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Author Topic: Brilliant concept, but will it work?  (Read 3236 times)

Offline tinman

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Re: Brilliant concept, but will it work?
« Reply #30 on: April 25, 2018, 03:56:38 PM »
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

Free Energy | searching for free energy and discussing free energy

Re: Brilliant concept, but will it work?
« Reply #30 on: April 25, 2018, 03:56:38 PM »

Offline Low-Q

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Re: Brilliant concept, but will it work?
« Reply #31 on: April 25, 2018, 04:33:11 PM »
[/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

Offline tinman

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Re: Brilliant concept, but will it work?
« Reply #32 on: April 25, 2018, 04:47:50 PM »
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

Free Energy | searching for free energy and discussing free energy

Re: Brilliant concept, but will it work?
« Reply #32 on: April 25, 2018, 04:47:50 PM »
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Offline broli

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Re: Brilliant concept, but will it work?
« Reply #33 on: April 26, 2018, 10:20:29 AM »
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  ;) .

Offline vikram_gupta11

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Re: Brilliant concept, but will it work?
« Reply #34 on: April 26, 2018, 02:36:23 PM »
It will not work due to water pressure

Free Energy | searching for free energy and discussing free energy

Re: Brilliant concept, but will it work?
« Reply #34 on: April 26, 2018, 02:36:23 PM »
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Offline ramset

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Re: Brilliant concept, but will it work?
« Reply #35 on: April 27, 2018, 03:52:11 PM »
a better path forward would be

how can we make it work ....despite water pressure issues ??

http://news.mit.edu/2013/hydrophobic-and-hydrophilic-explained-0716
and  there are OTHER  attributes of water not really shared by gravity .

here some test Data
 
https://arxiv.org/ftp/arxiv/papers/1304/1304.1485.pdf

Foil shapes ....which involve boundary layers .. laminar flows  Etc etc
maybe the tubes have a useful [less drag]foil in the Middle..on the downside.

and then there is "ION charge" which can be stored in the fluid and put to work ?
attraction /repulsion ....

a very deep rabbit hole where many things need a good looking at...
IMO

respectfully submitted

Chet K
and a Big PS on foils
Brad reminded ...
the foils which drove the google test unit strait into the wind at 2 times windspeed

against the wind


prior to that....you were nutty to even consider such ??
things like FRICTION were given as reasons for much expected failure......

now its old news and well understood  ::)

lets be nutty....as Hoppy says
its what we do in the asylum...

until there is a discovery!!!

Offline Low-Q

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Re: Brilliant concept, but will it work?
« Reply #36 on: April 27, 2018, 11:34:47 PM »
It will not work due to water pressure
Yes, you're right. That's what we figured out. Not that I didn't knew in advance that this device could not work, but now we've figured out the technical reason. That horizontal pressure gradient was not obvious at first untill we analyzed the shape on the pipes inside and between the wheels.
And it doesn't help making a notch in the wheel to make pipes with equal surface on front and rear, because that notch would have an area facing forward, and will be that extra surface that prevents the decive from running.

Free Energy | searching for free energy and discussing free energy

Re: Brilliant concept, but will it work?
« Reply #36 on: April 27, 2018, 11:34:47 PM »
Sponsored links:




Offline Low-Q

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Re: Brilliant concept, but will it work?
« Reply #37 on: April 27, 2018, 11:52:43 PM »

I was thinking: How is it possible to make those front and rear surfaces equal? What about deviding these pipes into portions of solid material, like hockey pucks, with some space between them and angle parallell to the wheels, so the total surface for each puck is equal on front and rear. If each wheel are 15° angled, and angle each portion so the front and rear surface is the same?
But then we struggle with the sideways pressure to force those pucks in and out of the water...


Vidar

 

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