12 times more output than input, dual mechanical oscillation system !

[johnny874]

@Cloxxki . No worries , no offence taken . Raymond Head admits that he is not the worlds best mathematician , and indeed has asked for help on at least one of his videos . But the great thing is that he gets stuck in and builds things . I have come to the conclusion that even those who claim to be professional mathematicians  are still confused about this device . Mathematical proofs of the OU of the 2SO are a dime a dozen .But where is the self runner . At least Jovan Marjanovic , the mathematician has built small wooden models .At this time , I am fairly sure that the 2so is not overunity . I am not sure if it might be using a rotating pendulum , and so, rather than try to calculate , I intend to do practical tests . I had a sudden burst of energy today , and made a start on my new rotary model . I work on a zero budget using 100% recycled parts . I found a nice piece of timber yesterday whilst out cycling , and that inspired me to get started. I also did an experiment today . I inverted my bicycle and added a 1 KG weight to the front wheel . I put the bathroom scales under the handlebar . At bottom dead centre , the scales showed a weight gain of about 4 Kg . Watch this space .
Regards to all , Ken .

   Neptune,
>>   I also did an experiment today . I inverted my bicycle and added a 1 KG weight to the front wheel . I put the bathroom scales under the handlebar . At bottom dead centre , the scales showed a weight gain of about 4 Kg . Watch this space . <<

  NOT POSSIBLE ! You posted the math yourself that shows that centrifugal force is relative to mass. If you noticed, the mv^2 was Liebniz's calculation for momentum. As such, for a 1kg weight to have 2 kg's of force, the math you said supports you states that it would have to fall 9.87 meters. To have the force of 4 kg's, it would have to fall 2 x 9.87 meters or 19.74 meters to have the force you measured.
I think this is why some like myself have taken the time to learn how math and science either supports or does not support our efforts. It's a time saver and leads to better idea's.

[Cloxxki]

   Neptune,
>>   I also did an experiment today . I inverted my bicycle and added a 1 KG weight to the front wheel . I put the bathroom scales under the handlebar . At bottom dead centre , the scales showed a weight gain of about 4 Kg . Watch this space . <<

  NOT POSSIBLE ! You posted the math yourself that shows that centrifugal force is relative to mass. If you noticed, the mv^2 was Liebniz's calculation for momentum. As such, for a 1kg weight to have 2 kg's of force, the math you said supports you states that it would have to fall 9.87 meters. To have the force of 4 kg's, it would have to fall 2 x 9.87 meters or 19.74 meters to have the force you measured.
I think this is why some like myself have taken the time to learn how math and science either supports or does not support our efforts. It's a time saver and leads to better idea's.

Jim, without knowing the turning speed of the wheel, you can't say it's impossible to read 4kgs.
If the scales would have datalogging with infinitely high resolution and refresh rate, I can tell you over the duration of one full cycle, it would register exactly 1kg providing consistant cycles.   
In Neptune's case, the rest of the setup (bike) probably weighed enough to keep it grounded. A lightweight setup would just bounce, or when fixed to the scale, register negative weight for part of the cycle.

Take a simple 1kg weight, suspend it from a spring fish scale, and press some clay onto the gauge. Give it some really good swings either in the horizontal or vertical plane, and read the gauge's clay impression. Be prepared to be impressed, end suffer a tennis arm injury in the process.

[Cloxxki]

Neptune, I think scientists and pseudo-scientist think too hard about the 2SO.

As I see it now, a physics teacher in college should be able to present the 2SO to students, and ask them to express it in formulae. I am tempted to do it myself, and consider I would have to dig up each and every part-formulae from the internet, by utter lack of knowledge.
It really is not that complicated, and for initial math approach, the crossbar could be enhanced to have the pivots on either ends be restricted to move only in a vertical plane, with constant crossbar length.

Important starting point is the CW. Is it supported or suspended freely. weight of pendulum and CW, and lenth of pendulul en the 2 pivots from the central crossbar pivot.

I think it only gets complicated when the professor hints students the answer will prove significant overunity and conservation of energy as a result will not get them a pass for the exam. Then they need to get into fuzzy math. Answer are to range from 3.75 to 12x overunity, it turns out :-)

[johnny874]

Jim, without knowing the turning speed of the wheel, you can't say it's impossible to read 4kgs.
If the scales would have datalogging with infinitely high resolution and refresh rate, I can tell you over the duration of one full cycle, it would register exactly 1kg providing consistant cycles.   
In Neptune's case, the rest of the setup (bike) probably weighed enough to keep it grounded. A lightweight setup would just bounce, or when fixed to the scale, register negative weight for part of the cycle.

Take a simple 1kg weight, suspend it from a spring fish scale, and press some clay onto the gauge. Give it some really good swings either in the horizontal or vertical plane, and read the gauge's clay impression. Be prepared to be impressed, end suffer a tennis arm injury in the process.

   Cloxxki,
If he spun the bicycle wheel by hand to achieve a velocity of 9.87m/s, it is on him to make that statement. He did not. A weight falling while attached to a bicycle wheel with a radius of 25cm could not reach such a velocity.
And if he wishes to be critical of people who use math, his "experiments" need to be accomplished adhering to some type of standard.

                                                                                                                 Jim

[Cloxxki]

   Cloxxki,
If he spun the bicycle wheel by hand to achieve a velocity of 9.87m/s, it is on him to make that statement. He did not. A weight falling while attached to a bicycle wheel with a radius of 25cm could not reach such a velocity.
And if he wishes to be critical of people who use math, his "experiments" need to be accomplished adhering to some type of standard.

                                                                                                                 Jim

Hi Jim,

Let's start by seeing the example as it is. 1G we get for free, the scale is placed level. That's 1kg right there.

Then, the CF part.
I am going to presume a high speed spin on the wheel. 5m/s at 0,25m from the axle (not a very big wheel, weight sitting against wheel). I am expecting violence from that, having some cycling experience.


Wiki gave me:

Fc= (m*v²)/r
Fc= (1 * 5²)/0.25 = 100N
100N ~ 10kg.
Plus the one for a vertically placed scale, makes ~11kg of scale pressure.

I tend to mess up math, so please correct me.

Please note the difference between KE from an earth's free fall and the CF from a weight taking a sharp turn.
Consider that a weight falling to a little wheel at 5m/s downward, and existing 5m/s upward, accelerated vertically in the time it takes to make half a revolution. Wet finger in the wind math: pi*0.25=0,8m on the wheel during half revolution, or 0.16s. Average (!) vertical acceleration is 10/0.16=60m/s or >5x G extra on top of the 5m/s it had.

My gut tells me the peak pressure at top bottom, in fact identical to the constantly spun weight, will amount to twice the acceleration because often 2's end up in useful formulae. I'd have to dig it up, though, but there'd offer the 11 figure I found above for you. :-)

Calculating back the velocity Neptune's wheel was spinning for you is a matter of reversing the formula. I am getting ~2.73m/s right now.

Fc= (1*2.73²)/.25=7.5*4=30N (~3kg). Add the 1kg for the level scales and there's your 4kg peak pressure.