I did this experiment quite a while ago; but it is easy enough to remember so I will just present it from memory.

I took a board about ½ inch thick by 2 inches wide and 1.1 meters long. I drilled a hole in the center of the board at 55 cm. I cut the hex head off of a ¼ inch bolt and fastened the bolt to the board through the hole. I placed the smooth end of the bolt in a hand held, variable speed, drill.

I slowly began to rotate the board with the drill: it rotated uniformly and smoothly with no wobble. Now lets make some observations.

One: the mass on both sides of the board are equal.

The torque on both sides of the board are equal.

Third: the energy used to rotate the left side and the energy used to rotate the right side are equal.

Now lets continue the experiment by drilling a hole at 10 cm from the point of rotation on the right side of the board. And then place a  half inch bolt, 4 inches long, through the hole fixing the bolt at the center of its mass.

Now slowly begin to rotate the board and bolt; you will quickly notice a distinct wobble. 

The mass on both sides of the board are no longer equal.

The torque on both sides of the board are not equal.

The energy used to rotate the left side and the energy used to rotate the right side are (*probably) not equal.  *The drill is flying around so who know where the energy goes.

Now lets continue the experiment by drilling a hole at 50 cm from the point of rotation on the left side. And then place a quarter inch bolt through the hole and fasten it at the center of mass. This quarter inch bolt should have one fifth the mass of the ½ inch bolt.

I slowly began to rotate the board with the drill: it rotated uniformly and smoothly with no wobble. Now lets make some observations.

One: the mass on both sides of the board are not equal.

The torque on both sides of the board are equal.

Third: the energy used to rotate the left side and the energy used to rotate the right side are equal. The energy on both sides was equal when we just had the board and we had no wobble; and now we are back again to no wobble so surly the energies are equal again.

The small bolt is moving 5 times as fast as the large bolt and it has five time as much energy; but the drill treats the two sides as if they were identical. The board with no weight has no wobble; and the board with these two weights has no wobble. The same amount of energy is used to accelerate the large and small bolt and the small bolt has 5 times as much energy.

The Law of Conservation of Energy is violated by this experiment.

A 10 kilogram mass moving 2 m/sec can give half of its Newtonian momentum to a 1 kilogram mass moving 10 m/sec. The energy loss for the 10 kg is 15 J and the energy gain for the 1 kilogram is 150 J.

A 10 kilogram mass moving 1 m/sec can give all of its Newtonian momentum to a 1 kilogram mass moving 20 m/sec. The energy loss for the 10 kg is 5 J and the energy gain for the 1 kilogram is 250 J.

Math: ½ mv²:  ½ * 10 kg * 2 m/sec * 2m/sec = 20 joules:    ½ * 10 kg * 1 m/sec * 1m/sec = 5 joules:    ½ * 1 kg * 10 m/sec * 10 m/sec = 50 joules:       ½ * 1 kg * 20 m/sec * 20 m/sec = 200 J          ½ * 1 kg * 30 m/sec * 30 m/sec = 450 joules:

Math: mv:   10 kg * 2 m/sec = 20;   10 kg * 1 m/sec = 10;  1 kg * 10 m/sec = 10;  1 kg * 20 m/sec = 20 ;  1 kg * 30 m/sec = 30;

I was in my lab and I saw a piece of extruded aluminum 362 mm long with 29 holes evenly spaced along its length. The holes were slightly elongated down the center line of this bar shaped aluminum. So I thought I would put some numbers to the above experiment.

I fixed a 1/4 bolt in the center hole: and then put this bolt in a hand held variable speed drill.

The bar accelerated smoothly and uniformly as I slowly turned the drill on. 

Then at two holes out (25mm) from the center; I placed a ¼ bolt nut and washers with a mass of about 115 grams.

The drill wobbled distinctly as I turned the drill on.

Then I balance the bar with the two attached bolts; with a third bolt at 14 holes out (175mm) on the other side. This was to the nearest washer (one nut; two washers and one bolt). This came to about 17 grams. And then I placed this back onto the drill.

The bar accelerated smoothly and uniformly as I slowly turned the drill on.

The drill accelerated 17 grams at 175 mm just as easily as it accelerated 115 grams at 25 mm. The drill did not balk at producing more energy on one side than the other side. And all other parts of the bar being equal: If I had ramped up the drill to 380 rpm you would have had .0575 joules on one side and .4165 joules on the other side.

