Idea I had since elementary school

[preoccupied]

That's right alexf, This idea extends distance without additional force because it rolls down a ramp. Is that what you mean?

[preoccupied]

I'm looking for someone to calculate weights on an axle for me. I need to know the torque. It's for the radiusshift11 animation and the following animations like it. The calculations start at the heavy weight on the L shaped lever all the way down and the long lever on top all the way up. They will come into balance somewhere. I need to know where that is. Then the next calculation is two new weights on this axle. Jim_Mich made calculations and they can be referred to in the post and I think Jim_Mich calculations are correct but I don't know for sure. The two new weights are the long lever all the way up and another heavy weight all the way down. The long weight all the way up has extra torque to spare so it will push the first heavy weight up some and its own heavy weight. I want to know where that comes into balance for all the weights. Then repeat the process and calculate with a couple more long lever weight and heavy weight added IF the first heavy weight doesn't reach horizontal position. Once the first heavy weight reaches horizontal position then the calculations are complete because a weight can be reloaded. I think only one weight needs to be reloaded to cause perpetual motion.

[path_finder]

http://members.cox.net/riley101/virginiahammercam.gif
from this web site:http://www.newpath4.com/index.html

[Richard]

Im wondering if the experiance you observed with the right angle jack handle was as follows.

By pulling the jack handle on the inside a torque is established outward from the axis...and by pulling the jack handle on the outside force is wasted in opposistion to the axis?

I am facinated with the cascadeing leverage and will continue to follow....

the greater torque is established after the handoff through to reset and is so ratched to be applied in that direction...would that be a correct assumption on my part?

[preoccupied]

Richard, The ratchet needs to be connected to the long lever when it falls and the short lever when it rises. The second machine connects to the same ratchet so that it receives force from the long lever falling on both machines.

I haven't proved anything about the right angle bar yet. I must measure the torque using a torque wrench but so far the person I know hasn't found his wrench yet. I think my experience is legitimate though and I might be happy with my experiment measuring the torque.

[Richard]

Preoccupied..please be patient with my interest vs. my comprehension.

we are discussing ratcheting; this is a directional influence on your cascadeing levers. What is lock and release direction of the ratcheting?

Again..It is my impression that pe is (or should) be built to the point of hand off and then released to its greatest potential at reset.

[preoccupied]

The ratchet catches counterclockwise and the release is clockwise for both axles from the view of the animation. After the weight is transferred to the L shaped lever it will go up to a ramp that rolls down to a long lever on top and that is the most potential on the device, like you said.

[preoccupied]

If there was some kind of catch and release gear system then a new kind of improvement could be applied to radiusshift11 animation. Two machines can be connected to each other but then the second machine disconnects with the first machine after it gives it some further distance. The first machine stays where it is but tries to push back. The reason the first machine stays where it is is because of a stationary ratchet that allows the lever to go one direction but not the other. When the machines disconnect the ratchet that prevents rising movement would have to disconnect also. There are two ratchets. One ratchet that is always on which catches as the lever falls but allows the lever to rise which turns on an axle and one that catches as the lever rises but allows the lever to fall which is stationary and doesn't turn. A third machine starts anew to the first machine which is lower but because only two machines are connected now a third machine's resistance is not part of the situation and the first weight should be able to be reloaded. The disconnected machine should be lower than the third machine and the third machine and the disconnected machine should connect and come into balance and then disconnect again bringing the first disconnected machine low enough to be reloaded by the newly reloaded weight when it reaches the long lever if it connects to the machine of the newly reloaded weight. This is mechanically possible but it is more complex than just connecting machines together because machines have to disconnect and reconnect. If this can be attempted it would be more close to reloading that one weight necessary to power the device using gravity alone.

[Richard]

preoccupied

I'm thinking maybe we can achieve all with not more than two

No doubt I'll reflect much on the questions you raise..and then seek more feed back from you.

Can I assume that in this cascadeing system...your ultimate goal is to spontaneously harness all energy potential in a (reverse) mass freefall?

