Gyroscopic Precession

[MachJok]

Hello All

I am working on a ship with a gravity wheel
exhibit a.

and wanted to calculate the effects of gyroscopic precession on the ship (no masses yet still very early dev stage). The ship so far is 1077 feet long, the gravity wheel has an external diameter of 200' and an internal diameter of 180'. According to spincalc.com the angular velocity to simulate 1g is ~32.5°/sec.

I tried to look up gyroscopic precession via wikipedia
exhibit b. [ame="http://en.wikipedia.org/wiki/Precession"]Precession - Wikipedia, the free encyclopedia[/ame]
and found the equation to be a bit confusing for my application. Precessional Velocity = (omega)_p = mgr/(I_s*(omega)_s).

m*g is what I am confused about. Do I replace g with my rotational acceleration (yaw acceleration) since the frame of reference is the ship?

That leads into the next question. Would the precession be significant enough to account for in game or should I let "otto" "compensate" for it?

Thanks for the help in advance!

RedMeansGo

 

[tblaxland]

Precessional Velocity = (omega)_p = mgr/(I_s*(omega)_s).

m*g is what I am confused about. Do I replace g with my rotational acceleration (yaw acceleration) since the frame of reference is the ship?

No, that equation is for torque induced precession (for example, gravity imparting a torque on a spinning top).

You want to look at the equation for torque-free precession (there is no external torque being applied to your vessel when it is free falling in space): [math]\omega_p = (I_s/I_p) \omega_s (1-\cos(\alpha))[/math]

 

[MachJok]

Thanks! I may be confusing terminology but in my case the axis of rotation and the axis of symmetry are aligned correct? If so:

[math]
\omega_p = (I_s/I_p) \omega_s (1-\cos(\alpha)) 
I_s = I_p , \alpha = 0 [/math] making [math] \omega_s (1-\cos \alpha) = 0
[/math]

which makes the entire thing 0 but that makes no sense
if alpha = 90

[math]
\omega_p = (I_s/I_p) \omega_s
[/math]

Thanks for the help!

 

[tblaxland]

I may be confusing terminology but in my case the axis of rotation and the axis of symmetry are aligned correct?

Well, yes, but try actually building something so precise. You will always have some small angle between the angular momentum vector and the axis of symmetry and therefore always some precession. I can't tell you how accurate you are able to build your spaceship. Its fiction so you get to make that bit up

 

[MachJok]

Well, yes, but try actually building something so precise.

Valid point! I am shooting for best case scenario...heck the thing is powered by handwavium so why not?

So I start off with [math] 1) \omega_p = (I_s/I_p)\omega_s(1-\cos(\alpha))[/math]

because it is a disk and according to the wiki "They should be the same, due to the symmetry of the disk." Making
[math] 2)I_s = I_p [/math]
thus leaving
[math] 3) \omega_p = 1 \times \omega_s(1-\cos(\alpha)) [/math]
becoming
[math] 4) \omega_p = \omega_s(1-\cos(\alpha)) [/math]
[math]\alpha[/math] is the angle between the axis of symmetry and axis of precession. If the moment is applied to the lateral axis the precession would be realized in the longitudinal axis. ie if the ship yaws, the wheel will try to precess and in turn roll the ship. so then
[math] 5)\alpha = \pi/2 [/math]
then taking 5) and putting it in 4)
[math] 6)\omega_p = \omega_s(1-\cos(\pi/2)) [/math]
which becomes
[math] 7)\omega_p = \omega_s(1-0) [/math]
simplified
[math] 8)\omega_p = \omega_s [/math]
That is to say, the Precessional rate = the rotational rate in all circumstances regardless of the force applied...Maybe I fubar'd something but that doesn't seem right.

Any ideas?

Thanks for the help!

 

[tblaxland]

If the moment is applied to the lateral axis the precession would be realized in the longitudinal axis. ie if the ship yaws, the wheel will try to precess and in turn roll the ship...
...Maybe I fubar'd something but that doesn't seem right.

Any ideas?

If the ship yaws, you are applying a torque and you need to consider the torque-induced model: [math]\omega_p = \frac{\tau}{I_s\omega_s}[/math]

Where [math]\tau[/math] is the torque that your vessel applies to the gyroscope. This should be the same as the torque applied by the thrusters (or whatever is doing the torquing) if the gyroscope is at the centre of mass.

Sorry if I misunderstood what you were trying to calculate.

Note that as soon as you start applying torque, [math]\alpha[/math] will be non-zero and you will also get some torque-free precession.

FWIW, precession (all types) is calculated by integrating Euler's equations for rigid body dynamics. This is the formulae that Orbiter integrates for any vessel (all vessels are considered rigid bodies). The formulae presented on the Wikipedia precession page make useful predictions but only under a small set of restricted conditions.

EDIT: Highly recommended reading on this topic: http://faculty.ifmo.ru/butikov/Applets/Gyroscope.html and http://faculty.ifmo.ru/butikov/Applets/Precession.html