steering losses ?

[Kussris]

Does anyone knows how to calculat the steering losses ?

[Vanamonde]

I do not, but I know where to ask, and so this thread has been moved to Gameplay Questions.

[worir4]

I think mechjeb can do this.

http://forum.kerbalspaceprogram.com/threads/12384-PART-0-90-Anatid-Robotics-MuMech-MechJeb-Autopilot-v2-4-2

[LABHOUSE]

It can (although it was just to a 1000th of a degree when I had mechjeb).

[Starwhip]

Steering losses of what?

Why does everyone know what we're talking about?!

[Yasmy]

Steering loss is speed lost due to steering, i.e., not thrusting in the direction of motion.

If you thrust prograde, your change in speed is the delta-v spent, ignoring gravity and drag losses.

If you thrust at an angle to prograde, the change in your speed is less than delta-v.

(Though the magnitude of your change in velocity is actually still equal to delta-v.)

If you thrust at angle alpha away from your prograde vector, your steering loss is proportional to (1-cos(alpha)).

This means that instead of your speed increasing by dv, your speed will increase by approximately dv * cos(alpha).

For alpha < 8 degrees, your steering losses are less than 1%, since (1-cos(8)) = 0.0097.

Your total steering loss is the time integral of thrust/mass * (1-cos(alpha)). The time integral of an acceleration is

a delta-v. Note that thrust, mass and your steering angle alpha are generally all functions of time.

Edited March 5, 2015 by Yasmy

[Xannari Ferrows]

You never take steering losses. You will always use 10 Km/s if you have that much Delta-V. If you want to thrust in the axis along your trajectory, but are moving off center, you aren't taking any losses. Just taking that energy and using it to change your trajectory.

[Yasmy]

You never take steering losses.

I understand what you mean by "take", but I disagree with never.

You will always use 10 Km/s if you have that much Delta-V.

Correct. Steering losses are not delta-v losses. No delta-v is lost, as I said in my first post.

Imagine this scenario. My rocket has a stability problem. Instead of pointing prograde when commanded to point prograde, it precesses around prograde like a top. On average it points prograde, but it never actually points prograde. I don't lose any delta-v, but I do uselessly spend some of my delta-v budget by not pointing where I want the rocket to point. That is a steering loss. You need to expend more delta-v to perform a maneuver than a rocket with better attitude control would have to expend.

Anyone who's ever built a floppy rocket in KSP has experienced steering losses, and anyone who played KSP before the improvements to ASAS has experienced steering losses.

If you want to say that intentional non-prograde burns are not steering losses because those burns were intentional use of delta-v in the non-prograde direction, I would argue that you are redefining the standard definition of steering loss. Is it the word "loss" that you object to? In this case (intentional non-prograde burns) I would agree that you don't "take" a loss. You don't lose anything. But you still have a non-zero steering loss. It's just not relevant.

Edited March 6, 2015 by Yasmy
grammmmmar

[Kussris]

Thanks for the reply!

I love to building rockets in KSP an see wath happens when I launch them, but i was wondering if I could calculate the rockete before launch so that the next Kerbal dose not get lost in space.

So the program "AscentComputron" calculates much, but might have a bug in the DeltaV diagram in terms of the steering losses. So I like to calculat them myself.

I suspect that the speed of the rocket in flight should be = DeltaV - drag losses -gravity losses - steering losses

[Red Iron Crown]

I suspect that the speed of the rocket in flight should be = DeltaV - drag losses -gravity losses - steering losses

I don't think so. If you insert yourself into an eccentric orbit your speed will be different at various points even though you've spent the same dV to get there.

[emerson]

By "steering loss" you mean losses due to trying to keep a wobbly rocket on course as it drifts barely controllable about your target vector? Then, yes, it's a thing - any offset vector from your intended direction of travel is fuel (dV) wasted. There probably is an equation to account for stiffness, flexion and vibration in a real rocket. Do you really need to account for it in KSP, probably not. V(rocket)=dV-D-G-(wobbly rocket*pilot compensation), (estimate through testing) might be the order of the day.

Edited March 7, 2015 by emerson

[rkman]

I think the margin of error of a manually steered ascent is similar to typical steering loss during ascent; in the order of a few 10's of m/s. Personally i doubt it's worth the effort to calculate it.

