New modification to my drawing

[preoccupied]

I took my previous drawing and added more right angle attachments and weights creating a large fan of weights rather than two weights. I noticed that the weight at the end of the square was in a unique position. Then I drew it with just one weight at the end of the square and it looks like it has promise.

I'd like some help calculating the more complicated part where the top weight move halfway along their path. That is what the yellow part of the drawing is supposed to represent. I think that the angle against the rim of the wheel is about 11.25 degrees so sin(11.25)*1.4distance= 0.27 and then the other lower yellow square looks like it's about 1 unit of force but I'm not sure. So I guess it's about 1.3 and this might be one of the most resistant parts of the turn because earlier in the turn the pivot on the lower yellow square would have been very mild.

I want to be kind and humble about this discovery or art rather than my usual rash self. Thank you for any of your input.

[Tarsier79]

Hi P.

It is good you realize even though the weight position looks good, the extra lift for the flip costs more than just positional torque.

That is difficult to calculate. There are ways to do it....

Anyway, I wouldn't. For a start, the flip mechanism is a little wasteful. You lift the end weight up to vertical, then it flops down to about the 90 degree position on the little wheel/ramp. So: Whatever you get from OB, you will lose in the lift. When you get the flop you lose some height so the design is about 80% efficient just by eye.

[preoccupied]

The flip mechanism might be wasteful like you said. If you look at what I've drawn here the weight could unload on the red line on the left and use a ramp to the unload position on the right. Look at the red lines.

I think that it would be cool if the flip mechanism would work because it would be a more smoothly operating machine but this ramp being used as a short cut to the positions has much more potential mathematically. It still has about 0.7-0.8 force I think verses the 1.7 driving the wheel. The top weight in red is less than 0.7 counter force and the weight below it is applying just a little bit at that point because the pivot point isn't very great yet. It might be close to 1 against 1.7. This looks very promising now that I've added a short cut.

[preoccupied]

I drew some blue triangles to try to calculate exactly what is happening. the blue triangle on the lower weight on the left it shows the distance from the pivot point which should show the force pushing against the peg which is about 0.3 and the angle of the triangle should show the force pulling down on the wheel which would reduce the distance from the axle about 0.2 I think because 11 degrees it looks like it might be and 11sin is about 0.2. so 1-0.2=0.8 0.8+0.3= 1.1. Then we have the triangle for the top weight and because its pivot is hanging on the other side of the axle it is only applying weight against the peg against the wheel. It looks like it could be approaching 45 degrees. sin(45)=about 0.7 and that means it might be 1.4distance/2*0.7=0.49. So it's probably just about 1.7 just like the driving force of the wheel at maximum. This makes mechanical sense due to conservation of energy. So I might not know the exact numbers or how to calculate it but I am on the right track because I am seeing conservation come up. The thing that I notice is the weights are loaded at 1.7 driving the wheel before unloading the weight on the left on the ramp. The early momentum might reload the weight over and over again.

What I mean is that at the start of the turn it will turn because the weight at the top right exists before the red weight reaches its position. The question is will it be able to use the extra starting momentum to reach the reload position if its relatively the same force altogether? I think that because it starts to turn that it might reload. It's like special it like starts with one weight higher at the beginning of the calculation.

[SHADOW]

Votre idée est intéressante si il y avait une phase de distribution le long de la ligne rouge donc neuf cages si huit poids.
Mais je ne vois pas comment vous immobilisez le guide excentrique.
J.B

Your idea is interesting if there was a distribution phase along the red line so nine cages if eight weights.
But I don’t see how you can stop the eccentric guide.
J.B

[Tarsier79]

Hi P.

Just one thing, your triangles point should be at the axis of rotation of that particular part, not to the opposite corner of the lever. Also, even though you now have a more efficient overbalance, count how many weights are on the left compared to the right.

[preoccupied]

Hi P.

Just one thing, your triangles point should be at the axis of rotation of that particular part, not to the opposite corner of the lever. Also, even though you now have a more efficient overbalance, count how many weights are on the left compared to the right.

There are six weights on the left.

I added some squares parallel because they would balance out. After drawing three drawings here shown I noticed that it was the shape of a stork's bill so I'm trying to fully utilize the Stork's bill in the design for these three drawings. I think that the best way to manage the Stork's bill version of this design is to have the black square rotate around the peg and the Stork's bill not actually touch it just rotate with it. So the reload action is the same as the original drawing but now there is a stork's bill that flips over. An extra weight on one end is needed to launch the stork's bill shown at the top on the right. I'm looking at the preview of my images right now and I think that it would be hard to reset the stork's bill. The stork's bill weight located on the top right might have to be heavier to cause it to reset reasonable on the bottom left.

[preoccupied]

For the stork's bill I like this version with the stork's bill driving weight on the right most diamond on the top right because it moves to the right further, as it also needs to be displaced some because it's a heavier weight.

