# It has begun... [Debunked]

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Yo guys,

Long time no see. I came back to see how all my people doing. Guess what? Prepare yourself with alot of mind. It's going to be on the basis of first come first serve once this is release. I've worked it all out "the correct current behavior of nature" . It's in mathematic proof, easily done experimentally, and only require integration calculus to actually work it yourself and verify. Well, for sneak peak, it will tear the first law of thermodynamic into half (literally) and connect all science phenomenas down to high school level. Probably can reduce quantum mechanic works into ... I don't know, a few thousand times?

Well, I'm trying to figure out how to do this. I emailed my professor and ..1st of all, hopefully he doesn't call in a psycho analyzing team, which I think I gave him alot of respect and made good grade on his exams. 2nd, I don't even know if he can do much even after he agrees. If that fails or even not, I guess I have to use the people in here to help me. Hopefull I didn't make any enemy and everyone loves me. lol

So what do you say guys? Help me, help you? Well, this process going to take a while and this is my favorite science forum (since other forums keeps kicking me out for some reason ) so you guys will be the leading team. But seriously, will you willing to help me ? I promise this prove is 100% legit and it will brings us to the age of truth. Are you with me?

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This entire post is irrelevant. Theories are judged upon the merit of the evidence supplied, not in regards to how much time you put into them or who else agrees with you.

Be serious and put up your theory so we can start going over the process and the proof; this post has nothing to contribute other than setting up one big appeal to authority, which is a logical fallacy, and is unacceptable anyways.

~moo

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self congratulatory praise is no recommendation, so just post it and lose the nonsense!

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Um... This is not a theory, and with your negative attitudes, I'm not exactly sure how far can we get. I need atleast 5 people who have positive attitude before I post it.

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do you also need them to sacrifice a goat?

seriously, if you sent a letter off to a peer review journal saying ' i have a theory of everything that works in all cases, but i'm not going to tell you it' you'll just be laughed at and likely not even get a response.

there is a very simple rule which should be applied to cases like this the rule is as follow:

Put up or Shut up.

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Um... This is not a theory, and with your negative attitudes, I'm not exactly sure how far can we get. I need atleast 5 people who have positive attitude before I post it.
What you're asking is getting a negative response because it's like asking someone to take a job or buy a house without checking it out first. You wouldn't become my business partner and invest your time and money without seeing my plan first, would you? Can you please post what you have so it can be checked out?
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Here's a question - how can you expect us to believe you've calculated some sort of 'Theory of Everything' when you can't even type coherent sentences?

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(since other forums keeps kicking me out for some reason )

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I need atleast 5 people who have positive attitude before I post it.

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I understand now.... I'm not the one who going to save this world... you guys are. I beg you... save this world.

In circular motion, 1 unit of force/work/energy in the centrifugal direction will amplify by a factor of 2 in the tangent direction.

This is the proof:

A ball mass M tie to a string is in circular motion. The tension force on the string is m(v1^2/r) where v1 is the tangent velocity. Work is being done by the string to pull the mass inward. Say that the radius is now half the original radius. Applying conservation of angular momentum L=mvr, where L is the angular momentum. v2 is now twice the orginal v.

The initial energy of the ball is 1/2mv1^2, the final energy of the ball is 1/2m(2v1)^2 which is 2mv1^2. So the energy in the final position is 4 times the energy of the initial position, which shows the different of 3 times the original energy.

The work done by the string is tension force F multiply by the differential radius distance. Tension force F= mv^2/r . Work = mv^2/r (dr). V is a function of angular momentum, rewriting v=L/mr . Therefore mL^2/(mr)^2 (1/r)x dr is the work. The integration becomes (mL^2/(m^2*r^3) )dr from r to 1/2 r . Since L and m are constants, the integration only in volve 1/r^3 which equals to -1/2r^2 +C . The final form is -mL^2/2(m^2)r^2 + C from r to 1/2r . When r is at infinity, the tension would be zero, thent he work is zero, therefore constant C is zero. Solving by substitude the value r and 1/2 r gets -mL^2/2m^2((1/r)^2 - 4/r^2) = 3/2mL^2/(2m^2r^2). The reason I haven't cancel out the m because now one can see L^2/(2m^2r^2) is just v1^2 . Simplifly this to 3/2*mv1^2 . This is the work done by the string. Compare the ratio of work done by the string and the energy gain tangent direction 3mv1^2 / (3/2)*mv1^2 = 2 .

Special notes: 3/2mv1^2 , the number 3/2 have nothing to do with kinetic energy. I only use momentum/force/distance to derive this. It just represent the amount of work done by tension. So the ratio is just an amount comparison and independant of what means by kinetic energy. To my believe, the kinetic energy equation should be E=mv^2 as originally state by Lebniz.

This means that each unit of work/force/energy input by the tension will result in 2 units of work/force/energy in the tangent direction. The reason I write work/force/energy because I'm not sure what work/force/energy should be define anymore.

If one remember Coriolis acceleration 2wv where w=angular velcocity= v/r. Then coriolis accelration is 2v^2/r. So anytime a force apply to move the mass inward at velocity v, the tangent force is 2v^2/r , which is twice the tension force v^2/r. Thrust is Ft, so for the same amount of thrust provide by the radial direction, twice the thrust achieve in the tangent direction.

