Roll up a cable

neufneufneuf

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 :

neufneufneuf

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.

neufneufneuf

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 ?

neufneufneuf

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.

neufneufneuf

The sum of forces on a barometer is 0 ?

neufneufneuf

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.

neufneufneuf

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:

neufneufneuf

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.

neufneufneuf

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.

neufneufneuf

If I roll up a cable like that :



Surface S1 = S2 but the direction are not the same, no ?

neufneufneuf

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:

neufneufneuf

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.

neufneufneuf

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.

locum-motion

Does 999 ever get fed up of talking to himself?
Does anyone else pay any attention to his ramblings?

neufneufneuf

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 ?

neufneufneuf

Or with this cycle ?

neufneufneuf

The fixed grey tube increase height of water, so increase pressure at bottom. External pressure don't change lateral force because grey tube is at right and at left. Inside container, at bottom there is more right force than left but this depend of the diameter of the tube, like tube has more volume inside water, the height of water is greater.

alias no.9

neufneufneuf wrote: »

The fixed grey tube increase height of water, so increase pressure at bottom. External pressure don't change lateral force because grey tube is at right and at left. Inside container, at bottom there is more right force than left but this depend of the diameter of the tube, like tube has more volume inside water, the height of water is greater.

Where's the energy recovery or more to the point where is energy captured?

neufneufneuf

When the container move like arrow, the weight is higher due to the presence of grey tube inside water, sure there is a right force at bottom and a left force at top due to pressure of water, but if water is higher the (left force - right force) don't depend of the high of water, it's a differential force, it depends of h' not h. But weight come from pressure of water and this depend of the total height of water, not a differential force, no ?

imagine grey shape like an helix, the length of helix is higher, so the volume is higher too

SOL

Okay look, I know you think that you keep finding fundamental flaws in physics, and I'm fine with you putting them up here for discussion but they are all the same topic so you have to keep it in the one thread rather than starting loads of new ones...

neufneufneuf

okay, I'll do that.

neufneufneuf

If I put a grey shape like triangle the section can be the same than a helix but with the helix the volume is around 2piR*s with s the section of cable. In the contrary, the triangle's volume is smaller. Helix is fixed, don't turn, don't move, it is only there for adding volume in water.

Look at this image, sections s1 and s2 are the same, volume are the same but with a helix the seection is the same but the volume is higher. If the volume is higher, the weight is higher too, no ?



there is a limit, the triangle shape give 4R and the helix 2piR, so the limit is 0.63

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If I take a container with a weight W1 (container and water inside). With the helix, I add the volume Vh:




with:

N: number of turns
s: section (area) of helix
p: pitch of helix
R: radius of helix

So the weight is now of W1+Wh, with Wh=Vh*g*rho (rho is density, and g acceleration of gravity). The weight works with sin(a), so the force from additional weight is Wh*sin(a)

Forces Fl and Fr:

Fl-Fr = hp*s/cos(b)

With hp the vertical distance between Fl and Fr and b the angle of the axis of helix.

hp = N*p*sin(a)

so

Fl-Fr = Nps*sin(a)/cos(b)*rho*g, with b angle of cable with axis.

Fl-Fr works with cos(a) so the work is Nps*sin(a)/cos(b)*rho*g*cos(a)
The total is :




angle b = P/Lh with Lh the lenght for one turn of the helix

Numerical application:

a=10°
s=0.0001 m²
N=1
R=1
P=1


Total = 1.104-1.101 = 0.003 N



With another values:

a=45°
s=0.0001 m²
N=1
R=1
P=1

Total = 4.49-3.22 = 1.27 N

I think I need to take in parameter the angle a inside the sin for delta, because the helix has a slope but it's possible to imagine the container like a parallelogram so this calculates is good ?

like that ?



pb with sign, it's 1.96-1.86 = 0.1 N

I think I need to divide by cos(a) for have the good angle with helix and plane, so the result is :



and the result is 1.96-1.80 = 0.16 N

no, the angle a must be retrieve from the sin() so it's:






but it's not 0, it's:



the angle 'a' must not be greater than angle 'b' for have only one entry

And the limit seems to be 2pi not 0.63

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With a vertical helix it's 0 for the sum.

But here with a lateral movement it's not zero. I found an error it's the angle 'b'. So with the good angle I found not 0:

neufneufneuf

I found this:



it's not 0. I can find 0 if I delete angle 'a' inside cos(atan()) and divided by cos(a), but it's strange to do that because if I increase 'a' at one moment 's' is perpendicular to the surface, so the angle must be 90°. It seems the angle of helix/vertical surface is not always the same, so I need to integrate ? The sum must be the same at each point, no ?

Yes, I think angle change at each point, so the force due from gravity to move down container is not constant.

it's that, I found 0 on one pitch:



but this would say the energy of the system oscillate around 0, it's strange, no ?

For a part of time a system can give energy and recover later ?

If I take a conic helix or conic spiral, the section can be the same but the volume is lower (even in vertical mode):

http://www.mathcurve.com/courbes3d/heliceconic/heliceconic.shtml
http://www.mathcurve.com/courbes3d/spiraleconic/pappus.shtml

neufneufneuf

I understood the volume is the integral of the section.

I have another idea: if I take this helix with gas inside it. The helix is closed to itself, no end. I put it inside a container where there is vacuum, I can change the pitch to increase diameter 'd' and keep volume of helix constant, no ? 'c' will change too but change the pitch don't need energy, in the contrary change diameter giver energy I think because there is no end.

neufneufneuf

2 densities. 2 containers have always the same relative position. System (2 container) move like red arrow. Black stem is fixed. F is an extra force, no ?

Note, that container at density 0.8 is posed on container at density 1. The height of water inside container 1 is higher due to the presence of container 0.8.

neufneufneuf

it's easier to understand like that I think. The container where density is 1 has less surface at bottom, this must be equal to the extra volume due to the stem, but the extra volume goes at part inside liquid at density 0.8. I think it must be exist one position and slope for the stem where the additional volume for density 0.8 is not the same at density 1. The lack of surface of containers between d1 and d0.8 move down container d0.8 but the distance is a mean for all surface of the container. In the contrary, the stem is at a precise position, not a mean.

Container d0.8 must slope at left a little due to the lack of pressure (containers can move like they want until obtain equilibrium state, but there are gaskets). The lack of surface at bottom decrease weight, but the volume adding by black stem must add weight in the same proportion, but I think it can't due to the slope and the mean.

neufneufneuf

With fixed black stems and triangle moves from right to left? Triangle has weight (not drawn) and it is stable inside water.




At final step, it's possible to move down grey rectangle because there is less up force, when all rectangle is there, move to the left and move up and repeat cycle.




or use this shape for triangle:

neufneufneuf

An idea with half torus inside water. All the system is under gravity. I compute forces F1 and F2 and they don't depend of radius (R), only of height (h) in the integrale. So this would say I can suppress bottom surfaces (grey color) so the weight is reduced too. Normally, surfaces must be compensated by volume of air, but here it's lower volume