Center of Gyration

[jim_mich]

I thought I'd look at "Center of Gyration".

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Center of Gyration...
That point in a rotating body at which the whole mass might be concentrated (theoretically) without altering the resistance of the inertia of the body to angular acceleration or retardation.

My trusty old Machinery's Handbook has three pages filled with 24 frames containing drawings with over 50 formulas relating to the calculation of "Center of Gyration" of rotating bodies of different shapes.

First a quick observation: A wheel is a balanced pendulum, whereas a pendulum is an unbalanced wheel.

Center of gyration is a theoretical point. If you were to squeeze all of the mass of a pendulum down to a tiny single point and locate that point so that it swings just like the original pendulum, then that point would be the Center of Gyration.

For instance suppose that you have a two inch diameter (one inch radius) steel ball hanging on a weightless spider web from an overhead support. Suppose the ball's center is ten inches from the support. A two inch diameter steel ball will weigh 1.1796 pounds. Now suppose we have an aluminum ball that weighs the same 1.1796 pounds. It will need to be 6.5641 bigger in volume in order to weigh the same. An aluminum ball 3.7447 inch diameter will weigh 1.1796 pounds.

You might think that these two metal balls would swing the same speed. Their mass is the same and their swing radius is the same. But the smaller steel ball will swing faster! Air resistance has nothing to do with it. Even in a vacuum the small ball will swing faster. The reason is the "Center of Gyration" is different for the two balls. And it is the Center of Gyration that determines the swing speed.

To determine the Center of Gyration for a spherical ball shaped pendulum the formula is...

L = SquareRoot(a^2 + 2/5 * r^2)

'a' is the pendulum arm length to the center of the ball.
'r' is the radius of the ball.

For the 2.000 inch steel ball L = 10.02 inches.
For the 3.7447 inch aluminum ball L = 10.07 inches.

The smaller steel ball swings faster because it has a shorter radial Length to its Center of Gyration.

My gut feeling keeps telling me that this is part of the Bessler wheel secret.

[Fletcher]

Jim, are you able to describe that with a drawing & some arrows ? I just usually think of a sphere as condensed down to a point mass on the end of a rod (windage aside).

What would be wrong with that approach ?

[ken_behrendt]

Jim...

I have never heard of that effect that you are describing! I knew that the moment of inertia of a rotating body will vary depending upon how its mass is distributed about its center of rotation, but I did not think that would alter the period of oscillation of a pendulum using that body just so long as the value used for the length of the pendulum was the distance between the pivot and the center of the body.

I assume that with the effect you are describing, one must then use the corrected value of L in the expression that determines the period of oscillation of the pendulum one makes from bodies with various mass distributions about their centers. It appears that the effect can be ignored when the radius, r, of the pendulum weight is small compared to the length of the pendulum arm, a.

However, since I am not a believer that swinging pendula were used inside of Bessler's wheels, I can not accept that this effect would have been the secret of his wheels. But, it is something to keep in mind when one is using pendula with short arms compared to the radius of the weight.


ken

[rlortie]

Ken,

You may not see this as a possibility by not believing there were pendulums involved. But as Jim states a pendulum is not unlike a wheel.

Now if two round weights of different size work In pairs as some interpret Bessler to be saying. Two that never find, but are always seeking equilibrium due to the simple fact that they have different radius. This radius would cause the center mass to always be out of balance, not unlike the two pendulums describe above. You could look at it as two unlike pendulums on each end of the rod. Neither being pinned to the wheel.

Patrick's reciprocator may be a good example if one end weight was of different size than the other. This would show that it is all about point of fulcrum and fulcrum relation to wheel.

Ralph

[jim_mich]

I used 'L' for the radial Length to the Center of Gyration whereas this picture from my Machinery's Handbook uses 'k'.

A simple way to think about this is to visualize a radial arc drawn through the weight. In order to get half the mass inside the arc and half the mass outside the arc then the radius of the arc must be farther out. But that alone is not enough.

Think of breaking down the mass into individual atoms. Then square the radial distance of each atom from the pivot. The sum of all the squares inside the radius must equal the sum of all the squares outside the radius. The point where the sum of the squares balances is the radius of gyration.

Each shape requires a slightly different formula in order to calaculate where that point is at. That is why Machinery's Handbook has so many different formulas.