[Richard]

heres a thought.

using two of your systems....place an upsidedown bicycle between the two.

the assossiative value of the mental picture and real potential value is not without merit

If you calculate this to work....we can move on to the actual handoff.

richard

[preoccupied]

Lets say the length of each individual square is 2 and the length from the square to the top pivot is two also. sqrt(40) would be the length from the bottom left corner of the square to the pivot if the square is vertical to the pivot, if the square is level horizontally, because the sides of the triangle are 6 and 2. sqrt(8) is the length of the top left corner of the square to the pivot because the sides of the triangle are 2 and 2. sqrt(40) or 6.32 - sqrt(8) or 2.83 = 3.49. 3.49 should be the minimum distance of the longest lever that can be pushed on the other side of the top pivot from the top part of the square.

In the animation lets say ten newtons are required for shortest part of the longest lever on the other side of the top pivot to push. How much input force would be needed? F2*D2=F1*D1. F1=10N, D1=2.83, D2=3.49, and F2 is something. F2*3.29=10N*2.83. F2=8.6N so output force over input force 10n/8.6n = MA = 1.16.

The ratio between the size of the circles needed would give me the numbers for the gear ratio. The driving circle circumference is 3.14(sqrt(32)=5.65)=17.74. The driven circle circumference is 3.14(sqrt(24.36)=4.93)=15.48. Driven gear over the driver gear = 15.48/17.74=.87. The driving square is larger and turns .87 times for every one turn of the driven gear. I think 1/.87=1.15 is the amount of MA needed to have the input of an identical device be the output of this device. 1.15 is about equal to 1.16MA so this calculation is correct.

Does having four lever systems create a bonus of some kind? I think it would create 4x the MA and have the same gear ratio. So is that over unity? Can somebody correct me if I'm wrong? IF 4x the mechanical advantage is achieved then a device that can restart itself can be created because that mechanical advantage can be transferred back to itself after time has passed doing actions. Additional mechanical advantage can be achieved by adding the extra mechanical advantage to another square and levers.

[Tarsier79]

Can somebody correct me if I'm wrong?

Consider yourself corrected.

A 2:1 lever creates mechanical advantage at the cost of distance... Is this overunity?

[preoccupied]

Tarsier79, the distance lost is 0 because it is going to an axle. In the math the mechanical advantage broke even so there isn't actually mechanical advantage with one lever operating. But I'm saying that if there are four levers operating would it be 4x the force to the one gear needed to turn them?

MA=1.16 for one operating lever from the four levers connected to the square. The gear ratio=1.15 so mechanical advantage would be nothing. But with 4 levers operating 1.16x4=4.64MA. 4.64-1.6gear ratio = 3.44MA. Am I doing the math correctly? I'm not sure if my math is correct. But the central idea is that 4x the force is outputted from the same gear necessary to operate the device. The MA should be 1.16x3 because the first 1.16 breaks even with the gear ratio (I think). Any math corrections would be appreciated. I would like explanations why this wouldn't work using math and correct my math where it is wrong.

[Tarsier79]

Ignore the 4 levers, I think they are confusing the situation.

Take a single lever (Because every other lever is just a replication of any other lever), and follow its path and speed in any of the four positions. the lever creates a 2 sided shape. If there is a weight on the end of the lever, and in half the shape it is driven, and the other half it drives, both halves will be equal... IE a gain of 0. No matter what number you multiply 0 by, you always still get 0.

Another way to look at it, is look closely at the left set on your animation, 1 is falling, while 3 rising... and the speed difference between them (about 3x I guess)....

Get the point? This also goes for any number of the same type of any mechanisms. If 1 doesn't produce an OU effect, then neither will an infinite number of the same mechanisms. Simply examine 1 only for your answer.

[AB Hammer]

preoccupied

Looking at your squarelever1.gif. You would have to have 4 separate systems due interference by crossing rods. But that is not the concern for when you look at change1.gif there is no advantage. Each of your rods extend and is replace by each corner but this would counter balance as well not to mention the friction to do this style of action.