Can Hydraulic Pressure Convert into Kinetic Energy?

[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.

 

[anorlunda]

It appears that you guys are having fun with this thread. Excuse me for jumping in late. It triggered memories. I played a lot with these things as a child. So much that I emptied my mother's sewing box of all her spare buttons. She got mad when she found out because for the first time in her life she had to go to the store and buy buttons. From those memories, I think I can point to the dominant energy loss in such a system and the exact likely failure location.

First, I believe that the dominant energy loss in the system is the bending and stretching of the fibers in the string/rope/cable. Think of the pulleys on a crane as in the picture below. It appears that the diameter of those pulley sheaves is about 15 inches. Why not smaller diameter? Because the smaller the diameter for bending, the greater the energy losses and the shorter the lifetime. Of course, the materials and braided/twisted structure, and the lubrication of the strands influence the quantitative results, but the qualitative statement that bending and stretching dominate energy losses remains true IMO. Twisting a bundle of strands necessarily bends and stretches.

For failure mode, refer back to @OmCheeto 's diagram from #36.

The holes in the button lie on a circle with radius $R$. The strands (two strands in the diagram, but often 4 strands with buttons) form a twisted bundle with radius $r$.

$r<<R$

At distance $D$ from the button on each side, the strands must depart from the bundle and fan out to go through the holes in the button. In @OmCheeto's diagrams $D$ is measured axially from the place where the strands cross to the button.

As $D$ approaches zero, further twisting becomes impossible and (at the final instant) nearly 100% of the inertial energy must go into stretching the strands in that tiny fan-out region. From my boyhood memories, failure occurs either when the strands break in the fan out region, or when the button fails as the strands cut through the material making the multiple holes merge into one hole in the center.

Thanks for sharing a topic fun for engineers.

 

[RonL]

It appears that you guys are having fun with this thread. Excuse me for jumping in late. It triggered memories. I played a lot with these things as a child. So much that I emptied my mother's sewing box of all her spare buttons. She got mad when she found out because for the first time in her life she had to go to the store and buy buttons. From those memories, I think I can point to the dominant energy loss in such a system and the exact likely failure location.

First, I believe that the dominant energy loss in the system is the bending and stretching of the fibers in the string/rope/cable. Think of the pulleys on a crane as in the picture below. It appears that the diameter of those pulley sheaves is about 15 inches. Why not smaller diameter? Because the smaller the diameter for bending, the greater the energy losses and the shorter the lifetime. Of course, the materials and braided/twisted structure, and the lubrication of the strands influence the quantitative results, but the qualitative statement that bending and stretching dominate energy losses remains true IMO. Twisting a bundle of strands necessarily bends and stretches.

For failure mode, refer back to @OmCheeto 's diagram from #36.

The holes in the button lie on a circle with radius $R$. The strands (two strands in the diagram, but often 4 strands with buttons) form a twisted bundle with radius $r$.

$r<<R$

At distance $D$ from the button on each side, the strands must depart from the bundle and fan out to go through the holes in the button. In @OmCheeto's diagrams $D$ is measured axially from the place where the strands cross to the button.

As $D$ approaches zero, further twisting becomes impossible and (at the final instant) nearly 100% of the inertial energy must go into stretching the strands in that tiny fan-out region. From my boyhood memories, failure occurs either when the strands break in the fan out region, or when the button fails as the strands cut through the material making the multiple holes merge into one hole in the center.

Thanks for sharing a topic fun for engineers.

Thanks for your input,
From the comments near the end of your post, I tend to agree that the fan area is the main action points, which also leads me to think of the twisted lines as a spring of sorts, I'm thinking the continuous linear force is keeping the spring like energy release very nearly consistent until the last twist unleashes. So if this is in any way close to being correct, I'll try "spring energy storage" and see if there might be any information to work on.
Any ideas about my question based on a six second time and 48 twist in the cable set ? I kinda think that as much as 50% of the energy will transfer in that last second, I just don't know how to set up a graph or flow chart, that shows a steady force feeding into a progressively increasing flywheel speed.