[Kussris]

...

Your total steering loss is the time integral of thrust/mass * (1-cos(alpha)). The time integral of an acceleration is

a delta-v. Note that thrust, mass and your steering angle alpha are generally all functions of time.

Thanks! Now I understand how to calculate the steering losses. I have the steering direction (from the Ascent Komputron) in which

the rocket is steered by a KOS script and I can log the prograde vector during the flight. That difference should be the angle alpha.

I think the margin of error of a manually steered ascent is similar to typical steering loss during ascent; in the

order of a few 10's of m/s. Personally i doubt it's worth the effort to calculate it.

By "steering loss" you mean losses due to trying to keep a wobbly rocket on course as it drifts barely controllable

about your target vector? Then, yes, it's a thing - any offset vector from your intended direction of travel is fuel (dV) wasted.

There probably is an equation to account for stiffness, flexion and vibration in a real rocket. Do you really need to account for it in KSP,

probably not. V(rocket)=dV-D-G-(wobbly rocket*pilot compensation), (estimate through testing) might be the order of the day.

You're probably right that it is neglectable but I found it interesting how to calculate the steering losses.

[Jovus]

Steering losses are not necessarily negligible in KSP. The normal definition also includes losses due to non-optimal insertion into your intended trajectory. The standard 45-degree turn at 10km in KSP is exactly such a non-optimal insertion, and the steering losses involved when compared with optimal insertion can be several hundred m/s

[Xannari Ferrows]

Steering losses are not necessarily negligible in KSP. The normal definition also includes losses due to non-optimal insertion into your intended trajectory. The standard 45-degree turn at 10km in KSP is exactly such a non-optimal insertion, and the steering losses involved when compared with optimal insertion can be several hundred m/s

Those aren't steering losses. They're gravity losses. Going completely vertical until 10 kilometers, as opposed to turning right away, means you spend more energy fighting gravity, while gaining no lateral velocity.

[Red Iron Crown]

Those aren't steering losses. They're gravity losses. Going completely vertical until 10 kilometers, as opposed to turning right away, means you spend more energy fighting gravity, while gaining no lateral velocity.

Jovus is referring to what happens immediately after the 45 degree turn, when the rocket is thrusting far away from the prograde direction. It's the divergence of the thrust vector from the velocity vector that causes steering losses.

[Kussris]

Hi !

I am very interested in the subject, but I have not understood it jet.

When the steering loss is "thrust/mass * (1-cos(alpha))", how big is it by alpha=90 degrees so the thrust is normal to the velocity?

Say thrust/mass = 10 [m/s²]; alpha = 90 degree

steering losses is = ???

[ElWanderer]

Hi !

I am very interested in the subject, but I have not understood it jet.

When the steering loss is "thrust/mass * (1-cos(alpha))", how big is it by alpha=90 degrees so the thrust is normal to the velocity?

Say thrust/mass = 10 [m/s²]; alpha = 90 degree

steering losses is = ???

10m/s per second.

If you're pointing to normal, you're not increasing or decreasing your orbital velocity. Not at first, anyway (a large enough burn will pull your orbit around and change the angle). But if you're pointing to normal that's likely to be deliberate (i.e. to change inclination) so it's not really a steering loss.

If you were to launch vertically and keep going up until above the atmosphere, then pitch-over 90 degrees, that would count as steering losses. They would be significant as it's a really inefficient way to reach orbit.

[Kussris]

10m/s per second.

If you're pointing to normal, you're not increasing or decreasing your orbital velocity. Not at first, anyway (a large enough burn will pull your orbit around and change the angle). But if you're pointing to normal that's likely to be deliberate (i.e. to change inclination) so it's not really a steering loss.

If you were to launch vertically and keep going up until above the atmosphere, then pitch-over 90 degrees, that would count as steering losses. They would be significant as it's a really inefficient way to reach orbit.

Thanks for the reply!

The total steering loss is 10 [m/s²] and the max. acceleration in this moment is also thrust/mass = 10 [m/s²].

So, 10-10=0 ?

Does this mean that the velocity vector does not change in direction nor in magnitude?

[maccollo]

I don't think so. If you insert yourself into an eccentric orbit your speed will be different at various points even though you've spent the same dV to get there.

