Retroburn for landing

[LostOblivion]

Hi,

If putting a dummy maneuver node just over the place I want to land, say, on the Mun, and given an initial orbital velocity in m/s and retrograde acceleration in m/s^2 (taken from KER), what time to maneuver should I start my burn in order to kill my horizontal speed just over the maneuver node?

Not so obvious is that this is different from the time it takes to decelerate to a stop, that is t=v/a.

Assume circular orbit and constant acceleration.

Edited August 13, 2015 by LostOblivion

[selfish_meme]

You should be able to see time to 0m/s, that and time to node should be close but not exact, your time to 0m/s will be affected by the muns gravity, so you need to add in some pork. Also you will stop faster than the predicted node, bringing the impact point closer than the node. Good luck and if you can figure it out please post an explanation video for dummies like me who end up eyeballing it and making really dV wasteful approaches.

[Sharpy]

A bit complex, strongly depending on TWR. In gravity-free environment, that's exactly half the burn time away from the node. With gravity, substract the gravitational acceleration from your deceleration.

KER provides "distance to suicide burn", but it neglects horizontal speed, so unless you're coming in very steeply that will actually be suicide burn. Besides, you'll never get the node exactly on the surface (especially that as your trajectory changes with horizontal speed, altitude of the landing spot changes with terrain.) Personally, I perform at least three burns, down to 100m/s around 1000m, 10m/s at 100m and 1m/s at 10m.

[psychopoak]

just watch the master doing exactly that :

there is all the key to do what you plan to do, coming with the calculation you will have to do.

[kreutzkevic]

just watch the master doing exactly that

the best part

[match]

In map mode with focus on your ship, click on the info button (the "I" button). Listed among other details is burn time to 0 m/s.

[LostOblivion]

Still no one has answered the question, though. :-)

Burn time to stop is easy.

[p1t1o]

Are you looking to be at v(total)=0m/s over the node or v(horizontal)=0m/s over the node?

The maths gets complicated if you are burning retrograde w.r.t your velocity vector, but if you are just zeroing your horizontal component (Thrusting at a constant pitch angle rather than "retrograde", assuming that your burn is not so long as to cover a significant arc of orbit), it should still be the t=v/a answer right? Or am I missing something?

Thats what I do anyway, if landing in vacuum, a horizontal burn to kill horizontal velocity, and a vertical drop to the landing site, I don't think there is any dV wasted that way, is there?

[strigon]

Thats what I do anyway, if landing in vacuum, a horizontal burn to kill horizontal velocity, and a vertical drop to the landing site, I don't think there is any dV wasted that way, is there?

Sorry, but you absolutely are; you know how a gravity turn is the most efficient way into orbit? (if not, there are plenty here who can explain how to do it)

Well, the same maneuver but in reverse is more or less the most efficient way to land.

When you do that, you are basically burning along the 2 shorter sides of a right triangle. A gravity turn is a burn along the hypotenuse, which is far more efficient.

[selfish_meme]

A suicide burn to stop at 0m/s on the ground is most efficient, I know some of the guys in my T-prize challenge could do it using nodes and KER, the rest of us plebs had to use mechjeb to finish the burn accurately enough

[LostOblivion]

What I'm after is to use a maneuver node as a target above my desired landing site, so I get its time to node information. If t=v/a gives me, say, 2 minutes and I start my burn when the time to node is -2 minutes, I will obviously stop before the target, because the time to node assumes only acceleration due to gravity, not from my burn, which will slow the spacecraft down. For a circular orbit, the constant orbital speed can be used to estimate the time to node (which the game does, only some more accurate orbital equations instead), which won't be constant if I start burning. Thus, the time to node cannot be given by t=v/a, but is seen as a sort of construed marker for when I am to start my burn.

I'm not very good at algebra, but I might find out myself.

[p1t1o]

Sorry, but you absolutely are; you know how a gravity turn is the most efficient way into orbit? (if not, there are plenty here who can explain how to do it)

Well, the same maneuver but in reverse is more or less the most efficient way to land.

When you do that, you are basically burning along the 2 shorter sides of a right triangle. A gravity turn is a burn along the hypotenuse, which is far more efficient.

I would have though that would be the case if you were descending into an atmosphere (the opposite of conditions at launch) but I am assuming a descent in vacuum in this case.

As far as I know, gravity turns are more efficient due to lower aerodynamic forces?

[LostOblivion]

Found a solution. It's actually t=v/2a!

Example:

If I travel at v=560 m/s and my engines provide a=6.5 m/s^2 of acceleration, I should start burning when the time to maneuver node is t=v/2a=560/(2*6.5)=43 seconds, or -43 seconds.

