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the need for symmetry that I expressed was not just to cancel rotational inertias but to also prevent gyroscopic action which would be problematic for a system suspended from a maneuvering aerial vehicle.
The angular momenta don't need to be colinear or otherwise placed in any particular way to cancel. They have to be equal and opposite in direction. If there is no angular momentum, there is no gyroscopic effect. Possibly it helps you to think about bent sticks to explain this. I failed.
Registered Member #2529
Joined: Thu Dec 10 2009, 02:43AM
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Ok, I've been trained in the theory of this, and I know real world examples.
Uspring is correct, provided the two angular momentums are equal and opposite, you can spin it up and nothing will happen.
If they're not equal and opposite, then the main body will start to spin in the opposite direction.
This effect is used in space craft, look up 'momentum wheels'.
To be equal and opposite Iw has to be equal and opposite (where w rotation vector which is along the rotation axis and has a length equal to the angular rate, I is the moment of inertia of each rotor.)
This can happen when the spin axes are parallel (they can be colinear, but don't have to) and the spin directions opposite, and the spin rate and angular momenta are correctly in proportion.
So, yeah, continuous rotation unless you momentum dump, by exerting an external torque at start up and during shutdown to stop it, or unless the parts counter rotate and have the right ratio of moment of inertia and spin speeds.
edit: if there's a non rigid frame then the axes will no longer cancel. If you add the resulting vectors you end up with a net angular momentum vector, and the rest of the body will happily start performing that. That is used for stuff also, there's a weird design for a device for balancing a bicycle-type thing on a wire using a feedback loop based on this.
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The important part is the counter-rotation and the equal moments of inertia and spin speeds. As I understand it, this cancels all external forces and moments irrespective of the relative positions of the two flywheels. (for qualifications about proportionality and spin axes etc. see Bigbad's post above)
The obvious way of mounting a generator directly to an engine has both of them rotating in the same direction, so there will be a gyroscopic effect and a torque reaction when the system accelerates or decelerates.
To achieve counter-rotation you would need gears or a twisted belt.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
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Steve Conner wrote ...
The important part is the counter-rotation and the equal moments of inertia and spin speeds. As I understand it, this cancels all external forces and moments irrespective of the relative positions of the two flywheels. (for qualifications about proportionality and spin axes etc. see Bigbad's post above)
The obvious way of mounting a generator directly to an engine has both of them rotating in the same direction, so there will be a gyroscopic effect and a torque reaction when the system accelerates or decelerates.
To achieve counter-rotation you would need gears or a twisted belt.
Equal MOI and spin speeds is sufficient, Steve, but it's not necessary. All that's required is that the angular momentum products, MOI*w, are equal and opposite.
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I don't think anyone has replied with a 'definitive' answer here, but the more I think about it, I think the losses, whether they be friction losses or copper losses, don't affect the 'angular motion' thing, so you are left with 'net' power produced on one side and 'net' work done on the other, so they 'must' cancel.
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Joined: Thu Feb 09 2006, 08:29AM
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Ash Small wrote ...
I don't think anyone has replied with a 'definitive' answer here, but the more I think about it, I think the losses, whether they be friction losses or copper losses, don't affect the 'angular motion' thing, so you are left with 'net' power produced on one side and 'net' work done on the other, so they 'must' cancel.
That's the best I can come up with so far.
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. It may be through lack of diagrams, but all the relevant equations have been presented, total angular momentum = sum_over_n(In.wn) where Ix is the moment of inertia of the xth component on that axis of rotation, and wx is the xth component's angular velcoity. Set that sum equal to zero, and Bob's your uncle.
Angular momenta equal and opposite is necessary and sufficient for there to be a) no external gyroscopic effect exhibited by the system and b) no external torque reaction when the internal components of the system speed up and slow down.
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.
Note that angular momentum is a 3 vector, it has a direction and magnitude but *no position*, unlike axis of rotation which is a 3 vector passing through a 3 point. You do not need axes of rotation co-linear to cancel the angular momentum vectors, just as when you bend a stick slightly between thumb and middle finger of each hand, you don't need both hands in the same place to cancel the two couples so that the stick doesn't accelerate. The rigid frame of the engine/generator system does the same job as the stick, transferring forces from one component to another as required by the torques.
The power generated by the engine, dissipated in various losses, friction, windage, noise, heat, and output by the generator are completely irrelevant. Total energy will be conserved of course, chemical in the fuel eventually ending up as heat in outputs and losses, and some for a while store in kinetic energy energy of the components, but that sum is totally unrelated to the momentum question, which is only sum which affects external gyro and torque reactions. Do you remember doing collisions between billiard (pool) balls in high school maths or physics, when you had to take account of coefficient of restitution (look it up on wikipedia if you've not come across the term)? If so, you did seperate sums for the momentum and the energy balance. Whether the collision was entirely elastic (no losses) or entirely sticky, the momentum sum was the same.
When doing system sums, use the right variables. If you want to see what the system costs, sum over the component costs, if you want to see how much energy is stored, sum over the component stored energies. If you want to find the linear momentum of the system, sum over the mass.velocity products of the individual components. And if you want to find the total gyroscopic effect, or torque reaction effect then sum the angular momentum vectors.
Sometimes, things are up for debate. Sometimes, answers are correct. I am not being arrogant, I am the first person to say that I don't know, or am uncertain, if that is the case. However, on this topic I am correct. Anybody that has trouble with that would do themselves a favour by trying to understand how the simplicity of my approach is the correct one. I know it's not always easy, I had a helluva time getting my head round special relativity, accepting the authority of the teachers, just going with the equations. Accept my authority, let go of irrelevant things like friction, power, axis of rotation, and follow the angular momenta, it's all you need. Start by accepting the result, even if you don't fully understand it.
As a hint, you can derive all of the angular results, including gyroscopic action, from linear mechanics. I did, for the fun of it, a long while ago when I was in college, which is perhaps why I feel comfortable with the angular equations. Most people freak out at a gyroscope, Laithwaite even trashed his career on one.
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Question About Hybrid (IC + generator) and Rotation.