Guess I thought this would be more simple than it seems to be.

 

[OmCheeto]

It appears that you guys are having fun with this thread.
...

It's not just fun, it's actually quite beautiful, IMHO.

The plot from my curve fitted equation, of position, vel, and acc:

maximum values set to 1​

actual max amplitudes

pos: -100.53 radians
vel: 19.321 radians/sec
acc: 5.987 radians/sec^2​

Guess I thought this would be more simple than it seems to be.

If it were, someone would have properly analyzed the physics by now.
From my earlier googling, the "button/buzzer" type whirligig may be about 2500 years old.

Per wiki, regarding "Buzzer (whirligig)"; "American Indians used the buzzer as a toy and, also ceremonially, as to call up the wind. Early Indian buzzers were constructed of wood, bone, or stone, and date from at least the Fourche Maline Culture, c. 500 B.C."​

[ref]

 

[RonL]

It's not just fun, it's actually quite beautiful, IMHO.

The plot from my curve fitted equation, of position, vel, and acc:

maximum values set to 1​

actual max amplitudes

pos: -100.53 radians
vel: 19.321 radians/sec
acc: 5.987 radians/sec^2​

If it were, someone would have properly analyzed the physics by now.
From my earlier googling, the "button/buzzer" type whirligig may be about 2500 years old.

Per wiki, regarding "Buzzer (whirligig)"; "American Indians used the buzzer as a toy and, also ceremonially, as to call up the wind. Early Indian buzzers were constructed of wood, bone, or stone, and date from at least the Fourche Maline Culture, c. 500 B.C."​

[ref]

It does look nice, a little like an electronic readout of a system of tank circuits
Your research is interesting and surprising, like most of my ideas they seem to fit the 1800's pretty well, however this one seems to be, as you said, a bit novel.
I looked in my shop and found just about everything I might need to setup a system fitting it around a 5" flywheel, 1" shaft and bearings, weight about 3 pounds (but my wife thinks there are other things that be a little more important) she is usually right I'm going to hold back a little and see if anything turns up in this thread. Really want to get that flywheel speed estimated.

 

[OmCheeto]

It does look nice, a little like an electronic readout of a system of tank circuits
Your research is interesting and surprising, like most of my ideas they seem to fit the 1800's pretty well, however this one seems to be, as you said, a bit novel.
I looked in my shop and found just about everything I might need to setup a system fitting it around a 5" flywheel, 1" shaft and bearings, weight about 3 pounds (but my wife thinks there are other things that be a little more important) she is usually right I'm going to hold back a little and see if anything turns up in this thread. Really want to get that flywheel speed estimated.

Glad to hear you've decided on a 3 lb vs 220 lb prototype.
I'm pretty sure you'd have lost [at least] a finger, otherwise.
My 0.03 lb prototype really hurt my fingers when I wasn't paying attention.
Leverage, time, and a suitable energy storage device can make a painful, if not deadly, combination.

 

[OmCheeto]

Gads.
I've done lots of maths, and this is not looking good.

notes:

ME = mechanical energy = kinetic energy + potential energy
Potential energy turned out to be less than 1% of the energy in the system, so I threw that out.
ME losses were ≈ 75% in less than a full cycle.
conclusion: this is a lousy losey system​

ps. I learned some very strange and wonderful things yesterday, in my quest to solve this problem:

The physics behind "twisted ropes" are as weird and perhaps older than this toy:

https://www.sciencenews.org/article/physicists-untangle-geometry-rope

"Ropemaking in ancient egypt. Tomb of Akhethotep and Ptahhotep, about 2300 BC.
...​

the intrinsic geometry behind the art of laying rope is not something you have to know or be aware of, just the instructions which have been passed down through generations." [ref: from the original paper]​

From this I decided: RonL, if you do build one of these devices, DO NOT let the length of the twisted bundle reduce to some amount referenced in the above paper, as you will have just created a new and novel "rope making device".

