Can Hydraulic Pressure Convert into Kinetic Energy?

[OmCheeto]

I'm moving like cold molasses , but I'll get there

ps. if I hooked a small electric motor to the wheel axle, would watts, in any way show a value equal to the moment of inertia ?

Not only no, but no.

 

[RonL]

Ok I now have a trifilar pendulum completed and it works great, but I'm not sure what I'm supposed to be doing. Based on your comments above, are you saying to measure one cycle or several over a time period ? I guess also I need to convert all my numbers to metric ?
I used 135 Kevlar thread, it swings almost like a Foucault Pendulum

ps. This seems to be the most documented, yet unexplained thing I have ever found

 

[Charles Kottler]

Ignoring elasticity, the twisted cord forms a spiral about the centre line with a radius of half the cord thickness. A piece of cord of length L when hanging straight will form the hypotenuse of a triangle of height H and base 2Πr*N after N rotations, so the new length will be:

H = ((2Πr*N)² + N²)^1/2

If the point of separation (the point of the fan) is unrestricted then I imagine the overall change of length due to the change in the angle of the fan would match the change due to the twisting, but this is just a guess at the moment.

As tested many times on a swing with long ropes, tying the cord together at a fixed point will virtually eliminate the jerk as the cords straighten.

 

[OmCheeto]

Ignoring elasticity, the twisted cord forms a spiral about the centre line with a radius of half the cord thickness. A piece of cord of length L when hanging straight will form the hypotenuse of a triangle of height H and base 2Πr*N after N rotations, so the new length will be:

H = ((2Πr*N)² + N²)^1/2

If the point of separation (the point of the fan) is unrestricted then I imagine the overall change of length due to the change in the angle of the fan would match the change due to the twisting, but this is just a guess at the moment.

I'm having trouble interpreting this equation. Do you have a link that explains it?
For instance, when r <<<<< N, H = N
which doesn't make any sense.

From my interpretation, the hypotenuse length should be constant.

I would do the maths, but I have to be at a birthday party in 30 minutes. Perhaps tomorrow.

As tested many times on a swing with long ropes, tying the cord together at a fixed point will virtually eliminate the jerk as the cords straighten.

I wish I'd thought of that 3 weeks ago. I think it would have made things quite a bit simpler. Thanks!

 

[RonL]

At last I found a detailed look at how to use a trifilar pendulum, if one likes "pocket Tops" this might be addictive the you tube link at bottom, led me to what follows......,

How to measure the moment of inertia
« Reply #18 on: July 24, 2016, 02:07:47 PM »
I have not studied mathematics nor physics, so I beg your pardon if what I have written here is not perfect. I hope there are not important errors.

The aim of the trifilar pendulum here is to measure the radius of gyration, (indicated here with the letter "r") of a spinning top. The radius of gyration is a kind of average distance of the mass from its axis of rotation, or, the distance from the axis of rotation where all the mass of the top could be concentrated without changing its moment of inertia.

The radius of gyration of a full cylinder is always 7.071/10 of its geometrical radius. That of a full sphere 6.324/10, that of a full cone 5.477/10. The radius of gyration of an imaginary pipe with all its mass points at the same distance from the axis of rotation, is 10/10 of the geometrical radius, the two radii coincide.
As a sample, the radius of gyration of a cylinder with a diameter of 12 inches, is 12/2 x 0.7071 = 4.2426 inches.

But in the case of more complex objects it is more difficult to calculate the radius of gyration.
We can use a trifilar pendulum to measure it; in fact the duration of the oscillations of the pendulum is proportional to the radius of gyration.

I made my pendulum with a wooden plate, (diameter mm 85, weight 5.6 grams), and some fishing line, (diameter mm 0.30, but for tops lighter than 100 grams I suggest diameter mm 0.15). Three lines are fixed to the plate in three equidistant points at mm 40.5 from the center of the plate.
The lines are fixed at their opposite ends to another plate, to which the first plate is suspended by the three lines. The distance between the two plates is mm 1540.
The upper plate is very little, and the three lines are fixed here at a distance of only mm 1.8 from the center of the plate.
You can use different measures of course, but this will affect the oscillation time. If the oscillating movement lasts for a longer time, the reading will be more accurate, because the angle of oscillation is more stable, not decreasing too rapidly.

