing gravitational field and the changing ship mass are negligible when calculating delta-v while falling from 1 km up. The ship losing mass over time doesn't actually matter at all, since that's equivalent to raising the thrust in a constant-mass ship. If you want, you can keep the ship at the same acceleration as it loses mass by continuously lowering the throttle.

My point was that the most efficient landing is one where you use the highest thrust possible at the latest time possible. That is true whether or not the mass of the ship decreases (which is equivalent to increasing your maximum thrust) or the gravitational field increases (which means you lose a higher proportion of your effective acceleration at lower thrusts).