Previous quote: “we can assume the attached velocity at 6 o’clock is the same as the suspended velocity of the same quantity of mass dropped the same distance.�
Note the word suspended; By suspended I meant as in (or as in a mass suspended from a string wrapped around the wheel) the ribbon which I have been using in the last few experiments that is wrapped around the wheel. I should have said “the same velocity that would be caused by a suspended mass�
If a suspended mass (on the end of a ribbon wrapped around the wheel) of 1 kg accelerates a 4 kg rim mass (nearly all the mass is in the rim) balanced wheel the acceleration will be 1/5 * 9.81 m/sec² or 1.962m/sec².
If a suspended mass of 1 kg accelerates a 40 kg rim mass wheel the acceleration will be 1/41 * 9.81 m/sec² or .239m/sec².
If a suspended mass of 1 kg accelerates a 1 kg rim mass balanced wheel the acceleration will be 1/2 * 9.81 m/sec² or 4.9m/sec².
If a suspended mass of 1 kg accelerates a 1000 kg rim mass (nearly all the mass is in the rim) balanced wheel the acceleration will be 1/1001 * 9.81 m/sec² or .00980m/sec².
This math is F = ma.
I should not have made it sound like the velocity of an overbalance wheel is an assumed quantity, it really isn’t. A one kilogram mass that is dropped straight down in freefall (9.81m/sec²) for one meter will have a velocity of 4.429 m/sec and it will have a momentum of 4.429. A one kilogram pendulum bob that has dropped down one meter has a 4.429 m/sec velocity and 4.429 units of momentum.
Near 6 o’clock the force applied to the bob (in the direction of travel) in a pendulum is small but the time over which the force acts is proportionally greater. So in a simple pendulum you end up with the same velocity as that of the freefall mass. Much like suspending the mass near the axis in your previous post, the time over which the force acts is longer but the force itself is proportionally small, and you will be getting the same overall velocity change.
The angles over which the force acts in a pendulum are identical to the working angles of the force in the wheel.
A one kilogram mass on a string that is suspended over a frictionless pulley (just off the edge of the table) with the string attached to a nine kilogram block on a frictionless plane will accelerate at a rate of .981m/sec² (1/10th of 9.81). This is because only 1 out of 10 kilograms is under gravitational acceleration.
The same is true for the balanced nine kilogram wheel with one kilogram of overbalance. The same quantity of force that was available to you in the simple pendulum is now available to you in the wheel. And the force is going to accelerate the other nine kilogram just like the suspended mass over the pulley accelerated the block. Now we have 1/10 the acceleration rate that got us 4.429 meters per second which will now give use 1.4007 m/sec. This is accepted Newtonian physics. Even though the acceleration is 1/10 the final velocity is greater than 1/10 and the final momentum is greater than 4.429. The velocity is determined by this formula d = 1/2v²/a, where v is the square root of (2 * d * a) which in this case is (2 * 1m * .981m/sec²) = 1.4007m/sec. That puts momentum at 10kg (all the overbalance wheel is moving 1.4007 m/sec) times 1.4007 m/sec for 14.007 units of momentum.
Now you put the 14.007 units of momentum into a yo-yo de-spin device and transfer all of the motion to the small overbalance mass of 1 kilogram. If (a very small if) Newton’s Three Laws of Motion apply to this event then the one kilogram must be moving 14.007 m/sec. At 14.007 m/sec the one kilogram will rise (d = 1/2v²/a) 10 meters, and it was only dropped one meter. All measurements that I have taken confirm Newtonian physics, and that it applies to this event.
I intend to do an experiment of mass attached directly to the wheel, but I am quite sure I will be checking bearing resistance not Newtonian physics. Well; but indeed, I will be checking both.
I don’t see anything that is left to speculation, NASA and RCA both performed the de-spin stops. And I have accomplished stops of disks, cylinders, and wheels. I have timed the spheres and pucks in these stops and have confirmed Newtonian physics. The wheel wrapped with a ribbon will give Newtonian accelerations even if an in wheel attachment of the mass won’t. So we know that the original input motion is achievable.
energy producing experiments
Moderator: scott
re: energy producing experiments
I placed a mass on the end of the red ribbon which was wrapped around the big wheel
I taped a photo gate flag (hex wrench) to the big wheel. I placed the two photo gates at a fixed distance from each other and in a position that the flag (after acceleration) interrupted the photo gate beams.
I raised the mass 296mm above the point where the flag was half way between the photo gates. After being dropped the mass accelerated the wheel and the flag which crossed the photo gates in .0272, and .0272 seconds, in two different runs.
I then taped the same mass to the inside of the rim of the wheel and raised its position 296 mm above the same photo gate position. After release the flag crossed the distance between the photo gates in .0269 and .0269 seconds.
So if the mass is suspended from the circumference with a ribbon or fixed to the wheel, it is the distance the mass drops that determines the final velocity of the wheel. The velocity is also dependent upon the wheel inertia and the dropped mass of course.
I taped a photo gate flag (hex wrench) to the big wheel. I placed the two photo gates at a fixed distance from each other and in a position that the flag (after acceleration) interrupted the photo gate beams.
I raised the mass 296mm above the point where the flag was half way between the photo gates. After being dropped the mass accelerated the wheel and the flag which crossed the photo gates in .0272, and .0272 seconds, in two different runs.
I then taped the same mass to the inside of the rim of the wheel and raised its position 296 mm above the same photo gate position. After release the flag crossed the distance between the photo gates in .0269 and .0269 seconds.
So if the mass is suspended from the circumference with a ribbon or fixed to the wheel, it is the distance the mass drops that determines the final velocity of the wheel. The velocity is also dependent upon the wheel inertia and the dropped mass of course.
re: energy producing experiments
Absolutely right Pequaide - people with wm2d can test this for themselves - make a center pivoted circle any radius you want - change the mass of the circle for different tests - attach an small ancillary weight to the rim at say 1 o'cl [so that it has some torque] - let the wheel rotate until the weight is at 6 o'cl.
Drop an exact same mass from the same vertical height as a control experiment.
Measure the velocity of the rim weight traveling in a radius circle, measure the velocity of the control falling vertically under gravity, & measure time.
Assuming the big circle has virtually no or little mass & that there is little or no pivot/axle friction then at any vertical height from the start height to 6 o'cl both the rim weight & the control will have the same velocity/speed at the same height - they get there in vastly different amounts of time though.
As soon as you introduce some pin friction & play with the mass of the big circle you find things change IINM - what is sure is that even with a big mass for the circle, although slow to get underway the rim weight travels up to a similar height vertically, due to momentum/inertia - in fact if you turn on air resistance it swings higher than the lighter circle mass comparison - that's because the force gravity sets to give a constant acceleration is proportionally greater than the air resistance force opposing it so it has a lesser proportional effect & it swings higher.
It's a long time since I did these experiments on wm2d so I reserve the right to be forgetful or confused ;)
Drop an exact same mass from the same vertical height as a control experiment.
Measure the velocity of the rim weight traveling in a radius circle, measure the velocity of the control falling vertically under gravity, & measure time.
Assuming the big circle has virtually no or little mass & that there is little or no pivot/axle friction then at any vertical height from the start height to 6 o'cl both the rim weight & the control will have the same velocity/speed at the same height - they get there in vastly different amounts of time though.