[preoccupied]

Actually I don't think there should be two weights. I think that one weight at the end of one lever, constructed at the end a stork's bill is all that is needed. I made the line of the lever in question thicker to show emphasis.

[preoccupied]

To address the flipping of the square I think that an extra peg should be put there, drawn in green in. It will take more to position the weights correctly but the Stork's bill pushes the weights REALLY FAR from the Axle.

The benefit that I see from this adaption is that there is two weights driving one weight for the beginning of the turn on the left of the peg and there is a huge difference in the distance of the weights. The weights on the right are super far and the weight being positioned by the peg is closer to the axle. The initial momentum is HUGE. What's not drawn but you can probably imagine is that the weights on the bottom would be hanging really low but they are about in the position handing downward below their pivot point. I am using the black square against the peg to position the stork's bill. Don't forget that.

[preoccupied]

The initial momentum would look like this position drawn here in all black.

The loading phase would have one of the low hanging weights move sideways to the left on the peg but because there is two fully extended stork's bills it should give it time to contract back into the wheel. I just want to put emphasis on how long this stork's bill can be and there is a fully extended stork's bill always ready to drive the wheel while one on the left contracts and one on the right extends. If I were to roughly think about the math it looks like about +25 is driving the wheel on the right and -3 on the left at the beginning and then it should be like +20 vs -10 on the left at maximum while stork's bill is contracting on the left. What do you think fellas?

[preoccupied]

I put a second weight to help contract the stork's bill. I think this helps bring it into the wheel some on the left side.

In the red position there is more than enough torque to turn the wheel. The purple part starts the red position because it needs to flip over. Once it flips over and expands there is plenty of extra torque in the red. The green looks like it might have a small amount of keel. The bottom green weights on the left and right might be pretty close to balanced. It looks like top green weights are sitting on the peg at about 11 degrees maybe so it's probably around 0.5 with both weights. It looks like the bottom left green weights are about 4 distance from the axle. the angle is about 45 degrees to 4*0.7*2=5.6 and plus 8 for pushing on the peg = 13.6, and this makes it about 14.1 on the left side. If the length of the right side's furthest green weight is about 7 then it's basically balanced at about 14 force on each side of the wheel. I think that it might be closer to 8 on the right side's green lever if the stork's bill is really flat or if it's generous this would work as it's drawn. The starting momentum should give it a little bit of a kick from the red positions.

[preoccupied]

Hello folks,
Now I've drawn just a Stork's bill with one peg at the top to flip the square.

This is just a new way to use the Stork's bill? Is this Bessler's Wheel? Further in the turn the top position will be flipped right and will begin extending at as steep of as a 45 degree angle I think. Then for part of the turn there will be 3 entire stork's bill's extended on the right side. The trick I think is the right angle that helps unravel the stork's bill and the squares that have to be flipped on the axis of a circle.

sincerely,
Jon

[preoccupied]

Here is two more drawings of the stork's bill version of my drawing.

There is consistently two long stork's bills vs two short stork's bills. I know a lot of people have been looking for the Stork's bill solution with all of of the like in the toy's page. I want to know if this is similar to anybody else's drawing, please. Any help is appreciated.

[preoccupied]

In my upper picture in the previous post I want to calculate the force over the peg and the weights on the left side of the axle. On the very top there is a weight pushing against the peg. I think that it's directly connected to the peg very close so it should be the distance upwards plus the distance of the weight to the left. 1+1=2. Then the weight below the on the left is about 3 units away from the axle. 2+3=5. Then the weight below that is 6 units left, making 5+6=11 total. On the right it's about 9+7=16 I think.

The bottom picture the issue is how well does the stork's bill retract on the left with the size of the right angle weight. I know that in the upper picture that it should be like I drew it for the weight right angle lifting upwards along the 45 degrees in the bottom left of the picture. I'll prove that Maybe? It has a ramp of 45 degrees so that's about 0.7 of a force to lift up it. Then the leverage on flat stork's bill was extremely high but when it's square it's 45 degrees another 0.7. Then it's 0.7*1=0.7 force and that's the limit because going up the ramp requires 0.7 force. I don't know if I did that right. I think that I could have larger right angles and whatever it takes really to make the design work, if that's all it takes is a larger right angle that is pretty simple.

It looks like the left middle most weight on the bottom picture has about 22.5 degree angle. That should be 0.38 up the ramp. So it should be able to push the bill to a 22.5 degree position. That is about where I drew it I think. I don't know for sure. If the stork's bill is more difficult to move because it's longer like if it has 0.38*0.38*0.38*0.38=0.02 or something then I would need a bigger right angle. I'm not sure it's like that in this situation. This might be the first time I've tried to calculate stork's bills as far as I can remember.