Can Hydraulic Pressure Convert into Kinetic Energy?

[OmCheeto]

In respect to berkeman and his hard task of watching this thread, I will request that all comments revolve around combustion of a fuel and the efficiency of the flywheel operation.
Thanks
RonL

And ignore how the wirligig works?
Strange little contraptions. I spent half an hour trying to find the "physics" behind their operation, and couldn't find anything.
So I built one out of an old AOL CD and some string.
Now the internet says a CD weighs about 0.02 kg, and I had to apply an equivalent force of 2 kg to keep the CD cycling. (I'm using a fish scale to measure the forces)
That's a factor of 100.
Scaling that up to your 100 kg flywheel gives me an equivalent force of 10,000 kg. (98,000 Newtons)
One place on the internet says that a 4" diameter gasoline driven piston applies the equivalent of 2860 kg (6300 lbs) of force near the top of its stroke, [ref]
So that's equivalent to 3.5 small block pistons.
Seems like a lot.
Although I was willing to build and test a CD version, I'm not willing to upscale to another model, as I'm not seeing any advantage in adding complexity to what appears to be a gas driven generator.

ps. Might make a fun chest muscle exercise machine.
pps. I'm still not sure how a wirligig works. Although, obviously there's torque involved. I'm assuming because of the 100:1 ratio of flywheel mass to equivalent force required, it's a function of the diameter of the string.

 

[RonL]

Your cylinder could be pneumatic or hydraulic, for which control of pressure can be easier than controlled combustion. Open a valve, the pressure drops; open another valve, you're connected to the compressor outlet.

In my mind I tend to see a 5 to 10 second cycle, not a lot of linear take up by the cables, so a short stroke.
Power will be determined basically by piston diameter, I think keeping the pressure steady (or increasing) as the cables unwind until that last twist, at which point a full release and the flywheel transfers energy to one generator half, save what is needed to twist the unloaded cable set in the other direction. This action would exhaust the chamber and have the pistons close together and ready for the next power stroke.

 

[RonL]

@ Om, I'm always amazed at how you test things, I hope your simple test was representing 50% of what was going on and that the CD disc was wasting the other half. If you had pulled harder would that have resulted in much more wasted energy ? I'm delighted you took time to test my idea in any detail.
Thank you
ps. Flywheel operation is the wirligig, at least in my mind

 

[OmCheeto]

@ Om, I'm always amazed at how you test things,

Sometimes it's easier to test things, rather than analyze them theoretically. This is especially true if you don't know how something works.

I hope your simple test was representing 50% of what was going on

I don't fully understand what is going on, so I have no idea.

and that the CD disc was wasting the other half.

Odd. I didn't include that part of my experiment. How did you know that?
From Jack Action's equation in post #17: $E =$ $\frac{1}{2}I\omega^2$ + $\frac{1}{2}kx^2$
I came up with that the energy imparted to the disk was 0.64 joules.
But I had supplied 1.2 joules. (work = force * distance)
I assumed that I was either missing something, or the test device was so poorly designed: Just me, the whirligig, and the fish scale
that there was no way I was going to get any meaningful values.
I did throw out Jack's "spring" part of the equation, as, well, there was no spring.
It's possible that "I" was the spring, but being a very poor spring, all springy energy from me was lost.

If you had pulled harder would that have resulted in much more wasted energy ?

Still not seeing where you are getting that there was wasted energy, from my post #31.

I'm delighted you took time to test my idea in any detail.

A real detailed analysis would require a VERY fancy rig.
Try operating your whirligig with you eyes shut. Without visual cues as to when to apply and release forces, I can't do it.
It would probably be much easier to figure out the math.
And I have no idea how to do twisted rope torque maths.
And given that Google says; "No results found for "twisted rope torque equation"."
I'm guessing it's not a well researched subject.

Thank you
ps. Flywheel operation is the wirligig, at least in my mind

This is starting to remind me of "Nobody knows how bicycles work!"
Yes, your flywheel is a whirligig, but I can't analyze this problem until I know how a whirligig works.