As soon as you introduce some pin friction & play with the mass of the big circle you find things change IINM - what is sure is that even with a big mass for the circle, although slow to get underway the rim weight travels up to a similar height vertically, due to momentum/inertia - in fact if you turn on air resistance it swings higher than the lighter circle mass comparison - that's because the force gravity sets to give a constant acceleration is proportionally greater than the air resistance force opposing it so it has a lesser proportional effect & it swings higher.
It's a long time since I did these experiments on wm2d so I reserve the right to be forgetful or confused ;)
re: energy producing experiments
Thanks; Fletcher
If a mass is dropped the same distance; off the circumference, or fixed inside the rim the final velocity will be the same.
If a mass is dropped the same distance; off the circumference, or fixed inside the rim the final velocity will be the same.
- Attachments
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- photo gates and flag
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- inside rim of wheel
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- circumference drop
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georgexbailey
- Enthusiast
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- Joined: Wed Dec 10, 2008 1:49 pm
re: energy producing experiments
Once the 10kg wheel has a momentum of 14.007 units, what would happen if you place a 1kg mass at 6:00? Would the wheel have enough momentum to carry it to 12:00? Is the yo-yo de-spin device the only way to transfer the momentum to the 1kg?Now you put the 14.007 units of momentum into a yo-yo de-spin device and transfer all of the motion to the small overbalance mass of 1 kilogram. If (a very small if) Newton’s Three Laws of Motion apply to this event then the one kilogram must be moving 14.007 m/sec. At 14.007 m/sec the one kilogram will rise (d = 1/2v²/a) 10 meters, and it was only dropped one meter. All measurements that I have taken confirm Newtonian physics, and that it applies to this event.
Obviously the answer to that question is No - otherwise any out-of-balance wheel would self rotate until it flys apart. I used to balance tyres on a tyre balancing machine a long time ago - trust me, this never happened!
This is all looking like confirmation that the Laws of thermodynamcs still work. If a mass dropped a specific height always accelerates to the same velocity, regardless of what tricks we pull - then obviously they all aquire the same momentum. And it's going to require all of the momentum, and then some, to get the weight back up the same height.
This is all looking like confirmation that the Laws of thermodynamcs still work. If a mass dropped a specific height always accelerates to the same velocity, regardless of what tricks we pull - then obviously they all aquire the same momentum. And it's going to require all of the momentum, and then some, to get the weight back up the same height.
Pequaide - thanks for that brilliant explanation. I think I understand what you are saying ...
This makes me ask a fairly dumb question ...
If we build a massive balanced water wheel, with a lot of mass built into the rim simply to bulk up the mass ...
And then inject a relatively small flow of water at the top to overbalance it ....
From what I understand you are saying, the momentum of this wheel is going to be far more than the energy required to pump that water back up again ...
So what is stopping us from attaching a generator to the wheel to create electricity, and using a small portion of this electricty to run an electric pump ...
This is the type of 'perpetual motion' scheme that is laughed at as being ridiculous - although I believe I have seen pictures of similar designs that allegedly work. Most people assume these are just utube hoaxes, or deluded people making a fool of themselves.
But you've got me wondering now ...
This makes me ask a fairly dumb question ...
If we build a massive balanced water wheel, with a lot of mass built into the rim simply to bulk up the mass ...
And then inject a relatively small flow of water at the top to overbalance it ....
From what I understand you are saying, the momentum of this wheel is going to be far more than the energy required to pump that water back up again ...
So what is stopping us from attaching a generator to the wheel to create electricity, and using a small portion of this electricty to run an electric pump ...
This is the type of 'perpetual motion' scheme that is laughed at as being ridiculous - although I believe I have seen pictures of similar designs that allegedly work. Most people assume these are just utube hoaxes, or deluded people making a fool of themselves.
But you've got me wondering now ...
Could somebody please direct me to the equation for calculating the final velocity of a known mass falling a known vertical distance to earth under gravity acceleration.
I know g is approximately 9.8 meters per second per second - so I could calculate the velocity of a 1 kg mass after 1 second, 2 seconds, 3 seconds, etc ...
It's been many years since I studied calculus and stuff, and i've forgotten it all by now.
It seem to me that I need a way to calculate the distance elapsed - because I need to know when the mass hits the bottom.
If I graphed time on the X axis, and velocity on the Y axis - is distance traveled the area under the graph?
It's been a long time ...
I want to play around with the numbers to see if there really is this unexpected gain in momentum. It seems to me that you (pequaide) are saying that when we attach a falling weight to another moveable weight that isn't falling, then the force of gravity is divided equally amongst the total weights. That I can understand.
What I don't understand is the numbers you are getting for the final velocities. You are seeming to suggest that when a falling weight is being used to accelerate some non-falling weight, the falling weight will take longer to fall, and the final velocity will be smaller - but not proportionally smaller???? Allegedly higher velocity than expected?
IF TRUE, this is the point where free energy is gained, therefore I need to be able to crunch these numbers myself to test the theory.
EDIT - sorry, I think you've already given this ...
I know g is approximately 9.8 meters per second per second - so I could calculate the velocity of a 1 kg mass after 1 second, 2 seconds, 3 seconds, etc ...
It's been many years since I studied calculus and stuff, and i've forgotten it all by now.
It seem to me that I need a way to calculate the distance elapsed - because I need to know when the mass hits the bottom.
If I graphed time on the X axis, and velocity on the Y axis - is distance traveled the area under the graph?
It's been a long time ...
I want to play around with the numbers to see if there really is this unexpected gain in momentum. It seems to me that you (pequaide) are saying that when we attach a falling weight to another moveable weight that isn't falling, then the force of gravity is divided equally amongst the total weights. That I can understand.
What I don't understand is the numbers you are getting for the final velocities. You are seeming to suggest that when a falling weight is being used to accelerate some non-falling weight, the falling weight will take longer to fall, and the final velocity will be smaller - but not proportionally smaller???? Allegedly higher velocity than expected?
IF TRUE, this is the point where free energy is gained, therefore I need to be able to crunch these numbers myself to test the theory.
EDIT - sorry, I think you've already given this ...
Need time to think about this ...The velocity is determined by this formula d = 1/2v²/a, where v is the square root of (2 * d * a) which in this case is (2 * 1m * .981m/sec²) = 1.4007m/sec.
re: energy producing experiments
The distance (or displacement formula; s) formula is s = 1/2at². I think I have seen d used instead of s so I went with that, who would think of s being distance. So now we have d = 1/2at², v = at or t = v/a and t² = v²/a², substituting v²/a² for t² the distance formula is now d = ½ v²/a. So lets check and see if it is correct. After one second in free fall the distance dropped is 4.9 meters (s = 1/2at²) and the velocity is 9.81m/sec (v = at) so lets plug in 9.81m/sec velocity in (d = ½ v²/a) instead of time in (s = 1/2at²): d = ½ * 9.81m/sec * 9.81m/sec / 9.81m/sec² = 4.9m, yep it works.
Lets double check it at two seconds. After two second in free fall the distance dropped is 19.62 meters (s = 1/2at²) and the velocity is 19.62m/sec (v = at) so lets plug in 19.62m/sec velocity in (d = ½ v²/a) instead of time in (s = 1/2at²): d = ½ * 19.62m/sec * 19.62m/sec / 9.81m/sec² = 19.62m, yep it works.
