extending levers

[preoccupied]

When the weights are 45 degrees I think they are deadlocked and won’t move so a pendulum will pull on it just a little bit to get it to start moving. Once it is moving the weight that is on the edge of the circle will be farther out than the weight that is on the straight edge.

I’ve seen people draw a circle and put a straight edge to block the weight on one side and I don’t think that will work because pushing against the straight edge takes extra force. Since the angle of the turn within the outside 45 degrees stays longer for longer it is more stable to overbalance something, I think. So if a wheel can be overbalanced by adding a straight edge it might only be able to be done on the turn of the 45 degrees on the long end like in the animation.

I don’t know how to calculate this. I know that extra resistance is at the straight edge but I don’t know what that resistance is. If this were to be built the weight falling on the edge of the circle would be hanging on a platform on the end of the lever and the weight lifting on the straight edge would be on a platform rolling up the straight edge. A wheel would be moving on the lever and another wheel on the straight edge for the platform for the straight edge.

Sorry Jim_Mich - that was a bunch of stuff that wasn't explained in the previous animation.

[Tarsier79]

A pendulum and a wheel have many of the same characteristics.

As much energy you use from the pendulum to get over the dead spot in the wheel, you will have to give back to the pendulum to get it to reset.


http://www.youtube.com/watch?v=u9w4W2z1X_o

I made this sim to show you which weight has more leverage, the vertical moving weight, or the weight fixed to the perimeter of the wheel. As you can see, the weight on the right, at the bottom has more leverage, because it isnt at a fixed radius to the wheel at any point, and gravity acts in the direction it is moving, compared to it acting at an angle compared to the weight with the fixed radius. This particular vid doesn't show it, but through this 90 degrees of rotation, there is 0 sum advantage. So even with a pendulum, you still need a little extra from somewhere.

[AB Hammer]

preoccupied

I believe it should be called the double D rocker idea. The first thing I see is that in each full cycle just from placement it balances out twice but then one has to go up the back side of the D and the other down the outer curve. A weight strait up and down has more than the curve if these are in a double D ramp. This shows you are thinking and now you need to start building to learn the effects you are facing.

Alan

[preoccupied]

I believe in measuring twice and nailing once. I would never build random experiments unless I can do all the math first and it is viable.

I improved on the design and I think this could be bessler’s clue “weights came to be placed together arranged one against the other�. The weights pull on each other by a string to pull the next weight out of balance. I don’t know what all the forces are but a possibility is having the pendulum fall the full distance connected to the weights and then the weights can push the pendulum all the way up on the end of the swing because there will be like 3-4 weights on the falling end and one pendulum on the opposing end. The animation wasn’t drawn as if the pendulum swings with the weights and the pendulum waits for the weights to finish falling instead. But if the pendulum were swinging with the weights it would be pushed up back into position as the weights are falling.

To calculate this we need to know how much torque is necessary to make one weight come out of position on top by the pendulum. How much torque the weight has when it is falling and if it can pull another weight out of balance. If the torque produced by the weight falling can pull another weight out of balance then if the weight of 3-4 falling weights can push the pendulum back into place then this would be perpetual motion and might be Bessler’s wheel because it is based off of the clues as I perceive them on the Bessler.com website clue section.

[Tarsier79]

If 1 mechanism is 0 sum, so with friction is a non runner, why would adding more mechanisms improve the system? Lets connect 100 mechanisms... 0 x 100 = 0. Maybe add just a few more? See my point? This goes for any less than ideal mechanisms.

What do you believe is the advantage of having the second lever in this updated mech?

[Grimer]

I believe in measuring twice and nailing once. I would never build random experiments unless I can do all the math first and it is viable.

I improved on the design and I think this could be bessler’s clue “weights came to be placed together arranged one against the other�. The weights pull on each other by a string to pull the next weight out of balance. I don’t know what all the forces are but a possibility is having the pendulum fall the full distance connected to the weights and then the weights can push the pendulum all the way up on the end of the swing because there will be like 3-4 weights on the falling end and one pendulum on the opposing end. The animation wasn’t drawn as if the pendulum swings with the weights and the pendulum waits for the weights to finish falling instead. But if the pendulum were swinging with the weights it would be pushed up back into position as the weights are falling.

To calculate this we need to know how much torque is necessary to make one weight come out of position on top by the pendulum. How much torque the weight has when it is falling and if it can pull another weight out of balance. If the torque produced by the weight falling can pull another weight out of balance then if the weight of 3-4 falling weights can push the pendulum back into place then this would be perpetual motion and might be Bessler’s wheel because it is based off of the clues as I perceive them on the Bessler.com website clue section.

I've being thinking about this and I think I can see a variation on your design which might well work.

At present you have the weights taking a closed path, one on the left of the axis and one on the right.

Unfortunately the closed paths are in opposite directions. One is clockwise. The other is counter-clockwise. If both were going around the same way so that we had a Vesica Pisces on each side then the wheel would have to turn as a whole to preserve angular momentum.

