We were throwing golf balls with the plywood wheel a few days ago. One full wrap of the circumference with the tether appeared to stop the wheel, so that would give it a rotational inertia of about 1800 grams. I could not positively say the wheel stopped because the tethered golf ball was throwing 51 meters into the neighbors hay field; and finding the tethered ball was no small task. Greater attention was placed on the flight of the ball than the motion of the wheel. We were throwing by hand so there was no after throw drive and the wheel indeed stopped but the bearing are, industrial, thick grease bearings and they would stop a slow moving wheel somewhat quickly. We were fortunate to find the throws and we were not too comfortable out there trampling the neighbors hay. Using F = ma experiments with hanging masses off of the circumference yielded 1900 grams for the wheel’s rotational inertia. So I feel comfortable in saying that the rotational inertia of the 19 inch plywood wheel is less than 2 kilograms.
Roughly timed experiments were done with a stopwatch but I decided to repeat the experiment using the photo gates. And now the added mass of 20.45 kilograms at the shaft or .9 kilograms at the circumference is now no longer a small portion of the total inertia of the wheel. The added mass is now roughly half the inertia of the system.
The objective is simple: If the same mass (suspended from the shaft) can accelerate a wheel with 40.9 kilograms, at the .75 inch shaft, at the same rate of rotation as 1.8 kilograms, at the circumference, then the Law of Conservation of Energy is false. For example: At 5 RPS the 40.9 kilograms is moving .3 m/sec and the 1.8 kilograms is moving 6.75 m/sec for 1.84 joules for the 40.9 kg and 41.00 joules for the 1.8 kg.
I have to make sure everything is free of mishap so I delayed on the photo gate setup; and I began trials with the stopwatch.
I set (45 pound) 20.45 kilograms bar bell masses on both sides of the .75 inch shaft; I suspended them with 3/16 inch nylon rope. I placed 7.5 more pounds on the clockwise side so that the wheel would rotate.
The dropping mass took about 2.02 second to drop one half rotation of the wheel.
I then replaced the 20.45 kilogram mass with a bolt, washer, and nut arraignment that had a total mass of 899.2 grams that was placed on the circumference. It took about 1.94 second to drop the same distance.
I then added 5 pennies to the mass on the circumference; (2.67 gram * 5) 13.35 grams. This gives us an opposite side mass of 912.55 grams. It took about 2.03 second to drop the same distance.
The wheel has been sanded some so its diameter is down to 18.86 inches and the .75 inch diameter with the stretched rope (rope center) measures .8685 inches. So 18.86 inches / .8685 inches = 21.7
The mass relationship is 20.45 kg / .91255 kg = 22.4. 21.7 / 22.4 3% difference
By using the same force of 7.5 lbs over the same distance the experiment shows that it takes the same amount of energy (newton * meters = joules) to cause the 20.45 kilograms to move 0.03456 m/sec as it does to make .91255 kilogram move .75 m/sec. But ½ * .91255 kilogram * .75 m/sec *.75 m/sec is .2566 joules and .5 * 20.45 kg * .03456 m/sec *.03456 m/sec = .0122 joules. Therefore the Law of Conservation of Energy is false.
Note: that everything in the two arraignments (2 - 20.45 kilogram at .75 or one 20.45 kilogram mass at the .75 inches shaft and one .912255 kilogram mass at the circumference) other than the motion of the added masses is the same; bearing resistance is the same, energy of the wheel is the same, same dropped mass, same distance, etc.
Here is a clue: look for a throw.
energy producing experiments
Moderator: scott
re: energy producing experiments
We have a few new people looking on, so I would like to nutshell it again.
There are at least two ways to make major amounts of free energy. The current experiment demonstrates one. And this method, like the other, is very simple. The current experiment could be referred to as a double Atwood's. I am going to round the numbers for math simplicity but it is exactly the same as the experiment.
Take a 3 cm shaft and mount it in pillow blocks or dry ice.
