Minimizing Delta-V in landing burn

[staplic]

I've been looking for a formula or a guide on exactly when to initiate a deceleration burn while landing on a planet that will get you to a safe landing speed and minimize the delta-V involved in the burn. I've been experimenting with this in game a bit too, but I'm finding it difficult to find that window in the first place from which to fit a model. If I start burning before the window, my speed will drop to 0 then I have to shut down (or I start going up again) and then I have to burn again later to kill off the speed I gain from falling. Conversely if I burn to late my speed doesn't fall enough and a blow up on impact. Without a formula I've been erring on the side of going early, but as I move on to progressively more difficult objects to land on (Tylo for instance) I'm trying to conserve as much delta-v as possible.

Edited October 11, 2013 by staplic

[Blue]

Tylo requires approximately 5km/s of delta-V if I recall correctly, for a landing and launch-to-orbit on its surface. The rule of thumb is "if it can work as a rocket SSTO on Kerbin, it can land on Tylo"- which is a surer thing than the number I just gave, because I'm not sure if I do actually remember.

Regarding a landing profile I'm not sure if there actually is a tried-and-true mathematically ideal landing equation or somesuch, since I've never seen one nor has the idea ever seem to've been presented. This is all assuming no atmosphere is present, since aerobraking is remarkably simple to do, and fuel-cheap.

The words I work by, have been recommended by various people at various times:

1. Be in orbit around the target body. Eccentricity of your orbit should be low.

2. Burn retrograde at the periapsis at a proportionate throttle level so that your lateral velocity decreases to a controllable stable level in an appropriate time for the descent. (On the mun, this can mean getting it down to 50~20m/s before reaching 1000m from the surface. It varies widely, since on Gilly you can land with reckless abandon using exclusively RCS, and on Tylo you'll likely be burning at full throttle until the last 250m from surface contact.)

3. While decreasing your lateral velocity by controlling the throttle, control your vertical velocity by changing the angle of thrust (changing the angle of your rocket) up or down relative to the surface. (On the Mun, 30% throttle at a 5~8° upward angle from retrograde or thereabouts is enough to get a descent within a few minutes, descending at a velocityless than 100m/s vertically.)

4. Once your speed is less than 150m/s relative to the surface and/or you're within 1000m of the surface, in most circumstances you can eyeball the landing from there on out.

[Geschosskopf]

NOTE: Assuming airless worlds so parachutes aren't an option.

Landing delta-V isn't so much a mathematical formula as it is a piloting and ship design function. It depends almost entirely on landing trajectory, which is a function of the planet's gravity, your rocket's TWR, the mission requirements, and what you as pilot feel comfortable with. At a bare minimum, you'll have to burn at least as much as your starting orbital velocity but it will always be a bit more than this.

The most fuel-efficient landing trajectory is just enough burn to deorbit, then freefall in a long arc, and finally a hard burn started as low as possible that kills both your horizontal and vertical speeds just as you reach the ground. But the shape of the descent arc depends greatly on the situation. You can vary 2 things about the shape: starting altitude (and thus speed), and steepness of descent, and there's also the landing site's altitude to consider. The stronger the gravity and/or the weaker the rocket's TWR, the longer you'll have to burn so you need to start higher or come in shallower. This means that the best trajectory is different for rockets with different TWRs, even at the same planet and trying to land at the same spot.

[capi3101]

There are really far too many variables involved that are not set until landing is underway - Blue and Geschosskopf have already gone over them.

General kinematic equations could give you an estimate on how long you could expect to burn, but those don't take into account the change in your lander's mass as you burn fuel, and then you're still left with one major unknown: the actual elevation of the surface you're landing upon.

[Blue]

Unless you're Scott Manley in which case you can land however the hell you wish and no one will think less of you for it; only more.

[Kosmo-not]

Landing is very simple. You start off with a periapsis approximately 1km above terrain. Upon reaching periapsis, start your landing burn. Control your vertical speed using pitch. Your goal is to be very close to the ground by the time you kill your horizontal velocity.

This works for craft with any amount of TWR, and is necessary for craft with low TWR.

Below is a video showing me landing Mulbin's Munbug. That thing has an insanely low TWR.

[Specialist290]

As the others have stated (directly or otherwise) in previous posts, the ideal landing trajectory from orbit for a given craft is essentially the exact reverse of the ideal takeoff trajectory from the surface to that same orbit. In practice, though, unless you're using some sort of autopilot, you're more likely than not going to want at least a quarter again as much delta-v (I usually aim for half again, myself) for landing as for takeoff, just to give yourself a bit of a safety margin, since without resorting to an autopilot your actual landing will be less than ideal simply by virtue of the fact that most of us tend to slow down our descent earlier than we really need to so we have a little more time to react to unforseen circumstances.

