Can Hydraulic Pressure Convert into Kinetic Energy?

[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....

 

[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....

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

 

[OmCheeto]

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)

I am of a differing opinion.
But, opinions ain't science, so I'll wait for me to get started, and fix my graphs and stuff from the last few days.

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

The twisted section is the key here, IMHO. And I can find very little, as in zero, quantitative analysis on it.
The most promising title so far: Catapult Physics
yielded little more than a picture. But such pages do direct me to terminology which comes in handy for googling. For example: Bundle Torsion.
But if you go to Wolfram|Alpha's page on "Bundle Torsion", it's blank!

And the internet is rife with "Bundle Torsion" stuff, relating to nothing we are discussing:

Spacetime tangent bundle with torsion
Abstract

It is demonstrated explicitly that the bundle connection of the Finsler spacetime tangent bundle can be made compatible with Cartan's theory of Finsler space by the inclusion of bundle torsion, and without the restriction that the gauge curvature field be vanishing.​

Ok then. Not only are we on the verge of warp drive, we've stumbled upon the basis of the "cloaking device".

ps. Sorry about all the jokes, but all this serious maths is making me a bit crazy. And finding humor along the way to our destination, makes it a bit more bearable.

pps. I would like to posthumously thank whoever it was that found "e", and its role in calculus: f'(e^ax) = ae^ax

 

[RonL]

I don't mind the humor at all, my attempts at humor generally fail, so I tend to remain my sixth grade self
Your last pps. is completely foreign to me and over my head, but I feel I understand almost fully a "Jacobs Brake" bringing an 18 wheeler from 60 mph down to 40 or less in a matter of seconds ( an almost complete waste of compressed air as a best example) sure saves a lot of wear and tear on the standard air brake system. But I have gone off topic a little.

The drawing depicts a force, compressed air, being multiplied at the tension arm and transmitted through the twisted cables, converting force to spin at the flywheel.
The main question in my mind is, after the spin has started each twist is passing the same force to a faster spinning flywheel, until that last twist unleashes. How fast will the wheel be turning. ( why is this not a piece of cake for anyone that did their homework )
I do find the links you shared, very interesting

I think I understand that your measurements using just gravity, are probably as close to accurate as numbers where force has been scaled up. I have a 12' shop ceiling, that I can hang two 40 pound pipe end covers from, that have 12 one inch bolt holes, a digital tachometer to measure speed.

Is your method of calculating, something you can explain here or in a PM ?

ps. I think I found a promising search area. "heavy lift cranes and twisted cables" there seems to be a lot of links (it's just a thought)

 

[OmCheeto]

...
The main question in my mind is, after the spin has started each twist is passing the same force to a faster spinning flywheel, until that last twist unleashes. How fast will the wheel be turning.

All in due time. All in due time.

( why is this not a piece of cake for anyone that did their homework )

I'm finding that I'm still really bad at homework.
Along with displacing that decimal point, I also discovered that I boogered the "potential energy" portion of the problem.
argh!

I do find the links you shared, very interesting

I think I understand that your measurements using just gravity, are probably as close to accurate as numbers where force has been scaled up. I have a 12' shop ceiling, that I can hang two 40 pound pipe end covers from, that have 12 one inch bolt holes, a digital tachometer to measure speed.

Is your method of calculating, something you can explain here or in a PM ?
...

Calculating is easy, once you've figured out:

1. You erroneously used the mass of 10 CDs for your 1 CD experiment
2. Your equation for potential energy was lopsided. (I just found out that I needed the absolute value)
3. Other things that pop up.
​

My latest graph...

Current confidence level that anything is correct, after a weeks worth of botching just about everything: 10%​

 

[RonL]

@ OmCheeto, Starting to look a little like Mardi Gras don't you think ? Just don,t give up

ps. If I understand anything about your graph, I would wonder how LITTLE extra force will be used to keep it at the full cycle.

 

[OmCheeto]

@ OmCheeto, Starting to look a little like Mardi Gras don't you think ? Just don,t give up

ps. If I understand anything about your graph, I would wonder how LITTLE extra force will be used to keep it at the full cycle.

Argh. I found another mistake.
I can't really help you until I filter out all the bugs.
In the interest of "start over and see what else isn't right" I compared my raw data to my curve fitted data. It doesn't look too bad.

Although "displacement" displays as smooth as a baby's butt, "acceleration", without the averaging, is incomprehensible.

But the overall character of the data looks like it matches the fitted curves. And the maximum values are moderately close.

