I constructed a three disk system from a 5/16 inch steel shaft and cut plywood disks. The ID of the bearing and the OD of the shaft were so similar that they could not move one upon the other. I could press fit the bearing on the shaft but that meant the bearing had to stay on the end of the shaft. I chose a shaft length of 7.25 inches and then I press fit the bearings on both ends. The 6 inch disk (all diameters) and the three inch disk are glued together and are ¾ of an inch from one bearing. The 1.5 inch disk is ¾ inch from the other bearing. The 6 inch disk is on the inside of the 3 inch disk. The bearings were placed on lumber.
By placing proportional amounts of inertial mass at the three diameters the results favored mr but the data was not by any means conclusive. So I thought I would try something else.
I placed 2000 grams at the 6 inch location, that is 1000 grams draped over both sides. I then increased the mass on one side, at the six inch diameter, by 22 grams. The disk made an average rotation in 3.63 seconds.
I then removed the 22 grams and left the 2000 grams at the 6 inch location and placed 44 grams on the 3 inch disk. The disk made one rotation in 3.60 seconds.
I then placed 88 grams at the 1.5 inch location with the 2000 grams remaining at the 6 inch disk. The disk made one rotation in 3.60 seconds.
Obviously the applied force is an mr or Fr relationship. Also obvious is the fact that the resistance mass or inertial mass is only important because it applies a resistance force if you try to move it. I can not believe that the applied force is Fr and inertial mass is some other physics.
I wonder if the applied inertial masses at the different disks causes a flexing of the 5-16 inch shaft, and that the flexing causes the bearing to have more resistance. This seems more logical than to have one set of Laws for the applied force and another set of laws for the inertia.
The data for the changing of inertial masses is not good for mrr either; in fact it is too fast for mrr.
The next step is probably a 3/8 inch shaft; and try it again. I have done levers before; they work, but the thing has to spin.
The other wheels preformed well with inertial mass changes, I think it is the shaft. Of course the bearings are open and just setting on 2 inch by 6 inch lumber. But I have them wedged. ??
My partial summary of pequaide's "energy producing experiments" thread
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Wubbly
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re: My partial summary of pequaide's "energy producing
I ran your numbers through the Atwoods 3 mass spreadsheet and you should have seen a time of about 3.63 or 3.62 seconds for all of the three experiments.
This would have been completely consistent with the known laws of physics.
You are using different mass, radius combinations to represent the same small input torque into your system.
0.022 kg @ 3" radius (0.0762 meters),
0.044 kg @ 1.5" radius (0.0381 meters),
0.088 kg @ .75" radius ( 0.01905 meters)
Most of your system mass is in your two 1 kg masses @ 3" radius (0.0762 meters). Since you are accelerating a much larger system mass with a small input torque, you should expect to see about the same angular acceleration in all the experiments. This is what the Atwoods spreadsheet predicted and is what you saw.
This would have been completely consistent with the known laws of physics.
You are using different mass, radius combinations to represent the same small input torque into your system.
0.022 kg @ 3" radius (0.0762 meters),
0.044 kg @ 1.5" radius (0.0381 meters),
0.088 kg @ .75" radius ( 0.01905 meters)
Most of your system mass is in your two 1 kg masses @ 3" radius (0.0762 meters). Since you are accelerating a much larger system mass with a small input torque, you should expect to see about the same angular acceleration in all the experiments. This is what the Atwoods spreadsheet predicted and is what you saw.
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re: My partial summary of pequaide's "energy producing
Wo; is there more here?
A short hand method for finding acceleration is to divide the accelerating mass by the accelerated mass. And then multiply this by 9.81 m/sec/sec.
For mr this would be 88 * 1.5 inch / 2000 * 6 inch + (88 * 1.5) * 9.81 m/sec/sec or .1067 m/sec/sec
For mr this would be 44 * 3 inch / 2000 * 6 inch + (44 * 3) * 9.81 m/sec/sec or .1067 m/sec/sec
For mr this would be 22 * 6 inch / 2000 * 6 inch + (22 * 6) * 9.81 m/sec/sec or .1067 m/sec/sec
The measured acceleration of this experiment is .0737 m/sec/sec
For mrr this would be 88 * 1.5 inch * 1.5 inch / 2000 * 6 inch * 6 inch + (88 * 1.5 * 1.5) * 9.81 m/sec/sec or .0269 m/sec/sec
For mrr this would be 44 * 3 inch * 3 inch / 2000 * 6 inch * 6 inch + (44 * 3 * 3) * 9.81 m/sec/sec or .0536 m/sec/sec
Both are below the actual acceleration.