 

[RonL]

@Om, I don't have but just a moment, wanted to say I looked at your reference and liked it, I might have missed something, but when the 6300 pounds of force on the crankshaft was mentioned, the increased pressure of the combustion explosion was not mentioned.
His numbers in general helped me to think about where to start, in the evaluation of my machine.
Waiting for berkeman to give me some guidance

 

[OmCheeto]

@Om, I don't have but just a moment, wanted to say I looked at your reference and liked it, I might have missed something, but when the 6300 pounds of force on the crankshaft was mentioned, the increased pressure of the combustion explosion was not mentioned.

I think it was:

...Very consistently, the explosion pressure in an internal combustion engine rises to between 3.5 and 5 times the compression pressure. Since our example engine had a compression pressure of 120 PSIA, this results in a momentary explosion pressure that peaks at around 500 PSIA. (We are going to slightly cheat here and call it 515 PSIA to simplify the following math!)

Since the piston is 4" in diameter, the top surface of it is just PI * (4/2)2 or around 12.6 square inches. Each of those square inches experiences the 500 PSI(G) pressure (Pascal's Law), so the total force then instantaneously applied to the top of the piston is 12.6 * 500 or around 6300 pounds.
...

[bolding is the author's]
His "120 psi" jives mathematically with my recollection of small block Chevys:

...
The 8.0 compression ratio means that the 15 PSIA beginning mixture, is now at about 8.0 times that pressure, or around 120 PSIA. (Technically, not precisely, because of some really technical characteristics of what happens when gases are compressed isentropically.)
...

And since I can never remember what any of those "gassy science" terms mean:

...
In thermodynamics, an isentropic process is an idealized thermodynamic process that is adiabatic and in which the work transfers of the system are frictionless; there is no transfer of heat or of matter...

and

An adiabatic process is one that occurs without transfer of heat or matter between a thermodynamic system and its surroundings.
...

His numbers in general helped me to think about where to start, in the evaluation of my machine.
Waiting for berkeman to give me some guidance

While we are waiting for berkeman to beat us like ugly stepchildren, I thought I'd share a doodle of a kind of free body diagram I just whipped up:

This image may, or may not, display all of the forces involved.

ps. From my recollection of college physics classes, this device was never used as a homework problem in my textbook, and I'm starting to understand why.

 

[256bits]

When I said practical applications of the whirliwig, there has been one since the 1600's or earlier, but of a different design.
An example is given in this picture, which includes the clock escapements using a spring.
The whirligig is the balance and the balance spring, or in physics terms,a rotating mass and torsion spring, along with the barrel spring. The whirlygig is given a tiny boost in energy from the barrel spring through the gearing and from the escapement as it ticks away. What it does is change the θ of the balance spring on each tick ( or is it tock ) so that the balance spring has a little bit more torsion, which it can then exchange with the rotating balance wheel during the cycles. The extra energy is just enough to overcome friction to keep the thing ticking and tocking.
One may notice that this would be a good mechanical way to keep time if we attach some markers to one or more of the gear wheels.

 

[256bits]

A bifilar pendulum can be made by attaching two wires to a mass.
A torsional displacement of the mass, in this case a bar is shown, will result in the strings being displaced from the vertical. Doing so will raise the mass vertically, and work will have been done on the mass. The potential energy of the mass can be converted to kinetic by allowing the mass to rotate from the torque produced by the horizontal force from the wires. Ideally with no friction and other damping forces the system will act continuously back and forth converting potential energy of the mass to kinetic and vice-versa.

The shorter the wires, distance x, the stiffer the "spring" action.
The longer the wires, the softer the "spring" action.

Distance d can also be changed.
A smaller d will have a softer spring effect and vice-versa.

Reason being for both cases, the for torsional displacement θ, the linear displacement of the wires from the plane of the top and bottom bars is minimized with larger x ( length of the wire ), and minimized with smaller d ( distance between wires ).

Note that as the bottom bar is displaced, the tension in the wires increases.

Note also that if the distance between the wires, d, is made smaller and smaller towards zero ( 0 ) an approximation of the restoring spring force approaches that of the torsional effects of one ( 1 ) wire being twisted and exerting a torque upon the mass.