[jim_mich]

Suppose I take four weights and attach them to the four corners of a parallelogram mechanism as in the drawing below. The colored spot in the center is where this mechanism attaches to a wheel. A number of these mechanisms are attached to the wheel. Now any of you familiar with weights will quickly see that the center of gravity of each mechanism is always located on that colored spot. The mechanism can be squeezed or elongated causing the weight distribution to change. When this happens the center of gyration changes, which causes the radius of gyration to change. This will cause the mechanism to exert its weight at varying radius of gyration distances from the wheel axle. It is like moving the weights in and out. If the weights get squeezed narrow sidewise at the bottom then their radius of gyration moves inward. If they get elongated sideways at the top their radius of gyration moves outward. You can measure to the center of each weight and tell that the combined distances are greater on one side than the other.

Things are never simple. Inertia likes to stick it nose into things that are rotating and other things conspire to make the wheel balance. But maybe a way can be found to use this little trick?

[Michael]

I'm sorry but I am having problems seeing what is going on here. If your saying that if two weights of different sizes that weigh the same and both have the same lever lengths from their centers to the pivots will swing at different rates, excluding air resistence than I would like to see a working model because I don't agree. Respectively.

[jim_mich]

Michael,
If the weights are rigid to their 'rod' so that they must rotate as they swing, then two pendulum weights of the same mass but different shape will swing at different speeds. This is because it is not the lever length that determines their speed, it is the distance from the axis to the center of gyration that determines the radius of gyration which is what determines the swing frequency. Text books simplify matters by using a point mass and then usually show a circular pendulum weight. This gets the idea across to new students but confuses them as to the real way that pendulums work.

The same principle applies to two cylinders of equal diameter and mass that roll down an incline at differing speed becuase one is hollow and made of heavy material while the other is solid and made of a lighter material. Both weigh the same. Both are the same diameter. When a cylinder rolls it is actually pivoting about the contact point. Again Center of Gyration comes into play. The solid cylinder with more of its mass near the center will roll (swing) faster.




Some quick numbers taken from the cad drawing that I threw together real quick like for my last post... Looking at a single mechanism with four weights, each weight is 4 inches apart, so each of the six 'rods' are 4 inches long. They are drawn at 70 degrees and 110 degrees. The colored connection point is 8 inches from the axle. The weights are located at...

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Right side weights at 3 o'clock...
X = +9.365	Y =  0.000	R = 9.365
X = +8.000	Y = +3.759	R = 8.839
X = +6.632	Y =  0.000	R = 6.632
X = +8.000	Y = -3.759	R = 8.839
-----------------------------------------
   +32.00 	          	   33.68

Left side weights at 9 o'clock...
X =-11.759	Y =  0.000	R = 11.759
X = -8.000	Y = +1.368	R = 8.116
X = -4.241	Y =  0.000	R = 4.241
X = -8.000	Y = -1.368	R = 8.116
-----------------------------------------
   -32.00 	          	   32.232

With wheel turned 45 degrees.

Right side weights at 1:30
x =  2.999	Y = 8.315	R =  8.839
x =  4.689	Y = 4.689	R =  6.632
x =  6.624	Y = 6.625	R =  9.368
x =  8.315	Y = 2.999	R =  8.839
------------------------------------------
    22.627	         	    33.678

Left side weights at 10:30
x = -4.689	Y = 6.624	R =  8.116
x = -2.999	Y = 2.999	R =  4.241
x = -8.315	Y = 8.315	R = 11.759
x = -6.624	Y = 4.689	R =  8.116
------------------------------------------
   -22.627	         	    32.232

When the X coordinates are summed they show the wheel to be balanced. But when you sum the Radii they are not equal. The weights on the right side average about 4.5 percent farther out from the axle. These four mechanisms like I've shown are balanced when in their curent position. If you turn the wheel 45 degrees as I've shown in the attached drawing it will still be balanced. But because the center of gyration is different then the weights want to swing (turn) at a different speed.

I'm not saying this works as there are other things to be considered. This is just an idea in its early stage of developement.

[AgingYoung]

Jim,

I thought the example you gave of the less dense solid pipe rolling faster than the hollow yet denser pipe would be an easy exercise with wm2d. You would have to construct the hollow pipe with the polygon. It would be informative but also a pain. It would be informative to see if wm2d would make the appropriate calculations. I don't think there has been an example of a way to test the accuracy of wm2d that is as obvious as this one.