A short hand method for finding acceleration is to divide the accelerating mass by the accelerated mass. And then multiply this by 9.81 m/sec/sec.
For mr this would be 88 * 1.5 inch / 2000 * 6 inch + (88 * 1.5) * 9.81 m/sec/sec or .1067 m/sec/sec
For mr this would be 44 * 3 inch / 2000 * 6 inch + (44 * 3) * 9.81 m/sec/sec or .1067 m/sec/sec
For mr this would be 22 * 6 inch / 2000 * 6 inch + (22 * 6) * 9.81 m/sec/sec or .1067 m/sec/sec
The measured acceleration of this experiment is .0737 m/sec/sec
For mrr this would be 88 * 1.5 inch * 1.5 inch / 2000 * 6 inch * 6 inch + (88 * 1.5 * 1.5) * 9.81 m/sec/sec or .0269 m/sec/sec
For mrr this would be 44 * 3 inch * 3 inch / 2000 * 6 inch * 6 inch + (44 * 3 * 3) * 9.81 m/sec/sec or .0536 m/sec/sec
Both are below the actual acceleration.
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re: My partial summary of pequaide's "energy producing
pequaide wrote:
Your measured acceleration matches the theoretical value predicted by the Atwoods acceleration equation almost exactly. Your experiment follows the known laws of physics.
Or you could use the Atwoods acceleration equation.A short hand method for finding acceleration is to divide the accelerating mass by the accelerated mass. And then multiply this by 9.81 m/sec/sec.
Your measured acceleration matches the theoretical value predicted by the Atwoods acceleration equation almost exactly. Your experiment follows the known laws of physics.
re: My partial summary of pequaide's "energy producing
I have a question.
Lets start with 1 kilogram of inertial mass at 3 inches (that would be 500 grams draped on both sides) with one kilogram of drive mass at 3 inches.
What would be the difference between; moving the inertial mass to 1.5 inches and doubling it; or moving the drive mass to 6 inches and halving it?
Lets start with 1 kilogram of inertial mass at 3 inches (that would be 500 grams draped on both sides) with one kilogram of drive mass at 3 inches.
What would be the difference between; moving the inertial mass to 1.5 inches and doubling it; or moving the drive mass to 6 inches and halving it?
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re: My partial summary of pequaide's "energy producing
pequaide wrote:
a = f/m = (1 kg x 9.81) / [(2 x .5 kg) + 1 kg] = 4.905 m/s² (linear acceleration of the driver mass)
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In your second example where you move your "balanced mass" to one half the radius and double the mass, you now have twice the mass, but the radius change would make it four times easier to rotate. So it's mass would feel like 2 kg x 0.5² = 0.5 kg to the driver mass. So your acceleration equation would be:
a = f/m = (1 kg x 9.81) / [(2 x 1 kg) x 0.5² + 1 kg] = 6.54 m/s² (linear acceleration of the driver mass)
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In your third example where you keep your balanced mass at (2 x .5 kg) @ 3", and move your driver mass to 6" and cut your driver mass in half to 0.5 kg, it would be similar to example 2. Your "balanced mass" is at half the radius as your driver mass, so your driver mass would find it 4 times easier to rotate the "balanced mass". So your acceleration equation for the driver mass would be:
a = f/m = (0.5 kg x 9.81) / [(2 x 0.5 kg) x 0.5² + 0.5 kg] = 6.54 m/s² (linear acceleration of the driver mass) but since your "balanced mass" is at half the radius, it's linear acceleration would be half of this number, and the angular acceleration of the pully would be half that of experiment 2.
