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Question About Hybrid (IC + generator) and Rotation.

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Uspring

Wed Jan 21 2015, 11:10AM

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

Dr. Slack wrote:

But I will admit I'm having a hard time though to get anybody else to see the sense and simplicity of it.

I think the hard part to grasp is, that angular momentum can be described as a vector. To be precise it is a pseudo vector (axial type) i.e. it doesn't change sign under parity inversion like normal (polar) vectors do.
More formally it is a second order tensor, though one which shows many of the properties a vector has. While I agree to that what you say technically, I have a different opinion on its simplicity. It's not really intuitive.

Dr. Slack

Wed Jan 21 2015, 12:05PM

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

Well, my mathematics stops pretty hard somewhere in the middle of vector calculus, so I can't comment on whether angular momentum is *really* a second order tensor. However, I do know that if you use a consistent convention ( I think we used right hand rule in high school and college), you can handle rotation and angular momentum as 3 vectors, and do all the things you'd need to do to calculate angular momentum, add them by putting the origin of one on the head of another, and multiply them by scalars. Anything else is complication for its own sake. For instance, scalars are special cases of complex numbers, which are themselves special cases of quaternions, but we handle each as simply as we can in any specific application. We don't teach complex numbers to 10 year olds, and there's no need to introduce tensors when dealing with rotation.

Vectors handles gyroscopes just fine for instance, as the applied torque vector, times time to give angular momentum, adds at right angles to the angular momentum of the flywheel, to change its direction, without changing its magnitude. Precession, described correctly in direction and magnitude. That is as far as you have to go into the maths.

I will admit that possibly few people have met rotation as a vector. But if you look on wikipedia, there is some sort of introductory treatment of rotation and angular momentum as vectors. Needless to say, it is not all that easy to follow, few wiki articles are for the uninitiated, but it is a start, and shows the concepts and conventions that you need.

Maybe I'll read through pseudovector on the same site for my further education!

Uspring

Wed Jan 21 2015, 01:50PM

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

...so I can't comment on whether angular momentum is *really* a second order tensor.

The usual kind of vector describes e.g. a shift in spatial coordinates, which amounts to adding some values to each coordinate. The "vector" w describes rotation in space, which can't be accomplished by adding some values to the coordinates but involves a matrix operation. The angular momentum, being the product of w and the angular moment of inertia, inherits this property, i.e. being a matrix, which is the stuff second order tensors are made of.

Sulaiman

Wed Jan 21 2015, 03:39PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Please don't get me wrong, I don't want to argue, just understand;
imagine a rigid frame with four identical gyroscopes arranged in parallell in it like this

when all four gyros are stationary
or if only gyros A and D spin in opposite directions at the same speed
or if only B and C spin in opposite directions at the same speed
I can imagine there being no nett gyroscopic forces

but if only gyros A and B (or C) spin, in opposite directions at the same speed
there is still no nett gyroscopic action ?

i.e. distance from pivot or centre of mass makes no difference ?

afterthought .... maybe clearer if I'd just used three gyros.

P.S the Laithwaite lectures are fascinating and some boggle my mind

going to re-watch them now to see if I can get a better feel for this stuff.

Dr. Slack

Wed Jan 21 2015, 04:02PM

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

If A spins left, and D spins right at the same speed, there will be no nett gyro effect on the system. The frame itself will experience an almighty internal stress when the frame is turned, as A and D still experience gyro forces in their own right.

If A spins left and B spins right, the external situation is exactly the same. Two gyros on parallel but non-coincident axes are spinning (the same case for A and D spinning) with their angular momenta adding to zero. Where on the frame does not matter. Still no nett external gyro effect. The frame has mass attached to it in various places, some of that mass happens to be stationary wheels C and D. But the symmetry of the frame doesn't affect the total angular momentum.

If A spins left, and B, C and D all spin right at 1/3rd of A's speed, again there will be zero total angular momentum, and zero external gyro effect. It really is just adding up the angular momentum as vectors.