If we twist the mass through and angle θ so that the wires cross and wind upon one another, then it becomes similar to that of d=0 at the bottom. With each wind, the length of the wire from the wind to the attachment becomes less, increasing the tension in the wire.

that is all preliminary analysis for what it is worth.

http://physics.dorpstraat21.nl/images/expts/bifilar%20pendulum1.png

 

[RonL]

The barrel springs or as called by my overhead door parts list "clock springs" might be able to perform a service in my drawing, but that might come later.
I did overlook the paragraph that mentioned the explosion.

As long as the power source is a secondary supply, we can consider, compressed air, hydraulic or fuel combustion as a pressure force to push the pistons apart and work the tension arms.
I'll come back in a bit and put some information out in an attempt to find a start point.

 

[256bits]

This image may, or may not, display all of the forces involved.

I also thought there should be a torque from the width of the wire or string.

 

[RonL]

I also thought there should be a torque from the width of the wire or string.

Are you talking about the distance from centerline to point of cable attachment ?

I'm putting the drawing here for ease of access.

 

[256bits]

Are you talking about the distance from centerline to point of cable attachment ?

No. I am in agreement. The same as you. As the string winds it is displaced from the centre line of the twist as you had shown.
I think that is what you had shown.

 

[RonL]

I'll throw some things out, they are estimates and are intended to be indicators of how things might need to be adjusted for better performance.

1. Consider the pistons to be 4" and a stroke of 12", that will produce a chamber volume of 4" X 24" (approx 1/3rd cubic foot)

2. Compressed air @500 psi, HP per cubic foot required by a two stage compressor .303, or a three stage compressor .289

3. The tension arm 48" long (push rod center to pivot center.

4. Cable mount center, to be 12" above pivot center.

5. Cable cluster each side made of, three 1" X 120" steel cables, attachment of each cable at flywheel and tension arm to be 12" radius.
An estimate of twists in the cables 60 ~ 70 (flywheel revolutions )

6 flywheel weight and diameter to be adjusted, based on above items and how much energy can be transferred and applied.

Hopefully this can show what goes on in the wirligig scaled up machine. let me know if this is workable.

 

[RonL]

The hardest thing for me to comprehend in my mind is, the strong torque as the cable starts to unwind and then transitions to speed by the time it moves past that last twist. Every twist should represent a different set of calculations, I feel sure that somewhere there is a computer that can make it simple ?

 

[RonL]

Had company over the weekend, but typical me, I think of things that might be better, before I can get started on a basic prototype.
I looked at a piece of rope in my shop and realized the number of twist will be more like 27~30 on each side of the flywheel, this will depend how little of an angle from twist center to the 12" radius attach point.
Second the longitudinal growth length will have a requirement that might move the cable center point to the 24" center of the tension arm.

My thoughts at this point are, a 6300 pounds push on each tension arm and a force at the cable center-line of 12,600 pounds
I'm pretty much lost about how to calculate the torque applied at each of the 120 degree cable attachment points to the cable center ( this also brought up the thought of having a 4" tube that the cable twist around ? it seems that more torque could be applied to spinning the flywheel and less stress on the twisted cable)

I would like to hear from anyone, questions or comments, if none I'll probably make a very crude mock-up on Tue. or Wed.

Thanks and hope everyone had a good weekend.

 

[OmCheeto]

Thanks and hope everyone had a good weekend.

No...

this also brought up the thought of having a 4" tube that the cable twist around

That was my thought!

 

[RonL]

No...
That was my thought!

Sorry your weekend wasn't the best,
One other thing I didn't say was the thought of those cables flying outward at that last twist, leads me to think some form of flexible band to hold the cables close to the center area might be a good thing.
Later.

 

[RonL]

After letting my mind operate in free fall mode for a couple of days, the startling thought sprung up I am thinking the twisted cables and an electric motor perform the exact same function, converting pressure into spin.
It will be nice if the twisted cables are a tad more efficient ?
I think any calculations between attachment points of the cables will yield a net of zero.