In answer to your earlier question:

Is your method of calculating, something you can explain here or in a PM ?

Here are the values and equations for most everything.

CD buzzer-button whirligig
mass(m)                  0.0155
radius(r)                0.06
moment of inertia(I)     0.0000279 = 0.5 * m * r^2
time(t)                  collected from video
displacement(θ)          100.53×SIN(0.2513×t+4.7124)×e^(−0.06×t)   curve fit from data collected from video
height(h)                ABS(7.83×10^(−7)×(θ)^2 − 0.0002(θ) + 0.0002)  curve fit from data
potential energy(PE)     mgh
velocity(vel)            −(6.0318×SIN(0.2513×t+4.7124)−25.2632×COS(0.2513×t+4.7124))×e^(−0.06×t)   derivative of displacement
kinetic energy(KE)       0.5 * m * vel^2
mechanical energy(ME)    KE + PE
acceleration(acc)        −(5.98673×SIN(0.2513×t+4.7124)+3.03158×COS(0.2513×t+4.7124))×e^(−0.06×t)  derivative of velocity
torque(τ)                I * acc
power                    vel * torque      (peak was 1.51 milliwatts)
rpm                      displacement(θ)/(2*pi) * 60

Here's a brief explanation of the numbers in the displacement equation:

displacement(θ) = 100.53×SIN(0.2513×t+4.7124)×e^(−0.06×t)

100.53 is the maximum amplitude
0.2513 determines the frequency
4.7124 determines the initial offset
e^(-0.06*t) is the amplitude damping factor over time​

I'm afraid I couldn't remember how to introduce a damping factor for frequency.
Perhaps tomorrow.

 

[OmCheeto]

...
I'm afraid I couldn't remember how to introduce a damping factor for frequency.
...

Yay!

position = (a)sin(e^(bx)kx+c)e^(gx)
where

a = original amplitude
b = frequency damp
x = time
k = original frequency
c = offset
g = amplitude damp
e = sinh(1) + cosh(1) ​

and

f(u) = sin(e^(bx)kx+c)
f(v) = e^(gx)
d/dx (uv) = du/dx(v) + dv/dx(u)
d/dx f(u) = ake^(bx)*(bx+1)*cos(kxe^(bx)+c) <---- not sure if the notation is correct, but I think I know what I'm doing
d/dx f(v) = ge^(gx)
d/dx f(x) = a((sin(e^(bx)kx+c))(ge^(gx))+(e^(gx))(ake^(bx)*(bx+1)*cos(kxe^(bx)+c)))
∴
angular velocity = ae^(gx)(gsin(e^(bx)kx+c))+(ake^(bx)*(bx+1)*cos(kxe^(bx)+c)) <---- "definite maybe" that I got this right​

But, I guess this means, that I have to start all over, again...

ps. I'll check and see if the graphs match.
and... no. The flavor is right, but everything is wrong.

I should probably stick to addition and subtraction.

 

[RonL]

At least you know enough to be able to make a choice, my brick wall was Cortisone Coordinates eleven years ago, I was taking a CC course in pre-algebra. I took a job that made a schedule for classes impossible. But I had always struggled with great difficulty, any kind of advanced math.

 

[OmCheeto]

At least you know enough to be able to make a choice, my brick wall was Cortisone Coordinates eleven years ago, I was taking a CC course in pre-algebra. I took a job that made a schedule for classes impossible. But I had always struggled with great difficulty, any kind of advanced math.

Our lives sound eerily similar.
And like my graphs, the paths are completely different.
hmm...

But that's neither here nor there.

It may be next week before I get back to the problem, as I have been invited to the coast. And although I don't want to go, I need some time to meditate.

ps. I'll take my hammer along, in case anyone tries to ruin my weekend again, like last week.

pps. My latest, and hopefully not my last, whirligig graph:

 

[RonL]

Om, I made a few test today, but I had no way to absorb energy from the flywheel so I had to be extra careful not to spin things too fast.
The flywheel weighs 4 pounds and 7 ounces. It is 5" in diameter and is 1/2" thick. The axle part is a 1" diameter and 6" long.
I used a two strand Kevlar thread on each side (400 pound test) about 1/8" thick.
I suspended it like you did the CD disk, from a 2X4 and applied energy with a 2 X 2, the distance between them was 50" and the string measure on each side 22"
Cycle time was right at 17 seconds, 85 twist to full stop and the digital tach shows 975 to 1025 rpm, the variation had to do with my unsteady applying of pressure by hand, which I would estimate at about 2 pounds pushing down on the 2 X 2.