[Blue]

your actual landing will be less than ideal simply by virtue of the fact that most of us tend to slow down our descent earlier than we really need to so we have a little more time to react to unforseen circumstances.

Due to the fact that landing "without wiggle room in the numbers" is more like a fluke than a plan due to terrain other variables. I've only ever seen it done once where someone brought their velocity to 1m/s (from a high velocity) at the same moment as they contacted the ground.

[Specialist290]

Pretty much, yeah; that was the main "unforseen circumstance" I had in mind. I have to admit that I was rather surprised on my very first Mun landing attempt when my lander hit the ground well before the altimeter reached zero

[capi3101]

That one nails everybody on their first attempt, and it's only later that folks learn about the IVA radar altimeter...

[rhoark]

The question of how to get 0 velocity at touchdown with a continuous burn is trickier than I expected at first. At least a second order nonlinear ODE, if not worse.

Simplifying down to a purely vertical descent and keeping constant TWR instead of constant throttle, you can pull acceleration out of the rocket equation and integrate it to tell you that vertical velocity at the start of the descent burn should be

(g w)/(h+r)^2-(g w)/r^2

where,

g = gravitational parameter

w = TWR

h = altitude

r = planetary radius

then if you throttle down as you go to keep TWR constant, you should hit sea level with 0 velocity.

I haven't tried this.

[staplic]

Since there wasn't an equation I decided to try and develop one, and I've almost got it, but unfortunately my mathematica software is maxing out my CPU trying to simplify this in closed form so I'll just post the fragments up here to see if someone else can simplify it (or point out an error in it, as is always a possibility):

First I computed the total minimum delta-v needed for touchdown using conservation of energy:

Start with Potential Energy in orbit (a height h above the surface) and subtract from it the potential energy at the surface:

(-μMm)/〖(ÃÂ+h)〗^2 -(-μMm)/ÃÂ^2 = Net Potential Energy

where

μ = gravitational parameter (G in our universe)

M = Mass of planet/moon (referred to as 'the body')

m = mass of rocket

ÃÂ = radius of the body

This simplifies to:

(mÕ*(2ÃÂh+h^2))/〖(ÃÂ+h)〗^2

where:

̉ۢ = acceleration due to gravity at the surface of the body.

this must be equal to the Kinetic Energy gained = 0.5m v^2. The m's cancel and since you started with 0 vertical speed the final v is your total vertical delta v, combine this with the horizontal speed you have to kill and

delta-v = √((2Õ*(2ÃÂh+h^2))/〖(ÃÂ+h)〗^2 +vh^2 )

where:

vh = horizontal speed

Using this we can then calculate the burn time:

t= (m*I/T)*(1-â…‡^((-v)â„I))

where

t= time

I = specific Impulse of your rocket

T = Thrust of your rocket

v= delta-v calculated above.

In a similar method to how we derived the total delta-v we can find our vertical velocity as a function of vertical height x where we will start our burn this velocity Y is:

Y=(ÃÂ*√(2Õ(h-x)*(h+x+2ÃÂ)))/((ÃÂ+h)*(ÃÂ+x))

where all variables are as previously defined.

Lastly since we want our velocity to be zero at touchdown we know that:

dx = Y+at dt

with a = (T/(m-nt))-̉ۢ

where n is the mass flow rate m-dot. notice a is a function of time so integrating this function we get:

x = (Y-I)*t-0.5̉ۢ*t^2 + (m*I^2/T)*Ln[(m*I)/(m*I-T*t)]

in theory solving this last equation for x (recall Y is a function of x as well) and plugging in the values calculated above you can solve for x in terms of h, rho, phi, I, T, and m, all of which are known to you at the time you start your descent. When attempting to solve for x you run into a fairly nasty looking fourth order polynomial to solve and then work your way through each of the four solutions to figure out which one yields a positive real result (and if there are two which one actually works). Like I said earlier my mathematica kernal timed out and shut down when I attempted to have it simplify[solve[]] this, but if we get it in a simplified closed form I wouldn't think it would be too difficult to have a computer crunch the numbers.

Edited October 5, 2013 by staplic
derivation error in x

[numerobis]

tavert did a bunch of math regarding kosmo-not's landing approach; might want to run with that?

[tavert]

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

You don't want to do a vertical burn, you want to burn at the necessary angle to keep your trajectory horizontal. Since the burn takes finite time, the delta-V is more than you would expect purely from an energy calculation.

[numerobis]

Meanwhile, I still haven't figured out how to optimally run an atmospheric ascent...

[tavert]

Did you read through this thread http://forum.kerbalspaceprogram.com/threads/46194-I-need-someone-help-me-do-some-math-for-launch-optimization ?