I believe Coriolis recognize this and has been forced by society to change kinetic energy to 1/2mv^2. It is an act of greed by us.

Please... this is important for you, for me, for all of us. I'm just a guy who wants to live a happy life and I know what I have yesterday,today, and tomorrow isn't comes from me, it comes from all of you.

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A ball mass M tie to a string is in circular motion. The tension force on the string is m(v1^2/r) where v1 is the tangent velocity. Work is being done by the string to pull the mass inward.

No work is performed in circular motion. The force is always normal to the path, so $\mathbf F \cdot d\mathbf l \equiv 0$.

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I think he means that work is being done to decrease the radius of the circular path, not to keep it in the same path.

Thank you Cap'n.

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I think he means that work is being done to decrease the radius of the circular path, not to keep it in the same path.

So right.

In that case the tension on the string is greater than that needed to maintain circular motion. If the tension were just F=mv^2/r the ball would not move inward. Assuming that the force is still mv^2/r leads to a nonsense result:

This means that each unit of work/force/energy input by the tension will result in 2 units of work/force/energy in the tangent direction.

Try to patent this idea. (Hint: The Patent Office rejects all claims to perpetual motion machines without prejudice.)

Edit

Well, I sure do feel stupid.

The work done in making a ball undergoing circular motion move from radius of r to r/2 is indeed $\frac 3 2 \frac {L^2}{mr^2}$. To see this is so, imagine that the force is infinitesimally greater than the force needed to maintain circular motion. That infinitesimal extra force will pull the string in very slowly, and will not contribute to the work (it is infinitesimally small).

How about the kinetic energy? The kinetic energy at some radius r is $\frac 1 2 \frac {L^2}{mr^2}$. Assuming angular momentum is conserved, the kinetic energy at half that radial distance will be $\frac 1 2 \frac {L^2}{m(r/2)^2} = 2\,\frac {L^2}{mr^2}$.

The difference between the kinetic energy at radius r and radius r/2: $\Delta KE = \frac 3 2 \frac {L^2}{mr^2}$

which of course is equal to the work performed.

Edited by D H
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Figure Skaters have been doing this for centuries, when you want to increase the spin rate pull your arms and leg in.

there`s nothing New here.

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DH

Thank you for following the math closely. I know it takes you time and effort, but I hope you enjoy it. Actually you're right. It boils down to

3(KE)/(3/2)mv1^2 = ?

If one say KE=1/2mv^2, then the ratio is one. If one say KE=mv^2, then the ratio is 2. I know KE=mv^2, but I have a hard time finding more clues to prove this so. I know for a fact this is not a question of logic anymore. It's either you believe or not believe. Through Lebniz(ke=mv^2), Eistein (E=mc^2), the earth, the moon, the solar system, cold fusion, electromagnetism, lift, and all science phenomena can be explain with ease, but the logic blockage is here and only here can one access all others. I know it's unfair to ask a person to make a decision weather it's 1/2mv^2 or mv2. I can only gives you similarity so you can self realize. Must this be the question this era must face....."to believe or not to believe". I know god is rolling dice on me. I had the proof yesterday and doesn't remember I assumed mv^2 is KE.

My best clue is that if one treat kinetic energy as a force when it collide, then as proven earlier, the kinetic energy at r/2 is 4 times the initial kinetic energy. The intial force centrifugal force mv^2/r is 8 times of that when it's at r/2 . The ratio is 8/4 which is 2. Sigh... this is meant to be, why is he making this so hard.

YT,

yes, it's nothing new. The skater just doesn't know that the amount of work/energy he use to pull himself in putout twice the amount of work/energy as rotation. Or does he...

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Let me save you a lot of trouble: http://tutor4physics.com/workenergy.htm

There's no "maybes" and "perhapses" in the formula for Kinetic Energy, Potential Energy, Work or any of those formulas.

Read 'em, Use 'em, stop trying to invent the wheel on your own whim, and if you must, then at the very least prove why you're using what.

~moo

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If you are going to try to reinvent the wheel, at least make sure that it is better than the previous ones. The same applies to scientific theories.

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I don't know what to say. I figure it out now. It's not that who's right and who's wrong. It's logic. Logic is not truth. Logic is creativity. I can keep 1/2mv^2 and still can make 2 parts of energy for 1 part and that still doesn't violate the first law of thermodynamics. I see what Godel's paradox talking about now. It's not the world that needs to be save... it's me that need saving.

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So, to sum up:

You thought you invented a perpetual motion machine, which you call a "theory of everything" (?) that will usher in a new "age of truth" that somehow, everyone else missed despite involving only a simple application of high-school level classical mechanics. However, it turns out you just left out a "1/2" coefficient in your equations.

If this is for real (as opposed to you just trying to mess with people), then I think there are some badly needed lessons you could draw from this experience.

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I don't know what to say. I figure it out now. It's not that who's right and who's wrong. It's logic. Logic is not truth. Logic is creativity. I can keep 1/2mv^2 and still can make 2 parts of energy for 1 part and that still doesn't violate the first law of thermodynamics. I see what Godel's paradox talking about now. It's not the world that needs to be save... it's me that need saving.
I congratulate you for trying, I congratulate you for your noble ideals, and most of all I congratulate you for being able to see where you went wrong. That is rare in the Speculations section, where so many theorists refuse to admit they made a mistake.

Don't stop trying.

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