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31-01-2014, 15:27   #1
neufneufneuf
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Gears and worm drive with differential pressure

I understood that when 2 gears turn, when the junction is P to 0 this need energy 2PdV with V the volume of one tooth. So if I use a worm drive and 8*15 gears like image shows. There are 7 junctions so the sum of energy from this 8 junctions is -2PdV for each row. 2 gears that in touch with worm drive give nothing and need nothing, this for each row. n row need -n2PdV but the worm drive give 3*2nPdV due to the number of 3 starts for the thread. The sum is 2*2nPdV ?




Maybe because the worm drive can be like this :

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Last edited by neufneufneuf; 31-01-2014 at 18:06.
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31-01-2014, 20:05   #2
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Maybe it's easier to move out vaccum object from one thread with N starts. When I turn one turn of the thread, I need to give a torque = T/N with T the torque for one start. But the volume out is always the volume of one turn with one start. Imagine the very thin walls for thread. In one turn the torque is divided by N but the volume of object with vacuum in it is always the same than with one start.

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01-02-2014, 09:07   #3
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If the thickness of thread is like gear, when I turn one round I move up not one tooth but 2 teeth. But gear give PV for one tooth, next it's blank so not PV. Here the sum of energy is 0, but with thickness of worm drive like near 0, in one turn I move one tooth not 2, the sum of energy is the same ?

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02-02-2014, 22:10   #4
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Problem with sum of torque

I try to find the sum of torque for the object below:




The water must give the same torque than the solid, but I don't find 0:



The difference is very small but must be at 0

For a radius of circle of 1 m, with a thickness of 1 m, water give a torque of 1666 Nm. But the solid give a torque higher. Where I'm wrong ?

I found the center of gravity of solid at 0.7766320 m maybe this is wrong ? For me, it's 0.5 m from the square, added to the right at 4/3/pi from the part of disk with a weight of pi/4.
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03-02-2014, 09:17   #5
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Sum of forces on barometer

The sum of forces on a barometer is 0 ?

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04-02-2014, 11:45   #6
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Wheel turn on the ground and roll a tube

Hi,

The wheel:

http://en.wikipedia.org/wiki/Cycloid

roll up a tube with pressure inside it. If there is no pressure at external, there is a work give from the tube, because the red trajectory seems to give a work ?





The force F seems to work:



Sure, there is a difference of surface inside/outside circle. But with a big radius for the wheel: 100 m, and a tube of 1cm the difference of surface is only of 6.28 cm and this don't depend of the radius. If I take a radius of 1000 m the difference of surface is always of 6.28 cm. But the works is not the same with a radius of 100 m or 1000 m, no ? The force from F seems to be constant, but the force from surface seems to change with the percentage of roll on wheel, at start to roll up, F seems to give more than difference of surface, at the middle the sum of force seems to be lower. Is it possible to compute the sum of force at each time ?

I think it's logical if I think with the lost of red surface. When the wheel turn and move, it's like a very small rotation about the point "Point of rotation". The red surface want to turn anticlockwise the wheel but if this surface deseappear, it could say the wheel want to turn more clockwise, sure the difference is very small.

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06-02-2014, 21:21   #7
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4 cases with different pressure

I imagine this new system for recover energy, I don't know if I can use the formula PV when I move down pressure.




It's possible to try with others values like 4-4-1-1 for example or 1-1-0-0

It's not easy to create small pressure, but it's possible to create cases with small pressure, for example for pass from 0.5 to 0.1, I can create 5 cases with 0.1 from one case, I use later these 4 cases. Energy needed for 5 cases is 0.7*4=2.8 PV and it's possible to do 5 cycles with 0.725 Pv won in each cycle, so it's 5*0.725=3.625 PV and recover energy from cases where there is 0.5 P in it. So like this in theory, the sum is not 0 ?



I done a mistake:

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Last edited by neufneufneuf; 07-02-2014 at 10:27.
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10-02-2014, 18:01   #8
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sum of torque wiith gears

Gears under differential pressure from water give a sum of torque at 0 ? Like it's possible to place gaskets like I need, (solution 1 or 2 for examples), there is a solution where the sum is not 0 I think.

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11-02-2014, 11:00   #9
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Wheel and tube inside water

If I turn the wheel and roll up the tube around the wheel, like pressure is not the same with proof, the sum of force on tube is not zero ?



the wheel is compressed with the external pressure apply to the tube, energy is recover later. Maybe like that no need water. Or turn wheel like a bike's wheel with right end of tube fixed.

Last edited by neufneufneuf; 11-02-2014 at 11:40.
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11-02-2014, 13:42   #10
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Roll up a cable

If I roll up a cable like that :



Surface S1 = S2 but the direction are not the same, no ?
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Last edited by neufneufneuf; 11-02-2014 at 19:27.
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12-02-2014, 08:11   #11
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one end move in one turn with length of L1=2*pi*sqrt(R²+(p/2/pi)²) (helix length), other end move with length of L2=2*pi*R and in addition the movement is not in the same axis than force.

the sum of energy for one turn is P*S*L1-P*S*cos(a)*L2, with P the pressure, S the surface of cable.



or if this is the solution, it's not 0:



Another solution but it's not 0:

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Last edited by neufneufneuf; 12-02-2014 at 10:23.
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12-02-2014, 20:22   #12
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If I change the pitch of the helix, at right the angle is bigger than at left, no ?



and in water, the pressure is not the same at top than at bottom, even there is a torque from each turn of cable.

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Last edited by neufneufneuf; 12-02-2014 at 22:00.
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13-02-2014, 07:17   #13
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I understood, like the cable move at left in the same time, the movement is exactly at the same angle than at right. But for the length, at right I move 2piR but at left I move more it's 2pi*sqrt(R²+(p/2pi)²), it's the length of the helix, no ?

If I imagine the wheel turn and move to the right at velocity of pitch for one turn, the cable move at L1=2pi*sqrt(R²+(p/2pi)²) but the angle decrease the force. At right the end of cable have full angle and the lenght is L1 too so there is a difference of force I think.

Last edited by neufneufneuf; 13-02-2014 at 11:37.
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13-02-2014, 18:44   #14
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Does 999 ever get fed up of talking to himself?
Does anyone else pay any attention to his ramblings?
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14-02-2014, 13:29   #15
neufneufneuf
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Increase are where pressure is lower

If I want to increase area where there is 0.5 bar I can insert another ball. I need to decrease pressure at 0.5 bar in step 2 (this can be recover later) and at step 3 I can increase the area without energy, no ?



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Last edited by neufneufneuf; 14-02-2014 at 14:28.
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