To use the pendulum first we need to calibrate it.
The suspended plate itself has its radius of gyration, we can start calculating it; the plate is a cylinder with diameter mm 85;
85/2 x 0.7071 = mm 30.05 , radius of gyration of the plate.
Let's see how much is the oscillation period of the plate.
I choose to make it oscillate for an angle of 45 degrees:
different angles will not affect much the oscillation period, (unless very large or very little angles), but readings will be more accurate if you make it oscillate always from the same angle.
The plate, (as also could be expected for whatever other object with a radius of gyration of mm 30.05 in this pendulum, but I better explain this later) oscillates 10 times in 36.42 seconds, (time, "t")
The ratio t/r is: 36.42/30.05= 1.212 (fixed number)
So, to know the radius of gyration of an object, I could simply divide its oscillation period for the fixed number:
36.42 seconds/1.212 = 30.05 (mm, radius of gyration).

But it is more complicated:
If you put an object on the plate and make it oscillate, you will measure an average radius of gyration given by the object together with the plate: you need to subtract the effect of the plate to have the correct data.
To do so we first need to calculate the moment of inertia of the plate:

The formula for the moment of inertia ("I") is:
r x r x m = I , where m is the mass of the plate, 5.6 grams.
30.5 x 30.5 x 5.6 = 5057 (gram-square millimeter, rounded off)
5057 is the moment of inertia of the plate.

Now I put an object (weight 82.6 grams, I want to calculate the radius of gyration of this object) on the pendulum and I clock the oscillation period; it oscillates 10 times in 32.17 seconds.

The weight of the oscillating mass (the object together with the plate) is 88.2 grams.

The radius of gyration of the object together with the plate is: 32.17 seconds/1.212 fixed number= 26.54 mm

The moment of inertia of the object together with the plate is: 26.54 mm x 26.54 mm x 88.2 grams = 62,126

Now we know the moments of inertia of the plate alone and of the object together with the plate.
I subtract the first from the second to know the moment of inertia of the object alone:
62,126 ( I tot) - 5057 ( I plate) = 57,069 ( I object)

To know the radius of gyration of the object alone;
57,069 ( I object) / 82.6 (m object) = 690.9 (r x r)
Square root of 690.9 = 26.28 mm , radius of gyration of the object alone.

So, now we know how to measure and calculate the radius of gyration and the moment of inertia of a spinning top or whatever other object using a trifilar pendulum.

But still the readings of the oscillation periods are not very accurate, we have a problem.
In fact fishing line is a bit elastic, and lengthens differently depending on the weight of the object put on the pendulum. When the lines are a bit longer, the pendulum oscillates a bit more slowly, and the readings will be misleading.
It is possible to calibrate the pendulum for different weights, I did so, using the various aluminum cylinders you have seen in the video.

Example:
An aluminum cylinder weighing 494 grams, diameter mm 69.5, oscillates in the pendulum 10 times in 33.55 seconds; I want to calculate the fixed number for this weight:

The radius of gyration of the cylinder is:
69.5/2 x 0.7071 = mm 24.57 (r)

Its moment of inertia is:
24.57 (r) x 24.57 (r) x 494 (m) = 298,220 ( I )

Moment of inertia of cylinder and plate together:
298,220 ( I cylinder) + 5,057 ( I plate) = 303,277 ( I tot)

Mass of cylinder and plate together:
494 (m cylinder) + 5.6 (m plate) = 499.6 (m tot)

Radius of gyration of cylinder and plate together:
303,277 ( I tot) / 499.6 (m tot) = 607.04 (r x r)
Square root of 607.04 = mm 24.64, (r tot)

Fixed number:
33.55 (t) / 24.64 (r tot) = 1.361 (fixed number for 499.6 grams).

