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The spoon,Bored Chemist,Both of you are not understanding is that you are taking it in wrong direction.Let me clarify all things.The scale length is 2 meter.There is a 2 or 4 meter long tube.Let's take tube length 2 meter.This 2 meter long tube is mounted on left arm of seesaw in middle so it's 1 meter part is above fulcrum and 1 meter is below fulcrum.A 10 kg.ball is resting at the bottom of the tube.Now counterweight.After calculating distance from fulcrum the counterweight will be 14.4 kg.to balance the seesaw as the distance of ball is 1.44 meter from fulcrum as the ball is in rest position in the tube.Now the seesaw is in balanced position. But I will take counterweight 20 kg.due to this excess counterweight the seesaw will be in tilting position at 90 degree angle.Now when I lift up the seesaw then I will not have to calculate the input energy of the 1 meter distance of tube which is mounted below fulcrum as I have added mass in counterweight.Now calculate input using mgh formulaMgh=10*10*1=100 JouleBut calculate output as ball will fall down from 2 meter heightMgh= 10*10*2=200 JouleBall will again fall down from 2 meter height at the time of reversingMgh=10*10*2=200 JouleSo total output is 400 Joule but input is 100 Joule.The very interesting point is that it is not important that the device must complete one cycle as ball has completed one cycle.See the link.you can see momentum in the video.//www.youtube.com/watch?v=TQr-MR2yj_U

The spoon,What I told you is very much correct.there will be no need of lifting the ball from 2 meter height as I have added the mass in counterweight.so the input will be required to lift the 1 meter only.It is logically correct.if I will lift up the ball from 2 meter then why I will increase the counterweight?counterweight must be 10 kg .but no ,counterweight will be increased to balance the seesaw .so there will be no need of lifting up the seesaw.

The spoon,See the sketch carefully .in the sketch the tube is mounted in middle with left arm so the ball is resting at the bottom of tube.the 1 part of tube which is above fulcrum will be not calculated as so the height will be only 1 meter but ball will fall down 2 meter.you can consult with any one on this point.the 1 meter part of tube mounted above fulcrum has nothing to do with input but output..

What I have not proven?these two video are sufficient to prove Overunity in this mechanism.If there is no overunity then the formulas of mgh and 0.5 mv^2 are wrong.

The main purpose is to get more output than input.

What I have not proven?

Bored chemist,The main purpose is to get more output than input.But tell me one thing if it move continuously then can I get more output than input.No.the output will be same in both condition if it moves continuously or it stop and will need to push it up again and again.

Like I said, you can't challenge me on time because I am the ''master'' on time and space and if Einstein was here today he would concede to me.