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Registered Member #2431
Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
if i have a piston engine rotating, putting out 746 watts. And a Generator taking out 746 watts (thers some loss obviously), and these two were counter-rotating, and hanging lets say from a balloon or zeppelin, would there be a continuous rotation attempting to assert itself?
i was wondering if this would be complicated by the piston having low mass and 15,000 rpm, and the generator having high mass but 3,000 rpm.
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
thoughts only, no practical knowledge;
1) the mass of the pistons is irrelevant as the motion is reciprocating (linear) with vibration being the only concern.
2) if you have an arrangement where rotational inertias are symmetrically cancelled about the centre of mass then the system should behave as a simple solid mass e.g. crankshaft with similar flywheels at each end, gearing/coupling at each end reversing the direction of rotation similar generators at both ends, each with rotational inertia of half of the engine+flywheels rotational inertia x gear ratio
then for constant or accelerating/decelerating speed there should be no nett torque on the system and no gyroscopic forces.
you could simplify this to a single generator with appropriate gearing and rotating masses but adding 'dead' weight seems inappropriate for aerial use.
for weight considerations lighter flywheels 'geared up' to higher speed would be appropriate
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
your closed system hanging freely from a balloon will behave as a single entity, with a total mass, total angular momentum, total linear momentum etc.
This means that if you make the angular momenta of the engine and generator equal and opposite, then the total system angular momentum so gyroscopic reaction will be zero. There will still be bearing loads on the engine and generator with respect to their mountings in the frame.
The system will only try to rotate if you are shedding angular momentum somewhere, so perhaps an off-centre engine exhaust.
If you have balanced the engine and generator momenta, then there should not even be a brief twist as they change speed, the total is zero at any speed.
Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
If the motor is at the front and the genny at the back compared to a centre pivit point it should rotate as the motor is going right the genny is going left, but if they were both at one end from centre then no they will cancel
Edit I would have the setup on the y axis if z axis is front to back, with the distance back based on centre of anti gravity, for a zapplion also based on centre of pressure, if the aircraft was going to travel quickly forward, I wouldhave it slightly further back from those points, with the x axis centred.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
There is no discussion, unless you want to invent new physics as Saywer is trying to do with a reactionless drive. If no mass goes over the side, there is no torque to rotate it. If the sum of the angular momentum vectors of all the inf=dividiual components adds up to zero, there is no angular momentum so no gyroscopic tendency for the entire system.
No if you start introducing 2nd or third order effects like vibration inducing a small back and forth motion of some part of the structure that induces a net aerodynamic force, like the fin of a diver, then go ahead, but I don't think that's what you or anyone else on the thread is talking about. Let's put this mythical motor generator set into a vacuum, as well as hang it from a balloon, remove earth's magnetic field, solar radiation, birds that might land on it, and then we're back to simple Newtonian/classical physics where momenta, both linear and angular, add as vectors.
Registered Member #2431
Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
Dr. Slack wrote ... Let's put this mythical motor generator set into a vacuum, as well as hang it from a balloon. . .
Your dam right its mythical, so dont any of you go jumping to conclusions on what i might be using for either...
So, if I=(1/2)mr^2 equals rotational inertia.
Then, IC = I=(1/2)mr^2 Generator = -I=-1(1/2)mr^2 -I + I = 0 The net rotation is zero, right?
Im thinking of a thin diameter cylinder shaft, and then a hollow large cylinder. But when looking at the calculus and its final result (I=(1/2)mr^2) i dont see where speed (RPM) rotational speed matters. or does the definition of mass include a time interval? it seems like that a^2 bit does.
im thinking of a SMPS with active PID feedback so as to avoid wild oscillations should a load change suddenly occur. taking to much power out, or suddenly cutting back and over revving could cuase unpredictable behavior either way...
The equation for the moment of inertia I is correct. But you want the angular momentum A to cancel, not I. The angular momentum is given by A=2*pi*f*I, where f is the rotational frequency. There are 2 f's and 2 I's, i.e. for the motor and the generator.
If the angular momenta cancel, there will be no particular rotational effect.
If they don't and the rotational frequencies don't change, you'll have gyroscopic effects. They don't matter, ias long as you don't try to turn the whole thing by an external force.
If the angular momenta don't cancel and you revv up and down, you'll see some torque on the payload.
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