Other fixed numbers I obtained with different cylinders:
5.6 grams: 1.212
100 grams: 1.293
200 grams: 1.333
300 grams: 1.349
400 grams: 1.357
500 grams: 1.361

Still there is some lack of precision in clocking manually the oscillation time, with errors until about 1 %, but making the average of more timings more accuracy is achieved.

With this pendulum I can now measure and calculate the radius of gyration and the moment of inertia of my spinning tops:

Example:
My top Nr. 20 weighs 269 grams and oscillates 10 times in 31.19 seconds.

Total oscillating mass: 269 (m top) + 5.6 (m plate) = 274.6 grams.

Fixed number for 274.6 grams: 1.344

Radius of gyration of top and plate together:
31.19 (t) / 1.344 (fixed number) = 23.21 mm

Moment of inertia of top and plate together:
23.21 (r tot) x 23.21 (r tot) x 274.6 (m tot) = 147,928 ( I tot)

Moment of inertia of the top alone:
147,928 ( I tot) - 5,057 ( I plate) = 142,871 ( I top)

Radius of gyration of the top:
142,871 ( I top) / 269 (m top) = 531.12 (r x r)
Square root of 531.12 = 23.05 mm ( r top)

That's all.
(see how to use the value to calculate: How much energy you can put into your spinning top ?
« Last Edit: July 25, 2016, 07:07:23 AM by Iacopo »ps. I'm not sure if I violated ethics here if yes I'll remove it.

 

[RonL]

@Charles Kottler , Thanks for looking in

 

[Charles Kottler]

Having slept on this is has become a bit clearer (I think...). The angle between the cord and the ground probably remains constant along the entire length. The formula I suggested was derived from the length of a line following a spiral down a cylinder for exactly one turn, but was wrong - it will have covered height H and horizontal distance 2*Pi*r, or 2*Pi*r*N for N turns.
So for a cord of length L and thickness 2r,
L^2 = H^2 + (2*Pi*r*N)^2
so H = Sqrt( L^2 - (2*Pi*r*N)^2 )

In the setup you have above, the angular momentum must come from the drop in height so by measuring the maximum speed you should be able to calculate the efficiency of the system.

 

[RonL]

Having slept on this is has become a bit clearer (I think...). The angle between the cord and the ground probably remains constant along the entire length. The formula I suggested was derived from the length of a line following a spiral down a cylinder for exactly one turn, but was wrong - it will have covered height H and horizontal distance 2*Pi*r, or 2*Pi*r*N for N turns.
So for a cord of length L and thickness 2r,
L^2 = H^2 + (2*Pi*r*N)^2
so H = Sqrt( L^2 - (2*Pi*r*N)^2 )

In the setup you have above, the angular momentum must come from the drop in height so by measuring the maximum speed you should be able to calculate the efficiency of the system.

Sorry I'm late getting back, I have been trying to work on Om's equation and the methods that I linked. I'll post what I have come up with after this reply.

Based on your post, my question is, should your equation be used four times, one for each section formed when the cables cross in the center above and below the flywheel ? The flywheel is moving upward at the same time the lower 2 X 2 is being pulled up at the same rate.
In my drawing both tension arms move and the flywheel unit stays stationary.
Also because of energy losses, I have revised the system (in my mind) to have a single direction of flywheel motion. (the drawing need not be changed in any way).
Thanks for your interest and I hope to eventually show or explain how it will have more efficiency than first thoughts bring to mind.

 

[RonL]

@OmCheeto , Michelob sends their thanks, their bottom line is looking better and my wife is starting to get concerned Old school to Metric is about to strain my sanity, if in fact I had any in the first place .
I'm going to put my numbers down and if anything really looks wacky, one might see any error without me trying to explain what I did.

