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December 19 - 25, 2023

Perpetual-Motion Wheel Mechanics

I've made some drawings. You can follow if you load them, they will be on separate tabs so that you can click from one to the other. The first shows how the wheel looks with only a little mercury at the bottom of tubes. The next two drawings have the tubes half filled with water for us to explore at this time so see what happens to the power level. The first of two drawings of water-filled tubes has a vertical spoke at the 12 o'clock position, and the second drawing has the same spoke travelled clockwise by 22 degrees, a total of 1/16th the wheel. Both drawings have had the tips of spokes removed to give a clear view of the tubes, which to-scale are 48 inches long here where the wheel is not quite 10 feet as showing between 9 and 3 o'clock:

Ten feet is from outer tube to outer tube, and black lines are 1 foot apart. Or, make the wheel any size your wish. I've got tubes in two different colors to help distinguish them (draw software won't allow a change of which items go in front of which, sorry). Horizontal measurements are needed for figuring the power level of each tube, but only in terms of gravity bite, not from the power of momentum.

Over the past two weeks, I made two blunders, I think. Trying to figure the mechanics of this wheel is complicated and deceptive, just try it. Try to figure out, on your own, how much advantage the do-work side has over the go-against side. If you love challenges, you've come to the right challenge. The first thing I did was to find how many total inches the tubes were from the vertical axle line, then subtracted the go-against-work inches from the do-work inches. That's a correct thing to do, but then I went and blundered, I think, by reducing those totals according to the angle of the spokes.

Where spokes are at a 45-degree angle, I'd reduce the inches to .5; if the spokes were at a 22.5 degree angle, I'd reduce the inches to .25. If the spokes were horizontal, I'd not make any reductions, and I'd assign 0 inches to vertical spokes. This is wrong in two ways. First, vertical spokes have tubes that produce some power, meaning they can't be assigned 0 force.

The second problem is the slippery snake. This week, I first realized that it's not the angle of the tubes that should reduce the inches, but the angle of the tubes when one draws a line from the center axle through the center of the liquid. I went and did over all the calculations, combining the gravity-bite factor with this angle factor, and found the following:


Spoke 1: (18.6" / 60") x (27-degree / 90-degree) = .09 force
Spoke 2: (48.6 / 60) x (85 / 90) = .77
Spoke 3: (37.8 / 60) x (52 / 90) = .36
Spoke 4: (6" / 60) x (7 / 90) = .007

Total force = 1.23 force (maximum 4)

Go-Against-Work Side

Spoke 5: (18.6" / 60") x (27 / 90) = .09 force
Spoke 6: (30.6 / 60) x (57 / 90) = .32
Spoke 7: (37.8 / 60) x (80 / 90) = .56
Spoke 8: (21 / 60) x (32 / 90) = .12

Total = 1.09 (maximum 4)

This formula is excellent because the result figures are being compared to water centered at the wheel rim at 3 o'clock, which has a value of 4 per wheel side, which is why I have "maximum 4" in the brackets. In other words, this formula allows one to know WHERE on the wheel the advantage is to go, a requirement for knowing how much power the advantage has, which is the very goal of this math.

[Insert December 27 -- Late last week, when this update was written, I thought I realized that it's wrong to combine the inch factor with the angle factor because the inch factor already does it all on its own. For example, if the water could be at the rim at 3 o'clock and at 90 degrees to the axle, it forms the brackets as: (60 / 60) x (90 / 90) = 1 force (maximum force). That is, 3 o'clock at the rim has a value of 1 force.

Then, as the water moves to the bottom of the wheel, the angle (from the axle) gradually changes exactly in proportion to the decrease of inches to the vertical axle line. When the water gets to the rim at a 45-degree angle from the axle, the inches from the axle line drop from 60 to 30, and this is half as much, even as 45 degrees is half of 90. The formula here is: (30 / 60) x (45 / 90) = .25 force, but after accepting that formula for a couple of weeks previous to this update, I rejected it while writing this update. Then, the week after, I remembered why it was necessary, and so I think it should be reinstated. It should be given a pardon-me; I was wr-wr-wrong again.

I remembered that we need the angle factor because Tubes 3 and 7 demand it. Both are the same distance from the axle line, yet Tube 3 has water at a 45-degree angle and cannot, therefore, receive gravity benefit as well as water at Tube 7, for the latter's is on the 9 o'clock line.

The formula above is done correctly with multiplication because, when two facets of the same force are combined, they should be "added" together with multiplication. The formula in the brackets above prove that multiplication is correct for combining gravity-bite due to axle-line distance and gravity bite due to angle from the axle. [End insert]

I thought that, when we calculate the inch difference in the first pair of brackets, it has already figured in the angle difference, and so, I'm thinking, we should not do the second pair of brackets. The reason I included the angle factor is that gravity has half the bite on the water when it's travelling at a 45-degree angle, as compared to water travelling straight down to gravity when it's at 3 o'clock. This 45-degree direction of travel is when the water is 30 inches from the axle line, and so this reduction in gravity bite has already been included when measuring the water's distance from the axle line i.e. we shouldn't calculate it a second time using the angle factor [wrong, I think that's wrong].

The math tells us that the wheel in this position, with vertical spokes (see drawing above), has gravity biting into the do-work side roughly as much as gravity bites into the go-against-work side. But this is when the wheel is stationary. Things change when the wheel begins to spin with momentum.

Compare spoke 3 to spoke 7. In an analogy, view both spokes as one beam, with a man standing at the extreme left end of the beam, and a second man of the same weight hanging down on a rope tied to the top of the extreme end of the beam on its right side. The beam will remain balanced because gravity has the same bite into both men, and it doesn't matter that one man is hanging beneath the beam while one man is standing on it.

Having said that, Spoke 3 can possibly have more power, even though both are the same distance from the vertical axle line, WHEN THE WHEEL IS SPINNING. The water hanging down from the spoke gives Tube 3 more torque power because further from the central axle.

[However, Tube 7 has the angle factor in its favor. I should have added this too

I just can't get this work done right. I'll need a few days, I think, due to complications before putting out the work I've been struggling with for about a week. As it's Thursday now (Dec 28), I may as well put it all into the next update, out Monday at noon. I still don't know how much power this wheel can produce. I'm determined to find out.]


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