George: if you leave the one kilogram attached to the wheel ‘no’ the 14 units of momentum will not be enough to return the one kilogram to12 o’clock, but if you separate the one kilogram from the wheel and give it all the momentum then it will rise 10 meters. Separate the overbalanced mass from the wheel and transfer all the momentum to it. The cylinder and spheres separates the mass and transfers all the motion to the mass, and energy is made.
A greendoor question: If I graphed time on the X axis, and velocity on the Y axis - is distance traveled the area under the graph?
Pequaide's answer: Yes.
If a suspended mass of 1 kg accelerates a 0 kg rim mass balanced wheel the acceleration will be 1/1 * 9.81 m/sec² or 9.81m/sec². Just brainstorming here this is free fall. At the end of a one meter drop the velocity will be 4.429 m/sec and its momentum will be 4.429 units of momentum, and its energy will be (1/2mv²) 9.81 joules, and the mass will rise 1 meter. This is all the energy that is needed to reload the system, less a little friction.
If a suspended mass (on the end of a ribbon wrapped around the wheel) of 1 kg accelerates a 4 kg rim mass (nearly all the mass is in the rim) balanced wheel the acceleration will be 1/5 * 9.81 m/sec² or 1.962m/sec². At the end of a one meter drop the velocity will be 1.98 m/sec and its momentum will be 9.90 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 9.90m/sec its energy will be (1/2mv²) 49.05 joules, and the mass will rise 5 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
If a suspended mass of 1 kg accelerates a 40 kg rim mass wheel the acceleration will be 1/41 * 9.81 m/sec² or .239m/sec². At the end of a one meter drop the velocity will be .6917 m/sec and its momentum will be 20.05 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 28.36m/sec its energy will be (1/2mv²) 402.1 joules, and the mass will rise 41 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
If a suspended mass of 1 kg accelerates a 1 kg rim mass balanced wheel the acceleration will be 1/2 * 9.81 m/sec² or 4.9m/sec². At the end of a one meter drop the velocity will be 3.13 m/sec and its momentum will be 6.26 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 6.26m/sec its energy will be (1/2mv²) 19.62 joules, and the mass will rise 2 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
If a suspended mass of 1 kg accelerates a 1000 kg rim mass (nearly all the mass is in the rim) balanced wheel the acceleration will be 1/1001 * 9.81 m/sec² or .00980m/sec². At the end of a one meter drop the velocity will be .14000 m/sec and its momentum will be 140.14 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 140m/sec its energy will be (1/2mv²) 9819 joules, and the mass will rise 1001 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
This math is F = ma. And d =1/2 v²/a, and (1/2mv²)
Lets double check it at two seconds. After two second in free fall the distance dropped is 19.62 meters (s = 1/2at²) and the velocity is 19.62m/sec (v = at) so lets plug in 19.62m/sec velocity in (d = ½ v²/a) instead of time in (s = 1/2at²): d = ½ * 19.62m/sec * 19.62m/sec / 9.81m/sec² = 19.62m, yep it works.
George: if you leave the one kilogram attached to the wheel ‘no’ the 14 units of momentum will not be enough to return the one kilogram to12 o’clock, but if you separate the one kilogram from the wheel and give it all the momentum then it will rise 10 meters. Separate the overbalanced mass from the wheel and transfer all the momentum to it. The cylinder and spheres separates the mass and transfers all the motion to the mass, and energy is made.
A greendoor question: If I graphed time on the X axis, and velocity on the Y axis - is distance traveled the area under the graph?
Pequaide's answer: Yes.
If a suspended mass of 1 kg accelerates a 0 kg rim mass balanced wheel the acceleration will be 1/1 * 9.81 m/sec² or 9.81m/sec². Just brainstorming here this is free fall. At the end of a one meter drop the velocity will be 4.429 m/sec and its momentum will be 4.429 units of momentum, and its energy will be (1/2mv²) 9.81 joules, and the mass will rise 1 meter. This is all the energy that is needed to reload the system, less a little friction.
If a suspended mass (on the end of a ribbon wrapped around the wheel) of 1 kg accelerates a 4 kg rim mass (nearly all the mass is in the rim) balanced wheel the acceleration will be 1/5 * 9.81 m/sec² or 1.962m/sec². At the end of a one meter drop the velocity will be 1.98 m/sec and its momentum will be 9.90 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 9.90m/sec its energy will be (1/2mv²) 49.05 joules, and the mass will rise 5 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
If a suspended mass of 1 kg accelerates a 40 kg rim mass wheel the acceleration will be 1/41 * 9.81 m/sec² or .239m/sec². At the end of a one meter drop the velocity will be .6917 m/sec and its momentum will be 20.05 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 28.36m/sec its energy will be (1/2mv²) 402.1 joules, and the mass will rise 41 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
If a suspended mass of 1 kg accelerates a 1 kg rim mass balanced wheel the acceleration will be 1/2 * 9.81 m/sec² or 4.9m/sec². At the end of a one meter drop the velocity will be 3.13 m/sec and its momentum will be 6.26 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 6.26m/sec its energy will be (1/2mv²) 19.62 joules, and the mass will rise 2 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
If a suspended mass of 1 kg accelerates a 1000 kg rim mass (nearly all the mass is in the rim) balanced wheel the acceleration will be 1/1001 * 9.81 m/sec² or .00980m/sec². At the end of a one meter drop the velocity will be .14000 m/sec and its momentum will be 140.14 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 140m/sec its energy will be (1/2mv²) 9819 joules, and the mass will rise 1001 meters. 9.81 joules is all the energy that is needed to reload the system, less a little friction.
This math is F = ma. And d =1/2 v²/a, and (1/2mv²)
Pequaide - thank you so much. I'm definately going to brush up my physics now and see if I can draw up some designs to exploit this effect.
I have a vague idea forming, based on various ideas from several threads here ...
I'm wondering if the Bessler secret may have been inside the weights. Apparantly he always removed these; handled them with gloves; allowed others to touch them - but not the ends; someone thought they were thin-walled cylinders; Bessler said something about the weights being the power source, gaining motion from their swinging or something to that effect ...
I'm wondering if these weights were actually a heavy weight on an axel, supported inside an thin tube cyclinder. (This would require bearings in the ends - and if anyone could see the ends, this would probably be a dead giveaway).
Imagine a 2kg mass of lead on an axle, mounted in roller bearings inside an outer tube. How would these behave if used inside a 'conventional' gravity wheel (which we know can only approach unity)?
I believe they would behave like solid weights, but the outer tube could roll - and being a much lighter mass than the inner 2kg weight, they could roll very easily, and the 2kg mass would remain stationary on account of inertia.
Now imagine the heavy mass was connected to the outer cylinder with a short leather leash that could coil up a certain distance. I believe this would allow the cylinder to roll a certain distance (picking up momentum) and then suddely it would reach the end of the leash (like Bessers 'dog' or the 'cat snatching the fat rats tail').
Once the coil was wound, the heavy mass should roll too, since it's effectively locked to the outer cylinder. Then - if the outer cylinder was suddenly stopped - I wonder where all that momentum would go to? I haven't thought this through very far - but i'm wondering if there is a situation where the heavy 2kb mass could be spinning while the outer tube is stationary - unwinding the leash, until suddenly it reachs the end of the leash, and the outer cylinder is suddently given all the momentum of the heavy spinning core ...
That's where your maths becomes very interesting to me ... could we get these special Bessler weights to roll uphill with more energy than they obtained falling or rolling down ...
