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Hey!

I was trying to create a cfg file for a reaction wheel and I wanted it to be as realistic as possible. I've already determined it's mass with it's molar mass, assuming it was made of steel and lead. but here's the problem: I'm only 16 and I haven't already learned how to determine an object's torque knowing its mass and its rotation speed. and I don't understand anything to raw formulas. does anyone has the time to explain me? Thanks :)

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What kind of "wheel" they have inside and what happens to it is an interesting theological argument, as reaction wheels in KSP don't behave anything like real reaction wheels, but since you ask:

For pushing things, mass is all you have to worry about, for rotating things, you have to worry about how far it is from you too.  The farther it is, the harder it is to do, like a lever.  This is why torque units describe the force it takes for you to pivot a weight on a stick.  One foot-pound of force is the force needed to pivot a one pound weight held by the end of a one-foot stick.

So that's what force is.  For mass, you have I, angular mass or moment of inertia.  For a weight on a stick that's just

I = m * r^2

where m is weight and r is length of stick.  For more complex things like a disc, it's modelled as the sum of an infinite number of tiny weights on sticks, but there's lots of ways to cheat.  Suppose your disc has a hole in the middle, with all of its weight on the rim, a thin ring in other words.  Then you're just back to I = m * r^2 again.  r is radius, which is length from the center of the ring to the edge.

So your reaction wheel is a thin ring 2*r wide (because radius is half diameter), weighs m, and has T torque applied to it for t amount of time, how fast does it end up spinning (@)?

The formula is T = I * w, where I is angular mass, substitute that:

T = m * r^2 * w

So:

w = T / (m * r^2)

...which makes sense on first look:  Stronger torque makes it faster, bigger weight and/or wider wheel make it slower.  Now what's missing?  Time (t).  Just multiply it by time to see how far the wheel actually turns.  The longer you push, the further it moves, and the faster it goes.

@ = (T*t) / (m * r ^ 2)

@ because nobody's got that stupid theta symbol on their keyboard.  It's an angular measure, and angles in physics mean radians, so you'll get out a number from this where 3.14159 means 180 degrees.

I won't go into momentum, and if you go into it yourself you'll quickly figure out why KSP reaction wheels are magic:  To provide constant force, they have to be constantly speeding up!  Yet KSP reaction wheels do that all the time without complaint.

This is half barely-remembered college facts and half wikipedia, so if there's anything blatantly wrong, I tried.

Edited by Corona688
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19 minutes ago, Corona688 said:

What kind of "wheel" they have inside and what happens to it is an interesting theological argument, as reaction wheels in KSP don't behave anything like real reaction wheels, but since you ask:

For pushing things, mass is all you have to worry about, for rotating things, you have to worry about how far it is from you too.  The farther it is, the harder it is to do, like a lever.  This is why torque units describe the force it takes for you to pivot a weight on a stick.  One foot-pound of force is the force needed to pivot a one pound weight held by the end of a one-foot stick.

So that's what force is.  For mass, you have I, angular mass or moment of inertia.  For a weight on a stick that's just

I = m * r^2

where m is weight and r is length of stick.  For more complex things like a disc, it's modelled as the sum of an infinite number of tiny weights on sticks, but there's lots of ways to cheat.  Suppose your disc has a hole in the middle, with all of its weight on the rim, a thin ring in other words.  Then you're just back to I = m * r^2 again.  r is radius, which is length from the center of the ring to the edge.

So your reaction wheel is a thin ring 2*r wide (because radius is half diameter), weighs m, and has T torque applied to it for t amount of time, how fast does it end up spinning (@)?

The formula is T = I * w, where I is angular mass, substitute that:

T = m * r^2 * w

So:

w = T / (m * r^2)

...which makes sense on first look:  Stronger torque makes it faster, bigger weight and/or wider wheel make it slower.  Now what's missing?  Time (t).  Just multiply it by time to see how far the wheel actually turns.  The longer you push, the further it moves, and the faster it goes.

@ = (T*t) / (m * r ^ 2)

@ because nobody's got that stupid theta symbol on their keyboard.  It's an angular measure, and angles in physics mean radians, so you'll get out a number from this where 3.14159 means 180 degrees.

I won't go into momentum, and if you go into it yourself you'll quickly figure out why KSP reaction wheels are magic:  To provide constant force, they have to be constantly speeding up!  Yet KSP reaction wheels do that all the time without complaint.

This is half barely-remembered college facts and half wikipedia, so if there's anything blatantly wrong, I tried.

oooh, thanks a lot :) , I think I understood... but I can't figure out why the wheel has to accelerate. if a mass attached to an axle itself attached to another mass, in a universe without any external forces applied to it, and if the first mass is turning constantly at a fixed speed, why would the mass b turn then deccelerate? why would mass a need to accelerate to keep a constant turning speed on mass b?

thanks tjt, I'll check that :)

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5 minutes ago, Tomato29 said:

oooh, thanks a lot :) , I think I understood... but I can't figure out why the wheel has to accelerate. if a mass attached to an axle itself attached to another mass, in a universe without any external forces applied to it, and if the first mass is turning constantly at a fixed speed, why would the mass b turn then deccelerate?

It wouldn't.  I'm talking about constant force, like a moon lander landing tilted, preventing itself from falling over with a reaction wheel.  The reaction wheel only exerts force on the lander when it accelerates.  So the wheel has to constantly speed up to hold it in place.

Take my math above with a grain of salt, I'm making a table and think I had some errors.

Edited by Corona688
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Just now, Corona688 said:

It wouldn't.  I'm talking about constant force, like a moon lander on a tilted hill holding itself up with a reaction wheel.

Take my math above with a grain of salt, I'm making a table and think I had some errors.

oh okay, I see now. well I'll take your math because I don't understand anything to raw formulas, and you explained to me perfectly :)

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If you know any Newtonian mechanics, rotational mechanics actually translates fairly easily, with torque in place of force and inertia in place of mass.  The trickiest part is figuring out what the inertia is for a given shape and mass, and dealing with the stupid untypeable symbols.

Symbol Linear Rotational Example Example
Distance d θ

d = V * t

Distance is velocity times time.

θ = w * t
Velocity V ω

V = a * t

Velocity is acceleration times time.

ω = α * t
Acceleration a

α

"fish"

   
Force F

T

F = m * a T = I * α
Mass m I    

So, momentum in linear motion is m*v.  Ergo in rotational motion that's I*ω .

Meaning, my equation above actually gave ω not θ, and needs to be multiplied by t again to produce "distance" i.e. θ

Edited by Corona688
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