Can Hydraulic Pressure Convert into Kinetic Energy?

[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 ?