Can Hydraulic Pressure Convert into Kinetic Energy?

[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.

 

[OmCheeto]

Sorry about my absence this week, but

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

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

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

 

[RonL]

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

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

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

 

[RonL]

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

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

 

[Nidum]

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

+1

 

[RonL]

+1

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

Naturally a lot of details need to be considered.