The first derivatives balance out. The second derivatives balance out and we are left with the third derivative - I think.

It's late here so I draw some diagrams tomorrow to explain what I mean.

[preoccupied]

I am following the clue that eight weights were heard falling for each turn of the wheel. I see area A plus area B equal zero but I see area B equal not zero. If more extra torque is in area B than area A, torque can be transferred to area A and create over unity. I don’t think the extra torque placed in area A is equal to area B and I think area B has more torque. When a weight is falling more torque exists than is necessary to move the weight.

So if there wasn’t eight weights and only two it would be able to run I think but it would stop eventually because the input energy would stop. The pendulum would cause the weight to go into area B and fall. But the pendulum would lose function after the first few swings. With eight weights I think the extra torque from multiple weights can push up the pendulum.

[Tarsier79]

Agreed, there is more torque in B than A, in the position I have shown in red.

This is due to X being greater than Y. At the top of Y, you have to get the weight to lift by the amount of Z1, and at the bottom of X it must be lifted by Z2.

So positive torque (X), - negative torque (Y+Z1+Z2) = 0

[preoccupied]

For 4 weights on each side the distance separated from each other should be about the distance of area A so that the first weight taken out of area A gets to apply torque in the falling direction to the other weights. Things that need to be calculated are, amount of torque on each weight in each position as weights fall on the curved side. Therefore we can know where the weights are because they are separated by about the length of area A. How much force is necessary to move weights out of area A from weights in area B? I think the pendulum should turn with the swing of the weights like a lever in a 2:1 gear ratio so that half a swing gets to the point where about 3 weights are in the B area and then the weights in the B area push the pendulum back up. So force of the Pendulum at its positions would be calculated with force of the weights on the levers in their positions. After we can calculate the torque on all the weights in their positions and the pendulum we can experiment with distances of the strings. The first string should be the distance of area A but the next strings could be shorter and shorter so that more weights can be in area B at the same time but I don’t know if that would be helpful or not but it might help restart the pendulum. Anyways lets do the calculations under the assumption that weights only apply force with the pendulum after they are being pulled by the string and then they move along with the pendulum equally after that. I guess a partially teethed gear could allow this to happen. What I mean is that the pendulum pushes the weights but the weights don't push against the pendulum until they are being pulled by the strings or are in position to be useful.

[preoccupied]

Does anybody know where I can find MT drawings to look at? I saw them in the past but now I can't find anywhere that has them.

[jim_mich]

Click the Wiki button at ther top of the forum page. Then click the Maschinen Tractate button. Then choose which group number of drawings.

[preoccupied]

I've been wanting to say that the weights are pulled by strings like is seen in MT 9. MT 9 is drawn funny like part of the strings are longer than the others and if the wheel were to turn it would look differently. I think that was for animation purposes by Bessler. I think pulling a weight can compound usefulness in area B. I also think adding more weights can compound usefulness in area B. I need to calculate but I need some help with that because I am uneducated in physics.

[Tarsier79]

I am also not great at physics. A diagram from Wiki is below.

Also below is how WM2D calculates the forces, hopefully this is correct. Because your weights and gravity are a constant, just assume it is 1, and ignore this part of the formula.

Use Trig to solve X1, as you have done before.

To solve X2, it is adj/cosθ


sum X1's - sum of x2's = resultant force.


Good luck!

[preoccupied]

I actually don’t know Trigonometry. I mean I used to know math but I forgot everything. I’m using an online trig calculator to get my numbers.
http://www.carbidedepot.com/formulas-trigright.asp

I see that you drew that the right angle goes on the end for the vertical incline. I’m going to try to use that to calculate all the math using the online trig calculator.

I want the pendulum to pull the weight into the 90 degree area at a 1:1 gear ratio and see if how that works…
Pendulum plus weight 5+3.54=8.54 and counterweight 7.07
Pendulum plus weights 0+3.54+5+3.54=12.08 and counterweight 7.07+3.54+7.07=17.68
The weights by themselves cannot produce extra force to push the pendulum back up. They just have the extra force from falling to go into place after the pendulum inputs the energy to get it started. It looks like the straight edge might have more extra torque than the curved edge. Maybe the pendulum can push the straight edge when it is in the center and the straight edge can help push the pendulum back into place somehow.

EDIT Now that I look at this I want to cut out one of the 7.07 somehow. I think a ramp could be used and the weight could fall to the end of the positive torque at the end of area B and roll over to the vertical incline. That way the pendulum causes the weight to fall through the first area A but the weight would not have to travel through a second area A. I don't know if this would help however that is what I see when I first look at those numbers.

[Tarsier79]

To tell the truth, I use an online trig calculator as well. Much easier than a calculator.

Yes you can see the reason for the right angle(rhs) on the wiki diagram.

Swapping the weights between the two paths in the positive torque area is an interesting concept, though I will have to consider it further before I comment further.

Cheers