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If something has an SAS torque of "20", what does that actually mean?


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It means it can apply that much torque. Torque is a rotational force, it does two useful things: in space it can be used to rotate your spacecraft, in atmosphere you're mostly interested in stopping it from rotating. So a part with high torque could either rotate your ship faster in space, or it could resist a larger force trying to destabilise your rocket in atmo.

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It's in kN/m, measures the force it can enact to maneuver a craft (rotate/pitch/yaw).

The actual maneuver speed of a craft will also be influenced by its mass and where the torque is, related to the center of mass (basically the lever).

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It means it can apply that much torque. Torque is a rotational force, it does two useful things: in space it can be used to rotate your spacecraft, in atmosphere you're mostly interested in stopping it from rotating. So a part with high torque could either rotate your ship faster in space, or it could resist a larger force trying to destabilise your rocket in atmo.

I get that, I was kinda hoping for a 'if your craft weighs X, a torque force of Y will move it Z fast' kind of thing. I know what it does and how it does it, what I'm unsure of is how much it does. I get really frustrated building probes because I can never work out if the RGU torque will be enough to make it relatively manoeuvrable. So I put on a reaction wheel and it's super jittery. I just would like to find/figure it all out before I get to space, really.

It's in kN/m, measures the force it can enact to maneuver a craft (rotate/pitch/yaw).

The actual maneuver speed of a craft will also be influenced by its mass and where the torque is, related to the center of mass (basically the lever).

So a rough calculation is doable, based on vessel mass and torque available then.. I remember reading a post here about how the placement of the SAS units doesn't actually matter and they all provide equal force. Any ideas how that fits into this?

HarvesteR himself
that "there should be a law that requires all programmers to document units".

Maybe we should remind him of that time to time :)

Heh, totally agree. That said, I do rather like the whole 'I have 45 units of fuel remaining'.. It might be that the units KSP measures things in is actually called Units :P

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I get that, I was kinda hoping for a 'if your craft weighs X, a torque force of Y will move it Z fast' kind of thing. I know what it does and how it does it, what I'm unsure of is how much it does. I get really frustrated building probes because I can never work out if the RGU torque will be enough to make it relatively manoeuvrable. So I put on a reaction wheel and it's super jittery. I just would like to find/figure it all out before I get to space, really.

You can use the debug menu to simulate a space environment. Go to the pad with your design, hit [mod]+[F12] and click the "hack gravity" button. This will give you zero g on the pad, and you can test how your design will perform in space. Don't forget to unhack it before you try a live launch!

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I remember reading a post here about how the placement of the SAS units doesn't actually matter and they all provide equal force. Any ideas how that fits into this?

Hm, it's very possible I'm wrong about that the position within the craft matters; it's an assumption I made and which didn't get contradicted while playing.

Then again, most of the time it's "the torque source is at the top of the craft and the longer the rocket, the harder it is to maneuver"... so it might just be the additional mass that makes them clunky ;)

Either way, for me a Large Inline SAS seems to be enough most of the cases for the first stages while I still have large tanks; after that, I'm fine with the command module's torque and can't remember adding more support.

Except RCS you can't do without when docking anyway.

I've rarely gone further than the Mun though.

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I remember reading a post here about how the placement of the SAS units doesn't actually matter and they all provide equal force. Any ideas how that fits into this?

I concur that all SAS units of the same type provide equal force.

However positioning has an effect upon how this force is utilized.

This force can not only be used to induce a rotation but also a movement.

Imagine a long rocket, say 20 tanks and at both ends there is a SAS module (do not forget energy and command module).

Rotating this ship with SAS will bend it, so that it looks like a very slim "S".

Bended objects in KSP will try to return to their original form. It is similar to a spring that oscilates until it is in it's original position.

This means that there is some kind of friction that reduces the oscillation.

So less than 100% of the physical energy generated by torque is used for the rotation. Some is lost in this fricion. I have however no idea about the magnitude of this loss.

For positioning I had the following impression: "the farer away from the CoM SAS modules are placed the more the parts bend."

So I try to place them as near to the CoM as possible.

Another thing to keep in mind is that a SAS unit can apply three times its torque at the same time (one for each rotation direction) but it uses also three times the energy from batteries.

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Rotating this ship with SAS will bend it, so that it looks like a very slim "S".

Bended objects in KSP will try to return to their original form. It is similar to a spring that oscilates until it is in it's original position.

This means that there is some kind of friction that reduces the oscillation.

So less than 100% of the physical energy generated by torque is used for the rotation. Some is lost in this fricion. I have however no idea about the magnitude of this loss.

I agree that some torque is lost on bending the ship, but once the ship is bent to the limit when bending force and its counterforce balance, all the torque goes to rotating it. Unless there is something horribly wrong with Unity physics simulation (which ofcourse is an option).

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I tend to put a small one in an upper stage or the payload, as it's useful for manuevers once you get out of atmo. For stability on lower stages a set of four fins weighs less than the large SAS.

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Actually, SAS isn't affected by leverage in KSP. A SAS unit at CoM does the same as one 10m from CoM.

I beg to differ.