Still thinking, but i'm throwing this idea into the ring for the good thinkers here to dissect ...
Thanks!
I have a vague idea forming, based on various ideas from several threads here ...
I'm wondering if the Bessler secret may have been inside the weights. Apparantly he always removed these; handled them with gloves; allowed others to touch them - but not the ends; someone thought they were thin-walled cylinders; Bessler said something about the weights being the power source, gaining motion from their swinging or something to that effect ...
I'm wondering if these weights were actually a heavy weight on an axel, supported inside an thin tube cyclinder. (This would require bearings in the ends - and if anyone could see the ends, this would probably be a dead giveaway).
Imagine a 2kg mass of lead on an axle, mounted in roller bearings inside an outer tube. How would these behave if used inside a 'conventional' gravity wheel (which we know can only approach unity)?
I believe they would behave like solid weights, but the outer tube could roll - and being a much lighter mass than the inner 2kg weight, they could roll very easily, and the 2kg mass would remain stationary on account of inertia.
Now imagine the heavy mass was connected to the outer cylinder with a short leather leash that could coil up a certain distance. I believe this would allow the cylinder to roll a certain distance (picking up momentum) and then suddely it would reach the end of the leash (like Bessers 'dog' or the 'cat snatching the fat rats tail').
Once the coil was wound, the heavy mass should roll too, since it's effectively locked to the outer cylinder. Then - if the outer cylinder was suddenly stopped - I wonder where all that momentum would go to? I haven't thought this through very far - but i'm wondering if there is a situation where the heavy 2kb mass could be spinning while the outer tube is stationary - unwinding the leash, until suddenly it reachs the end of the leash, and the outer cylinder is suddently given all the momentum of the heavy spinning core ...
That's where your maths becomes very interesting to me ... could we get these special Bessler weights to roll uphill with more energy than they obtained falling or rolling down ...
Still thinking, but i'm throwing this idea into the ring for the good thinkers here to dissect ...
Thanks!
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georgexbailey
- Enthusiast
- Posts: 26
- Joined: Wed Dec 10, 2008 1:49 pm
re: energy producing experiments
http://ntrs.nasa.gov/archive/nasa/casi. ... 006811.pdf
See the PDF page 9, first sentence at top of page. "During this operation angular momentum is transfered to the weights and the wires."
At least this confirms a part of pequaid's theories.
Very interesting.
GB
See the PDF page 9, first sentence at top of page. "During this operation angular momentum is transfered to the weights and the wires."
At least this confirms a part of pequaid's theories.
Very interesting.
GB
re: energy producing experiments
It is the fact that the rocket stopped rotating is what I would like to point out. I came to a different conclusion as to what form of momentum is conserved.
Place a thick immovable post in the center of a frictionless plane and start a puck rotating around it on the end of a fixed string. As the string wraps around the post the puck maintains its same linear Newtonian momentum, while its angular momentum decreases. This is proof that angular momentum is not a conserved quantity.
Place a thick immovable post in the center of a frictionless plane and start a puck rotating around it on the end of a fixed string. As the string wraps around the post the puck maintains its same linear Newtonian momentum, while its angular momentum decreases. This is proof that angular momentum is not a conserved quantity.
re: energy producing experiments
try googling Ballistic Pendulums re: momentum & energy
http://en.wikipedia.org/wiki/Ballistic_pendulum
http://webphysics.davidson.edu/physlet_ ... istic.html
http://hyperphysics.phy-astr.gsu.edu/Hbase/balpen.html
http://hyperphysics.phy-astr.gsu.edu/hb ... alpen.html
http://en.wikipedia.org/wiki/Ballistic_pendulum
http://webphysics.davidson.edu/physlet_ ... istic.html
http://hyperphysics.phy-astr.gsu.edu/Hbase/balpen.html
http://hyperphysics.phy-astr.gsu.edu/hb ... alpen.html
re: energy producing experiments
I like the ones where over 95% of the kinetic energy is lost, yet no one can find the alleged heat.
re: energy producing experiments
When the tether (with the 3 kg mass on the end) unwraps from the rocket (NASA’s yo-yo de-spin device) the rocket’s rotation will be stopped by a certain length of unwrapped tether. If the rocket has a little greater mass it will take a little longer tether to stop it, a little less massive rocket and you will need less tether length. For those expecting angular momentum to be conserved: if you change the angular momentum of the rocket then the tether length used to stop the rocket will be changed to accommodate the change in the angular momentum of the rocket. If Newtonian physics is the controlling factor, then the linear momentum changes would be the controlling factor the tether length. I set up an experiment to test which one of these would be the controlling factor for tether length: is it linear Newtonian momentum or is it angular momentum.
The most common hollow cylinder that I have used is the 3 inch PVC pipe coupler (for a 3 inch I. D. PVC pipe). This coupler has a 3.5 inch I.D. and a 4.0 inch O.D. Once the cylinder is chosen the two determining factors for stopping a certain mass of cylinder are: the mass of the spheres and the length of the tether. The mass of the cylinder is increased by placing a 3 inch I.D. PVC pipe (this pipe has an outside diameter of 3.5 inches) into the lower half of the coupler. For a certain length of tether with 132 grams (total) for both spheres a certain mass of cylinder is stopped precisely when the tether string is at 90° to a line tangent to the surface of the cylinder, at this point the tether enters a slit in the cylinder that allows easier observation of this cylinder stop. For a certain cylinder mass the 132g spheres stop the cylinder when the spheres are straight out. The spheres can not stop the cylinder when it has a slightly greater mass, and the cylinder will still be moving forward while the tether line is in the slit. The spheres and tether will have the cylinder moving backwards while the tether is in the slit if the mass of the cylinder is to light. This perfect 90° stop is equivalent to a certain length of tether for the rocket with the 3 kilograms end mass, only a certain mass of rocket or cylinder can be stopped at this unique point.
To get the spheres to stop the coupler in the first setup; a 365.8g three inch I.D. pipe was originally added to the bottom half of the coupler with spheres attached to the top half. This 3 inch pipe was removed and replace with a 4 inch I.D. pipe that was placed on the outside of the coupler. A length of 4 inch pipe was cut so that it had an equivalent Newtonian rotational inertia, that when spun at a same rpm will have an equal linear Newtonian momentum. The formula for that inertia would be mass times radius of gyration times circumference .3658 kg * .96 * 3.5 = 4 in. pipe mass * .96 * 4.5 or 284.5g for the 4 inch pipe mass. If linear Newtonian momentum controls the cylinder and spheres phenomenon then you would need a 284.5.3g 4 inch pipe for the 90° stop to be maintained.
The angular momentum formula is mass * radius * radius * radians per second. The radians per sec (I will use 2 radians per second) for both pipe is the same for both the 3 and 4 inch pipes so that drops out. So a 3 inch pipe mass of .3658 kg * 3.5 * .96 * 3.5 * .96 * 2 radians per second = new 4.0 inch pipe mass of X * 4.5 * .96 * 4.5 * .96 * 2 radians per second or 221.3g. If angular momentum controls the cylinder and spheres phenomenon then you would need a 221.3g 4 inch pipe for the 90° stop to be maintained.