Unfortunately, Laithwaite ended up thinking you could make a space drive with gyros, which you can't. One of his demonstrations at the Royal Instution used a small girl to hold a heavy gyro at the end of a 3m pole. His thesis was that she wasn't strong enough to do that unless the gyro was essentially generating anti-gravity. Normal gyro precession was sufficient to explain what was happening. Take Laithwaite's gyro presentations with that pinch of salt in mind. Now, his 3 phase linear motors, he didn't ever go loopy on those!

Ash Small

Wed Jan 21 2015, 04:41PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Dr. Slack wrote ...

.
Au contraire, Ash, my answer is definitive. Perhaps you're confusing lack of concensus with not having a definitive answer. But I will admit I'm having a hard time though to get anybody else to see the sense and simplicity of it..

This is what I meant, lack of concensus, Neil.

Dr. Slack wrote ...

As long as, that is, we assume a nice clean theoretical system where these loss terms don't leak out and cause various second order thrusts that complicate things in the real world, cooling air exhaust from the generator blowing out in a jet for instance.

.

This is the bit that gets me: 'As long as we assume a nice clean theoretical system'.

If we assume that exhaust and cooling are 'controlled', and have no overall effect, we are left with friction and copper losses.

Friction in the IC engine just reduces the overall 'power' in the system. Only the power transferred to the generator actually has any effect.

All of the transferred power contributes to torque on the generator side, though. Friction in the generator bearings contribute to induced torque, and copper losses have already been converted from mechanical power to electrical power, so the torques on both sides are identical.

But then, they have to be, due to 'conservation of energy'.

Dr. Slack

Wed Jan 21 2015, 05:40PM

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

Ash Small wrote ...

Friction in the generator bearings contribute to induced torque,

Torque between what and what? Any torque due to this is internal, not external to the system.

Anyhows, another way of looking at Sulaiman's multi-gyro.

Let's say A is spinning right, B left, and C and D stationary.

Now impose a 1 radian per second rotation on the frame, at right angles to the spin axes. The rigid frame will impose that rotation rate on gyros A and B, which will each react with a torque. However, as their spins are opposite and equal, the torques will be opposite and equal. The rigid frame will experience two equal and opposite torques (remember the stick bent into a bow-shape where one hand is twisting it left, the other twisting it right?). The two torques will cancel in the frame (or the stick), yielding no net torque to the outside world.

If the frame is rigid, it does not matter where on the frame the torques act, symmetrical or not.

Take care if you have ever used a torque spanner. Some are calibrated assuming that you apply your linear force in a particular place. If you don't, or also apply a torque as well as a linear force, you will get the wrong result. This effect may be the cause of some of the doubt and uncertainty expressed here. Fortunately gyros only ever apply a pure torque through their bearings when they are rotated.

Sulaiman

Wed Jan 21 2015, 05:55PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Thanks !

D'oh !
you wrote it several times .."torque"
in my mind I kept thinking "force"

nevertheless, a new addition to my "To Do" list

#2,532 investigate gyroscopes

assuming that I live long enough to get the other 2,531 done :)

Patrick;
now you have a consensus of at least one knowledgeable member
and one newly enlightened convert.

Ash Small

Thu Jan 22 2015, 12:03AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Dr. Slack wrote ...

Ash Small wrote ...

Friction in the generator bearings contribute to induced torque,

Torque between what and what? Any torque due to this is internal, not external to the system.
.

I was just trying to 'balance' the two sides of the equation, so to speak. If all the 'internal' torques cancel, there can be no overall 'external' torque.

Maybe it's just a case of there being more than one way to skin a cat?

Uspring

Thu Jan 22 2015, 09:02AM

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

Here's an intuitive way of understanding gyroscopic forces: Imagine a top, made, e.g. from a circular disk, in front of you, the axis being vertical and spinning clockwise. The rim at the right will be moving towards you, the one on the left away from you.

Now tilt the tops axis away from you. The rim on the right will then have an upward velocity component and you need an upward force on the right to generate that. On the left you'll get a downward velocity component and you'll need a downward force there. In total this means, that you need a turning left torque on the axis to account for that. This torque appears only during the tilting motion and will be larger for faster spinning tops, since then the vertical components will be larger.

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