Suspend 20 kilograms off of both sides as in an Atwood's. The forces on the shaft or bar are balanced and the shaft will not rotate.
Place 3 extra kilograms on the right side. That makes 23 kilograms on the right and 20 kg on the left side of the shaft. The shaft will now rotate with the right side going down; and as we know from Atwood's experiments this will be an F = ma acceleration.
Allow the three extra kilograms of drop far enough so that it is moving one meter per second. This drop distance will be (ideally) less than .75 meters.
On your 3 cm shaft mount a low mass wheel that has a diameter of 60 cm.
Place one kilograms suspended on the left side on the circumference of the 60 cm wheel. Remove the 20 kilograms on the left side of the shaft.
The shaft will rotate at exactly the same rate as it did with two 20 kilogram masses on both sides. But now at the end of your .75 meter drop you have 23 kilograms moving 1 m/sec and 1 kilogram moving 20 m/sec. “I will now reiterate:I have done this exact experiments with several wheels, or double Atwood's. This is not speculation these are real experiments.� On one wheel experiment I did the equivalent of removing the other 20 kilograms and only the 3 kilograms was left in the center.
Now remove the other 20 kilograms by the 3 kilograms and place one kilogram on the right side on the circumference of the large wheel. Now at the end of the .75 meter drop you will have 3 kilograms moving 1 m/sec and 2 kilograms moving 20 m/sec.
The energy of 3 kilograms dropped .75 meters is .75 m * 9.81 N/kg * 3 kilograms = 22.07 joules.
Two kilograms moving 20 m/sec is ½ * 2 kg * 20 m/sec * 20 m/sec = 400 joules.
In the actual experiments you have to deal with air resistance and bearing resistance but the results are still very good. And in the real experiments the distanced the drive mass was dropped was only a few cm. But the results are the same. The fact that the two kilogram expanded Atwood’s on the large wheel rotates as easily as the two 20 kg on the shaft is one key to free energy.
There are at least two ways to make major amounts of free energy. The current experiment demonstrates one. And this method, like the other, is very simple. The current experiment could be referred to as a double Atwood's. I am going to round the numbers for math simplicity but it is exactly the same as the experiment.
Take a 3 cm shaft and mount it in pillow blocks or dry ice.
Suspend 20 kilograms off of both sides as in an Atwood's. The forces on the shaft or bar are balanced and the shaft will not rotate.
Place 3 extra kilograms on the right side. That makes 23 kilograms on the right and 20 kg on the left side of the shaft. The shaft will now rotate with the right side going down; and as we know from Atwood's experiments this will be an F = ma acceleration.
Allow the three extra kilograms of drop far enough so that it is moving one meter per second. This drop distance will be (ideally) less than .75 meters.
On your 3 cm shaft mount a low mass wheel that has a diameter of 60 cm.
Place one kilograms suspended on the left side on the circumference of the 60 cm wheel. Remove the 20 kilograms on the left side of the shaft.
The shaft will rotate at exactly the same rate as it did with two 20 kilogram masses on both sides. But now at the end of your .75 meter drop you have 23 kilograms moving 1 m/sec and 1 kilogram moving 20 m/sec. “I will now reiterate:I have done this exact experiments with several wheels, or double Atwood's. This is not speculation these are real experiments.� On one wheel experiment I did the equivalent of removing the other 20 kilograms and only the 3 kilograms was left in the center.
Now remove the other 20 kilograms by the 3 kilograms and place one kilogram on the right side on the circumference of the large wheel. Now at the end of the .75 meter drop you will have 3 kilograms moving 1 m/sec and 2 kilograms moving 20 m/sec.
The energy of 3 kilograms dropped .75 meters is .75 m * 9.81 N/kg * 3 kilograms = 22.07 joules.
Two kilograms moving 20 m/sec is ½ * 2 kg * 20 m/sec * 20 m/sec = 400 joules.