First, SAS is a control function. It is the reason why when you try to turn your ship, all aerodynamic control surfaces, engine gimbal, and reaction wheels work in accord to meet that request. But I assume that by SAS you mean torque.

Second, while reaction wheel at CoM and reaction wheel at the end of the ship will eventually have the same effect on the ship, i.e. it will turn by the same amount around CoM, the way it gets there (bending) and stability and comfort of the operation may be very different.

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http://en.wikipedia.org/wiki/Moment_of_inertia is how "hard" it is to turn your rocket.

http://en.wikipedia.org/wiki/List_of_moments_of_inertia has some calculations. For rocket-shaped rockets the thin rod approximation is probably suitable. For any shape it shows that doubling the mass will double the moment of inertia and thus the torque needed for the same handling, while doubling the length will quadruple the torque requirement.

And this applies to however you're turning it: not just reaction wheels, but engine gimballing and RCS too.

And I'm pretty sure reaction wheel/SAS module placement does matter. With 16 reaction wheels in a somewhat loosely-attached payload, my rocket bent severely back and forth when I tried to make the gravity turn, and with a bit of deliberate input I made it snap. I admit I never tried 16 reaction wheels at the CoM.

I may try some experiments. If KSP's physics is right, then if I have a reaction wheel at the end of a long thin object it will take four times as long to make a turn (without SAS to avoid confusing things) than if the wheel is at the middle.

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I may try some experiments. If KSP's physics is right, then if I have a reaction wheel at the end of a long thin object it will take four times as long to make a turn (without SAS to avoid confusing things) than if the wheel is at the middle.

Wouldn't a reaction wheel (or RCS thruster or vectored engine) have more leverage the further away from the CoM it is? I.e. applies more torque to the craft?

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For RCS, yes, certainly if it's balanced (RCS at each end) and possibly even if it's not, because they're capable of a certain linear force that then gets used to create a torque. For reaction wheels I believe no, because they give torque directly, around the point they're located at.

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Wouldn't a reaction wheel (or RCS thruster or vectored engine) have more leverage the further away from the CoM it is? I.e. applies more torque to the craft?

To convince you, I made an experiment with torque wheels on a spring. I always let the pendulum stabilize before I made the screenshot. In all cases I was holding the Q key (used short interrupts to speed up the stabilization but I was holding it depressed while taking the screenshot and the pendulum was not rotating. First image is the machine. Second to fourth image are individual wheels turned on. Last two images are two and all three wheels turned on.

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Wouldn't a reaction wheel (or RCS thruster or vectored engine) have more leverage the further away from the CoM it is? I.e. applies more torque to the craft?

Actually no, since the SAS module tries to rotate in itself. The following picture shows the forces that RCS and SAS produce when turning counterclockwise.

JUUe9bU.png

Edit: @Kasuha nice test setup

Edited by mhoram
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Reaction wheel tests

Test ship:

13745558554_e91805165e_c.jpg

The four fuel tanks are full. We have a pair of reaction wheels at each end, and a pair in the middle. We also have two batteries in the middle to ensure we can power the reaction wheels fully.

I threw it up on a Skipper with fuel-hack, in case you're interested.

Method:

For each combination of reaction wheels, the following is done.

  1. SAS is disabled, and any residual rotation killed with a burst of timewarp.
  2. Full pitch or yaw in one direction is applied for 2 seconds.
  3. Letting the ship spin freely, the time taken to make one full turn is recorded using the in-game MET.

The RGU's torque is off at all times, to not confound things, and testing is all in space so air drag isn't a factor.

Results:

[table=width: 500, class: grid]

[tr]

[td]Configuration[/td]

[td]Time/s[/td]

[/tr]

[tr]

[td]2 in middle[/td]

[td]39[/td]

[/tr]

[tr]

[td]2 one end[/td]

[td]38[/td]

[/tr]

[tr]

[td]1 each end[/td]

[td]37[/td]

[/tr]

[tr]

[td]1 end, 1 middle[/td]

[td]38[/td]

[/tr]

[tr]

[td]1 end[/td]

[td]74[/td]

[/tr]

[tr]

[td]1 middle[/td]

[td]71[/td]

[/tr]

[/table]

Additional notes: During all turns, the ship rotated about its centre of mass, geometric centre, and root part (which are all in the same place). At no point did the batteries run anywhere near dry.

Conclusion:

The results speak for themselves. In terms of speed of turning, distance of the reaction wheels from from the Centre of Mass is not important. This matches what was reported by KvickFlygarn87 and the results from Kashua's torsion pendulum setup. It is I believe contrary to actual physics.

Halving the amount of torque approximately halves the time for the turn, as expected.

The way in which reaction wheels act to deform the ship remains to be investigated, as does whether KSP properly implements moment of inertia. The latter could be tested with a ship like mine but with two tanks full and two empty, shifting the fuel for different tests.

Edited by cantab
Fixed the typo in the 1 end, 1 middle result!
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To convince you, I made an experiment with torque wheels on a spring. *snip*

Nice demonstration. If I'm interpreting it correctly, a single wheel at any position deflects the rig by the same amount, and multiple wheels are additive. So it is different from RCS, in that the position is almost irrelevant. Cantab's results would seem to confirm it.

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