 

[OmCheeto]

Preliminary data from my experiment does not look good.
But we may have accidentally stumbled across the reason for quantum chirality.
That is, if you believe in my "Nylon String Theory".

Wound clockwise, things looked kind of normal.

turns    time (sec)     α = 2θ/t^2     torque = I * α
_1        3.32          1.140          0.00032
_2        5.21          0.926          0.00026
_4        8.08          0.770          0.00021
_8        9.73          1.062          0.00030
16       11.13          1.623          0.00045

Wound counter-clockwise, nothing looked normal.

turns  time    α      torque    notes
_1     n/a     n/a     n/a      would not turn
_2     n/a     n/a     n/a      made 1/2 turn & stopped
_4     n/a     n/a     n/a      made 2 3/4 turns & stopped
_8     15.7    0.408  0.00011  
16     14.05   1.019  0.00028

The nylon string is three stranded, and displays distinct properties when wound one way, versus the other.

I don't even know where to begin with these numbers...
The above experiment was done with 10 CDs weighing a total of 0.155 kg, yielding a moment of inertia of 0.000279

A second run with only a single CD yielded better results, but did not exhibit what I would call the "cog" effect of the counter-clockwise wound strings, which kind of ixnays this idea.

turns  time     α         torque         notes
_1     3.66     0.938     0.0000262     wound clockwise
_2     5.21     0.926     0.0000258  
_4     6.15     1.329     0.0000371  
_8     7.04     2.028     0.0000566  
16     7.53     3.546     0.0000989  

_1     5.98     0.351     0.0000098     wound counterclockwise
_2     9.60     0.273     0.0000076  
_4     8.54     0.689     0.0000192  
_8     8.06     1.547     0.0000432  
16     8.00     3.142     0.0000877

time: time to unravel
α: angular acceleration

 

[RonL]

I guess something along these lines would limit the twist problem, there are two types, dynamic and static. The static type does not have a tendency to stretch.

Are your test using just gravity ?

 

[OmCheeto]

...
Are your test using just gravity ?

Yes.
And after a good nights sleep, I've decided that this is a most complicated case of recreational mathematics.
Lots of things going on.
My angular accelerations and torques, from yesterday's experiment, are obviously all wrong.
They were based on constant acceleration, which is obviously not what was happening.
I did some more measurements this morning, regarding static distances and angles.

full     top    top twist   dist(adj)  radians     degrees
twists
_0      0.480     n/a              
_1      0.480     0.295       0.185     0.108       6.2
_2      0.480     0.325       0.155     0.128       7.4
_4      0.480     0.365       0.115     0.172       9.9
_8      0.480     0.425       0.055     0.340      20.0
16      0.480     0.456       0.024     0.695      39.8

I'm still trying to figure out the new free body diagram.

I'm guessing it's been too many years since I've studied this at university. (2016-1986 = 30 years!)

ps. I don't think choosing a rope design is going to help solve this problem, before we've figured out what's going on.

 

[RonL]

I have always considered steel cables only in a larger scale up machine, in the toy version the long and small diameter string or fishing line will hide any obvious loss due to the layering bias of how the line is formed. I believe a regular style three strand twisted steel cable, would show the same problems you found, but at a much greater loss.
I think I have seen steel cable in the same pattern as the ropes in the picture.

Thinking back to the toy button, it is force input by a persons arms (or wrist) and several cycles before the maximum speed of the button is reached, then it is a steady power applied and absorbed that keeps things consistent.

The simple thoughts that prompted my drawing, lead to the thermodynamics of compressed air and heat transfer in some form of hybrid system. So yes it presents a lot of engineering challenges to reach a smooth and steady mode of operation. IMHO

ps. I'm pretty sure what's going on with the toy version, is that with a method for measuring the energy input and the energy being absorbed, one would find they are "ALMOST" identical.
The energy to slow the button and set it up for reverse movement, represents the energy to harvest from. I hope that is correct, it seems right in my mind

 

[OmCheeto]

...
I think I have seen steel cable in the same pattern as the ropes in the picture.