Alterbaron got something working pretty well about a month ago, but there hasn't been much activity since.

[arq]

The ideal burn involves getting a low periapsis (preferably within a km or so above the surface) and waiting to get near peri to start killing lateral velocity and entering a vertical ascent. Vertical descent is expensive (whereas 'horizontal' descent, having a low periapsis, is nearly free), so an ideal profile will minimize that part.

Ofc on atmospheric bodies it's very different. But for some ships, it's actually not worth packing chutes. Parachutes are actually surprisingly heavy and hurt both your dV and TWR. On atmospheric bodies the air will usually slow you down to <150m/s or less without chutes, and sometimes the fuel to make that landing is less than the delta-v to drag several tons of parachutes across the kilometers/sec of delta-v to get there and back.

[Kasuha]

Assuming you can trust the map view (which is definitely not true in cases of small moons) and assuming you're landing on airless body, you can lower your orbit to intersect with terrain, then put a maneuver just above the terrain and bring your future apoapsis to that point using retrograde burn. That will tell you how much dv you will need, how long your burn will be at exactly that point and when do you need to do it. If executed correctly it will leave you low above the terrain with zero vertical speed at cost of minimum fuel.

Of course you cannot start braking when time to node is half the burn time as you would do in space - you'd crash into terrain. My experience is that starting the burn at 70% burn time left to node is about right.

If you can trust the map view. Because if the terrain is higher than the map tells you, you just crash.

[Deep silence]

My main parameter is the time the vessel spends near the planet (once in suborbital trajectory) : the more time its gravity accelerates the vessel, the more DeltaV you will need (same thing for take off, btw).

Thus, You need to have enough deltaV to zero your orbital speed, plus enough deltaV to cancel gravity effects during the orbital speed zeroing.

=> I think kosmo-not video is good and bad at the same time : he kills its orbital speed from the lowest point, but its vessel's TWR is too low : the maneuver takes too much time so he spends 50% of its deltaV to fight gravity (50% because the vessel has most of the time a 45° inclination).

My method is to get the most powerfull lander (yet avoiding a too heavy lander), and to slow down in one time from the lowest altitude possible. Of course, it is dangerous...

[tavert]

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

You'll see that too low of a TWR does require more delta-V, however since engines in KSP are fairly heavy it's easier to get more delta-V with a low-TWR craft than a high TWR craft. And past a TWR of ~2 relative to the body you're landing on, the engine mass easily outweighs the delta-V savings as you get into diminishing returns.

[staplic]

a note for anyone using the derivation I posted, i got ahead of myself a bit and forgot a factor of t in the integral deriving x. It's fixed now but the term (V-phi-I)t is actually (v-I)t-0.5phi t^2.

[numerobis]

(50% because the vessel has most of the time a 45° inclination).

A nit: actually it's less than 30% lost. At a 45 degree angle you have a cos(45) ~ 0.7 fraction of your thrust pointing horizontal.

[Kosmo-not]

My main parameter is the time the vessel spends near the planet (once in suborbital trajectory) : the more time its gravity accelerates the vessel, the more DeltaV you will need (same thing for take off, btw).

Thus, You need to have enough deltaV to zero your orbital speed, plus enough deltaV to cancel gravity effects during the orbital speed zeroing.

=> I think kosmo-not video is good and bad at the same time : he kills its orbital speed from the lowest point, but its vessel's TWR is too low : the maneuver takes too much time so he spends 50% of its deltaV to fight gravity (50% because the vessel has most of the time a 45° inclination).

My method is to get the most powerfull lander (yet avoiding a too heavy lander), and to slow down in one time from the lowest altitude possible. Of course, it is dangerous...

Can you propose a different landing method for that craft that uses less delta-v?

[Kasuha]

Can you propose a different landing method for that craft that uses less delta-v?

I watched that video and I think that if there was a hill in your path before you killed your horizontal velocity you'd be toast.

And I also think your method is not the most efficient as you're burning a lot in non-retrograde direction, moving your orbit along the surface instead of braking.

The least delta-v is used when you freefall on sub-orbital trajectory and then stop right on the surface after just one full throttle retrograde burn. That applies to any craft. You can use maneuver node to estimate it.

[Kosmo-not]

I watched that video and I think that if there was a hill in your path before you killed your horizontal velocity you'd be toast.

And I also think your method is not the most efficient as you're burning a lot in non-retrograde direction, moving your orbit along the surface instead of braking.

The least delta-v is used when you freefall on sub-orbital trajectory and then stop right on the surface after just one full throttle retrograde burn. That applies to any craft. You can use maneuver node to estimate it.

Make a lander that has a TWR of 1.5 - 2.0 (relative to Mun) and try both methods. Record your results (video would be awesome).