In the actual experiments you have to deal with air resistance and bearing resistance but the results are still very good. And in the real experiments the distanced the drive mass was dropped was only a few cm. But the results are the same. The fact that the two kilogram expanded Atwood’s on the large wheel rotates as easily as the two 20 kg on the shaft is one key to free energy.
re: energy producing experiments
The advantage of physical experiments:
Someone wanted to see some videos so I decided to repeat the three cylinder experiment shown on page three. This is where three cylinders with different masses and different radii are placed under the same cylinder and spheres experiment. The three cylinders have the same momentum when rotated at the same rate. The spheres throw out from the upper cylinder in the same manner no matter what added cylinder is placed on the bottom.
The repeat performance was successful. The cylinder stopping as the string came into the opening. When the stops occurred the different cylinders were within 3% of each other for momentum conservation. They were 26% away from each other for energy conservation. These timings were done with video tapes where the tape can be slowed.
372 grams were moving on a radius at 3.284 in. = 1,221.6
279.1 grams were moving on a radius at 4.257 in. = 1,188.1
The third was a combo of 219.2 grams at 3.77 in. and 112.8 grams at 3.284 in. = 1,198.4
It is always rewarding to do a good physics experiment but the important part came quite unexpectedly. I surveyed the three cylinders on the table and it occurred to me that the three different cylinders are like; 122.5 kg at a radius of .6875 inches, or 7.656 kilograms at an 11 inch radius, or 21 kilograms at a 4 inch radius. If the wheel had the same rotational speed all three would throw the missile in the same manner.
We know from the double Atwood’s experiments that the three (different masses at different radii) could be started with the same suspended mass dropped the same distance; ‘or’ they could be set into motion with the same energy. And the cylinder and spheres shows that all three arrangements throw the sphere or missile at the same speed. Its kind of like killing two birds with one stone.
Someone wanted to see some videos so I decided to repeat the three cylinder experiment shown on page three. This is where three cylinders with different masses and different radii are placed under the same cylinder and spheres experiment. The three cylinders have the same momentum when rotated at the same rate. The spheres throw out from the upper cylinder in the same manner no matter what added cylinder is placed on the bottom.
The repeat performance was successful. The cylinder stopping as the string came into the opening. When the stops occurred the different cylinders were within 3% of each other for momentum conservation. They were 26% away from each other for energy conservation. These timings were done with video tapes where the tape can be slowed.
372 grams were moving on a radius at 3.284 in. = 1,221.6
279.1 grams were moving on a radius at 4.257 in. = 1,188.1
The third was a combo of 219.2 grams at 3.77 in. and 112.8 grams at 3.284 in. = 1,198.4
It is always rewarding to do a good physics experiment but the important part came quite unexpectedly. I surveyed the three cylinders on the table and it occurred to me that the three different cylinders are like; 122.5 kg at a radius of .6875 inches, or 7.656 kilograms at an 11 inch radius, or 21 kilograms at a 4 inch radius. If the wheel had the same rotational speed all three would throw the missile in the same manner.
We know from the double Atwood’s experiments that the three (different masses at different radii) could be started with the same suspended mass dropped the same distance; ‘or’ they could be set into motion with the same energy. And the cylinder and spheres shows that all three arrangements throw the sphere or missile at the same speed. Its kind of like killing two birds with one stone.
re: energy producing experiments
The cylinder had just been dropped, so its downward motion is not great. But the vertical motion is blurred and the rotational or horizontal motion is not. In the frame by frame mode these black squares are directly below each other. This occurs because the cylinder has stopped rotating and the string or tether is in the long slit in the side of the cylinder. Shortly there after the string will come to the other end of the slit and the cylinder will begin accelerating in the same direction in which it was started.
If the add on cylinder is 6 grams too heavy the cylinder will be moving forward when the string is in the slit.
If the add on cylinder is 6 grams too light the cylinder will be moving backward when the string is in the slit.