Some keywords I was missing off the bat were; torsional harmonic oscillator
"physics of twisted rope" kind of got me started.​

pps. Other useful things I've learned:

e = sinh(1) + cosh(1)... I got very tired of looking up that number.​

ppps. Other things brought back:

d/dx ax^b = bax^(b-1)
d/dx sin(x)=cos(x)
d/dx cos(x)=-sin(x)
d/dx (uv) = du/dx(v) + dv/dx(u)​

pppps. I think I broke Wolfram|Alpha's "definite integral calculator".

But, it was a long equation:

find the definite integral of power from t = 29.3 to 36.5

given the equation:

P = 0.000279 * (−(5.98673×SIN(0.2513×t+4.7124)+3.03158×COS(0.2513×t+4.7124))×e^(−0.06×t)) * −(6.0318×SIN(0.2513×t+4.7124)−25.2632×COS(0.2513×t+4.7124))×e^(−0.06×t)​

 

[RonL]

@ Om, I really am thankful for all your input and research But I just don't understand how your measurements are calculated. The time is confusing to me, as I see complete cycles from zero to maximum and back to zero in about 10 seconds, the speeds seem to be very fast ( I would say close to 2000 rpm ) if I'm right, here is a calculator for showing my little 3 pound flywheel.
http://www.calculatoredge.com/mech/flywheel.htm
Guess my numbers don't show I used 48 oz. 5" dia. and 2000 rpm. (just guessing at speed, which is very momentary)
My searching has taken me to a book that is Greek to me " Vector Mechanics for Engineers" an 84 edition. (nothing I can see that looks close)

I think it's time to start the slow contemplating process for my three pound wheel ( don't hold your breath )
I,ll get back later.

 

[OmCheeto]

@ Om, I really am thankful for all your input and research But I just don't understand how your measurements are calculated. The time is confusing to me, as I see complete cycles from zero to maximum and back to zero in about 10 seconds, the speeds seem to be very fast ( I would say close to 2000 rpm )

Maximum angular velocity was 185.4 rpm @ t=4.5 seconds.

if I'm right, here is a calculator for showing my little 3 pound flywheel.
http://www.calculatoredge.com/mech/flywheel.htm

Uh oh. My numbers don't match.
Ah ha!
I used the moment of inertia of 10 disks vs 1 disk.
My "real" numbers may have been off for the last few days.

Guess my numbers don't show I used 48 oz. 5" dia. and 2000 rpm. (just guessing at speed, which is very momentary)
My searching has taken me to a book that is Greek to me " Vector Mechanics for Engineers" an 84 edition. (nothing I can see that looks close)

I think it's time to start the slow contemplating process for my three pound wheel ( don't hold your breath )
I,ll get back later.

That one error is having really nasty repercussions in my calculations.
But, then again, my graphs now indicate that I may have found a source of "negative energy".
and do you know what wiki says about that?

In [some] theories, negative energy is involved in wormholes which allow time travel and warp drives for faster-than-light space travel.​

And there you have it. The secret to warp drive is to move the decimal point one place in the wrong direction.

ps. Actually, I think this means that the potential energy is more significant than previously calculated. I'll try and fix this in the morning.

 

[OCR]

Guess my numbers don't show I used 48 oz. 5" dia. and 2000 rpm. (just guessing at speed, which is very momentary)

RonL ... I don't mean to butt in here, and it might be a fact that...

If you don't know what you're talking about, then I prefer you don't say anything about this at all.

But, I think you about have to take a screen grab to keep your numbers showing... I input the same numbers as I bolded in your quote. Is this what it showed... ?

BTW... nifty calculator.[COLOR=#black]...[/COLOR]

 

[RonL]

RonL ... I don't mean to butt in here, and it might be a fact that...

But, I think you about have to take a screen grab to keep your numbers showing... I input the same numbers as I bolded in your quote. Is this what it showed... ?

BTW... nifty calculator.[COLOR=#black]...[/COLOR]

View attachment 105557​

That is what mine showed, thanks and anyone is welcome to have a say especially if they can help show how the linear force is transferred through the twisted section. ( I feel there is very little loss there)

ps. I didn't mean to limit input by anyone, to just the twisted section