Brilliant concept, but will it work?

[tinman]

This design https://www.youtube.com/watch?v=wI7j6YYZ8-I change buoyancy in front and rear of the wheel.


One counterforce/countertorque is the difference in surface area on the front half side of each pipe and rear half side. Pressure will push more on the greater surface. Say you have 1 bar pressure at the top, and 2 bar pressure at the bottom, and the front side of the pipe is 100cm² and the rear side is 90cm², then the difference in force on the highest pipe is 10kg, but 20kg at the bottom. This difference in surface area must correspond to the angle between the wheels, and also corresponds to the difference in submerged volume. Do my conclusion seem right?


Vidar

I will make this post here,as i see you are working on it as well Vidar.

The upward force of each pipe is the weight of the displaced fluid.
With fresh water,if your tube displaces 1 liter of water,then the generated upward force is 1KG.

Once again ,the diagram shows my calculated results from my build so far.


Brad

[Low-Q]

[/size]I will make this post here,as i see you are working on it as well Vidar.The upward force of each pipe is the weight of the displaced fluid.With fresh water,if your tube displaces 1 liter of water,then the generated upward force is 1KG.Once again ,the diagram shows my calculated results from my build so far.Brad

Actually, the torque will be less than the upward force because the pipe doesn't move vertically. It is 30 degrees between the tubes.
4.1kg is correct
But then (right hand side):
3.06kg
1.75kg
0 at the bottom
Total 8.91kg angular force


Left hand side:
1,07kg
0,905kg
0,66kg
minus 2.64kg angular force.

in total 6,28kg angular force.

Add the torque from the 75mm diameter tube itself due to greater volume on right hand side than left hand side.

Then substract horizontal force, going to the left due to the greater surface area on the right hand side of the tubes.
How do this add up at the end?
Don't forget to calculate torque in NewtonMeter at 1 meter radius for every part you calculate.[/size]
I would guess, you end up in zero, but I haven't got time to calculate this yet.


Vidar

[tinman]

Actually, the torque will be less than the upward force because the pipe doesn't move vertically. It is 30 degrees between the tubes.
4.1kg is correct
But then (right hand side):
3.06kg
1.75kg
0 at the bottom
Total 8.91kg angular force


Left hand side:
1,07kg
0,905kg
0,66kg
minus 2.64kg angular force.

in total 6,28kg angular force.

Add the torque from the 75mm diameter tube itself due to greater volume on right hand side than left hand side.

Then substract horizontal force, going to the left due to the greater surface area on the right hand side of the tubes.
How do this add up at the end?
Don't forget to calculate torque in NewtonMeter at 1 meter radius for every part you calculate.[/size]
I would guess, you end up in zero, but I haven't got time to calculate this yet.


Vidar

The torque on the shaft from each tube will actually be greater than the upward force,due to the leverage distance between the vertical point of each tube in relation to the center of the shaft.

Torque = force x distance

If we take for example the top tube on the active side,with a lifting force of 4.1KGs,at a distance of 400mm from center of tube to center of shaft,the torque at the shaft is 16.08 newtons,or 11.86 ft-lb.

Hope that helps.


Brad

[broli]

Calculating the area of the counter acting cylinder part is not that difficult actually because the area is a triangle if you unwrap it. All you need is the circumference of he tube and the angle the wheels are positioned at. However something tells me this sum will tend to go to zero as well  .

[vikram_gupta11]

It will not work due to water pressure