The take up of line distance surprised me, a total of 13" at which the travel was limited by the scaffold frame and the fan of the string would absorb the rest of the spin motion.
A lot of numbers, I'll try to make better order of them later.
Hope you have a good week end.

 

[OmCheeto]

Om, I made a few test today, but I had no way to absorb energy from the flywheel so I had to be extra careful not to spin things too fast.

My system seems to absorb all the energy in my flywheel quite rapidly. hmmmm...

The flywheel weighs 4 pounds and 7 ounces. It is 5" in diameter and is 1/2" thick. The axle part is a 1" diameter and 6" long.
I used a two strand Kevlar thread on each side (400 pound test) about 1/8" thick.
I suspended it like you did the CD disk, from a 2X4 and applied energy with a 2 X 2, the distance between them was 50" and the string measure on each side 22"
Cycle time was right at 17 seconds, 85 twist to full stop and the digital tach shows 975 to 1025 rpm, the variation had to do with my unsteady applying of pressure by hand, which I would estimate at about 2 pounds pushing down on the 2 X 2.

Sounds like my original experiment:

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)

The problem with those methods was that applying pressure or force by hand is not really going to be useful.

So I redesigned my system. It looks very much like the system you've set up, so I've inserted some of your numbers.

The test mass would, like you and I applying force, be added and removed at the proper moments.
It would be added initially, and whenever the device stops.
It would be removed when the twisted bundle is no longer twisted.
My plan is to vary the test mass, until little noticeable change in full cycle time is noticed.

I probably won't use 130 CDs. Perhaps 20.

The take up of line distance surprised me, a total of 13" at which the travel was limited by the scaffold frame and the fan of the string would absorb the rest of the spin motion.
A lot of numbers, I'll try to make better order of them later.

"A lot of numbers"?
I'm up to 10 tabs on my spreadsheet.

hmmm...

number counts
__≈# tab name
__28 cd button wirligig
__85 cd button wirligig 2.0
_190 cd button wirligig 3.0
3800 cd button wirligig 4.0
_350 cd button wirligig 4.5
1200 fun with sines
_470 more fun with sines
_140 simplified damped oscillation
1100 whirligig 5.0
1100 whirligig 6.0
-------------------------------
8500 total numbers

Hope you have a good week end.

That would make a good thread.

 

[RonL]

That moment of cogging you mentioned, seems to be a real energy killer, I tried putting a few twist in the upper section before sliding the string over the 2 X 2 at the bottom, that eliminated the bump, as twist went from one direction to the other, but it seemed to cripple the input of power (not sure exactly what was happening).
I'm beginning to think, the piston chamber area needs to function more along the lines of a sterling cycle. But that is a long way off.

Still think my drawing is close to accurate, just need some very good energy absorbing controls that are precise and easy to adjust during operation.

I was very happy to see the speed of the flywheel hit that 1,000 rpm mark, That flywheel speed is the very heart of what I'm thinking, but my footing the bill for special prototyping is not very likely.

Basically our test setups are almost alike, mine is lifting close to seven pounds, thirteen inches and when it reaches the cross brace of the scaffold and the 2 X 2 can't go any higher, the string and fan area act like a spring, sending the flywheel in the opposite direction. ( that 400 pound test, Kevlar is pretty good stuff)

 

[RonL]

Here's a picture of my test setup.

I switched from a single string to a three string setup (for safety reasons), things changed to a smoother and more powerful feel when the twisting reached it's limit, but seemed to be a little slower.

 

[OmCheeto]

Here's a picture of my test setup.
...

I switched from a single string to a three string setup (for safety reasons), things changed to a smoother and more powerful feel when the twisting reached it's limit, but seemed to be a little slower.

Looks like a grand setup!

Have you calculated the moment of inertia of your disk yet?
I do believe it's a crucial element in knowing what's going on.

I found a site the other day, Dynamics: Trifilar Pendulum, that listed everything you need for the experiment to determine the value.

     R^2*g*(tau)^2
I = ---------------  * m
      4*(pi)^2*L

I  refers to moment of inertia through center of mass about the z-axis of the system
R  is the distance from the center of the disk to each string 
g  is acceleration due to gravity (approximated to by 9.8 m/s^2)
τ  is the period of oscillation
L  is the length of the strings
m  refers to mass

I collected and ran the numbers from my video from the other day, and was only off by 2 orders of magnitude, from what I had originally calculated.
I'd forgotten to factor in the "m". (0.0155 kg)

Original (I), based on physical measurements:

CD radius: 0.06 m
m: 0.0155 kg
moment of inertia (I): 0.0000279 (from I = 0.5 * m * r^2)​

Trifilar pendulum derived I:

I = 0.0000275 (first cycle period)
I = 0.0000302 (average cycle period)
I = 0.0000423 (final cycle period)​

I'm guessing the "first cycle" value is the one we want. But the "average" would have suited me just fine.