If the unique 90° stop is maintained with the 284.5g 4 inch pipe then the cylinder and spheres machine is conserving linear Newtonian momentum not angular momentum and not kinetic energy. The cylinder and spheres machine was spun with the 4 inch 284.5g pipe and the 90° stop was maintained. I will add the account of the actual experiment for those who may wish to work through the experiment, but I will first give you my conclusions.
After working with cylinder and spheres machines of many designs and sizes it becomes very apparent that the machines are extremely accurate at determining the quantity of mass in the cylinder. The cylinder and spheres machines can pick up changes in the cylinder mass as small as 6 grams. But it is not just the mass that they are measuring; it is either mv or 1/2mv² or it is angular momentum that the spheres, at the 90° cylinder stop, are measuring or being controlled by.
You can remove the 3 inch 365.8 g pipe which is a certain quantity of linear Newtonian inertia, (which upon spinning would become linear Newtonian momentum) and replacing it with an equal quantity of linear Newtonian inertia in the form of the 284.5 g 4 inch pipe. After changing the pipes and after spinning and releasing the cylinder and spheres the 90° stop of the cylinder remains the same. This proves beyond doubt that linear Newtonian momentum is what is conserved by the cylinder and spheres phenomenon.
The actual experiment.
The top of the cylinder and spheres experiment can be connected to different lengths and diameters of PVC pipe (1. a 3 in. pipe, 2. a 3 in. pipe with coupler attached, and a 4 in. pipe). The top is a 212.5g cylinder (coupler 3.5 in. I.D., both the pipe and the coupler have a ¼ inch wall) with two 66g spheres seated in it surface. The cylinder has a 3 inch I.D. PVC pipe, 1/2 inch high, inside the coupler that has a mass of 22.2g to help seat the spheres. It also has a 7.2g plastic strip crossing a diameter for a center hole, but its mass moves half as fast as the cylinder, I will give it a momentum of 3.6g time cylinder velocity.
Without adding mass to the cylinder (238.6g) the spheres (132g) stop the top cylinder before the tether strings reach 90° to a tangent line at the surface of the cylinder. Only by adding mass to the cylinder can you arrange the spheres to stop the cylinder when the spheres are at 90° to tangent.
A slit is placed in the cylinder behind the entry hole of the tether string to allow you to evaluate the motion of the cylinder at that point; the cylinder will maintain the same rate of motion for about 1/10 of a second as the string move across the slit. One and only one length (mass) of pipe makes the spheres stop the cylinder at 90°, too little cylinder mass and it stops before 90° and the cylinder is moving backward while the string is in the slit at 90°. Too much added mass to the cylinder and the cylinder will still be moving forward while the string is in the slit.
The first mass is added by placing a straight 3 inch I.D. pipe in the bottom of the top coupler. For this particular tether length and cylinder arrangement the mass of 3 in. pipe that stops the cylinder at 90° is about 365.8g.
After being spun the top cylinder is forced into a stop when the spheres swing out to 90°. This is accomplished when a mass of 365.8 g is added to the top cylinder in the form of a 3.0 in. I.D. (.25 in. side wall) PVC pipe placed in the bottom portion of the coupler. There is one and only one mass that will cause a perfect 90° (to tangent) stop.
I then replaced the 3.0 in. pipe with a 4.0 I.D. PVC pipe. I calculated its relative velocity (at the same RPS) to be about 4.25/3.25 that of the 3.0 pipe. Therefore to maintain the same momentum the four inch pipe mass will be 3.25/4.25 times 365.8, which is 279.7g. I then placed a 279.1g pipe on the bottom of the top cylinder and began video taping it while spinning and releasing it by hand. Once the appropriate mass is determined the cylinder will stop at 90° at any RPS, being hand held offers no real problem, in fact it proves that the RPS does not matter. And the hand held models are much more readily available for physics class demonstrations.
At 279.1g the system had a slight backward motion that indicates that it is a little light. I added 7.5g and the forward motion was greater than the previous backward motion. That means that the 279.1g 4.0 in. pipe was about 4g light, 283.1g/279.7g is about 1% off. This is the third object (added to the bottom of the top coupler) with different diameters that have the same rotational momentum, and they all cause the same 90° stop. If graphed this would be three points on the same line, a line that indicates that the cylinder and spheres experiment is driven by the Law of Conservation of Momentum.
To conserve kinetic energy the 4.0 in. pipe would have to have a mass of 214g, about 30% off. .5 * 214g * 4.25/3.25 * 4.25/3.25 = .5 * 365.8 * 1 * 1. I see no way that kinetic energy is being conserved in this experiment. That makes the Law of Conservation of Energy false.
The second added mass has two pieces: an added 3 inch I.D. (cut to adjust the mass) pipe with a 3.5 I.D. diameter coupler (219.2g). You can make calculations and construct a system that keeps the spinning momentum of the added mass (both coupler and pipe) the same as the straight 3 in. pipe. But if the momentum is kept the same the kinetic energy of the spinning larger diameter coupler is not the same.
So if the entire system maintains its 90° stop it is reasonable to assume that nothing has changed and that the momentum of the system is being conserved since the input momentum had not change and the apparent output had not changed either.
If the 3 in. connecting pipe for the added on 219.2g bottom coupler is cut at 73.2g; the momentum change from the 3 in. (I.D.) 365.8g pipe length would be 12% low. The momentum would be the same for this lower portion if the connecting pipe is cut to a length that gives you 112.8g.
If the connecting pipe for the coupler is cut at 112.8g; the kinetic energy change from the 3 in. (I.D.) 365.8g pipe length would be 11% high. The kinetic energy would be the same for this lower portion if the connecting pipe is cut to a length that gives you 73.2g.
This 11% and 12% change for this lower portion doesn’t mean that the spheres motion will be going up or down by 11% or 12%, that has to be calculated separately because the spheres are going to absorb all the motion of the connecting pipe whether it has a mass of 73.2g or 112.8g. The 11% and 12% only predicts that the 90° stop will be lost if you use the wrong one (73.2g or 112.8g) of the two masses. The spheres themselves would ideally have about 550% of the original energy at the 90° stop.
The 112.8g connecting pipe was used with the 219.2g coupler and the 90° stop was maintained, that means that kinetic energy is not what the spheres are responding to. The spheres are responding to momentum, and since there was no change in the momentum there was no change in the 90° stop.
The 11% change in Kinetic Energy is still well outside of the accuracy of the 90° stop method. I think I could detect a 6g change in the connecting pipe mass, and the 73.2g pipe is 39.6 grams off. If the connecting pipe only had a mass of 73.2g the method would be telling you to add more mass to get it to the stop at 90°, but by using 112.8g the 90° stop has been achieved and the method is not telling you to add or subtract any mass. The cylinder and spheres experiment responds to momentum.
Here again is the main point.
Two 66g spheres embedded in the surface of a 3.5 in. PVC pipe coupler.
The original momentum of this model is proportional to 212.5g * 3.75 in. (top cylinder) + 22.2g * 3.25 in.(seat pipe) + 365.8g * 3.25 in. (mass added pipe) + 132 * 4 in. (spheres seated on the surface of the cylinder) + 3.6 * 3 (plastic strip at dia.) = 2581. The final momentum would be the same and is held only by the sphere because the cylinder is stopped 2581 / 132 = 19.55. The final velocity of the spheres has increased 19.55 / 4 = 4.88 times if momentum has been maintained.