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You can see all of this by plugging numbers into the Atwoods 3 mass spreadsheet posted earlier. All of this assumes a massless pully and frictionless bearings. Once you add a real pully, it's mass contributes varying amounts to the system mass based on the radius from which you accelerate it. Also stiff bearings skew the numbers by slowing everything down, and if they are stiff enough they can make two accelerations that should be different actually look similar.
In your first example (2 x .5 kg) @ 3" with 1 kg driver mass @ 3", your driver mass and "balanced mass" are both at the same radius. The driver mass would not find it any harder or easier to rotate the (2 x .5 kg). Your 2 x .5 kg would feel exactly like 1 kg to the driver mass. So your acceleration equation would be:I have a question.
Lets start with 1 kilogram of inertial mass at 3 inches (that would be 500 grams draped on both sides) with one kilogram of drive mass at 3 inches.
What would be the difference between; moving the inertial mass to 1.5 inches and doubling it; or moving the drive mass to 6 inches and halving it?
a = f/m = (1 kg x 9.81) / [(2 x .5 kg) + 1 kg] = 4.905 m/s² (linear acceleration of the driver mass)
-------------------
In your second example where you move your "balanced mass" to one half the radius and double the mass, you now have twice the mass, but the radius change would make it four times easier to rotate. So it's mass would feel like 2 kg x 0.5² = 0.5 kg to the driver mass. So your acceleration equation would be:
a = f/m = (1 kg x 9.81) / [(2 x 1 kg) x 0.5² + 1 kg] = 6.54 m/s² (linear acceleration of the driver mass)
-------------------
In your third example where you keep your balanced mass at (2 x .5 kg) @ 3", and move your driver mass to 6" and cut your driver mass in half to 0.5 kg, it would be similar to example 2. Your "balanced mass" is at half the radius as your driver mass, so your driver mass would find it 4 times easier to rotate the "balanced mass". So your acceleration equation for the driver mass would be:
a = f/m = (0.5 kg x 9.81) / [(2 x 0.5 kg) x 0.5² + 0.5 kg] = 6.54 m/s² (linear acceleration of the driver mass) but since your "balanced mass" is at half the radius, it's linear acceleration would be half of this number, and the angular acceleration of the pully would be half that of experiment 2.
-------------------
You can see all of this by plugging numbers into the Atwoods 3 mass spreadsheet posted earlier. All of this assumes a massless pully and frictionless bearings. Once you add a real pully, it's mass contributes varying amounts to the system mass based on the radius from which you accelerate it. Also stiff bearings skew the numbers by slowing everything down, and if they are stiff enough they can make two accelerations that should be different actually look similar.
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Last edited by Wubbly on Wed Jul 18, 2012 1:03 am, edited 1 time in total.
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re: My partial summary of pequaide's "energy producing
pequaide wrote:
-----------------------
In Experiment 1, your 2000 gram mass is being accelerated at the same radius as your driver mass. The driver mass doesn't find it any harder or easier to accelerate the 2000 gram mass. So the linear acceleration equation for the driver mass would be:
a = (.022 kg x 9.81) / [(2.000 kg x 1) + (.022 kg x 1) + 0.95 kg] = .073 m/sec² => resulting in a drop time of 3.63 seconds which is what your experiment showed.
note that the 0.95 kg is the mass of the pulley (moment of inertia of the pulley divided by the radius squared at which you are accelerating it. This number also includes any bearing friction).
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In experiment 2, your 2000 gram mass is being accelerated at twice the radius as your driver mass, so the driver mass would find it 4 times harder to accelerate the 2000 gram mass. So the linear acceleration equation of the driver mass would be:
a = (.044 kg x 9.81) / [(2.000 kg x 4) + (.044 kg x 1) + 3.8 kg] = .036 m/sec² => resulting in a drop time of 3.62 seconds where your experiment showed 3.60 seconds.
But this is the linear acceleration of the driver mass (at half the 2000 gram mass radius), so the linear acceleration of the 2000 gram mass would be twice this at .073 m/sec². Note that the 3.8 kg is the mass of the pulley (moment of inertia of the pully divided by the radius squared at which you are accelerating it. This number also includes any bearing friction). Since the driver radius decreased, the pulley feels more massive to the driver mass and any bearing friction feels larger.
