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Shrad

Wed Mar 13 2013, 11:47AM

Registered Member #3215 Joined: Sun Sept 19 2010, 08:42PM
Location:
Posts: 780

with the lid on works too ;) but if you try to do the powerball with it, expect the heads to go clunk on the platters and the disk to stop ^^

Patrick

Wed Mar 13 2013, 07:44PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

ive tried this with a dremel+dvd, but it was dangerous and still didnt work, ill go get my grinder...

EDIT: i see the gyro forces, most definately, but its hard to see when i tip it, which edge is rising and which is falling.

Dr. Slack

Wed Mar 13 2013, 07:45PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Patrick wrote ...

im trying to figure out how gyroscopic precession works, but the wiki article is complicated, the flat spinning disk has two axis of rotation and im confused, is there a better explantation anyone can point to?

i thought the rotations (torque, and spin) were suposed to oppose each other (CW and CCW),the difference between the two then generating the straight vector out the prependicular "top" ? right? or am i crazy?

???

I don't know if this helps, but this is how I thought about it, when realising that the time integral of torque was limited for the gyroscopic effect.

Consider first a reaction wheel, to provide pitch torque. In the first diagram, the axis of the wheel runs across the craft from left to right. We are sitting at the back, looking forwards.

If the wheel is stationary, or runs at constant speed, there is no torque. If we want to change the rotation speed of the wheel, we apply a torque to it whose reaction which will pitch the nose of the craft up or down.

If we want to apply a continuous torque to the craft pitch, then that means we will apply a continuous acceleration to the wheel. At some point, the wheel will reach a maximum speed, limited by the wheel strength, the motor supply voltage, or both. At this point, we can supply no more torque to the craft.

The angular momentum of the wheel is a vector quantity. The only unique direction in a spinning wheel is along the axis, this is therefore the direction that the angular momentum vector takes. Conventionally, the right hand rule is used to disambiguate the two possible directions. Wrap your right hand round the axis with the fingers pointing in the direction of wheel rotation, your thumb points in the direction of the angular momentum vector.

The next two diagrams show the maximum speeds that can be obtained in either direction. Applying a torque changes the momentum vector.

Now we can change the momentum vector on the craft left-right axis by other means. Let's say we have a wheel like the second picture, vector pointing to the right. Now instead of changing the speed of the wheel, leave it running at exactly the same speed, but turn the axis. Take the right hand bearing and lift it up, while pushing the left bearing down. Keeping turning it until the W vector points up as in the 4th picture. If trying to make sure that the wheel rotation arrow is consistent from picture to picture is doing your head in, try applying the right hand rule to both pictures.

What has happened? The system had finite angular momentum along the left-right axis, and now has none. The up-down axis used to have none, and now has some. We know from thinking about the reaction wheel that changing the left-right axis angular momentum pitches the nose. So rotating this wheel's axis about the fore-aft axis also pitches the nose.

This is how a gyroscope works, by changing the amount of angular moment along one axis, by being turned about another. The position taken by the fourth diagram is the normal position for a propeller. Instead of making radical changes to point fully left or fully right, it will tend to make small movements to left or right. When it does, there will be a component of its fixed angular velocity that will be directed to left or right, which will cause the nose to pitch up or down.

What about the torque about the up-down axis? In practice there will be two counter rotating props which will move in opposite directions. The yaw axis torque changes will cancel, the pitch axis changes will add.

The limits of fully left and fully right are useful to demonstrate that there is a limit to the time integral of torque that a gyroscope can apply.

Patrick

Wed Mar 13 2013, 08:03PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

based on further grinder experience, i beleive the above force diagram is right. im still pondering Dr. Slack's comments...

Uspring

Thu Mar 14 2013, 11:55AM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

I believe you got it backwards, if I understand this correctly, meaning the red arrow to be a force and the blue arrows to mean the direction the gyro responds. The angular momentum (AM) behaves like a vector according to the right hand rule: The AM points to the direction of your thumb, while the circular motion is shown by your fingers. In your case, the thumb points downward. You then apply the red force, which means according to the right hand rule that you add AM in the direction pointing at you.

