Why are suicide burns better?

[Hannu]

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.

You are 50 years late. NASA have found it out yet. Landing profile of unmanned Surveyor Moon probes was very near of the suicide burn. They had a SRB (!) with dv of over 2 km/s and only the last 100 m/s was decelerated with adjustable engines and closed control loop. Unfortunately, some anti-cheat purist has disabled functionality of F5 and F9 -buttons of computers in Nasa's command centers, so they chose less risky strategy for manned Apollo landers.

[m4v]

You are 50 years late. NASA have found it out yet. Landing profile of unmanned Surveyor Moon probes was very near of the suicide burn. They had a SRB (!) with dv of over 2 km/s and only the last 100 m/s was decelerated with adjustable engines and closed control loop. Unfortunately, some anti-cheat purist has disabled functionality of F5 and F9 -buttons of computers in Nasa's command centers, so they chose less risky strategy for manned Apollo landers.

Doesn't prove anything, efficiency isn't the only factor or constraint when designing a project, especially in real life. They could have used that profile because of numerous different reasons.

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

What is the basis of that assumption? I have heard it a lot of times but never the justification. Remember that is impossible to do a "reverse ascend" since you can't reverse how your vessel loses mass: during take off you lose mass as you go up, but when landing the reverse (gaining mass as you go down) can't happen. Edited September 29, 2014 by m4v

[Red Iron Crown]

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

Why do you need a more powerful engine? With a smaller engine doesn't the suicide burn just start earlier?

[xtoro]

Why do you need a more powerful engine? With a smaller engine doesn't the suicide burn just start earlier?

Not if it's too small and the TWR is too low.... Take Tylo as the common example. Use a lander with Rockomax 45's and a low-ish TWR. From a stable X-by-X circular orbit, you can burn retrograde all the way down until you crash into the surface. You need a more powerful engine, or more engines, to do a burn like this. Or, don't waste dv on the horizontal, spend more on the vertical speed to keep you off the ground, and your horizontal will bleed away. You need an engine(s) large enough to fight the gravity and win.

Edited September 29, 2014 by xtoro

[Red Iron Crown]

Not if it's too small and the TWR is too low.... Take Tylo as the common example. Use a lander with Rockomax 45's and a low-ish TWR. From a stable X-by-X circular orbit, you can burn retrograde all the way down until you crash into the surface. You need a more powerful engine, or more engines, to do a burn like this. Or, don't waste dv on the horizontal, spend more on the vertical speed to keep you off the ground, and your horizontal will bleed away. You need an engine(s) large enough to fight the gravity and win.

Such a vessel would not be able to do a controlled landing no matter the technique. A suicide burn should be possible with any design with >1 TWR, the only difference would be how early the burn must be started.

[maccollo]

What is the basis of that assumption? I have heard it a lot of times but never the justification. Remember that is impossible to do a "reverse ascend" since you can't reverse how your vessel loses mass: during take off you lose mass as you go up, but when landing the reverse (gaining mass as you go down) can't happen.

The basis for that assumption is that Newtonian physics is reversible, so the same concepts apply.

Actually decelerating is simpler, because you have the benefit of having a TWR that's increasing while the effective acceleration towards the ground is also increasing, so it's easier to maintain a constant angle of attack, which allows you to aim closer overall to the horizon, which means to get the most vertical acceleration possible for the lowest possible loss horizontal acceleration.

Now as to why you should come in as horizontal as possible (within reason) there are multiple ways of looking at it. Let's say we're trying to land on a body and we're fighting a gravitational acceleration of 5 m/s. Our ship accelerated at 10 m/s. If we come in vertically our effective acceleration is our acceleration-gravity, so 5 m/s. However, if you come in horizontally we can aim 30 degrees above the horizon. Then our vertical acceleration is sin30*10 m/s, which is 5 m/s. The acceleration due to gravity was also 5 m/s, so that cancels out. Our horizontal acceleration however is cos30*10 m/s, or 8.66 m/s. So in this example coming on horizontally allows us to boost our effective acceleration by 73%.

Edited September 29, 2014 by maccollo

[metaphor]

I don't really understand the disagreement here. A constant-altitude burn is a suicide burn, at least in the way me and EdFred defined it (burning at max thrust at the last possible moment so that speed reaches 0 right as altitude reaches 0). It's just that your thrust to weight ratio dictates how high above retrograde you have to burn. And it's a special case of a suicide burn where your trajectory is tangent to the body's surface. If your trajectory intersects the surface at an angle, it's no longer a constant-altitude burn since you start out with a vertical component of velocity, but it's still a suicide burn. Both of these are the most delta-v-efficient way to land when starting with their respective initial conditions. (If you start out in high orbit though, the most efficient way to get down is to get your trajectory tangent to the body's surface and then perform the suicide/constant-altitude burn.)

It's not always done this way in practice since there's other considerations besides efficiency (can't compute the exact trajectory beforehand, safety issues, abort options, terrain obstacles, etc).

[Rusty6899]

Is your gravity loss not equal to the second integral of dg/dt? Or something like that? So a lower time will result in a lower gravity loss and therefore a lower delta-V expenditure. The best way to minimize the time taken is to have as short a burn as possible at the end of your descent.

[dzikakulka]

Is your gravity loss not equal to the second integral of dg/dt? Or something like that? So a lower time will result in a lower gravity loss and therefore a lower delta-V expenditure. The best way to minimize the time taken is to have as short a burn as possible at the end of your descent.

It's true but it is only valid when you're already on collision trajectory. Cancelling horizontal velocity first makes constant altitude burn (sometimes) better but offers close to zero landing site precision and very long burn that need to be executed well. Or you could combine methods and come up with what you want.

[xtoro]

Is your gravity loss not equal to the second integral of dg/dt? Or something like that? So a lower time will result in a lower gravity loss and therefore a lower delta-V expenditure. The best way to minimize the time taken is to have as short a burn as possible at the end of your descent.

A short burn at full thrust is not necessarily using less delta v than a longer burn at lower thrust...

[toril]

A short burn at full thrust is not necessarily using less delta v than a longer burn at lower thrust...

in a gravity well assuming the flight ends at the end of the burn yes it does. also assuming you don't want to leave a crater at your landing site lol. although you would probably be correct on minmus or any other similarly low gravity rocks.

[CavemanNinja]

(deleted because redundancy)

Edited September 29, 2014 by CavemanNinja