I'm starting to get the feeling that there's a reason I have never seen braided cables used in such a manner.
The more I think about it, the less I like it.

...
ps. I'm pretty sure what's going on with the toy version, is that with a method for measuring the energy input and the energy being absorbed, one would find they are "ALMOST" identical.
The energy to slow the button and set it up for reverse movement, represents the energy to harvest from. I hope that is correct, it seems right in my mind

I'm still not confident I know how these things work, so I can't say anything yet.
But after a bit more experimenting today, I made a few more discoveries.
I decided my latest free body diagram in post #51 is missing a force.
To twist the 3 cables, I have to maintain an additional torque on the flywheel, other than just the weight of the flywheel itself.

I also made a video of the disk spinning from 16 turns clockwise to unraveled, and attempted to analyze that data as well as I could.
I discovered that my camera records 4 frames per second.
So in the 8 seconds it took to unravel, I collected 33 data points.
Since I wasn't in the mood to measure everything to the degree, I divided the circle into 16 points, and converted those points to radians.
This made for an incredibly ugly and incomprehensible acceleration curve, so I relearned the most basic of calculus, and calculated the first and second time derivatives of the displacement, giving me angular velocity and acceleration.

Since the acceleration is decreasing, that means the force was decreasing. So the weight of the flywheel was not the only force involved. Since there was a slight wobble in the flywheel, I'm assuming some of that is windage loss.
So what does this tell me?
At the 6 second point, two full seconds before the 3 strings had unraveled, the net torques on the disk went from positive to negative, and I'm not sure why.

I may digitize some more of the video in the morning, as it's a bit tedious doing it by hand. But I recorded a full minute of oscillations, and it might be interesting.

notes for future reference:
displacement in radians = -0.1322*t^3 + 2.5043*t^2 + 0.9963*t - 0.1581

approx
time
(sec)
0.00 wound 16 times clockwise
8.00 unraveled
13.25 stopped (now wound ccw)
19.00 unraveled
24.50 stopped (wound cw)
30.25 unraveled
34.75 stopped (ccw)
39.00 unraveled
43.00 stopped (cw)
46.50 unraveled for the last time and continued oscillating as a trifilar* pendulum

*Ha! I thought I'd just made up a new word. It looks like it's been around for a while.
http://www.paultitchener.com/dynamics-trifilar-pendulum.html

 

[RonL]

I'm starting to get the feeling that there's a reason I have never seen braided cables used in such a manner.
The more I think about it, the less I like it.I'm still not confident I know how these things work, so I can't say anything yet.
But after a bit more experimenting today, I made a few more discoveries.
I decided my latest free body diagram in post #51 is missing a force.
To twist the 3 cables, I have to maintain an additional torque on the flywheel, other than just the weight of the flywheel itself.

I also made a video of the disk spinning from 16 turns clockwise to unraveled, and attempted to analyze that data as well as I could.
I discovered that my camera records 4 frames per second.
So in the 8 seconds it took to unravel, I collected 33 data points.
Since I wasn't in the mood to measure everything to the degree, I divided the circle into 16 points, and converted those points to radians.
This made for an incredibly ugly and incomprehensible acceleration curve, so I relearned the most basic of calculus, and calculated the first and second time derivatives of the displacement, giving me angular velocity and acceleration.

Since the acceleration is decreasing, that means the force was decreasing. So the weight of the flywheel was not the only force involved. Since there was a slight wobble in the flywheel, I'm assuming some of that is windage loss.
So what does this tell me?
At the 6 second point, two full seconds before the 3 strings had unraveled, the net torques on the disk went from positive to negative, and I'm not sure why.