The original kinetic energy of this model is proportional to ½ * 212.5g * 3.75 * 3.75 (top cylinder) + ½ * 22.2g * 3.25 * 3.25 (seat pipe) + ½ * 365.8g * 3.25 * 3.25 (mass added pipe) + ½ * 132 * 4 * 4 (spheres seated on the surface of the cylinder) + 3.6 * 3 *3 = 4595.9.
The final proportional velocity of the spheres has already been determine by momentum conservation and is 19.55. Therefore the final kinetic energy is ½ * 132g *19.55 * 19.55 = 25,225, 25,225 / 4595.9 = 549%
My original estimations of a 350% energy increase (from different machines) were based upon attempts to actually measure the velocity of the cylinder with embedded spheres (using a mechanical release) and then measure the spheres alone at the 90° stop. I used strobe light photograph, video tapes, and photo gates. This different diameters 90° stop method is an entirely different and second method that confirms in my mind that the cylinder and spheres conserves momentum and can increase energy by 300% or more. This will be the driving force of perpetual motion machines.
Math used for the Experiments
These 90° stop experiments prove that the quantity of motion lost by the cylinder is momentum. Newton’s Three Laws of Motion require that the momentum lost by the cylinder must be gained by the spheres, because the force in the tether string must be equal in both directions.
The cylinder stops with the spheres at 90° with any RPS used if the proper amount of mass at a certain diameter has been added to the bottom of the top coupler. Three proper amounts of mass at a certain diameter have been used to stop the top cylinder when the spheres are at 90° to tangent.
The cylinder will stop with any initial R.P.S. given it by the hand release method. I am going to use 2 rotations per second to get some actual numbers. I think 2 R.P.S. is within the ability of the hand release method.
A 3.0 inch inside diameter PVC pipe with a length just under 10 inches and with a mass of 365.8g gives the cylinder a nice stop when the spheres are at 90°. The effective rotational mass of the cylinder is at about a 3.25 in. diameter. Here is the math used for the quantities of motion mentioned.
Mass: 365.8g of the 3 inch straight PVC pipe
Momentum: (mv) 365.8g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .18973
Kinetic energy: (1/2mv²) .5 * 365.8g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .04920
A 3.5 inch inside diameter PVC pipe coupler with a mass of 219.2g combined with a 112.8g connecting 3.0 in. pipe gives the cylinder a nice stop when the spheres are at 90°. The effect rotational mass of the coupler is at about 3.75 in. and effective rotational mass of the connecting pipe at a 3.25 in. diameter. Here is the math used for the quantities of motion mentioned.
Mass: 219.2 for the coupler and 112.8 for the connecting pipe
Momentum: (mv) 219.2g * 1kg/1000g* 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .13118
Momentum: (mv) 112.8g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .058506
Total momentum for the 219.2g coupler and 112.8g connecting pipe equals .18969
Kinetic energy: (1/2mv²) .5 * 219.2g * 1kg/1000g * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .039255
Kinetic energy: (1/2mv²) .5 * 112.8g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .015186
Total kinetic energy for the 219.2g coupler and 112.8g connecting pipe equals .05459
Momentum changes by .18969/.18973 = 0.02%
Kinetic energy changes by .05459/.04920 = 11%
A 4.0 inch inside diameter PVC pipe with a mass of 283.6g gives the cylinder a nice stop when the spheres are at 90°. The effect rotational mass of the cylinder is at about 4.25 in.
Here is the math used for the quantities of motion mentioned.
Mass: 283.6g for the 4 inch PVC
Momentum: (mv) 283.6g * 1kg/1000g* 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .19262
Kinetic energy: (1/2mv²) .5 * 283.6g * 1kg/1000g * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .06533
Momentum changed from the 3.0 pipe was .19262/.18973 = 1.5%
The kinetic energy change from the 3.0 in. pipe was .06533/.04920 = 33%
When the tether achieves 90° it enters a slit, if the cylinder is stopped it remains stopped as the tether string crosses the slit. If the spheres stop the cylinder before 90° they will have started moving the cylinder backward and the cylinder will be moving backward while in the slit. If the spheres have not yet stopped the cylinder before they reach 90° then the cylinder will be moving forward while the string is in the slit.
The cylinder will be stopped only if it has a certain mass, add 6g and the cylinder will still be moving forward in the slit, subtract 6g from the proper mass and the cylinder will be moving backward. It is accurate within 2% of the added mass. Momentum conservation falls within this 2%, kinetic energy conservation fall outside the 2% at 11% and 33%. Kinetic energy is not conserved by the cylinder and spheres experiment.
If the momentum of the cylinder and spheres is conserved in the motion of the spheres alone (when the cylinder is stopped) then the systems has an over 400% increase in energy.
Momentum: (mv) 212.5g * 1kg/1000g* 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .12722
Momentum: (mv) 22.2 g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .011515
Momentum: (mv) 365.8g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .18972
Momentum: (mv) 132g * 1kg/1000g* 4.00 in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .084265
Total initial momentum = .41273
Final momentum will be .132kg * v = .41273 for a velocity of 3.127 m/sec
Kinetic energy: (1/2mv²) .5 * 212.5g * 1kg/1000g * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .03827
Kinetic energy: (1/2mv²) .5 * 22.2g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .002986
Kinetic energy: (1/2mv²) .5 * 365.8g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .04920
Kinetic energy: (1/2mv²) .5 * 132g * 1kg/1000g * 4.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 4.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS =.026896
Total Initial Kinetic energy = .117352
Final Kinetic energy will be .5 * 132g * 3.127 * 3.127 = .64535 .64535/.117352 =550% of the original energy.
Now; one could claim that the 90° stop still occurs but that the spheres are moving faster or slower, but under these conditions the Law of Conservation of Momentum would be false. You simply can’t pretend that both these two formulas are conserved.
This discrepancy between momentum and kinetic energy exists in ballistic pendulums, there kinetic energy is assigned an imaginary friend (heat) to hide the formula’s (1/2 mv²) incompetence.
In the cylinder and spheres experiment it is totally impossible to have this phantom friend of heat come to the rescue, because kinetic energy is increasing not decreasing.
The most common hollow cylinder that I have used is the 3 inch PVC pipe coupler (for a 3 inch I. D. PVC pipe). This coupler has a 3.5 inch I.D. and a 4.0 inch O.D. Once the cylinder is chosen the two determining factors for stopping a certain mass of cylinder are: the mass of the spheres and the length of the tether. The mass of the cylinder is increased by placing a 3 inch I.D. PVC pipe (this pipe has an outside diameter of 3.5 inches) into the lower half of the coupler. For a certain length of tether with 132 grams (total) for both spheres a certain mass of cylinder is stopped precisely when the tether string is at 90° to a line tangent to the surface of the cylinder, at this point the tether enters a slit in the cylinder that allows easier observation of this cylinder stop. For a certain cylinder mass the 132g spheres stop the cylinder when the spheres are straight out. The spheres can not stop the cylinder when it has a slightly greater mass, and the cylinder will still be moving forward while the tether line is in the slit. The spheres and tether will have the cylinder moving backwards while the tether is in the slit if the mass of the cylinder is to light. This perfect 90° stop is equivalent to a certain length of tether for the rocket with the 3 kilograms end mass, only a certain mass of rocket or cylinder can be stopped at this unique point.