-----------------------
In experiment 3, your 2000 gram mass is being accelerated at four times the radius as your driver mass, so the driver mass would find it 16 times harder to accelerate the 2000 gram mass. So the linear acceleration equation of the driver mass would be:
a = (.088 kg x 9.81) / [(2.000 kg x 16) + (.088 kg x 1) + 15 kg] = .018 m/sec² => resulting in a drop time of 3.62 seconds where your experiment showed 3.60 seconds.
But this is the linear acceleration of the driver mass, so the linear acceleration of the 2000 gram mass would be four times this at .073 m/sec². Note that the 15 kg is the mass of the pulley (moment of inertia of the pulley divided by the radius squared at which you are accelerating it. This number also includes any bearing friction). Since the driver radius decreased again, the pulley feels even more massive.
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So your experiment just proved mr². You just had to know where to look in the numbers to find it, and do the correct math.
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If you understood my previous post, it can be applied to your actual experiment from this post here: http://www.besslerwheel.com/forum/viewt ... 605#101605A short hand method for finding acceleration is to divide the accelerating mass by the accelerated mass. And then multiply this by 9.81 m/sec/sec.
-----------------------
In Experiment 1, your 2000 gram mass is being accelerated at the same radius as your driver mass. The driver mass doesn't find it any harder or easier to accelerate the 2000 gram mass. So the linear acceleration equation for the driver mass would be:
a = (.022 kg x 9.81) / [(2.000 kg x 1) + (.022 kg x 1) + 0.95 kg] = .073 m/sec² => resulting in a drop time of 3.63 seconds which is what your experiment showed.
note that the 0.95 kg is the mass of the pulley (moment of inertia of the pulley divided by the radius squared at which you are accelerating it. This number also includes any bearing friction).
-----------------------
In experiment 2, your 2000 gram mass is being accelerated at twice the radius as your driver mass, so the driver mass would find it 4 times harder to accelerate the 2000 gram mass. So the linear acceleration equation of the driver mass would be:
a = (.044 kg x 9.81) / [(2.000 kg x 4) + (.044 kg x 1) + 3.8 kg] = .036 m/sec² => resulting in a drop time of 3.62 seconds where your experiment showed 3.60 seconds.
But this is the linear acceleration of the driver mass (at half the 2000 gram mass radius), so the linear acceleration of the 2000 gram mass would be twice this at .073 m/sec². Note that the 3.8 kg is the mass of the pulley (moment of inertia of the pully divided by the radius squared at which you are accelerating it. This number also includes any bearing friction). Since the driver radius decreased, the pulley feels more massive to the driver mass and any bearing friction feels larger.
-----------------------
In experiment 3, your 2000 gram mass is being accelerated at four times the radius as your driver mass, so the driver mass would find it 16 times harder to accelerate the 2000 gram mass. So the linear acceleration equation of the driver mass would be:
a = (.088 kg x 9.81) / [(2.000 kg x 16) + (.088 kg x 1) + 15 kg] = .018 m/sec² => resulting in a drop time of 3.62 seconds where your experiment showed 3.60 seconds.
But this is the linear acceleration of the driver mass, so the linear acceleration of the 2000 gram mass would be four times this at .073 m/sec². Note that the 15 kg is the mass of the pulley (moment of inertia of the pulley divided by the radius squared at which you are accelerating it. This number also includes any bearing friction). Since the driver radius decreased again, the pulley feels even more massive.
-----------------------
So your experiment just proved mr². You just had to know where to look in the numbers to find it, and do the correct math.
-----------------------
Last edited by Wubbly on Wed Jul 18, 2012 6:07 pm, edited 1 time in total.
re: My partial summary of pequaide's "energy producing
Okay lets go through this again:
22 grams at 6 inches causes the same acceleration as 44 grams at 3 inches, or 88 grams at 1.5 inches.
22 is half the mass at twice the distance at 6 inches; as 44 grams is at 3 inches, and they have the same acceleration
Or 44g at 3 inches is half the mass at twice the distance as 88 at 1.5, and they have the same acceleration.