My trifilar pendulum works well, line length ............2159mm
Flywheel.....................1502.52 grams
Radius of string....................69.85mm
Radius of Flywheel..................63.5mm
Tau (as you described)...1.10s X 2...............2.20s
Flywheel moment of inertia (based on MOI link, I posted).........6,058,556.43 ??
Flywheel radius of gyration (based on MOI link, I posted).........63.49mm

Using the equation you posted, I got a number in the amount of...210,023.778
Other numbers I got were more like the weight of the moon
I wonder if the radius of gyration is supposed to come out so near to the radius of the object, that seems weird

Any way hope you have been well.

 

[Charles Kottler]

Based on your post, my question is, should your equation be used four times, one for each section formed when the cables cross in the center above and below the flywheel ? The flywheel is moving upward at the same time the lower 2 X 2 is being pulled up at the same rate.

I was trying to work out a formula to explain how the shortening of the cables relates to the number of twists. In your example you would use it once to measure the height change of the disk and twice that value for the beam at the bottom. In practise though, it would be much simpler to just measure the change in height directly as it is wound up . The radius of gyration for a flat disk is root 2 * r, or 0.707r.

 

[OmCheeto]

Sorry about my absence this week, but

[last Saturday]
I would do the maths, but I have to be at a birthday party in 30 minutes. Perhaps tomorrow.

someone kicked in my front door while I was at the party.
So I've been in a bit of a cranky, "I ain't got time for no maths right now..." mood.
But that's all fixed, and might make a fun new topic: "The physics of deadbolt locks".

ps. It may be a couple of more days before I get back to analyzing the numbers that have transpired over the last few days, as other "stuff" is also going on.

 

[RonL]

Sorry about my absence this week, but
someone kicked in my front door while I was at the party.
So I've been in a bit of a cranky, "I ain't got time for no maths right now..." mood.
But that's all fixed, and might make a fun new topic: "The physics of deadbolt locks".

ps. It may be a couple of more days before I get back to analyzing the numbers that have transpired over the last few days, as other "stuff" is also going on.

Sorry to hear about the break in, I too have been doing things in little spurts of time between the more important things.
One of my solutions, leave so much stuff outside that the thieves get so tired loading things, they never make it to the inside. RonL logic
I understand and take your time.

 

[RonL]

I was trying to work out a formula to explain how the shortening of the cables relates to the number of twists. In your example you would use it once to measure the height change of the disk and twice that value for the beam at the bottom. In practise though, it would be much simpler to just measure the change in height directly as it is wound up . The radius of gyration for a flat disk is root 2 * r, or 0.707r.

If I recall correctly I was surprised at the six and a half inch reduction in length above and below the disk, the disk moved up six and a half and the 2 X 2 came up the full total of 13 inches.
I'm having a number of thoughts after watching the pendulum actions, thinking that there will be some very consistent values, no matter what mechanical design comes to mind.
Something has come up, Ill get back later.

 

[Nidum]

It would help if you explained the real world purpose of using this toy as a machine.

+1

 

[RonL]

+1

I'm glad you peeked in,
I'll see if I can condense my thoughts to the smallest amount of words,...Can a flywheel be energized more effectively with pressure transfer through a twisted cable set, than an electric motor that has to administer it's power through an application of carefully applied Volts and Amperage ?

Naturally a lot of details need to be considered.

 

[OmCheeto]

+1

pfft!

I'm pretty sure the "wheel" was once considered a toy.

 

[RonL]

I think I should change the thought of my last post to more of a statement than a question. I have quite a few smoked speed controllers, but they involved different types of flywheel like situations. I have determined that the system I pictured above, responds to pressure force as well as I expected (maybe better) but I kept it very simple as there are no safety measures in place.
The basic machine as I have drawn, in my mind, might be useful as a small wattage design that falls in a 5 to 10 KW range that might service a typical household. It's success would be based around how well heat loss is controlled and what type of power system is put into the design.

If this method is as good as i believe,(it most definitely works) it will blend well with making use of a smaller motor/generator system.