I may digitize some more of the video in the morning, as it's a bit tedious doing it by hand. But I recorded a full minute of oscillations, and it might be interesting.

notes for future reference:
displacement in radians = -0.1322*t^3 + 2.5043*t^2 + 0.9963*t - 0.1581

approx
time
(sec)
0.00 wound 16 times clockwise
8.00 unraveled
13.25 stopped (now wound ccw)
19.00 unraveled
24.50 stopped (wound cw)
30.25 unraveled
34.75 stopped (ccw)
39.00 unraveled
43.00 stopped (cw)
46.50 unraveled for the last time and continued oscillating as a trifilar* pendulum

*Ha! I thought I'd just made up a new word. It looks like it's been around for a while.
http://www.paultitchener.com/dynamics-trifilar-pendulum.html

I hate to admit and am slightly embarrassed, to say I am completely illiterate at anything higher than simple math, so I have to rely on information that is pretty well documented as, "that's just about right" That statement has served me well in almost everything, so now I would like to ask a couple of questions.

Can your dusted off calculus make a graph from the following ? If yes...

As you found out, there should be some force applied, now looking at post #43, if we use the air pressure, piston value, then assume the force on the cables are twice the piston force, we can adjust anything needed to fit the required travel of the cable as it unwinds (lets use a length and size that gives 48 twist) and because of a set force we can use a time factor of six seconds. What that gives, is a set of numbers that I think will determine flywheel speed at the end of piston travel.
We have two pistons applying force and that force flows through and becomes a twisting force on each side of the flywheel, I don't have a clue how to calculate the pull force on the end of the cables into twist force...(my mind sees the crazy picture of three people spaced 120 degrees apart, pulling with equal force, three starter ropes that have been wrapped around the starter pulley of a lawn mower) whatever the force at the twisted cable center, it will be the same at the flywheel connections.
That hopefully leads to the numbers of, 6300 pounds force applied smoothly, over a distance of 12" (48...1/4"points for 48 twist of the cables) expended over the time of six seconds...(because the two pistons move in opposite directions at the same time, the force X 2 is applied to the flywheel)

The flywheel is turned the number of 48 turns, then locked in place. From a large source and adequate inlet, the air pressure of 500 psi is applied. everything is in a static mode ready for release. When released the first 1/4" is a slow movement, then each 1/4" of motion becomes progressively faster and the last is fastest, at which point the air inlet is closed. I think with no load this would be a micro-second answer.

The second question...would the six second time, be the deciding factor of what mass and diameter the flywheel needs to be, in order to absorb this much energy over this time duration ?
I hope this is not too much of a dysfunctional post thanks again

Thanks Om, for your interest and time so far.

 

[OmCheeto]

I hate to admit and am slightly embarrassed, to say I am completely illiterate at anything higher than simple math, so I have to rely on information that is pretty well documented as, "that's just about right" That statement has served me well in almost everything, so now I would like to ask a couple of questions.

Can your dusted off calculus make a graph from the following ?

In a word, no.

If yes...

As you found out, there should be some force applied, now looking at post #43, if we use the air pressure, piston value, then assume the force on the cables are twice the piston force, we can adjust anything needed to fit the required travel of the cable as it unwinds (lets use a length and size that gives 48 twist) and because of a set force we can use a time factor of six seconds. What that gives, is a set of numbers that I think will determine flywheel speed at the end of piston travel.
We have two pistons applying force and that force flows through and becomes a twisting force on each side of the flywheel, I don't have a clue how to calculate the pull force on the end of the cables into twist force...(my mind sees the crazy picture of three people spaced 120 degrees apart, pulling with equal force, three starter ropes that have been wrapped around the starter pulley of a lawn mower) whatever the force at the twisted cable center, it will be the same at the flywheel connections.
That hopefully leads to the numbers of, 6300 pounds force applied smoothly, over a distance of 12" (48...1/4"points for 48 twist of the cables) expended over the time of six seconds...(because the two pistons move in opposite directions at the same time, the force X 2 is applied to the flywheel)

The flywheel is turned the number of 48 turns, then locked in place. From a large source and adequate inlet, the air pressure of 500 psi is applied. everything is in a static mode ready for release. When released the first 1/4" is a slow movement, then each 1/4" of motion becomes progressively faster and the last is fastest, at which point the air inlet is closed. I think with no load this would be a micro-second answer.

The second question...would the six second time, be the deciding factor of what mass and diameter the flywheel needs to be, in order to absorb this much energy over this time duration ?

I think it would take me a week to decipher what you are saying here, so don't hold your breath.