The .5 kilograms at 6 is half the drive mass at twice the distance as 1 drive kilogram at 3 inches with the one inertial kilogram at three inches, so it is going to have the same acceleration.
When the inertial mass moves to 1.5 inches and doubles this gives you the same arrangements as .5 kilograms at 6 inches with 1 inertial kilogram remaining at 3 inches. When the inertial mass moves you will have one drive kilogram at 3 inches and 2 inertial kilograms at 1.5 inches. The size of the circle has nothing to do with it, it is the relative length of the radii and the masses at those radii.
If the ultra complex formula agreed with my results it is either a coincidence or someone plugged in the appropriate numbers. My real life experiment still had a large bearing resistance and would not match anything theoretical. It did not surprise me that the formula got my question wrong. The mr concept is is simple, it does not need a three page formula.
You have this lauded formula (mrr) that you never use. It must be nice having a formula that never has to be applied to any real life situation.
When the applied force that is causing the acceleration is at 1.5 inches and the inertial mass is at 6 inches the inertial mass has a 4 to 1 radius advantage. This is the concept we are talking about; the interaction of forces applied at different radii in a circle. Some of the forces are applied forces and some are inertia forces. Is the relationship mrr or mr; or you could use (Fr and Frr). But in a real life situation you want to back away from your formula. Well of course you do: because it does not work.
If 2000 grams has an mrr m*4*4 advantage over 88 grams at m*1*1 then the (.088 kg * 9.81N/kg) .863 newtons will act upon a mass made 16 times harder to move. Or F = ma; F/m*r*r = a; .863 N/32 kg = .0269 m/sec/sec.
But mr gives a four times harder to move; .863 N/8 kg = .1078 m/sec/sec
It is still the relationship of the radii that is used by both formulas. Even though the 2000 grams remains in the same place the radii relationships are 1 to 1, 1 to 2, and 1 to 4. Both formulas must use these relationships: one formula works and one does not.
You are assuming that if the 2000 grams at 6 inches does not change its radius then the quantity of torque needed to accelerate it at a certain rate will remain the same; which is correct. And you assume that the applied force must deliver the same torque for the acceleration to remain the same, this is true also. But how does the applied force deliver the same torque? If the radius of the applied force halves the mass (applied force) doubles. If the radius of the applied force is reduced to a fourth then the mass quadruples. This is also mr; it is like the truth is screaming at you, but you refuse to see it.
Oh I see your magic numbers .95 3.8, 15. your formula need a little help does it?
22 grams at 6 inches causes the same acceleration as 44 grams at 3 inches, or 88 grams at 1.5 inches.
22 is half the mass at twice the distance at 6 inches; as 44 grams is at 3 inches, and they have the same acceleration
Or 44g at 3 inches is half the mass at twice the distance as 88 at 1.5, and they have the same acceleration.
The .5 kilograms at 6 is half the drive mass at twice the distance as 1 drive kilogram at 3 inches with the one inertial kilogram at three inches, so it is going to have the same acceleration.
When the inertial mass moves to 1.5 inches and doubles this gives you the same arrangements as .5 kilograms at 6 inches with 1 inertial kilogram remaining at 3 inches. When the inertial mass moves you will have one drive kilogram at 3 inches and 2 inertial kilograms at 1.5 inches. The size of the circle has nothing to do with it, it is the relative length of the radii and the masses at those radii.
If the ultra complex formula agreed with my results it is either a coincidence or someone plugged in the appropriate numbers. My real life experiment still had a large bearing resistance and would not match anything theoretical. It did not surprise me that the formula got my question wrong. The mr concept is is simple, it does not need a three page formula.
You have this lauded formula (mrr) that you never use. It must be nice having a formula that never has to be applied to any real life situation.
When the applied force that is causing the acceleration is at 1.5 inches and the inertial mass is at 6 inches the inertial mass has a 4 to 1 radius advantage. This is the concept we are talking about; the interaction of forces applied at different radii in a circle. Some of the forces are applied forces and some are inertia forces. Is the relationship mrr or mr; or you could use (Fr and Frr). But in a real life situation you want to back away from your formula. Well of course you do: because it does not work.