 

[OmCheeto]

@OmCheeto , Michelob sends their thanks, their bottom line is looking better and my wife is starting to get concerned Old school to Metric is about to strain my sanity, if in fact I had any in the first place .
I'm going to put my numbers down and if anything really looks wacky, one might see any error without me trying to explain what I did.

My trifilar pendulum works well, line length ............2159mm
Flywheel.....................1502.52 grams
Radius of string....................69.85mm
Radius of Flywheel..................63.5mm
Tau (as you described)...1.10s X 2...............2.20s
Flywheel moment of inertia (based on MOI link, I posted).........6,058,556.43 ??
Flywheel radius of gyration (based on MOI link, I posted).........63.49mm

Using the equation you posted, I got a number in the amount of...210,023.778
Other numbers I got were more like the weight of the moon
I wonder if the radius of gyration is supposed to come out so near to the radius of the object, that seems weird

Any way hope you have been well.

Using your numbers, I come up with the following for "moments of inertia":

0.0040795 (m^2)(kg) when using the trifilar equation (from post #85)
and
0.0030293 (m^2)(kg) when using the geometric equation: I = 0.5 * m * r^2​

Which are neither like your two numbers, even accounting for decimal point shifting.
And unfortunately, neither of your MOI links work, so I have no idea what equations you used.

 

[RonL]

@OmCheeto Did you overlook post #95 ? I copied and pasted his comments and at the end of the you tube video it brings up his next video on balancing spinning tops ( I think it is very interesting )

 

[OmCheeto]

@OmCheeto Did you overlook post #95 ? I copied and pasted his comments and at the end of the you tube video it brings up his next video on balancing spinning tops ( I think it is very interesting )

Of course I saw your post.
I did not though, watch the "next video".

ps. I just verified Chuck's equation.

H = Sqrt( L^2 - (2*Pi*r*N)^2 )

I have no idea how that worked. But it did.

 

[RonL]

Having slept on this is has become a bit clearer (I think...). The angle between the cord and the ground probably remains constant along the entire length. The formula I suggested was derived from the length of a line following a spiral down a cylinder for exactly one turn, but was wrong - it will have covered height H and horizontal distance 2*Pi*r, or 2*Pi*r*N for N turns.
So for a cord of length L and thickness 2r,
L^2 = H^2 + (2*Pi*r*N)^2
so H = Sqrt( L^2 - (2*Pi*r*N)^2 )

In the setup you have above, the angular momentum must come from the drop in height so by measuring the maximum speed you should be able to calculate the efficiency of the system.

I'm afraid that I'm a bit lost as to what numbers to use.

The total L will be (6.5" above and below) for a total 13" of twisted cord.
The radius will be based on two cords twisted ? (each cord is about (.125") thick) or the disk ?
The disk is 5" or a radius of 2.5" (weight 53 oz)
The number of turns = 85
The maximum speed of the disk measured 1025 rpm (digital tach)

The pressure applied by hand to the 2 X 2 is guessed to be about 2 or 3 pounds of force. As I think I understand, revolutions per minute will be based on how much pressure is applied to the twisted cords (more pressure and even less turns can produce high speed). So without force, what efficiency is being calculated ?

Anyway if anyone got higher than 87%, we will use your numbers

 

[RonL]

Got tired of spinning my trifilar pendulum, so the thought came to mind of suspending from a plate attached to an electric motor of proper size, the motor can put in turns at the same time the object weight spins them out...a perpetual drop that never loses height...reading the watts consumed by the motor and knowing efficiency of it and bearings should tell me something.
Laying this on a horizontal and using sprag clutch bearings, will give a one direction spin of a flywheel and the ability to force an additional pressure spin as needed. How to use the flywheel, other than spinning it will get me in trouble.
I think this thread needs to go to the sea of dead threads I have learned a few things. THANKS

 

[OmCheeto]

...
I think this thread needs to go to the sea of dead threads

No!

I have learned a few things. THANKS

Me too!

Along with some long lost maths skills, things keep popping up: The energy stored in the cable/thread disappears if you remove the tension from the system. (?)

