Drawings by preoccupied

[Tarsier79]

The problem the way I see it, isn't resisting fall, but increasing PE.

[preoccupied]

Tarsier79, PE= potential energy?

Lets say there is one pound weight in the center and two one pound weights on the outside. The block and tackle connected to the center weight is 1024:1 or wrapped around 10 times (is that correct?). If the leverage is determined by the angle of the string then at 90 degrees you have infinite leverage and any amount of weight will move it from that position. Very little string is moved and I don't know how much (can someone tell me how to find out?), but whatever it is it will be 1024x more string moved into the block and tackle if it can move it.

[Tarsier79]

Think carefully. Why does the angle of the string determine leverage?

[preoccupied]

Does the angle of the string determine leverage in part because of how much string is pulled? So when there is infinite leverage there must be no string being pulled.

But my calculator says 1/(sin0)=cannot divide by zero so there must be infinite leverage near zero. When I punch in the numbers there is a huge advantage to the string near horizontal. 1/(sin1)=57.29, that's 57.29 times the weight in leverage at 1 degree, almost horizontal. I think that maybe between 0 and 1 degree it approaches infinity leverage near 0. If it exponentially increases into infinity then even a small amount of string movement from that angle might be more than naturally would occur with other mechanical devices. Unless the leverage is gains is times 90 the weight, then it is not infinite.

[Tarsier79]

I guess you are on the right track.

But simpler:

Leverage isn't ultimately decided by the string at all. Providing the weights are connected in some action/reaction system, it is decided by the Potential weight loss of both weights in the system. If one weight falls twice as far as the other, then I guarantee you the system it is attached to will ensure it also has twice the leverage as the other.

Think of a basic see-saw with a 1:2 leverage. the weight at "2" falls twice as fast.

Think of any block and tackle system. If one weight falls 2, 3 or 4 times quicker than the amount the other, then it will also have that much leverage in comparison.

Fact: Gravity pushes mass down.

Fact: Without energy applied to a system, total mass in a system will move downwards, the way gravity pushes it.

How then, do you overcome this to create a "Gravity engine"?

[pequaide]

If the ratio is 1024 to one then the smaller mass is moving 1024 times faster. Calculate the KE ½mv² of the two masses. It is not the distance that gives you energy it is the speed.

[preoccupied]

Using an online trig calc I calculated that the mechanism would pull less string than the leverage gained the smaller the angle got by almost 10 times. It is super inefficient idea. Unless compressing stuff using inefficiency can give material special energy, I found nothing useful. Where does the energy go anyway? If I were to try to compress something with this string exchange, I would lose energy it seems because it is super inefficient. Do I destroy energy? Maybe I'm too much of a noob.

0.0001523280439077 is string moved with 1/1sin=57.295, 57.295*0.0001523280439077=0.0087276352756916715
0.000001523089032 is string moved with 1/0.1sin=572.95, the string pulled gets smaller faster than the leverage gets larger
0.0000000001523088 is string moved with 1/0.001sin=57295.77
0.0000000000015232 is string movedw ith 1/0.0001sin=572957.7

[preoccupied]

When I was using drawing the classical overbalanced wheel with the levers hanging out and I used the ball system I had in mind that if I could get all the weights to be in the same spot on the lifted side of the wheel and can get them to separate on the falling side that I could get the wheel to turn. I couldn't get the balls to come together all on the lifted side in the pictures from hanging positions. I have in mind having the weights fall back together on the way down like at the top right in the ms painting... If I can get the weights to separate and then fall back together on the way down then they can be in the same spot on the lifted side. In my ms painting I drew a blue line where I want the tube to go and a regular tube where I think it would go. For this thought process I would be open to both of the weights on the lifted side being further from the axle as long as they are both in the same spot but I have not thought of anything.

[preoccupied]

I think this is a good ball system for this too. There is a sharper fall and the weight would be more likely to fly out with the tube at some velocity. The yellow line in the ms painting is meant to be more elastic and bounce the tube further out. Maybe a slightly warped board?

[raj]

In this PE, Seesaw, block and Tackle, leverage and Speed debate, I would disagree with Tarsier79 and agree with Pequaide.

I think I have found a way to show that leverage and speed do not need to be in the same ratio, which could be a solution to a possible PMM.

I have done a minor build test, and it was positive.

I am currently trying build a larger demonstration to prove it again to MYSELF.

This test result is only days away.

[preoccupied]

This is my first idea for a Bessler Wheel. The very first design I ever drew I mean. I am romantically falling in love with it again.

Bessler's Wheel!!!
http://www.youtube.com/watch?v=QYzVwdbFh-U

[justsomeone]

Pre, I like the cleverness of the design but you are trying to lift three with one.

[murilo]

worry,
the basic idea is CLEVER!
It can be useful for another set with greater number of elements.
The important reduction you show on radius is a goal.
Best!
M

[preoccupied]

The ramps in the chambers make the weights pretty closely balanced at the different positions. Here I have the four chamber painting as is seen in the youtube video and I ms painted a three chamber and a five chamber example.

I want to calculate where the weights are on the horizontal axis. I don't know how to find the exact positions of the weights when they are on the ramps in the chambers but I can find the positions of the weights once they are at the end of the foot. (the foot is the door with a ramp on it that looks like a foot). This is because I know how to use the online trig calc to look at triangles but I do not know how to locate positions on circles.

[AB Hammer]

First picture of your diagram. Note each of the hinged points. When in swing, is where the weight effects to. Now count the spacing of each weight from the axle. Also realize that the one at six effect is effected as soon as it swings.

Use my grid over the design
http://www.besslerwheel.com/forum/downl ... er=user_id

Then turn it on it's side and check if you have enough weigh above to continue the actions.
http://www.besslerwheel.com/forum/downl ... er=user_id

This is a simple test but will help with a lot of designs.

Alan

PS I thing I will make a video how to use the grid some time next month.