Clearly the upper cylinder is responding to the total momentum of the rational system. Or; Momentum throws the spheres and it is yet another experimental example of momentum conservation. All three add on cylinders have the same rotational momentum. I could give you pictures of the other two add on cylinders and the upper cylinder is behaving in the same manner.
What is not conserved is energy. I did this once with a 1 inch steel pipe (with a large mass and an equivalent momentum) in the center and the upper cylinder responded in the same way.
Force as in F = ma is used to start the rotational motion. This F * t relationship makes an equivalent quantity of momentum no mater what the balanced mass distribution of the wheel. And then when the wheel throws it throws its momentum into the missile.
If the add on cylinder is 6 grams too heavy the cylinder will be moving forward when the string is in the slit.
If the add on cylinder is 6 grams too light the cylinder will be moving backward when the string is in the slit.
Clearly the upper cylinder is responding to the total momentum of the rational system. Or; Momentum throws the spheres and it is yet another experimental example of momentum conservation. All three add on cylinders have the same rotational momentum. I could give you pictures of the other two add on cylinders and the upper cylinder is behaving in the same manner.
What is not conserved is energy. I did this once with a 1 inch steel pipe (with a large mass and an equivalent momentum) in the center and the upper cylinder responded in the same way.
Force as in F = ma is used to start the rotational motion. This F * t relationship makes an equivalent quantity of momentum no mater what the balanced mass distribution of the wheel. And then when the wheel throws it throws its momentum into the missile.
re: energy producing experiments
It appears that some are doing experiments but not posting their results here. So progress is being made.
re: energy producing experiments
I finally suspended 912 grams from each side of the 19 inch wheel. It is in the 2.03 second category.
So lets go over the experiment again (posted April 21 2012). The same accelerating force (7.5 pounds) can accelerate a 20.45 kilogram mass suspended from both sides of the .75 inch shaft (total mass of 40.9 kilograms; or the 7.5 pound can accelerate a 912 gram mass suspended from both sides (total mass 1.824 kilograms) on the circumference of the 19 inch wheel at the same rotational rate.
If the 7.5 pound force is dropped a sufficient distance to have the 40.9 kilograms moving 1 meter per second at the shaft then the same drop distance would have the 1.824 kilograms moving 21.7 meters per second at the wheel circumference.
A formula for energy is 1/2mv²
½ * 40.9 kg * 1 m/sec * 1 m/sec = 20.45 joules
½ * 1.824 kg * 21.7 m/sec * 21.7 m/sec = 429.45 joules.
Real world experiments prove that 430 joules can be produces just as easily as 20.5 joules; by the same force dropped the same distance. The other formula for energy is force times distance; N * m, joules.
Real world experiments prove that The Law of Conservation of Energy is false.
So lets go over the experiment again (posted April 21 2012). The same accelerating force (7.5 pounds) can accelerate a 20.45 kilogram mass suspended from both sides of the .75 inch shaft (total mass of 40.9 kilograms; or the 7.5 pound can accelerate a 912 gram mass suspended from both sides (total mass 1.824 kilograms) on the circumference of the 19 inch wheel at the same rotational rate.
If the 7.5 pound force is dropped a sufficient distance to have the 40.9 kilograms moving 1 meter per second at the shaft then the same drop distance would have the 1.824 kilograms moving 21.7 meters per second at the wheel circumference.
A formula for energy is 1/2mv²
½ * 40.9 kg * 1 m/sec * 1 m/sec = 20.45 joules
½ * 1.824 kg * 21.7 m/sec * 21.7 m/sec = 429.45 joules.
Real world experiments prove that 430 joules can be produces just as easily as 20.5 joules; by the same force dropped the same distance. The other formula for energy is force times distance; N * m, joules.
Real world experiments prove that The Law of Conservation of Energy is false.
re: energy producing experiments
One kilogram dropped one meter creates 14.7 units of momentum in a 10 kilogram (5 kg each side) Atwood’s.