If 2000 grams has an mrr m*4*4 advantage over 88 grams at m*1*1 then the (.088 kg * 9.81N/kg) .863 newtons will act upon a mass made 16 times harder to move. Or F = ma; F/m*r*r = a; .863 N/32 kg = .0269 m/sec/sec.
But mr gives a four times harder to move; .863 N/8 kg = .1078 m/sec/sec
It is still the relationship of the radii that is used by both formulas. Even though the 2000 grams remains in the same place the radii relationships are 1 to 1, 1 to 2, and 1 to 4. Both formulas must use these relationships: one formula works and one does not.
You are assuming that if the 2000 grams at 6 inches does not change its radius then the quantity of torque needed to accelerate it at a certain rate will remain the same; which is correct. And you assume that the applied force must deliver the same torque for the acceleration to remain the same, this is true also. But how does the applied force deliver the same torque? If the radius of the applied force halves the mass (applied force) doubles. If the radius of the applied force is reduced to a fourth then the mass quadruples. This is also mr; it is like the truth is screaming at you, but you refuse to see it.
Oh I see your magic numbers .95 3.8, 15. your formula need a little help does it?
re: My partial summary of pequaide's "energy producing
For those interested in proving for themselves without building an Atwoods.
http://www.youtube.com/watch?v=hKZ7Jywwwsc
Same experimental principle except stick coins or washers to a single CD [collared to axle] - test time with 2 X 1 unit masses opposing at 1 unit radius v's 2 x 1/2 unit masses at 2 units radius.
What is the fallacy revealed ?
N.B. resistance forces will be minimal as will be the actual mass of the disk & axle, so times will be indicative of the inertia = mr or mr^2 relationship.
P.S. run the disk without masses as a control baseline.
http://www.youtube.com/watch?v=hKZ7Jywwwsc
Same experimental principle except stick coins or washers to a single CD [collared to axle] - test time with 2 X 1 unit masses opposing at 1 unit radius v's 2 x 1/2 unit masses at 2 units radius.
What is the fallacy revealed ?
N.B. resistance forces will be minimal as will be the actual mass of the disk & axle, so times will be indicative of the inertia = mr or mr^2 relationship.
P.S. run the disk without masses as a control baseline.
re: My partial summary of pequaide's "energy producing
You invent numbers (.95, 3.8, 15) to keep your theory and ignore numbers (22, 44, 88; all 3.60sec) that destroy it. Then Fletcher tries to change the subject. You three have delayed a lot of good research.
Obviously you can not be reasoned with.
Obviously you can not be reasoned with.
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re: My partial summary of pequaide's "energy producing
pequaide wrote:
Nothing was invented or ignored. The person ignoring numbers is pequaide, specifically ignoring the pulley mass and the bearing friction. The pulley and the bearing friction are real, even though pequaide insists on ignoring them and pretending they do not exist in all of his experiments.
Perhaps I should have explained the 0.95 kg, 3.8 kg and 15 kg numbers came from. The moment of inertia (0.0055 kg m²) came from plugging numbers into the Atwoods spreadsheet until pequaide's observed time matched the theoretical time for his experiment 1. (note: I combined the bearing friction together with the moment of inertia number so I wouldn't have to add an additional number for bearing friction). The bearing friction makes the mass feel "heavier", so combining it in with the moment of inertia seems like a rational thing to do (more rational than just ignoring it like pequaide insists on doing again and again and again). Once I knew the moment of inertia for experiment 1, I left the number the same for experiment 2 and experiment 3.
The Atwoods acceleration equation contains a term in the denominator for (moment of inertia of the pulley divided by the (driver radius squared). This term was not invented. It comes from using F=ma and deriving the atwoods acceleration equation as described in this post here: http://www.besslerwheel.com/forum/viewt ... 6980#96980
For experiment 1, the Moment_of_Inertia divided by the driver radius squared comes out to (0.0055 kg m²) / (0.0762 m)² = 0.95 kg
The pulley mass and the bearing friction feel like 0.95 kg to the driver mass.
For experiment 2, the Moment_of_Inertia divided by the driver radius squared comes out to (0.0055 kg m²) / (0.0381 m)² = 3.8 kg.
