Question About Hybrid (IC + generator) and Rotation. / General Science and Electronics / Forums

Forums

4hv.org :: Forums :: General Science and Electronics

« Previous topic | Next topic »

Question About Hybrid (IC + generator) and Rotation.

previous 1 2 3 4 next last

Move Thread

LAN_403

Andy

Sat Jan 17 2015, 02:31PM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874

Dr slack not sure what you are meaning, as the centre of the object even away from gravity in space and complete isolated will rotate, as a centre point will be created based on its own gravity, unless you dont remove all input output varaibles, if the system is contained and isolated then you can transfer koules from the battery to roation, which will turn it like the pic below, if the input ouput parts swamp or flood the effect, it not because it can,t do it, say a compass needle compared to earth and a neo.

1421505075 4266 FT168370 Img 20150118 032510

Sulaiman

Sat Jan 17 2015, 08:49PM

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

I believe that it is not enough just to have cw and ccw rotational inertias equal
they also have to be balanced about the centre of mass
hence the arrangement that I suggested.

Patrick

Sun Jan 18 2015, 05:23AM

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

Sulaiman wrote ...

I believe that it is not enough just to have cw and ccw rotational inertias equal
they also have to be balanced about the centre of mass
hence the arrangement that I suggested.

if the mass1 X radius1 = mass2 X radius2 shouldn't that mean I + -I = net 0. even if its off center? or is the center off of rotation a lever?

Dr. Slack

Sun Jan 18 2015, 10:31AM

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

Please guys, don't invent new physics, don't invent new difficulties, it is really, *really* simple.

Take a closed system, comprising perhaps a rigid frame with an IC engine geared to a generator. You can represent its linear momentum by a single 3-vector (x,y,z or i,j,k vector), and its angular momentum by another 3 vector, a single vector. Its linear intertia acts *as if* all its mass was concentrated at a point called the centre of gravity, this point does not have to be co-incident with any physical point of the body, think of 'C' shape for instance. It is *fully described* by a very small number of parameters.

To find the system momentum, vectorially add all the component momenta. This goes for both linear and angular momentum.

The angular momentum of a rotating body is I.w, where I is moment of inertia on that axis, w is the rotation speed. If the engine and the generator are geared such that we.Ie = -wg.Ig, then their sum will be zero, regardless of the magnitude of any w.

Now if the frame is non-rigid, and this may be where the draught of conceptual doubt is blowing up your trouser leg, the axes of we and wg will change a bit if the system is rotated by an external force, because their gyroscopic forces will not be fully resisted by the compliance of the frame. I haven't drawn a diagram yet to see whether they will continue to sum to zero, or whether their sum will now be finite due to them no longer lining up exactly, but I strongly suspect the latter.

Anyhow, the point I'm making is that a closed rigid system, seen from outside, just doesn't have enough external variables to express the internal distribution of contributors to momentum. In theoretical cosmology, the same sentiment is expressed as 'a black hole doesn't have hair'. The internal elements can be distributed arbitrarily, and the angular momentum will be exactly sum_over_n(In.wn). If you have designed the gearing and axis directions so that that sum stays zero, then the system will not demonstrate any external gyroscopic effect or generator speed change slew, regardless of generator speed.

<edit>

While prepping the veg for the mid-day meal, my wife noticed me staring into the middle distance. Here's the result of that stare ...

The assumption of the rigid frame holding the components is quite important. Consider that instead we were trying to describe the electrical charge of the system. Different arrangements of charged bodies within the system would give different electrical fields externally *unless* all the charges were within a conductive box that formed the boundary of the system. Then all that could be determined externally would be a total charge, regardless of how that was made up of internal charges. The conductive box does electrically what the rigid frame does mechanically, it combines all the components together and delivers a single figure to the outside world. The same will go for any conserved quantity.

It is the conductive screen that allows us to call an ensemble of charges a system. Similarly, it is the rigid frame that allows us to call a collection of mechanical components a system.

</edit>

Uspring

Sun Jan 18 2015, 12:33PM

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

Dr. Slack wrote:

it is really, *really* simple.

While what you write is correct, it is not obvious, at least to me. Consider the case of 2 momenta, equal and opposite, but not on the same axes. Then, by vectorial addition the linear momenta will cancel, but the net angular momentum is not 0.
Now think of 2 angular momenta, equal and opposite, but not on the same axes. Again, they will cancel, but won't there be some higher order moment left over? I think not, but I'm not sure, why.

