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Where do the input energy go? |
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| Jan14-13, 05:40 AM | #52 |
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Where do the input energy go?Personally, I think your assumptions are false. I.e. further input energy is not required to sustain rotation. You are simply asserting that assumption without any valid justification, and since that assumption leads to a contradiction it must be false. Please post a detailed analysis using Newton's laws for a frictionless piston showing this continuous energy input requirement. |
| Jan14-13, 05:54 AM | #53 |
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In a 'realy real', non-ideal mechanism, you have to allow the structure to be rigid (enough) to allow the loss-less transfer of KE between piston and flywheel. etc etc |
| Jan14-13, 06:58 AM | #54 |
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It does not really matter how heavy the flywheel is. I am sure the KE in that wheel is not changing anything regardin the pistons. Ofcourse, the heavier the flywheel is the longer it can transfer its KE to the pistons befor it stands still, if the motor which power it is released from it. I do question yor claim that this isnt a real thing. It is possible to build and test this in real life. So since conservation of energy must be conserved, there must be some place for this input energy to go - that isnt heat... Might the energy loss be "fed back" to the motor some how?
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| Jan14-13, 07:02 AM | #55 |
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| Jan14-13, 07:30 AM | #56 |
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We are assuming that the reciprocating mechanism is not powering anything or being driven - just running freely, with no added friction. If the flywheel had no MI (mass) then the piston (in an open cylinder with zero mass con rod) would move at uniform speed along the length of the bore as the flywheel could not affect it in any way. How could it do otherwise (Newton 1), as there would be no forces on it? That would, as I have already said, result in a discontinuity in velocity at the ends of motion - the piston would either stop with a jolt at the extremity of its motion or be pulled back due to nothing more than resilience in the con rod etc. and that is not 'allowed' in a real model. Also, the on-going mean KE in the flywheel is irrelevant in this argument as it doesn't change in a situation where there are no losses. If there is an inelastic collision at each end of piston travel then there is energy loss - of course. But there is a force on the con rod because of the presence of the flywheel which slows down and speeds up the piston and forces a (timewise) sinusoidal oscillation. That force times distance (integrated) represents work done to and by the flywheel. At the end of travel, the change of direction involves no motion so no work is done actually at that time. The energy is transferred from piston to flywheel and back again, leaving the mean KE of the flywheel unchanged. Thinking about a situation like this, one must avoid being subjective about it. There is an understandable feeling that the 'jerk' at the ends of motion is necessarily a reason for loss of energy but, although the acceleration is at a maximum at the extremes of motion (as in all SHM) the motion is smooth and continuous so no energy is lost if the bearings are all tight and there is no slop. |
| Jan14-13, 07:33 AM | #57 |
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A mass bouncing on a spring 'loses' Kinetic energy to the spring and comes to a halt each end of its motion but gets it back on the way down and up. |
| Jan14-13, 08:05 AM | #58 |
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Take a seesaw. Put a heavy weight on each side of the pivot, and power it up. How fast can this seesaw run with a given input work? I bet the seesaw will stop accelerating at a given frequency even - if the weights are perfectly balanced. The lighter the weights are, the higher the frequency will be with the same input work. @sophiecentaur said, that shorter stroking engines can certainly rev higher, indicates that there is higher efficiency in such a system. I want to add that a lighter piston, with same stroke as a heavy piston, performs better (Higher efficiency). I assume race cars use as light pistons as possible for the engine to perform as good as possible. So even if the friction would be the same in those two engines, the different KE in those pistons can't be neglectet. So I dare to claim that there is some wasted energy input we (Probably only myself) yet do not understand where is heading. You don't need to comment this anymore if you feel that I'm not "getting it" ( ). I will try this in practice with some of the brushless RC motors I have available - running a piston, or similar, with a flywheel, and measure the energy consumtion with light and heavy pistons.I will post my findings here when the temperature outside is acceptable (-20'C as I write)... Vidar. |
| Jan14-13, 08:39 AM | #59 |
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I have already demonstrated the analysis for your early gyroscope, but you are coming up with far more new systems than I am willing to analyze in detail. So the burden of analysis falls to you. From first principles we know that energy is conserved, so we know any assumption to the contrary is wrong. The details are up to you to show. |
| Jan14-13, 10:23 AM | #60 |
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If you try an experiment, the results will be meaningless because they will be swamped by practical errors. Just think about the following, ignoring the slight errors in geometry when a short con rod is used. Read it through with your mind switched to 'look and learn' mode. A piston being driven by a simple crank will be oscillating with simple harmonic motion (sinusoidal oscillation) and so will the same piston, bouncing up and down on a spring, which has been chosen to produce the same period of oscillation. The two graphs of displacement in time will be precisely the same. There is no loss of energy in the mass on spring so where will the energy loss come from when precisely the same motion is produced by another method? (The crank can be chosen to be as long as you like, to suit this argument - it's assumed to be mass-less). This is theory, which your argument is supposed to be based on. But your argument is based on a subjective view of the situation - which ain't Physics. |
| Jan14-13, 01:22 PM | #61 |
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I have tried to analyze:
I use a clock as reference. At 12 o'clock (Of the flywheel) the piston starts to accelerate. At 3 o'clock the piston has the highest possible velocity and KE. This KE is transfered back to the flywheel between 3 o'clock and 6 o'clock. The cycle repeats with acceleration from 6 o'clock to 9 o'clock. Maximum KE at 9 o'clock. KE is transferred back to the flywheel between 9 o'clock and 12 o'clock. The net energy is zero - zero loss. Can't believe I have not seen that. The similar will apply to the seesaw, and probably the gyro as well... |
| Jan14-13, 03:35 PM | #62 |
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Well done. You stuck at it and got there.
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| Jan18-13, 04:28 AM | #63 |
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I found an interesting and well explaining article about gyros:
"A gyro will resist any force that attempts to change the direction of its spin axis. However, it will move (precess) in response to such force; NOT in the direction of the applied force, but at right angles to it. The direction a gyro will precess also depends on the direction the gyro is spinning. Precession is actually the result of two forces: angular momentum (spinning force) and the applied force (torque). The direction of precession is always offset from the direction of the applied force. The offset is always in the direction of rotor spin. For example, when a force is applied upward on the inner gimbal, as shown in figure 3-8, the force may be visualized as applied in an arc about axis Y-Y. This applied force is opposed by the resistance of gyroscopic inertia, preventing the gyro from rotating about axis Y-Y. With the rotor spinning clockwise, the precession will take place 90º clockwise from the point of applied force. The gyro precesses about axis Z-Z in the direction of the arrow "P"." ![]() Vidar |
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