At half the radius, the pulley mass and the bearing friction feel like 3.8 kg which is FOUR times heavier than experiment 1. Also, the balanced mass feels FOUR times heavier to the driver mass.
For experiment 3, the Moment_of_Inertia divided by the driver radius squared comes out to (0.0055 kg m²) / (0.0191 m)² = 15 kg.
At one fourth the radius, the pully mass and the bearing friction feel like 15 kg which is SIXTEEN times heavier than experiment 1. Also, the balanced mass feels SIXTEEN times heavier to the driver mass.
When you INCLUDE the Moment_of_Inertia and the bearing friction into the equation, the theoretical predicted acceleration matches pequaide's observed values perfectly (as shown in the Atwoods 3 mass spreadsheet posted earlier). And I mean perfectly. OK, maybe he's off by a few hundreths of a second in two of his experiments, but that's pretty close for pequaide.
pequaide could not have come up with a better example to prove mr² than the experiment he performed. It demonstrated mr² perfectly. This is what happens when you don't ignore the BUS in your experiment, pretending it does not exist.
.
You invent numbers (.95, 3.8, 15) to keep your theory and ignore numbers (22, 44, 88; all 3.60sec) that destroy it.
Nothing was invented or ignored. The person ignoring numbers is pequaide, specifically ignoring the pulley mass and the bearing friction. The pulley and the bearing friction are real, even though pequaide insists on ignoring them and pretending they do not exist in all of his experiments.
Perhaps I should have explained the 0.95 kg, 3.8 kg and 15 kg numbers came from. The moment of inertia (0.0055 kg m²) came from plugging numbers into the Atwoods spreadsheet until pequaide's observed time matched the theoretical time for his experiment 1. (note: I combined the bearing friction together with the moment of inertia number so I wouldn't have to add an additional number for bearing friction). The bearing friction makes the mass feel "heavier", so combining it in with the moment of inertia seems like a rational thing to do (more rational than just ignoring it like pequaide insists on doing again and again and again). Once I knew the moment of inertia for experiment 1, I left the number the same for experiment 2 and experiment 3.
The Atwoods acceleration equation contains a term in the denominator for (moment of inertia of the pulley divided by the (driver radius squared). This term was not invented. It comes from using F=ma and deriving the atwoods acceleration equation as described in this post here: http://www.besslerwheel.com/forum/viewt ... 6980#96980
For experiment 1, the Moment_of_Inertia divided by the driver radius squared comes out to (0.0055 kg m²) / (0.0762 m)² = 0.95 kg
The pulley mass and the bearing friction feel like 0.95 kg to the driver mass.
For experiment 2, the Moment_of_Inertia divided by the driver radius squared comes out to (0.0055 kg m²) / (0.0381 m)² = 3.8 kg.
At half the radius, the pulley mass and the bearing friction feel like 3.8 kg which is FOUR times heavier than experiment 1. Also, the balanced mass feels FOUR times heavier to the driver mass.
For experiment 3, the Moment_of_Inertia divided by the driver radius squared comes out to (0.0055 kg m²) / (0.0191 m)² = 15 kg.
At one fourth the radius, the pully mass and the bearing friction feel like 15 kg which is SIXTEEN times heavier than experiment 1. Also, the balanced mass feels SIXTEEN times heavier to the driver mass.
When you INCLUDE the Moment_of_Inertia and the bearing friction into the equation, the theoretical predicted acceleration matches pequaide's observed values perfectly (as shown in the Atwoods 3 mass spreadsheet posted earlier). And I mean perfectly. OK, maybe he's off by a few hundreths of a second in two of his experiments, but that's pretty close for pequaide.
pequaide could not have come up with a better example to prove mr² than the experiment he performed. It demonstrated mr² perfectly. This is what happens when you don't ignore the BUS in your experiment, pretending it does not exist.
.
Last edited by Wubbly on Wed Jul 18, 2012 8:09 pm, edited 3 times in total.
re: My partial summary of pequaide's "energy producing
I see you are calculating the acceleration at the new drive location, say one fourth. I think the calculations have to remain at the inertial location. Otherwise you are effectively getting rid of one of your radii by calculation acceleration at the smaller radius.