Solution (using the equations of motion found here):

v^2=v0^2+2a(r-r0)

v=velocity at node=0
v0=initial velocity=orbital speed
a=acceleration from engines (assume constant)
r=position at node=0
r0=initial position


r0=v0t (assume circular orbit)
t=time to node


v^2=v0^2+2a(r-r0)
0^2=v0^2+2a(0-r0)
0=v0^2-2ar0
0=v0^2-2a(v0t)
0=v0^2-2av0t
v0^2=2av0t
v0=2at
v0/2a=t


t=v0/2a

Edited August 13, 2015 by LostOblivion

[Sharpy]

I would have though that would be the case if you were descending into an atmosphere (the opposite of conditions at launch) but I am assuming a descent in vacuum in this case.

As far as I know, gravity turns are more efficient due to lower aerodynamic forces?

Yes, the optimal vacuum landing is a suicide burn (which is a kind of reverse gravity turn but exceptionally short), but that is tricky and dangerous. The standard procedure of "reverse gravity turn" OTOH allows for very precise safe landings with little extra losses.

When launching from vacuum worlds, you too rise only above mountains, and accelerate horizontally as much as you can, pitching up only as far as needed not to drop.

[LostOblivion]

Oh! This actually means that I can do like I normally do. Place half my burn ahead and half after.

[Sharpy]

Found a solution. It's actually t=v/2a!

Example:

If I travel at v=560 m/s and my engines provide a=6.5 m/s^2 of acceleration, I should start burning when the time to maneuver node is t=v/2a=560/(2*6.5)=43 seconds, or -43 seconds.

And where's the gravitational acceleration? Over 43 seconds at 1.63 m/s² of Mun acceleration your vertical speed has grown by 70m/s from v0. Did you account for that?

And no, you can't happily substract 1.63 from a=6.5; It's not 1.63 at all times - it depends on horizontal velocity in a quadratic relation.

Edited August 13, 2015 by Sharpy

[LostOblivion]

And where's the gravitational acceleration? Over 43 seconds at 1.63 m/s² of Mun acceleration your vertical speed has grown by 70m/s from v0. Did you account for that?

It only accounts for horizontal speed, and gravitational acceleration is not horizontal, so it's accurate. However, it's not optimal considering you'll have to burn off the vertical speed gained, but you have to do that anyway. It will put you on target however, and being much better than t=v/a, this means I don't have to eyeball the landing anymore.

(Using mechjeb or other autopilots is not an option for me. )

Also, if you you want to avoid the "burn-horizontal-until-stop, then straight down at 70 m/s" thing, you can start a bit before 43 seconds to account for the loss of horizontal acceleration due due to pointing at the retrograde marker instead of the horizon.

- - - Updated - - -

I just realised, for those of you who use KER, this means setting a maneuver node that will kill your horizontal speed, and then looking at the "time to burn" field in the orbital menu. Burn a few seconds before this should make it pretty accurate. Not as accurate as any autopilot, of course, but landing manually in kerbal is all the fun!

- - - Updated - - -

Actually, you don't burn for 43 seconds. 43 seconds is just the construed marker of time to node of when you should start the burn. The burn time (for horizontal movement) is actually double that at t=v/a=560/6.5=86 seconds. This means even more vertical speed will have built up, so you have to compensate more in the vertical, and probably begin even further ahead of "time to burn".

Edited August 13, 2015 by LostOblivion

[Sharpy]

Yes, and that's why it's unoptimal - and why you should always take quite a bit more delta-V for landing than absolutely necessary.

Quite recently I was in a situation where I underestimated my fuel amount needed (KER counted the delta-V of carried rovers into the total, and the way they were mounted and the way I was unable to pilot both at the same time made them useless for landing), I did have to perform one perfect suicide burn on the Mun to land the carrier with the rovers in one piece.

Let me tell you no matter how precisely you estimate the time to surface at impact point of where your trajectory would bring you, the actual altitude of the landing site varies wildly depending on a fraction of second when you start the burn. I think I crashed into terrain past the edge of the crater good fifteen times, and crashed into the bottom of the crater another ten times before I got the timing good enough that I ran out of fuel only some 30 meters above the surface and with maybe 15m/s on the clock. The carrier got totalled but both rovers survived intact, if in awkward positions that required a lot of effort to set them both upright.

[xtoro]

If you're using maneuver nodes, you don't need KER or MechJeb or anything. Just adjust the maneuver node so it's right above ground and pull the retrograde adjustment until the AP is where the node is. And there you go, the info that shows next to your navball is all you need, you have your required m/s to come to a stop just above ground, and burn time, and time to node. So just make sure you start your burn while the time to node is greater than burn time. Done.