I think this is a grand thread, as it makes me "think".

ps. I think 256bits was spot on when he compared this to a clock mechanism back in post #37.

 

[Charles Kottler]

It's probably easiest to work it out backwards, and I'm very rusty with imperial units so will convert to metric. The kinetic energy of a rotating disk is:

E = (1/4) m r² * w² when w is measured in radians per second.

Using metric units, E will be in joules, m will be the mass of the flywheel in kg, r will be the radius of the flywheel in meters, and w will be the angular velocity in radians/second.

To convert rpm to radians per second, you'd take

(revolutions / minute) x (2 pi radians / revolution) x (1 minute) / (60 seconds)

i.e w (rad/sec) = (rpm)*(2*pi) / (60)

So, starting with the system at its' lowest point, all the energy is in the form of kinetic energy from the spinning disk.
m = 53oz = 1.5Kg
w = 1025rpm = 1025*2*pi/60 rad/sec = 107.3 rad/sec
r = 2.5 inches = 0.0635 meters
E = 0.25*1.5*0.0635*0.0635*107.3*107.3 = 17.41 joules

After 85 revolutions the disk will have been lifted 6.5 inches, or 0.165m, and the wooden bar at the bottom will have been raised twice this: 0.33m.
By this point we want all the rotational energy to have been used up (so that it will reverse direction and star unwinding again). The energy to raise a mass is the force required * distance covered, and the force is 1.5 * 9.8 (mass*gravitational acceleration), so
E_disk = 1.5Kg * 9.8m/s² * 0.165m = 2.43J

If a steady weight of 2.5Lbs (=1.13Kg) had been applied (including the weight of the lower beam) throughout the winding phase, the energy required to raise this would have been:

E_beam = 1.13Kg * 9.8m/s^2 * 0.33m = 3.65J

The total energy used to wind the system up is therefore 6.08J, and on release it generates 17.41J we have a super perpetual motion generator, giving out more than twice the input energy .

To get a more plausible answer you would need to accurately measure the weight on the bottom beam. Your disk also has a solid axle which should be treated as tall narrow disks above and below the main disk. I assume that the 1.5kg includes these, and from the look of it that would account for almost half the mass. This would have a much lower moment of inertia as the weight is more central and could well account for the discrepancy.

If you were to build this as a generator you would presumably be applying extra force in the expansion phase to build up the speed, and then use a dynamo to convert the kinetic energy to electricity as it winds up again. It would be interesting to see how the maximum rotational speed changes with an increase in applied force.

To get an estimate of the energy loss in the system, apply a fixed force and measure how much the maximum speed drops on each cycle. If possible do this with a wide range of weights on the beam. If you decide not to proceed with this, then I look forward to your next challenge!

 

[RonL]

No!

Me too!

Along with some long lost maths skills, things keep popping up: The energy stored in the cable/thread disappears if you remove the tension from the system. (?)

I think this is a grand thread, as it makes me "think".

ps. I think 256bits was spot on when he compared this to a clock mechanism back in post #37.

Glad that's how you feel

 

[RonL]

It's probably easiest to work it out backwards, and I'm very rusty with imperial units so will convert to metric. The kinetic energy of a rotating disk is:

E = (1/4) m r² * w² when w is measured in radians per second.

Using metric units, E will be in joules, m will be the mass of the flywheel in kg, r will be the radius of the flywheel in meters, and w will be the angular velocity in radians/second.

