BesslerWheel.com

[rlortie]

because Wubbly is to wobbly and Kaine is not Able! :-)

This is no longer a topic thread, it is a blog with very few interested members.

[pequaide]

I have succeeded with numerous machine designs, why people don't follow is beyond me.

I have been asking people who don't believe free energy can be made to stay off the thread, that would make it as private as I would like. Why aren’t they happy with their own threads; why do they bother attaching me? I have a non-read list; if they don't like what I am proposing then they can put me on their non-read list. Do it now.

When the two 45 pound (20.7 kg) drive masses are on the same side, the wheel will immediately resume rotating after it stops when the throw is made. I wanted to make sure that the throw did not cause a counter rotation; for that would be hard to detect without a video tape of the throw. So I placed the two barbell masses on the opposite sides of the .75 inch shaft. This would give the wheel the same mass but it would have no internal force applied for acceleration.

I then released the .250 kilogram mass at the same position as before (about 1:00) but I did the throw by hand. There was no backward motion of the wheel with the hand throws. The wheel stopped nicely as the tether loop slipped off the pin and the bag took off for the wall. I had guessed my mass and tether length correctly form a few 100 gram throws I had made.

But I noticed that the hand throws were wimpy compared to the mechanical throws (with the two barbell masses on the same side). As hard as I could try; it seems that the hand throws were half or two thirds the velocity of the mechanical throws. I have thrown over the tree tops by hand: about 80 feet. And 15 meters is only 49 feet. And the velocity of the mechanical throw can probably be doubled by letting it run more before the throw is made.

Pumpkin chucking is to restrictive: in a field I would use metal missiles not pumpkins. Check out the golf ball guys; I think they were up to 1100 feet.

[daxwc]

Ralph, "that's all I can stands, I can't stands no more"; Ralph you are not welcome on this thread, please do not post on this thread 8P 8P 8P

http://www.youtube.com/watch?v=h97kbv4mbsc

[pequaide]

Daxwc: I am fearful of missing the great contributions you may make to the field of science: but I will fight the feeling. You are on the non-read list. Please don't post on this thread.

[Wubbly]

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Kane and peq got different results because peq couldn't see the mr^2 relationship that is there.

Wubbly and peq got different results because wub found no extra energy in the increased momentum.

Kaine and I got similar results in that neither of us found any extra energy in an Atwoods.
.

[Wubbly]

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People have looked into the Atwoods and found no energy there.

People have looked into the C&S and found no energy there.

The fact that peq still claims he creates energy with these proves he is a fraud.

He fooled us for a while, but no longer.

This thread does not belong in a private forum. It belongs in the fraud section.

[rlortie]

Peq, wrote;

[pequaide]

I took the 19 inch diameter wheel outside today and threw the .250 kg BB bag as I did in the lab. This throw has a nearly flat trajectory with only a slight rise. The bag traveled 15 meters before it struck the ground. It was released above the ground at about 1.8 meters. So in the time it takes to drop 2.3 meters it travels 15 meters. This seems fairly consistent with the guess of 20 m/sec.

This is a drop of 4.4 cm for 41.4 kilograms. You are allowed 47 kilograms dropped 8 cm.

[Art]

When the 250 g is actually in flight I think your energy calculations are probably correct .

However when the weight transfers its energy on reaching the ground you can only ever recoup half of its momentum because of 1/2 MV^2 and Newtons 3rd Law.

The time it takes in flight is governed by the acceleration of gravity and the speed (distance travelled) is governed by the "leverage" produced at the driving wt . Of the energy produced by the 41 kg falling 4.4cm only half is transferred to to the 250 g and then only half of that can be recovered . I think the overall loss comes in at 25 %

If atmospheric pressure supplied the energy "free" then 25 % gain would be ok : )

_________________
Have had the solution to Bessler's Wheel approximately monthly for over 30 years ! But next month is "The One" !

[pequaide]

But we have more energy than you propose. Input energy is in Nm 41.4 kg * 9.81 N/ kg * .044 meters = 17.87 joules. Output energy is 1/2mv² =.5 * .250 kg * 20 m/sec * 20 m/sec = 50 joules.