One kilogram dropped one meter creates 7.67 units of momentum in a 2 kilogram (1 kg each side) Atwood’s.
So the force of gravity over a certain distance does not conserve momentum.
7.5 pound of force can create 20.5 joules of energy or 430 joules as shown in the 19 inch wheel experiment.
So the force of gravity over a certain distance does not conserve energy.
The two Atwood’s in the first two paragraphs product the same quantity of energy but the force of gravity works for different amounts of time. Gravity works for 1.497 seconds in the first Atwood’s and for only .782 seconds in the second.
So the force of gravity over a certain amount of time does not conserve energy.
Note that the time over which the force of gravity works in the two Atwood’s is proportional to the momentum produced.
So gravitational force is a force as in F = ma where a = v/t; and therefore Ft = mv. So time and momentum have a direct relationship, but time is free. So gravity places no restriction on whatever we choose to produce with it, and it is all done for free.
One kilogram dropped one meter creates 7.67 units of momentum in a 2 kilogram (1 kg each side) Atwood’s.
So the force of gravity over a certain distance does not conserve momentum.
7.5 pound of force can create 20.5 joules of energy or 430 joules as shown in the 19 inch wheel experiment.
So the force of gravity over a certain distance does not conserve energy.
The two Atwood’s in the first two paragraphs product the same quantity of energy but the force of gravity works for different amounts of time. Gravity works for 1.497 seconds in the first Atwood’s and for only .782 seconds in the second.
So the force of gravity over a certain amount of time does not conserve energy.
Note that the time over which the force of gravity works in the two Atwood’s is proportional to the momentum produced.
So gravitational force is a force as in F = ma where a = v/t; and therefore Ft = mv. So time and momentum have a direct relationship, but time is free. So gravity places no restriction on whatever we choose to produce with it, and it is all done for free.
re: energy producing experiments
Two 20.45 kg masses at .75 inches or two .906 kg masses at 19 inches are rotated at the same rate by the 7.5 pounds off of the shaft. I pictured then together to save space.
Above is a 19 inch plywood wheel mounted on a .75 inch shaft.
Above is a 19 inch plywood wheel mounted on a .75 inch shaft.
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re: energy producing experiments
The shaft and the wheel in the double Atwood’s could be looked at as two lever arms. If the shaft has a working diameter of .828 inch and the large wheel has a diameter of 18.8 inches then the lever advantage is 22.7 to one. We can suspend 20.45 kilograms from one side of the shaft and 20.45 kilograms from the other side of the shaft. We can then accelerate the two suspended masses with an additional 3.4 kg suspended from one side. We will get an F = ma relationship for the acceleration of the 44.3 (20.45 + 20.45 + 3.4) kilograms being accelerated by 3.4 kg (3.4 kg * 9.81 N/kg / 44.3 kg =.753 m/sec/sec). How close we get to a perfect F = ma will be determined by how good our bearings are and how small we can keep the rotational inertial of the wheel. We could put the motion of the wheel in as part of the F = ma equation but how we apply this motion to the equation is what is being discussed. So for now it would probably be best to ignore the inertia of the wheel.
The 3.4 kilogram accelerating the 44.3 kilograms will give us an acceleration of about .753 m/sec/sec. The 3.4 kilograms would have to drop .664 meter to give the system a velocity of 1 m/sec. That would give you ½ * 44.3 kg *1 m/sec * 1 m/sec = 22.15 joules of energy. This would be equal to the input energy of 3.4 kg * 9.81 N/kg * .664 meters = 22.14 joules.
What would the acceleration be if we replaced one of the 20.45 kilogram masses suspended from the shaft with a balancing mass of .9 kilograms suspended from the circumference of the wheel?
Would the accelerating mass of 3.4 kilograms back down the acceleration because it would be producing too much energy?
