Easy Gravity wheel

[Bizzy]

To explain this. The Ls on mine are tubes with capes on each end. There are heavy metal balls in each tube. At 3 oclock the ball rolls down the tube and impacts the endcap on the bottom of the vertical tube which gives is force to move the others. at 6 oclock the ball rolls backwards towards the bend in the L.
At 9 oclock the ball is rolling towards the center. I forgot to draw a small sping at the inside portion of the L to prevent any unwanted force on the left side of the wheel. at 12 oclock the ball is resting at the bottom of the L.
Bizzy

[wattsup]

Even if the vertical stem is heavier then the horizontal stem, if the union point is a pivoting point, here is what will happen.

1) The heavier vertical stem holds all its weight in its limited diameter.
2) Anything going outwards from the vertical stem is the horizontal stem.
3) Since the vertical is balanced, 100% of the horizontal will be considered as excess or off balance to the vertical, pushing the vertical to the right somewhat.
4) This means when the weight is to the right of the wheel it will be pushing the vertical to the left of the pivot point, but when the weight is on the left side of the wheel, it will also push the vertical again to the left somewhat, thus canceling any momentum potential.

Any modification of the weight ratios between the vertical and horizontal stems will just change the overall angle of the weight, but the distribution will be the same.

So in reality, the drawing is impossible.

But I love these mind teasers because always at first glance your mind wants to work it through as intended.

wattsup

[Pirate88179]

wattsup:

I can not disagree with your analysis.

Bill

[Bizzy]

Even if the vertical stem is heavier then the horizontal stem, if the union point is a pivoting point, here is what will happen.

1) The heavier vertical stem holds all its weight in its limited diameter.
2) Anything going outwards from the vertical stem is the horizontal stem.
3) Since the vertical is balanced, 100% of the horizontal will be considered as excess or off balance to the vertical, pushing the vertical to the right somewhat.
4) This means when the weight is to the right of the wheel it will be pushing the vertical to the left of the pivot point, but when the weight is on the left side of the wheel, it will also push the vertical again to the left somewhat, thus canceling any momentum potential.

Any modification of the weight ratios between the vertical and horizontal stems will just change the overall angle of the weight, but the distribution will be the same.

So in reality, the drawing is impossible.

But I love these mind teasers because always at first glance your mind wants to work it through as intended.

wattsup

Good morning
You are right that is the equilibrium portion that was bugging us earlier.
Thanks

Bizzy

[Bizzy]

Hi Bill
another dumb techinal question if you don't mind. I attached a PDF file successfully and it can be accessed other it is not showing up on the thread like gurji's original drawing.How can I make that happen?
Thanks
Bizzy