Gravity losses do not stop once you shut down the engine. It is simply the velocity you loose as you ascend a gravity well. If you're falling down to periapsis you get negative gravity losses.

[ElWanderer]

Thanks for the reply!

The total steering loss is 10 [m/s²] and the max. acceleration in this moment is also thrust/mass = 10 [m/s²].

So, 10-10=0 ?

Does this mean that the velocity vector does not change in direction nor in magnitude?

To be pedantic, I shouldn't have said the orbital velocity wouldn't change as that is a vector. If you burn to normal, the orbital speed (magnitude of velocity) won't change but the velocity vector will, as you will be going in a slightly different direction (i.e. you've changed inclination).

If you burn, you will change your velocity! Steering losses don't mean your burns are having no/reduced effect, it's that you will end up spending more delta-v than necessary to end up in the final orbit.

[Red Iron Crown]

Gravity losses do not stop once you shut down the engine. It is simply the velocity you loose as you ascend a gravity well. If you're falling down to periapsis you get negative gravity losses.

We are defining gravity losses differently, I think. Falling around an eccentric orbit does not change the total orbital energy of the vessel, so by my understanding no losses are occurring. Kinetic energy is simply being exchanged for potential energy.

I think of gravity losses as the delta-V expenditures spent overcoming gravity and not changing the vessel's velocity. If a vessel can accelerate at 10m/s^2 and is ascending straight up against 4m/s^2 of gravity, then net acceleration is 6m/s every second even though the vessel is spending 10m/s of delta-V. So every second of such a maneuver results in 4m/s of gravity losses.

[Nich]

I am going to have to agree with Red here.

Also you have to equate energy i am guessing but unforunitly dV does not convert to energy without knowing how it was spent. Work (energy) = intergul of F*distance however this has to include dV, gravity losses, aero losses, and stearing losses at the same time.

1=launch pad

2=orbit

m1g1H1 + 1/2*m1V1^2 + energy of dV = m2g2H2 + 1/2m2V2^2 - energy of stearing losses - energy of gravity losses - aero losses

[maccollo]

Falling around an eccentric orbit does not change the total orbital energy of the vessel, so by my understanding no losses are occurring. Kinetic energy is simply being exchanged for potential energy.

[...]

I think of gravity losses as the delta-V expenditures spent overcoming gravity and not changing the vessel's velocity.

This seems very limiting, because then you can't use deltaV minus drag, gravity losses and steering losses to calculate your current velocity at any arbitrary point of an orbit.

Also what happens if you have a cruise phase during an ascent, and then circularise 70 km higher? Does the velocity lost during this phase not count? After all it's just an orbit like any other, and no total energy losses are occurring. If it doesn't count then final velocity plus gravity, drag and steering losses wont match total delta V expended. If it does count then why does it stop counting once you have an orbit that doesn't intersect the ground? This way of defining gravity losses also means that gravity losses can be steering losses and vise versa, like when you are doing a constant altitude ascent.

I like to think of in the same way that mechjeb shows it. Gravity losses is velocity that is lost due to kinetic energy being exchanged for potential energy when ascending a gravity well. Steering losses are cosine losses.

Edited August 13, 2015 by maccollo

[Red Iron Crown]

This seems very limiting, because then you can't use deltaV minus drag, gravity losses and steering losses to calculate your current velocity at any arbitrary point of an orbit.

Nor should you, IMO.

Also what happens if you have a cruise phase during an ascent, and then circularise 70 km higher? Does the velocity lost during this phase not count? After all it's just an orbit like any other, and no total energy losses are occurring. If it doesn't count then final velocity plus gravity, drag and steering losses wont match total delta V expended. If it does count then why does it stop counting once you have an orbit that doesn't intersect the ground? This way of defining gravity losses also means that gravity losses can be steering losses and vise versa, like when you are doing a constant altitude ascent.

IMO there is no gravity loss when coasting (though there might be drag losses, of course). Not sure what is meant by a "constant altitude ascent", can you clarify?

I like to think of in the same way that mechjeb shows it. Gravity losses is velocity that is lost due to kinetic energy being exchanged for potential energy when ascending a gravity well. Steering losses are cosine losses.

I'm not really familiar with how MechJeb calculates it, to be honest. Are you saying that an idealized two burn ascent, with a horizontal instantaneous burn to set Ap and another at Ap to circularize, would show gravity losses?