Would the 3.4 kilograms detect the difference in torque between 20.45 kilograms at .828 inch as opposed to .9 kilograms at 18.8 inches? Is there a torque difference between 20.45 kilograms at .828 inch as opposed to .9 kilograms at 18.8 inches?
My data shows that the 3.4 kilogram drive mass will rotationally accelerate 20.45 kg on one side and .9 kg on the other side at the same rate as 20.45 kg on both sides. This would be with 20.45 kg at the .828 inch shaft and .9 kg at the 18.8 inch circumference.
F = ma is satisfied with this because .9 kg * 22.7 m/sec = 20.43 units of momentum and 20.45 kg * 1 m/sec = 20.45 units of momentum. At any rotational velocity the suspended mass at the circumference is moving 22.7 times faster than the suspended mass off of the shaft. I picked 1 m/sec and 22.7 m/sec.
My data also shows that two .9 kg masses at the circumference accelerates at the same rate as two 20.45 kg at the shaft.
But: ½ * 1.8kg *22.7 m/sec * 22.7 m/sec = 463.76 joules: and we only put in 22.14 joules. After a drop of .664 m for the 3.4 kg the shaft is moving 1 m/sec and the circumference is moving 22.7 m/sec.
The 3.4 kilogram accelerating the 44.3 kilograms will give us an acceleration of about .753 m/sec/sec. The 3.4 kilograms would have to drop .664 meter to give the system a velocity of 1 m/sec. That would give you ½ * 44.3 kg *1 m/sec * 1 m/sec = 22.15 joules of energy. This would be equal to the input energy of 3.4 kg * 9.81 N/kg * .664 meters = 22.14 joules.
What would the acceleration be if we replaced one of the 20.45 kilogram masses suspended from the shaft with a balancing mass of .9 kilograms suspended from the circumference of the wheel?
Would the accelerating mass of 3.4 kilograms back down the acceleration because it would be producing too much energy?
Would the 3.4 kilograms detect the difference in torque between 20.45 kilograms at .828 inch as opposed to .9 kilograms at 18.8 inches? Is there a torque difference between 20.45 kilograms at .828 inch as opposed to .9 kilograms at 18.8 inches?
My data shows that the 3.4 kilogram drive mass will rotationally accelerate 20.45 kg on one side and .9 kg on the other side at the same rate as 20.45 kg on both sides. This would be with 20.45 kg at the .828 inch shaft and .9 kg at the 18.8 inch circumference.
F = ma is satisfied with this because .9 kg * 22.7 m/sec = 20.43 units of momentum and 20.45 kg * 1 m/sec = 20.45 units of momentum. At any rotational velocity the suspended mass at the circumference is moving 22.7 times faster than the suspended mass off of the shaft. I picked 1 m/sec and 22.7 m/sec.
My data also shows that two .9 kg masses at the circumference accelerates at the same rate as two 20.45 kg at the shaft.
But: ½ * 1.8kg *22.7 m/sec * 22.7 m/sec = 463.76 joules: and we only put in 22.14 joules. After a drop of .664 m for the 3.4 kg the shaft is moving 1 m/sec and the circumference is moving 22.7 m/sec.
re: energy producing experiments
The wheel and shaft with 20.45 kilograms suspended from each side. The rotation is jamed by the screw driver; the shaft is under an unbalanced load from the 3.4 kilograms of drive.
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re: energy producing experiments
As you see from the photograph of the plywood wheel I have placed duct tape on the circumference of the wheel. That tape was not there for earlier tests. I used the tape to make sure that the cord between the two suspended masses was no finding a rough groove in the sanded plywood. From tape measurements this duct tape takes us back to 18.98 inches with the cord diameter. This increase drops the mass that is needed at the circumference to balance the 20.45 kilograms at the shaft to 896.4 grams.
The 20.45 kilograms mass, marked 45 pounds, was taken to a certified scale and its actual mass was 20.54 kilograms. So after the duct tape changes the new mass ratio is 896.4 g / 20.54 kg = 22.9 to one. Within experiment error; this mass ratio should also be the lever arm ratio.