If you now add up the original AM (pointing down) and the AM you supply (pointing to you) you get a tilt of the axis of rotation opposite the direction the blue arrows indicate. Or, you would need to apply a force according to the blue arrows to keep it from tilting that way.

You can view this in a different way. Assume you apply a force according to the blue arrows. That means adding AM pointing to the right, which would tilt the rotational axis like the red arrow indicates.

BigBad

Mon Mar 18 2013, 05:03AM

Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
Location:
Posts: 600

Yes, you've definitely got it backwards.

Actually precession is fairly easy to understand.

In the diagram you apply a torque on that side of the disk, consider the rotor tip of the rotor as it spins around.

You might think that the torque would just cause the disk to twist up on that side, but the rotation delays that twist by 90 degrees in the following way.

What happens is that after the tip passes the furthest point away from us (the neutral axis that the torque is being applied about) it will certainly feel the twisting force (much as normal) and start to rise but it will reach its maximum rise as it reaches the point closest to us, because it's going to be rising due to the force that whole half turn, but after it reaches the nearest point, the nearest neutral axis, it will only then start to fall.

So if you think about it, applying a force where you've drawn it will cause the rotor disk to tilt upwards on the nearest edge.

It's really just linear momentum, once it's risen up, it takes time to fall again, but it looks really strange.

OK, now if I say that quickly enough you'll believe that that's how it works, but you may have something nagging at you; if it's angling up with the closest edge, why doesn't that rotation cause a 90 degree rotation too?

The answer is that the precession is always associated with a slight wobble called nutation; which is imperceptible at high spin speeds, but is bloody obvious at low spin speeds. And you might think that that wobble is just irrelevant detail, but that wobble actually makes it all work, roughly speaking as the disk spins the torque causes the wobble, and the wobble ratchets up the rotation angle. (As I say it's easiest to see at low spin speeds).

The other thing is that you shouldn't assume that the rotor disk is completely rigid, in the real world, it's always slightly springy, and again the slight springiness is critical to it all working; the nutation needs that slight springiness; as long as there is any springiness at all (and all real world materials are springy) then it will precess in the way everyone is familiar with.

Hope this helps.

P.s. the maths to all this crap is really simple see:

which is alll true, but it hides the nutation that is one of the big secrets.

See also

(I don't think it explains it particularly well)

IRC the Feynman lectures get it spot-on; if you can get hold of them, he explains it better than I just did.

WaveRider

Mon Mar 18 2013, 02:45PM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

Some time ago, I put together a brief mathematical review of the symmetric top (gyroscope subjected to gravitational field) *here*. There is also a numerical simulation based on Octave which you might adapt to your needs (adding torque forcing-functions, etc.). I would be happy if you find it useful..

Cheers!

Patrick

Mon Mar 18 2013, 06:12PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Ah crap...

its possible i got it backwards, but its working better this way, so i dont want to change it till the new mechanics are ready... then well see for sure.

Waverider, ill look, but i dont know anything about Octave. And even less about lagrangian mathematics.

BigBad

Mon Mar 18 2013, 11:23PM

Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
Location:
Posts: 600

The proper maths for this isn't actually that difficult, you don't need Lagrangians. it's a single first year undergrad physics lecture, really just A-level maths.

You need to know about cross products and summing vectors, and that's about it.

basically it's just

T = dL/dt

where L is the angular momentum expressed as a vector and T is the torque expressed as a vector.

L is just the sum of:

L = r x mv

over the rotor.

and the Torque is just

T = r x F

where r is where you apply the force, and F is the force vector.

That's all the maths you need for your stuff.

WaveRider

Tue Mar 19 2013, 08:37AM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

I assumed Patrick has in mind to develop a physical model of his flying machine so he can simulate the effects of the rotors, coupled to the full airframe. This is much more than "just A-level maths" and is generally approached using Lagrangian techniques, especially if the plan is to add the control system to the model later.

This would be a fun problem that might actually yield results worthy of publication in a technical journal. I could help you through some of the details if this is what you want to do.

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