new perpetual motion machine, check out and comment

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bucky1andonly
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my machine

Post by bucky1andonly »

1 = spring O = pulleys [ ] = box 2 = metal bearing
the spring is connected to the pulley system on the right, now depending on how much weight the spring can launch will determine how many pulleys are required to lessen the force needed when the bearing rides the box down to compress the spring. the top part should be angled down towards the box and the bottom angled down toward the spring. when the box reaches the bottom the spring will catch on a lip on the underside of the spring, and when the bearing rolls over the spring and hits the wall, it will press in a little switch to release the spring shooting the ball upward and the box as well restarting the process over. now im sure my picture doesnt offer that much insight but ill see what i can do to get a real drawing if someone would like me to try. any input on why this wouldnt work will be much appreciated

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bucky1andonly
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re: my machine

Post by bucky1andonly »

the text picture was screwed up after submitting :P
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jim_mich
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re: my machine

Post by jim_mich »

Welcome bucky!

That's what happens to text pictures! You have two choices. You can edit your post and put the text picture inside a "code" box so it displays properly. Or make a paint picture and save it as a GIF or PNG file then upload the picture.

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ken_behrendt
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re: my machine

Post by ken_behrendt »

Welcome to the board, Bucky...

I've found that the easiest for me to include images with my posts is to make a quick sketch using MS Paint. I then save this as a .jpeg image file to a Desktop folder. Then, after I finish the text of my post, I left click Browse below the post text field and can then upload my sketch from my Desktop folder. I think there is a 250 KB maximum size limit on each uploaded image. This can make uploading an animation a challenge.


ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
bucky1andonly
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my machine picture, i hope

Post by bucky1andonly »

well, i followed the advice about getting my picture up
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bucky1andonly
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New perpetual motion machine

Post by bucky1andonly »

Ok, im trying once again, and this time I was able to draw it on mspaint, i found a way to open it finally. :)
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Fletcher
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re: my machine

Post by Fletcher »

Bucky .. I take it the ball rolls to the right & enters the elevator ? The balls weight causes the elevator to descend & in doing so, via the pulley system, resets the spring mech which will then theoretically toss the ball vertically so it can roll sideways again, closing the loop for the system ?

I think you will find that the energy required to toss the ball vertically is far greater than the energy gained via the pulley reset mech. Pulleys are just another form of leverage. You can double the number of pulleys so that it is easier to lift or tension something but it moves half the distance. This scenario is best calculated by using the Work Done formula's rather than Energy per se.

Work Done = Force X Distance.
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re: new perpetual motion machine, check out and comment

Post by ken_behrendt »

Bucky...

Interesting design...looks like a self-motive pinball machine!

However, I find myself in agreement with Fletcher. In such a design, if the ball is to force the elevator to full descend, then the amount of energy stored in the spring will probably be insufficient to "fire" the ball all the way back up to the top of the device again.

But, I think your design is interesting because you are using a spring to temporarily store energy to move a weight about. That is an approach that more members on this board might want to look into.


ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
bucky1andonly
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re: new perpetual motion machine, check out and comment

Post by bucky1andonly »

wouldnt the distance for the ball depend on just the spring itself, you could have 2 x 5cm springs with different release for just based on material used and how it was made, one would make it go higher than the other, but then you would need a heavier ball as well, or more pulleys, the spring itself i think wont be very tall, so the distance the elevator needs to go would be depending on how many pulleys (correct, i really dont know), but i think it really still comes down to the spring, just need some experiments with different springs and one constant weight.
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re: new perpetual motion machine, check out and comment

Post by Fletcher »

The spring must be compressed by the right amount to provide enough acceleration for the ball to travel vertically the required height (like a pin ball release). This requires work to be done at some time in the system to cock the spring.

The ball in the elevator must move downwards the same distance, so it starts with a certain amount of potential energy. PE = mgh. It is not allowed to just fall unrestricted in a gravity field (if it did it would give a Kinetic Energy of impact KE = 1/2mv2 on impact at the bottom of the fall).

The elevator is required to do work via a pulley linkage to compress the spring. This assumes there are no losses in vertical height between sides of the mech but in reality there is as the ball must be on a down slope to traverse across to it's next position.

A simpler way of imagining the energy losses that rob a system like this would be to visualize a ball falling & accelerating due to gravity (not in an elevator). It hits one side of an ordinary see-saw balance beam. The KE of impact is transferred into the see-saw depressing it downwards. Here the other side of the see-saw replaces the compression spring so that there is a direct transference of the energy from the fall via the fulcrum to flick another ball as high as it can go.

With no normal losses of friction etc it would be an energy balanced system but accounting for normal losses we come up short, not being able to shoot the second ball high enough to get to the same PE as we started with so that it can roll across to repeat the process.

I'll find the HyperPysics link page & post it here. It is an excellent source of information about forces & how to calculate work done etc.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
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