To convert rpm to radians per second, you'd take

(revolutions / minute) x (2 pi radians / revolution) x (1 minute) / (60 seconds)

i.e w (rad/sec) = (rpm)*(2*pi) / (60)

So, starting with the system at its' lowest point, all the energy is in the form of kinetic energy from the spinning disk.
m = 53oz = 1.5Kg
w = 1025rpm = 1025*2*pi/60 rad/sec = 107.3 rad/sec
r = 2.5 inches = 0.0635 meters
E = 0.25*1.5*0.0635*0.0635*107.3*107.3 = 17.41 joules

After 85 revolutions the disk will have been lifted 6.5 inches, or 0.165m, and the wooden bar at the bottom will have been raised twice this: 0.33m.
By this point we want all the rotational energy to have been used up (so that it will reverse direction and star unwinding again). The energy to raise a mass is the force required * distance covered, and the force is 1.5 * 9.8 (mass*gravitational acceleration), so
E_disk = 1.5Kg * 9.8m/s² * 0.165m = 2.43J

If a steady weight of 2.5Lbs (=1.13Kg) had been applied (including the weight of the lower beam) throughout the winding phase, the energy required to raise this would have been:

E_beam = 1.13Kg * 9.8m/s^2 * 0.33m = 3.65J

The total energy used to wind the system up is therefore 6.08J, and on release it generates 17.41J we have a super perpetual motion generator, giving out more than twice the input energy .

To get a more plausible answer you would need to accurately measure the weight on the bottom beam. Your disk also has a solid axle which should be treated as tall narrow disks above and below the main disk. I assume that the 1.5kg includes these, and from the look of it that would account for almost half the mass. This would have a much lower moment of inertia as the weight is more central and could well account for the discrepancy.

If you were to build this as a generator you would presumably be applying extra force in the expansion phase to build up the speed, and then use a dynamo to convert the kinetic energy to electricity as it winds up again. It would be interesting to see how the maximum rotational speed changes with an increase in applied force.

To get an estimate of the energy loss in the system, apply a fixed force and measure how much the maximum speed drops on each cycle. If possible do this with a wide range of weights on the beam. If you decide not to proceed with this, then I look forward to your next challenge!

There are some differences in the unit in the picture and the units I used in the trifilar suspension, mainly the shaft 1" X 6" weighing 19 ounces, it is centered and the measure above and below the disk would be 2-3/4".
Definitely there should be some more controlled and accurate measurements.
I don't think you are too far off the potential as I see it, the pressure converted to spin should far exceed what all but a ridiculously over-sized electric motor can supply, without burning itself up by consuming too many amps.
Thanks a lot for your time to show what you did, it should be a big help to me, if and as I transform this to a more horizontal layout and at a little larger scale.
I will be thrilled if the thread continues and especially if someone actually takes it to a level that I think is possible. Maybe me, but I think my time is about to have some serious limits imposed in the near future.

ps. I should amend my comment about too many amps, to "a machine of more serious size"

 

[RonL]

I will be out of touch for a few days

 

[RonL]

OK I'm back and I have come to some conclusions about how to think of the energy application of this system.
First a couple of links that, if not just interesting, can help show the value of the twisted cable.

This video shows a lot of mechanics, which I think will translate to the machine we have been discussing. I have no connections with the site, just found it on the google search.


And then his web site...
http://www.billetspin.com/home/

Without trying to explain too much, what I have struggled with so much is, what's going on with the twist in the cables ? I have come to think of this example...In the same manner as the thumb and index finger snap spin into the Top one time...The twisted cable performs the same function on two sides of the flywheel, as many times as there are twist, each twist is applying the pressure from the pistons to the ever increasing flywheel speed.
The energy in the flywheel grows at a greater rate than the displacement of piston travel.

I'm afraid I'm about to become an impulse purchaser of an expensive Pocket Top

 

[OmCheeto]

...

...

I don't mean to be rude, but that is one of the most worthless videos I've ever watched.

 

[RonL]

I don't mean to be rude, but that is one of the most worthless videos I've ever watched.

I have rolled a lot of eyes in my time, so that is OK I did not learn anything new, but I think what I saw was a tiny scale of how the twist of the thumb and finger compares to what the twisted cable does in the machine we have been discussing.
As a result of this thread I now have come to realize how to make the flywheel rotate in one direction and maintain speed, and make the cable rewind with electric motors. A piston and chamber design that is powered by flash steam.

I think the tops are beautiful and I only recently found out about this activity.