To get the spheres to stop the coupler in the first setup; a 365.8g three inch I.D. pipe was originally added to the bottom half of the coupler with spheres attached to the top half. This 3 inch pipe was removed and replace with a 4 inch I.D. pipe that was placed on the outside of the coupler. A length of 4 inch pipe was cut so that it had an equivalent Newtonian rotational inertia, that when spun at a same rpm will have an equal linear Newtonian momentum. The formula for that inertia would be mass times radius of gyration times circumference .3658 kg * .96 * 3.5 = 4 in. pipe mass * .96 * 4.5 or 284.5g for the 4 inch pipe mass. If linear Newtonian momentum controls the cylinder and spheres phenomenon then you would need a 284.5.3g 4 inch pipe for the 90° stop to be maintained.
The angular momentum formula is mass * radius * radius * radians per second. The radians per sec (I will use 2 radians per second) for both pipe is the same for both the 3 and 4 inch pipes so that drops out. So a 3 inch pipe mass of .3658 kg * 3.5 * .96 * 3.5 * .96 * 2 radians per second = new 4.0 inch pipe mass of X * 4.5 * .96 * 4.5 * .96 * 2 radians per second or 221.3g. If angular momentum controls the cylinder and spheres phenomenon then you would need a 221.3g 4 inch pipe for the 90° stop to be maintained.
If the unique 90° stop is maintained with the 284.5g 4 inch pipe then the cylinder and spheres machine is conserving linear Newtonian momentum not angular momentum and not kinetic energy. The cylinder and spheres machine was spun with the 4 inch 284.5g pipe and the 90° stop was maintained. I will add the account of the actual experiment for those who may wish to work through the experiment, but I will first give you my conclusions.
After working with cylinder and spheres machines of many designs and sizes it becomes very apparent that the machines are extremely accurate at determining the quantity of mass in the cylinder. The cylinder and spheres machines can pick up changes in the cylinder mass as small as 6 grams. But it is not just the mass that they are measuring; it is either mv or 1/2mv² or it is angular momentum that the spheres, at the 90° cylinder stop, are measuring or being controlled by.
You can remove the 3 inch 365.8 g pipe which is a certain quantity of linear Newtonian inertia, (which upon spinning would become linear Newtonian momentum) and replacing it with an equal quantity of linear Newtonian inertia in the form of the 284.5 g 4 inch pipe. After changing the pipes and after spinning and releasing the cylinder and spheres the 90° stop of the cylinder remains the same. This proves beyond doubt that linear Newtonian momentum is what is conserved by the cylinder and spheres phenomenon.
The actual experiment.
The top of the cylinder and spheres experiment can be connected to different lengths and diameters of PVC pipe (1. a 3 in. pipe, 2. a 3 in. pipe with coupler attached, and a 4 in. pipe). The top is a 212.5g cylinder (coupler 3.5 in. I.D., both the pipe and the coupler have a ¼ inch wall) with two 66g spheres seated in it surface. The cylinder has a 3 inch I.D. PVC pipe, 1/2 inch high, inside the coupler that has a mass of 22.2g to help seat the spheres. It also has a 7.2g plastic strip crossing a diameter for a center hole, but its mass moves half as fast as the cylinder, I will give it a momentum of 3.6g time cylinder velocity.
Without adding mass to the cylinder (238.6g) the spheres (132g) stop the top cylinder before the tether strings reach 90° to a tangent line at the surface of the cylinder. Only by adding mass to the cylinder can you arrange the spheres to stop the cylinder when the spheres are at 90° to tangent.
A slit is placed in the cylinder behind the entry hole of the tether string to allow you to evaluate the motion of the cylinder at that point; the cylinder will maintain the same rate of motion for about 1/10 of a second as the string move across the slit. One and only one length (mass) of pipe makes the spheres stop the cylinder at 90°, too little cylinder mass and it stops before 90° and the cylinder is moving backward while the string is in the slit at 90°. Too much added mass to the cylinder and the cylinder will still be moving forward while the string is in the slit.
The first mass is added by placing a straight 3 inch I.D. pipe in the bottom of the top coupler. For this particular tether length and cylinder arrangement the mass of 3 in. pipe that stops the cylinder at 90° is about 365.8g.
After being spun the top cylinder is forced into a stop when the spheres swing out to 90°. This is accomplished when a mass of 365.8 g is added to the top cylinder in the form of a 3.0 in. I.D. (.25 in. side wall) PVC pipe placed in the bottom portion of the coupler. There is one and only one mass that will cause a perfect 90° (to tangent) stop.
I then replaced the 3.0 in. pipe with a 4.0 I.D. PVC pipe. I calculated its relative velocity (at the same RPS) to be about 4.25/3.25 that of the 3.0 pipe. Therefore to maintain the same momentum the four inch pipe mass will be 3.25/4.25 times 365.8, which is 279.7g. I then placed a 279.1g pipe on the bottom of the top cylinder and began video taping it while spinning and releasing it by hand. Once the appropriate mass is determined the cylinder will stop at 90° at any RPS, being hand held offers no real problem, in fact it proves that the RPS does not matter. And the hand held models are much more readily available for physics class demonstrations.
At 279.1g the system had a slight backward motion that indicates that it is a little light. I added 7.5g and the forward motion was greater than the previous backward motion. That means that the 279.1g 4.0 in. pipe was about 4g light, 283.1g/279.7g is about 1% off. This is the third object (added to the bottom of the top coupler) with different diameters that have the same rotational momentum, and they all cause the same 90° stop. If graphed this would be three points on the same line, a line that indicates that the cylinder and spheres experiment is driven by the Law of Conservation of Momentum.
To conserve kinetic energy the 4.0 in. pipe would have to have a mass of 214g, about 30% off. .5 * 214g * 4.25/3.25 * 4.25/3.25 = .5 * 365.8 * 1 * 1. I see no way that kinetic energy is being conserved in this experiment. That makes the Law of Conservation of Energy false.
The second added mass has two pieces: an added 3 inch I.D. (cut to adjust the mass) pipe with a 3.5 I.D. diameter coupler (219.2g). You can make calculations and construct a system that keeps the spinning momentum of the added mass (both coupler and pipe) the same as the straight 3 in. pipe. But if the momentum is kept the same the kinetic energy of the spinning larger diameter coupler is not the same.
So if the entire system maintains its 90° stop it is reasonable to assume that nothing has changed and that the momentum of the system is being conserved since the input momentum had not change and the apparent output had not changed either.
If the 3 in. connecting pipe for the added on 219.2g bottom coupler is cut at 73.2g; the momentum change from the 3 in. (I.D.) 365.8g pipe length would be 12% low. The momentum would be the same for this lower portion if the connecting pipe is cut to a length that gives you 112.8g.
If the connecting pipe for the coupler is cut at 112.8g; the kinetic energy change from the 3 in. (I.D.) 365.8g pipe length would be 11% high. The kinetic energy would be the same for this lower portion if the connecting pipe is cut to a length that gives you 73.2g.
This 11% and 12% change for this lower portion doesn’t mean that the spheres motion will be going up or down by 11% or 12%, that has to be calculated separately because the spheres are going to absorb all the motion of the connecting pipe whether it has a mass of 73.2g or 112.8g. The 11% and 12% only predicts that the 90° stop will be lost if you use the wrong one (73.2g or 112.8g) of the two masses. The spheres themselves would ideally have about 550% of the original energy at the 90° stop.