 

[RonL]

I'm going to have to study a lot, as best I can.
This link helped a lot


This one is interesting, but I think I got a little more confused.


I need to explain that when I said three lines for safety, it is still a bifilar pendulum, as presented by 256bits , (three strands in the hole of the axle) The multi strands seem to soak up a lot of energy, as opposed to the single strand. A bit of a surprise there

I think the 1" axle will mess up my moment of inertia, or at least make it more difficult to calculate.
Your link was interesting, but well over my head at this time. It did help me a little in understanding what you are doing.

Later

 

[OmCheeto]

I'm going to have to study a lot, as best I can.
This link helped a lot


This one is interesting, but I think I got a little more confused.


I need to explain that when I said three lines for safety, it is still a bifilar pendulum, as presented by 256bits , (three strands in the hole of the axle) The multi strands seem to soak up a lot of energy, as opposed to the single strand. A bit of a surprise there

I think the 1" axle will mess up my moment of inertia, or at least make it more difficult to calculate.
Your link was interesting, but well over my head at this time. It did help me a little in understanding what you are doing.

Later

Even though I can now explain how to get the moment of inertia in about 60 seconds, I still love listening to Prof. Lewin.
I found the first one quite entertaining, as I guessed all the wrong answers. I probably don't have what I would call a "good feel" for it.
But I was relieved when Prof. Lewin in the second video said he couldn't remember any of the equations, and said it was ok for us not to bother memorizing them either.

To find the moment of inertia of your axle+disk, all you need are 4 measurements, which shouldn't take more than a minute to collect.
You need 2 lengths, 1 mass, and one time duration. So a ruler, scale, and stopwatch are the tools you'll need.

You will of course have to rig up a trifilar string setup.
This requires that the strings be all the same length, all be initially vertical, and equally spaced(120° apart).
I would recommend they be placed towards the outer edge of your disk-axle.
Place a mark somewhere on the disk.

And then, you start the experiment:

Twist the disk through one half a rotation, and let it go.
When the disk first stops, with the mark ≈180° on the other side, start the stop watch.
When the disk again stops, stop the stop watch.​

Multiply that time by 2, yielding (tau).
Plug all the numbers into the equation.
Write down the answer.

     R^2*g*(tau)^2
I = ---------------  * m
      4*(pi)^2*L

I  refers to moment of inertia through center of mass about the z-axis of the system

g  is acceleration due to gravity (approximated to by 9.8 m/s^2) (given)

R  is the distance from the center of the disk to each string
τ(tau)  is the period of oscillation
L  is the length of the strings
m  refers to mass

 

[RonL]

Thanks Om, I can get started tomorrow evening, the link you gave shows everything I need except the height, what dictates that ?

 

[OmCheeto]

Thanks Om, I can get started tomorrow evening, the link you gave shows everything I need except the height, what dictates that ?

By height, do you mean the length of the strings?
If so, anywhere between 2 and 6 feet should be plenty.
The shorter the string, the shorter the period.

From a rough estimate, I come up with the following:

tau    L(ft)
1.0    1.6
1.5    3.6
2.0    6.6

 

[RonL]

Checkups at the doctors, for my wife and me, have prevented me from doing hardly anything. Sitting here waiting to leave again, I found something that I haven't looked at yet, but think there is something here that we can draw on later,

http://www.mytreelessons.com/Pages/Rope Angle Leverage Calculator.htm

Have to go now, another appointment.

 

[OmCheeto]

Checkups at the doctors, for my wife and me, have prevented me from doing hardly anything. Sitting here waiting to leave again, I found something that I haven't looked at yet, but think there is something here that we can draw on later,

http://www.mytreelessons.com/Pages/Rope Angle Leverage Calculator.htm

Have to go now, another appointment.

This would be the basis of anorlunda's comment; "It's going to break here".

For failure mode...

So far, the tricks to make your invention not fail are:

1. Never exceed 120° (where according to your website, the horizontal force equals the vertical force, and kind of goes up from there.)
2. Don't make a rope:

​

ps. Science!

 

[RonL]

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 ?

 

[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.