I hope this is not too much of a dysfunctional post thanks again

Thanks Om, for your interest and time so far.

Well, all I can say is, this is fun.
I haven't solved a derivative in 30 years!
I actually passed a couple of classes back then called "Elementary Differential Equations with Boundary Value Problems".
I have no idea how. It's all greek to me now. It was actually very difficult back then. But no matter. What is past, is prologue.

And now I see that I have to re-learn how to find the derivatives of a sinusoidal function...

God help me...

 

[OmCheeto]

I believe what we have here, is a classic example of a discontinuous function.

1. whirligig vs trifilar pendulum maths
2. chirality of the winding direction frictional forces
3. there may be more​

This might make a grand mathematical problem for @micromass to post in the maths forum.

 

[RonL]

I believe what we have here, is a classic example of a discontinuous function.

1. whirligig vs trifilar pendulum maths
2. chirality of the winding direction frictional forces
3. there may be more​

This might make a grand mathematical problem for @micromass to post in the maths forum.

I can only hope he or anyone else can decipher what I tried to say.

ps. What I tried to say was a description of one side of the two identical sides. The air volume is only important when the need to evaluate efficiency comes around.

 

[OmCheeto]

I can only hope he or anyone else can decipher what I tried to say.

ps. What I tried to say was a description of one side of the two identical sides. The air volume is only important when the need to evaluate efficiency comes around.

As I've inferred previously, I'm not willing to consider your ideas, until we've figured out the basics.

ps. One of the big "googley" problems with this problem is that there are multitudes of different things called "whirligigs".
In all of my googling over the last week, I've yet to run across anyone willing to tackle the "button" whirligig. The closest I came was some guy who flapped his lips around for a bit, and at the end said; "See the yo-yo".

 

[RonL]

As I've inferred previously, I'm not willing to consider your ideas, until we've figured out the basics.

ps. One of the big "googley" problems with this problem is that there are multitudes of different things called "whirligigs".
In all of my googling over the last week, I've yet to run across anyone willing to tackle the "button" whirligig. The closest I came was some guy who flapped his lips around for a bit, and at the end said; "See the yo-yo".

And I thought the drawing, as crude as it might be, would show the basics.
The piston chamber does the exact same thing a persons arms do when they spin the button. Linear force converted to spin energy.
The spin speed of the flywheel becomes the most important thing, I'm pretty sure.

Reading in a book, by Tom Monroe, "Clutch and Flywheel Handbook" He explains the energy of a 30 pound, 11-3/4" diameter, flywheel, @ 1,000rpm the Kinetic energy is 612 foot pounds, and @ 5,000rpm the Kinetic energy increases to 15,289 foot pounds.
As I think we both know, the V-8 engine can accelerate that much in about one second or two, I'm not sure I understand if the potential energy in the flywheel is never exceeding the power that induced the speed ? I have hand cranked a Lister generator too many times in my life.

All that to say I think in my example of 1/4" increments of continually increasing the energy into the flywheel, the last of the twist would leave a speed of more than 48 or 480 revolutions per minute. The speed of spin and the speed of unwinding are two different things I believe.
Sure hope you don't get too bored with this,

 

[OmCheeto]

...
Sure hope you don't get too bored with this,

Not so much bored, as over my head.

I generated an equation for a damped sine curve that simulates the first 1 1/2 cycles of oscillation.

position = 100.53*sin(0.2513*t+4.7124)*e^(-0.06t)
From that I could extract:

angular vel = -(6.0318*sin(.2513*x+4.7124)-25.2632*cos(.2513*x+4.7124))*e^(-0.06*x)
angular acc = -(5.98673*sin(.2513*x+4.7124)+3.03158*cos(.2513*x+4.7124))*e^(-0.06*x)​

From maximum angular acceleration, 6 rad/sec^2, we should now be able to determine the maximum force on the strings, given that we've already determined the moment of inertia of the disk.
And from the change in velocity, we should be able to figure out the energy losses.
They seem to be pretty significant, as my device made only 3 full cycles.

perhaps tomorrow, as I'm typing one-eyed.