The 112.8g connecting pipe was used with the 219.2g coupler and the 90° stop was maintained, that means that kinetic energy is not what the spheres are responding to. The spheres are responding to momentum, and since there was no change in the momentum there was no change in the 90° stop.
The 11% change in Kinetic Energy is still well outside of the accuracy of the 90° stop method. I think I could detect a 6g change in the connecting pipe mass, and the 73.2g pipe is 39.6 grams off. If the connecting pipe only had a mass of 73.2g the method would be telling you to add more mass to get it to the stop at 90°, but by using 112.8g the 90° stop has been achieved and the method is not telling you to add or subtract any mass. The cylinder and spheres experiment responds to momentum.
Here again is the main point.
Two 66g spheres embedded in the surface of a 3.5 in. PVC pipe coupler.
The original momentum of this model is proportional to 212.5g * 3.75 in. (top cylinder) + 22.2g * 3.25 in.(seat pipe) + 365.8g * 3.25 in. (mass added pipe) + 132 * 4 in. (spheres seated on the surface of the cylinder) + 3.6 * 3 (plastic strip at dia.) = 2581. The final momentum would be the same and is held only by the sphere because the cylinder is stopped 2581 / 132 = 19.55. The final velocity of the spheres has increased 19.55 / 4 = 4.88 times if momentum has been maintained.
The original kinetic energy of this model is proportional to ½ * 212.5g * 3.75 * 3.75 (top cylinder) + ½ * 22.2g * 3.25 * 3.25 (seat pipe) + ½ * 365.8g * 3.25 * 3.25 (mass added pipe) + ½ * 132 * 4 * 4 (spheres seated on the surface of the cylinder) + 3.6 * 3 *3 = 4595.9.
The final proportional velocity of the spheres has already been determine by momentum conservation and is 19.55. Therefore the final kinetic energy is ½ * 132g *19.55 * 19.55 = 25,225, 25,225 / 4595.9 = 549%
My original estimations of a 350% energy increase (from different machines) were based upon attempts to actually measure the velocity of the cylinder with embedded spheres (using a mechanical release) and then measure the spheres alone at the 90° stop. I used strobe light photograph, video tapes, and photo gates. This different diameters 90° stop method is an entirely different and second method that confirms in my mind that the cylinder and spheres conserves momentum and can increase energy by 300% or more. This will be the driving force of perpetual motion machines.
Math used for the Experiments
These 90° stop experiments prove that the quantity of motion lost by the cylinder is momentum. Newton’s Three Laws of Motion require that the momentum lost by the cylinder must be gained by the spheres, because the force in the tether string must be equal in both directions.
The cylinder stops with the spheres at 90° with any RPS used if the proper amount of mass at a certain diameter has been added to the bottom of the top coupler. Three proper amounts of mass at a certain diameter have been used to stop the top cylinder when the spheres are at 90° to tangent.
The cylinder will stop with any initial R.P.S. given it by the hand release method. I am going to use 2 rotations per second to get some actual numbers. I think 2 R.P.S. is within the ability of the hand release method.
A 3.0 inch inside diameter PVC pipe with a length just under 10 inches and with a mass of 365.8g gives the cylinder a nice stop when the spheres are at 90°. The effective rotational mass of the cylinder is at about a 3.25 in. diameter. Here is the math used for the quantities of motion mentioned.
Mass: 365.8g of the 3 inch straight PVC pipe
Momentum: (mv) 365.8g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .18973
Kinetic energy: (1/2mv²) .5 * 365.8g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .04920
A 3.5 inch inside diameter PVC pipe coupler with a mass of 219.2g combined with a 112.8g connecting 3.0 in. pipe gives the cylinder a nice stop when the spheres are at 90°. The effect rotational mass of the coupler is at about 3.75 in. and effective rotational mass of the connecting pipe at a 3.25 in. diameter. Here is the math used for the quantities of motion mentioned.
Mass: 219.2 for the coupler and 112.8 for the connecting pipe
Momentum: (mv) 219.2g * 1kg/1000g* 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .13118
Momentum: (mv) 112.8g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .058506
Total momentum for the 219.2g coupler and 112.8g connecting pipe equals .18969
Kinetic energy: (1/2mv²) .5 * 219.2g * 1kg/1000g * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .039255
Kinetic energy: (1/2mv²) .5 * 112.8g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .015186
Total kinetic energy for the 219.2g coupler and 112.8g connecting pipe equals .05459
Momentum changes by .18969/.18973 = 0.02%
Kinetic energy changes by .05459/.04920 = 11%
A 4.0 inch inside diameter PVC pipe with a mass of 283.6g gives the cylinder a nice stop when the spheres are at 90°. The effect rotational mass of the cylinder is at about 4.25 in.
Here is the math used for the quantities of motion mentioned.
Mass: 283.6g for the 4 inch PVC
Momentum: (mv) 283.6g * 1kg/1000g* 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .19262
Kinetic energy: (1/2mv²) .5 * 283.6g * 1kg/1000g * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .06533
Momentum changed from the 3.0 pipe was .19262/.18973 = 1.5%
The kinetic energy change from the 3.0 in. pipe was .06533/.04920 = 33%
When the tether achieves 90° it enters a slit, if the cylinder is stopped it remains stopped as the tether string crosses the slit. If the spheres stop the cylinder before 90° they will have started moving the cylinder backward and the cylinder will be moving backward while in the slit. If the spheres have not yet stopped the cylinder before they reach 90° then the cylinder will be moving forward while the string is in the slit.
The cylinder will be stopped only if it has a certain mass, add 6g and the cylinder will still be moving forward in the slit, subtract 6g from the proper mass and the cylinder will be moving backward. It is accurate within 2% of the added mass. Momentum conservation falls within this 2%, kinetic energy conservation fall outside the 2% at 11% and 33%. Kinetic energy is not conserved by the cylinder and spheres experiment.
If the momentum of the cylinder and spheres is conserved in the motion of the spheres alone (when the cylinder is stopped) then the systems has an over 400% increase in energy.
Momentum: (mv) 212.5g * 1kg/1000g* 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .12722
Momentum: (mv) 22.2 g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .011515
Momentum: (mv) 365.8g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .18972
Momentum: (mv) 132g * 1kg/1000g* 4.00 in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .084265
Total initial momentum = .41273
Final momentum will be .132kg * v = .41273 for a velocity of 3.127 m/sec
Kinetic energy: (1/2mv²) .5 * 212.5g * 1kg/1000g * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .03827
Kinetic energy: (1/2mv²) .5 * 22.2g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .002986
Kinetic energy: (1/2mv²) .5 * 365.8g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS = .04920
Kinetic energy: (1/2mv²) .5 * 132g * 1kg/1000g * 4.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS * 4.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2 RPS =.026896
Total Initial Kinetic energy = .117352
Final Kinetic energy will be .5 * 132g * 3.127 * 3.127 = .64535 .64535/.117352 =550% of the original energy.
Now; one could claim that the 90° stop still occurs but that the spheres are moving faster or slower, but under these conditions the Law of Conservation of Momentum would be false. You simply can’t pretend that both these two formulas are conserved.
This discrepancy between momentum and kinetic energy exists in ballistic pendulums, there kinetic energy is assigned an imaginary friend (heat) to hide the formula’s (1/2 mv²) incompetence.
In the cylinder and spheres experiment it is totally impossible to have this phantom friend of heat come to the rescue, because kinetic energy is increasing not decreasing.