[Art]

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I think the problem may be in the two different formulae used to calculate the energy levels of the thrown weight and the driving weight .

The thrown weight is using the energy formula based on momentum ( 1/2 MV^2) and the driving weight energy is being calculated using the torque formula (Newton /metres).

The Units are the same (Joules) but the two formulae are not equivalent .

I think a conversion formula has to be used ie E = magnitude of the torque multiplied by the angle moved (in radians ).

[pequaide]

The meter in Nm is the distance over which the force acts not a lever arm length. Momentum is mv not 1/2mv². Both energy formulas are good formulas but energy is not a conserved quantity.

The two energy formulas are equivalent and it has nothing to do with torque. One is motion energy and the other is potential energy.

[pequaide]

Art: gravity exerts 9.81 newtons of force upon a one kilogram object. If the one kilogram mass is 1 meter above the ground then it has 9.81 N * 1 meter = 9.81 joules of energy; because the force can be exerted over a distance of one meter.

If the one kilogram objects is dropped 1 meter it will have a velocity of the square root of 2 times the distance dropped times the acceleration = 4.429 m/sec. Put 4.429 m/sec into the Kinetic energy formula and you get 9.81 joules.

So the two energy formulas do give you the same results.

So how can 41.4 kilogram of mass dropped 4.4 cm, which has 17.68 joules of energy, throws a .250 kilogram mass 20 m/sec for 50 joules.

The 19 inch wheel with the .75 shaft throws the .250 kilogram BB bag 5 meters in the lab. Using a stop watch; the time it takes to cover that distance is about .22 second. 5/.22 = 22.7 m/sec.

Just to make it clear: The 41.4 kilogram of mass dropped 4.4 cm, which is 17.68 joules of energy, throws a .250 kilogram mass 20 m/sec for 50 joules.

[pequaide]

I was a bit awe struck. I could not keep from laughing the first few times I did each of these steps.

I said I was going to change the 9 o'clock starting position and not let the 250 gram BB bag just hang there waiting for the release of the 41.4 kilograms of drive at the .75 inch shaft position. Well I implemented my plan. I took some 3 inch PVC pipe and made a trough for the 250 gram bags to slide horizontally into the 12 o'clock position so that it was no longer lifted into that position by the drive mass. I was just expecting things to remain the same with maybe a slight increase in velocity. Boy; but was I wrong.

The BB bag shot to the wall with distinctly more violence. The bag had possibly a 2 or 3 meters per second increase over the previous 20 m/sec. There is no question but that there is big time free energy here. I had not included the lift of the .250 kilograms of the bag in the input energy evaluation, I just ignored it for excess to cover margin of error.

The bag was maintaining its horizontal throw so I got real brave.

I placed the bag deeper down the horizontal tube so as to increase the drive rotation by about 30°. I calmed my nerves and let her fly. Again another 2 or 3 meter per second increase. This puts us at about 25 m/sec and above a comfortable velocity for the lab. I think a miss throw would now damage the walls or ceiling of the lab. A good throw hits a curtain suspended at the end of the lab.

The 30° increase it drive rotation puts the input energy at .053 meters * 9.81 Newton/ kilogram * 41.4kg = 21.5 joules. But the output energy is now at about 75 joules.

The throws are now to fast for a stopwatch; you can't pull the trigger on the watch that fast. But it is definitely a decrease from .22 sec.

[Art]

I think the momentum of the thrown weight has to be ½ MV ^2 because it is essentially falling in the gravity field.

And because the potential energy is acting through the radius of 0.75/2 inch I think it becomes a torque calculation right through until the 250g weight loses contact with the tether .

Also the force on the thrown weight acts through a radius to the same pivot point as the driving weight and that would have to be a torque calculation .

Which makes calculating whether energy is coming in mathematically too complicated for me .
( I’m OK on percentages though : ) and I’m full of admiration for those that can do it .

Just a thought :
There is as much gravitational force in 6 square inches of atmosphere as there is in 41kg of air ( or water or lead ).

And it can be easily thrown ( sort of )

And its everywhere .