I placed the 2000 grams (1000 g on each side) at the 1.5 inch diameter, and accelerated it with 22 gram on the 6 inch wheel. It did a complete rotation in 1.22 (average) second.
I then removed the 22 grams and placed 88 extra grams on the 1.5 inch diameter wheel. It did a complete rotation in 1.22 (first run) second.
No matter how you want to explain this; what the machine is doing is very obvious. 22 at 6 is equal to 88 at 1.5; maybe this is just too simple?
I placed the 2000 grams (1000 g on each side) at the 1.5 inch diameter, and accelerated it with 22 gram on the 6 inch wheel. It did a complete rotation in 1.22 (average) second.
I then removed the 22 grams and placed 88 extra grams on the 1.5 inch diameter wheel. It did a complete rotation in 1.22 (first run) second.
No matter how you want to explain this; what the machine is doing is very obvious. 22 at 6 is equal to 88 at 1.5; maybe this is just too simple?
Last edited by pequaide on Wed Jul 18, 2012 8:12 pm, edited 1 time in total.
Re: re: My partial summary of pequaide's "energy produc
Your theory is independent of an Atwoods & must apply to all situations [the rolling disk is another situation] - your theory is that rotational inertia is a mr relationship & not anything else - i.e. that it is just as easy to accelerate an object at 1/2m x 2r as 1m x 1r.pequaide wrote:You invent numbers (.95, 3.8, 15) to keep your theory and ignore numbers (22, 44, 88; all 3.60sec) that destroy it. Then Fletcher tries to change the subject. You three have delayed a lot of good research.
Obviously you can not be reasoned with.
Wubbly has explained himself clearly & you still don't get it.
Therefore a rolling disk test removes the concerns about bearing losses - it isolates the "mr relationship & not anything else" that is the crux of what you insist is reality.
Most people won't take the time to even build a simple rolling disk with a few coins stuck to it because they know both won't arrive at the bottom in the same time, thus disproving mr.
What do you think will happen in the rolling disk tests & how would you interpret the results ? - that's a rhetorical question to you pequaide because you just claim it is changing the subject, which obviously it is not.
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re: My partial summary of pequaide's "energy producing
If you ignore the pulley and its bearing friction, you get a perfect mr relationship.
If you include the pulley and its bearing friction, you get a perfect mrr relationship.
maybe this is just too simple?
So do you want to believe the analysis where a part of the experiment is ignored, or believe the analysis where the whole experiment is accounted for? Obviously pequaide has no problem throwing part of his experiment under the bus.
If you include the pulley and its bearing friction, you get a perfect mrr relationship.
maybe this is just too simple?
So do you want to believe the analysis where a part of the experiment is ignored, or believe the analysis where the whole experiment is accounted for? Obviously pequaide has no problem throwing part of his experiment under the bus.
re: My partial summary of pequaide's "energy producing
You mean those magic numbers you got by working backwards.
The same wheel moving at the same speed should have the same bearing resistance, that seems logical. But instead you work backwards to get your wanted numbers.
Did you notice that the CDs rolling down the slope are not rolling on their circumference, but the wooden axle alone is rolling on its circumference.
With the wooden axle attached to the CDs the wooden axle is moving slower; but how fast is the circumference of the CDs moving? The mass of the CD is outside the rolling surface and is moving quite fast. The rolling mass of the wooden axle alone is inside the rolling circumference and it is moving slower than the circumference. Were as the CDs mass is moving much faster than the rolling surface. If looked at properly the experiment would tell you the opposite of what you think is says.
The same wheel moving at the same speed should have the same bearing resistance, that seems logical. But instead you work backwards to get your wanted numbers.
Did you notice that the CDs rolling down the slope are not rolling on their circumference, but the wooden axle alone is rolling on its circumference.
With the wooden axle attached to the CDs the wooden axle is moving slower; but how fast is the circumference of the CDs moving? The mass of the CD is outside the rolling surface and is moving quite fast. The rolling mass of the wooden axle alone is inside the rolling circumference and it is moving slower than the circumference. Were as the CDs mass is moving much faster than the rolling surface. If looked at properly the experiment would tell you the opposite of what you think is says.