- #106
OmCheeto
Gold Member
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pfft!Nidum said:+1
I'm pretty sure the "wheel" was once considered a toy.
pfft!Nidum said:+1
RonL said:@OmCheeto , Michelob sends their thanks, their bottom line is looking better and my wife is starting to get concernedOld school to Metric is about to strain my sanity, if in fact I had any in the first place
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I'm going to put my numbers down and if anything really looks wacky, one might see any error without me trying to explain what I did.
My trifilar pendulum works well, line length ............2159mm
Flywheel.....................1502.52 grams
Radius of string....................69.85mm
Radius of Flywheel..................63.5mm
Tau (as you described)...1.10s X 2...............2.20s
Flywheel moment of inertia (based on MOI link, I posted).........6,058,556.43 ??
Flywheel radius of gyration (based on MOI link, I posted).........63.49mm
Using the equation you posted, I got a number in the amount of...210,023.778
Other numbers I got were more like the weight of the moon
I wonder if the radius of gyration is supposed to come out so near to the radius of the object, that seems weird
Any way hope you have been well.![]()
Of course I saw your post.RonL said:@OmCheeto Did you overlook post #95 ? I copied and pasted his comments and at the end of the you tube video it brings up his next video on balancing spinning tops ( I think it is very interesting )![]()
I have no idea how that worked. But it did.Charles Kottler said:H = Sqrt( L^2 - (2*Pi*r*N)^2 )
Charles Kottler said:Having slept on this is has become a bit clearer (I think...). The angle between the cord and the ground probably remains constant along the entire length. The formula I suggested was derived from the length of a line following a spiral down a cylinder for exactly one turn, but was wrong - it will have covered height H and horizontal distance 2*Pi*r, or 2*Pi*r*N for N turns.
So for a cord of length L and thickness 2r,
L^2 = H^2 + (2*Pi*r*N)^2
so H = Sqrt( L^2 - (2*Pi*r*N)^2 )
In the setup you have above, the angular momentum must come from the drop in height so by measuring the maximum speed you should be able to calculate the efficiency of the system.
No!RonL said:...
I think this thread needs to go to the sea of dead threads![]()
Me too!I have learned a few things. THANKS
OmCheeto said:No!
Me too!
Along with some long lost maths skills, things keep popping up: The energy stored in the cable/thread disappears if you remove the tension from the system. (?)
I think this is a grand thread, as it makes me "think".
ps. I think 256bits was spot on when he compared this to a clock mechanism back in post #37.
Charles Kottler said:It's probably easiest to work it out backwards, and I'm very rusty with imperial units so will convert to metric. The kinetic energy of a rotating disk is:
E = (1/4) m r2 * w2 when w is measured in radians per second.
Using metric units, E will be in joules, m will be the mass of the flywheel in kg, r will be the radius of the flywheel in meters, and w will be the angular velocity in radians/second.
To convert rpm to radians per second, you'd take
(revolutions / minute) x (2 pi radians / revolution) x (1 minute) / (60 seconds)
i.e w (rad/sec) = (rpm)*(2*pi) / (60)
So, starting with the system at its' lowest point, all the energy is in the form of kinetic energy from the spinning disk.
m = 53oz = 1.5Kg
w = 1025rpm = 1025*2*pi/60 rad/sec = 107.3 rad/sec
r = 2.5 inches = 0.0635 meters
E = 0.25*1.5*0.0635*0.0635*107.3*107.3 = 17.41 joules
After 85 revolutions the disk will have been lifted 6.5 inches, or 0.165m, and the wooden bar at the bottom will have been raised twice this: 0.33m.
By this point we want all the rotational energy to have been used up (so that it will reverse direction and star unwinding again). The energy to raise a mass is the force required * distance covered, and the force is 1.5 * 9.8 (mass*gravitational acceleration), so
Edisk = 1.5Kg * 9.8m/s2 * 0.165m = 2.43J
If a steady weight of 2.5Lbs (=1.13Kg) had been applied (including the weight of the lower beam) throughout the winding phase, the energy required to raise this would have been:
Ebeam = 1.13Kg * 9.8m/s^2 * 0.33m = 3.65J
The total energy used to wind the system up is therefore 6.08J, and on release it generates 17.41Jwe have a super perpetual motion generator, giving out more than twice the input energy
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To get a more plausible answer you would need to accurately measure the weight on the bottom beam. Your disk also has a solid axle which should be treated as tall narrow disks above and below the main disk. I assume that the 1.5kg includes these, and from the look of it that would account for almost half the mass. This would have a much lower moment of inertia as the weight is more central and could well account for the discrepancy.
If you were to build this as a generator you would presumably be applying extra force in the expansion phase to build up the speed, and then use a dynamo to convert the kinetic energy to electricity as it winds up again. It would be interesting to see how the maximum rotational speed changes with an increase in applied force.
To get an estimate of the energy loss in the system, apply a fixed force and measure how much the maximum speed drops on each cycle. If possible do this with a wide range of weights on the beam. If you decide not to proceed with this, then I look forward to your next challenge!
RonL said:...
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OmCheeto said:I don't mean to be rude, but that is one of the most worthless videos I've ever watched.
That's probably a good way to tie off the thread.RonL said:I think this thread needs to go to the sea of dead threadsI have learned a few things. THANKS