You sure don't want open the can of worms by discussing non-rigid frames. Every degree of freedom is an extra internal parameter, just think of a mass connected to the rest with a spring. Even rotating bodies inside the payload aren't really consistent with rigidity.

Sulaiman

Sun Jan 18 2015, 02:17PM

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

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.

gyroscope mathematics eludes me so I chose the simplest configuration that I could think of,
given that the rotating masses can not be co-incident.
I'm not even sure that the arrangement that i suggested WOULD cancel gyroscopic forces.

Dr. Slack

Sun Jan 18 2015, 03:17PM

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

Uspring wrote ...

Now think of 2 angular momenta, equal and opposite, but not on the same axes. Again, they will cancel, but won't there be some higher order moment left over? I think not, but I'm not sure, why.

Yeah, that one!. It used to do my head in at school when I was 14/15 ish and calculating the stability of a ladder standing on ground with friction, leaning against a frictionless wall, what are the critetira for it sliding down. But eventually, doing the sums over and over again, it gradually became clear. You can choose to take any axis, and you still get the same answer, as long as you include all the terms correctly.

If the angular momenta you refer to cancel, then the axes of rotation are parallel. From the point of view of angular sums, it doesn't matter where you take the actual rotation axis to be, the total angular momentum is
a) the angular momentum of the body about its c.of.g and
b) the angular momentum of the body's c.of.g about the axis you have chosen.

But, and it's a very important but, although the axis of rotation is a defined line is space, the angular momentum vector has a direction only, not any given position. This angular momentum can be considered to refer to any of the parallel axes you are considering.

In the specific case of this vehicle with an engine and generator on board, we are assuming the vehicle is not rotating, so the (b) line above is identically zero. All that's left is the (a) line, which is independent of any chosen axis of rotation, only depending on the body itself.

If you say 'whoa, the vehcile *can* rotate!', then sure, of course it can, but that then is a different problem, where you have to calculate the MOI of the total system, and worry about the system's angular velocity, momentum etc.

Exercise

Take a thin stick, one end in each hand, and bend it into a bow shape. Torque is a close cousin of angular momentum, and obeys similar rules about being considered to act on any arbitrary parallel axis. The torque at each end of the stick causes it to bend. Calculate the bending moment in the middle of the stick, and at the ends. There are four forces acting on the stick, two at each end. Take moments of all of these about several different axes of rotation. Your hands aren't in the middle, why is the stick bending there?

Uspring

Sun Jan 18 2015, 03:52PM

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

Getting somewhat OT here, but I guess, the way the stick bends depend on whether you just press the ends together, i.e having force components parallel to the stick, or bend it by two equal and opposite forces near to each other at each end. In the latter case I'd expect it to be bent everywhere, not only in the middle.

Here's a more OT exercise for you: Why is angular momentum a vector?

Sulaiman

Sun Jan 18 2015, 04:25PM

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

"take a thin stick, one end in each hand, and bend it into a bow shape. Torque is a close cousin of angular momentum, and obeys similar rules about being considered to act on any arbitrary parallel axis. The torque at each end of the stick causes it to bend. Calculate the bending moment in the middle of the stick, and at the ends. There are four forces acting on the stick, two at each end. Take moments of all of these about several different axes of rotation. Your hands aren't in the middle, why is the stick bending there?"
................................................. .....................................

there need be only one force at each end of the stick,
e.g. think of a bow with only tension in the string.

Dr. Slack

Sun Jan 18 2015, 07:30PM

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

I'm pointing at the moon, and you're looking at my finger! While you can make a bow shape in a stick with a bow-string, or by glue-lam of several thin sheets, neither was the point of the exercise. The point was to describe a thought experiemtn whereby two couples in two seperate places cancel each other out, and you can see their effect in the bend of the stick, to demonstrate that the rotational axes need not be co-linear for cancellation of torques. The only difference between torque and angular momentum is multiplication by a period of time, so all else is the same.

previous 1 2 3 4 next last

Moderator(s): Chris Russell, Noelle, Alex, Tesladownunder, Dave Marshall, Dave Billington, Bjørn, Steve Conner, Wolfram, Kizmo, Mads Barnkob