Gene

[amateur]

why "hollow" polygon?

Can you not do one of the following for equivalent results?
1) Select two different materials (plastic, wood, steel)
2) Select different masses for two different weights (thus having "custom" material)
3) Overlay smaller inset instances onto one of objects (2Dness of model I would think keep the "imbalance" in z-axis from being a problem, although if a problem [or using 3d modeler] could overlay "balanced" layers on each side of "host" object)

Just some thoughts..

[edit5]
make both circles the same mass, but leave with their radii, and they still swing at different rates... hmm, they "gain force by their swinging"? Does faster swing mean there is greater force here? Or is force just "consumed" by the rotation of the pendulum weights? What happens to "falling" bodies (g ) acceleration if they are rotating vs. not rotating?

[edit4]
Regarding AgingYoung's (next) response, I have "confused" issues - the above was directed toward the pipe example. The attachments deal with pendulums. The pendulum's were easier for me to try.

[edit]
Think the attached demonstrates wm has basic concept...
(think [at least] units must be changed before running, to english(pounds))
the lighter smaller sphere (in background in my incarnation) apparently outruns the larger heavier one (in front)
[edit2]
Note that single rope(s) attached to center's of objects does not produce phenomena - the single rope attachment allows the object to pivot relatively, and the object itself is not forced to rotate. (A key point in Jim's explanation that I almost missed...) I hope my attempt to use two ropes on each introduces differing pendulum lengths that would be explanation - hmm, I probably won't reattach, but will try full-half circle arc - think that will have symmetry with the way the ropes are attached.
[edit3]
Think 3rd try does have appropriate "similar" pendula...

[AgingYoung]

amateur,

It's not precise yet you can make wheels using overlays with more density at their middle (effectively less dense around the circumference). Also they do make it down the hill faster. Neat idea.

Gene

ps edit: I was talking about the example that Jim gave about the solid cylinder vs. the pipe rolling downhill; not the pendulum. I just finished looking at it with overlays and although it's not precise it is interesting. I don't think you can make a circle that's more dense near its circumference than in the middle with overlays.

[jim_mich]

Gene,
Yes, your CenterOfGyrationTest3 shows the smaller weight swinging faster!

To make a circle with different densities try this. Make a smaller circle square pinned to a larger circle. Set both circle's mass to a same value. Select all. Copy and past. Now change one inner circle so it is smaller. The masses should be the same on all four circles. Position both circles on a ramp. One circle will roll down the ramp faster.

[jim_mich]

Imagine a cube weight fixed onto the end of a pendulum rod. At first you might think that the distance from the pivot to the center of the weight is the dimension to use for calculating the pendulum's swing rate. That is NOT right! Imagine slicing the cube into eight cubes. Now measure the distances from the pivot to the centers of each cube. The average distance is longer than the original distance.

The formula for radius of gyration for a square (parallelepiped) is shown in the attached picture.

If the cube is 2 x 2 with its center at 10 from the pivot then the Radius of Gyration is 10.033 which is 0.033 more than might be expected. Think of it as the corners being a little farther out from the center.

[Fletcher]

Guys, perhaps more grist for the mill.

This appears to me to be a similar situation to the example of 2 pendulums set side by side. They are exactly equal in all respects (say massless connecting rod & spherical bob weight).

Except one is rigidly pinned (fixed) to the rod attachment point so that as it swings in an arc it must rotate its mass & orientation relative to the earth.

The second bob is pivot pinned to the connecting rod so that it can swing freely or you could spin it around by hand if you wanted to.

If dropped from the same height they both show different inertia & acceleration characteristics. The freely pivoted bob gives least inertial resistance IIRC.

There is a thread about it where they were looking for ways to take advantage of this difference by strategically locking & unlocking the sphere bobs thru each quarter of the swing.

Is this effect due to the center of gyration changing or something else ? If its something else perhaps they are complimentary.

[jim_mich]

Fletcher,

I think what you describe is complimentary. They both relate to inertia. By locking and unlocking a weight you are in effect changing its radius of gyration from fixed to infinite.

I have a strong feeling that inertia plays a big role in a working PM wheel. I think it is avantageous to study and learn as much as possible about inertia.