When comparing one arrangement against the other; I use the same half rotation of the wheel as the 7.5 pound force drives the wheel.
It takes the same time for the force to drive the wheel a half rotation with two 20.54 kilogram masses at the shaft; as it does with two .8964 kilogram masses at the circumference. Both half rotations were right on top of two second.
So we have the same input energy producing 41.08 kilograms moving .03275 m/sec for .022 joules of energy. Or you have 1.7928 kilograms moving .75 m/sec for .5042 joules. All other parts of the system have the same final energy it both arraignments, so this is obviously a violation of The Law of Conservation of Energy.
The momentum produced by the 3.4 kilograms dropping the same distance is the same in both experimental arrangements. The circumference masses have 1.7928 kg * .75 m/sec = 1.3446 units of momentum or the 41.08 kg * .03275 m/sec = 1.345 units of momentum. All other parts of the system have the same momentum in both arrangements; or F = ma is true.
The 20.45 kilograms mass, marked 45 pounds, was taken to a certified scale and its actual mass was 20.54 kilograms. So after the duct tape changes the new mass ratio is 896.4 g / 20.54 kg = 22.9 to one. Within experiment error; this mass ratio should also be the lever arm ratio.
When comparing one arrangement against the other; I use the same half rotation of the wheel as the 7.5 pound force drives the wheel.
It takes the same time for the force to drive the wheel a half rotation with two 20.54 kilogram masses at the shaft; as it does with two .8964 kilogram masses at the circumference. Both half rotations were right on top of two second.
So we have the same input energy producing 41.08 kilograms moving .03275 m/sec for .022 joules of energy. Or you have 1.7928 kilograms moving .75 m/sec for .5042 joules. All other parts of the system have the same final energy it both arraignments, so this is obviously a violation of The Law of Conservation of Energy.
The momentum produced by the 3.4 kilograms dropping the same distance is the same in both experimental arrangements. The circumference masses have 1.7928 kg * .75 m/sec = 1.3446 units of momentum or the 41.08 kg * .03275 m/sec = 1.345 units of momentum. All other parts of the system have the same momentum in both arrangements; or F = ma is true.
re: energy producing experiments
I transfer the videos into DVDs but the DVDs can not be transfers to the computer. The DVD can be stopped frame by framed and you can see the cylinder stop as all the motion is transferred to the spheres.
If you hit the pause button you can stop the motion at any certain point. Then double click the pause button to jump through every third or fourth frame or so. I don't know how many frames it jumps. It depends on how fast you can double click, I suppose.
This is an experimental example of a very large mass transferring its motion to a much smaller mass. The mass of the upper and lower cylinders are marked. The mass of the spheres are about 66.7 g each. This 133.4 grams for the spheres is a portion of the original mass in motion. If the motion transfer is F = ma (which is what I see in all my experiments) then there is a very large increase in energy.
You are not seeing a stop action camera stop the motion. If the spheres did not stop the cylinder you would see a total blur until the cylinder hits the cushion.
I should have labeled it; top cylinder 233.7 g, bottom cylinder 373 g, and spheres 66.7 g each.
If you hit the pause button you can stop the motion at any certain point. Then double click the pause button to jump through every third or fourth frame or so. I don't know how many frames it jumps. It depends on how fast you can double click, I suppose.
This is an experimental example of a very large mass transferring its motion to a much smaller mass. The mass of the upper and lower cylinders are marked. The mass of the spheres are about 66.7 g each. This 133.4 grams for the spheres is a portion of the original mass in motion. If the motion transfer is F = ma (which is what I see in all my experiments) then there is a very large increase in energy.
You are not seeing a stop action camera stop the motion. If the spheres did not stop the cylinder you would see a total blur until the cylinder hits the cushion.
I should have labeled it; top cylinder 233.7 g, bottom cylinder 373 g, and spheres 66.7 g each.