The point is that real world experiments do not show any trend toward conservation of energy. 

And it is important to note that the Newtonian momentum on both sides is equal.

Experiments conserve momentum not energy.

Now it is true that this is not a closed system; but double radius Atwood's experiments perform exactly the same way. They will accelerate smaller objects to higher speeds while conforming to  F = ma.  Not ½ mv²

Good stuff magneat. So you think Chas figured a way to use this for overunity benefit?

Quote from: Toolofcortex on January 09, 2020, 11:18:32 PM
Good stuff magneat. So you think Chas figured a way to use this for overunity benefit?

I am sure that it is.
Did you notice that YouTube is filled with mechanical OU generators made in Asia, Africa, Latin America and other "non-European" countries? there are already tens, if not hundreds of them, on YouTube, despite the fact that they are periodically removed.

imho, this is because the authors either taught physics from other textbooks, or self-taught.

"European" physics textbooks hammered into the heads of people from the 1st grade that the "Law of conservation of energy" is a "sacred cow", no one dares to doubt it. but there are many situations where the "Law of conservation of energy" conflicts with the "Law of conservation of momentum."

Continuing the topic: task
_____________________________________________________________________________
we write down the boundary conditions for the task:

we will consider only uniform, steady movements. transient processes, and hence accelerations at the moments of action of force pulses will not be considered. we are not at all interested in the magnitude of the force F and the duration of the force Δt. we are only interested in the result of the action (F * Δt) - a change in the velocity of the body ΔV.   
those. the "installation" operation algorithm is as follows:
we have a "body" of mass M, which moves uniformly with speed V1.
the "mechanism" periodically acts on the "body" by a force impulse (F * Δt = const).
transients (moments of acceleration) skip.
after the action of the force impulse (F * Δt), the "body" again moves uniformly, but with a different speed V2 = V1 + ΔV
i.e., in accordance with Newton's 1st law, at times when the body moves uniformly, we will have equal inertial reference systems (IRS).
__________________________________________________________________

for simplicity, consider linear motion.
the physics of the process will be the same as for the rotational motion.
there is a body of mass M, there is a certain "mechanism" ("mechanisms)" that is capable of
transmit to the body some impulse of an external force (F * Δt).
to simplify, let all momenta of force be the same (F * Δt = const)
using the "mechanism", the body accelerates to a certain speed V1, while the body
acquires kinetic energy Ek1 = 0.5 * M * V1 ^ 2
if you paint all the formulas:
body weight - M
initial body speed - V1
final body speed - V2
body speed gain - ΔV = V2 - V1
the energy expended by an external force to increase the speed of the body - ΔEst = 0.5 * M * ΔV ^ 2
important point: body speed increased from V1 to V2 (changed by ΔV)
those. for an external force there is no difference with what speed the body moves before and after its impact.
there is such a theorem - Impulse Theorem:
______________________________________________________________________________
The change in the momentum of a mechanical system over a certain period of time is equal to the geometric sum of the elementary impulses of external forces applied to the system over the same period of time.
______________________________________________________________________________

initial kinetic energy of the body - Ek1 = 0.5 * M * V1 ^ 2
kinetic energy of the body after an impulse of an external force - Eк2 = 0.5 * M * V2 ^ 2
increase in kinetic energy of the body:
ΔEk = Ek2 - Ek1 = 0.5 * M * V2 ^ 2 - 0.5 * M * V1 ^ 2 = 0.5 * M * (V2 ^ 2 - V1 ^ 2) = 0.5 * M * ((V1 + ΔV) ^ 2 - V1 ^ 2)) = 0.5 * M * ((V1 ^ 2 + 2 * V1 * ΔV + ΔV ^ 2) - V1 ^ 2)) = 0.5 * M * (2 * V1 * ΔV + ΔV ^ 2) = 0.5 * M * ΔV ^ 2 + M * V1 * ΔV
write again:
ΔEst = 0.5 * M * ΔV ^ 2
ΔEk = 0.5 * M * ΔV ^ 2 + M * V1 * ΔV
ΔEc> ΔEst
_______________________________________________________________________________________
where is the mistake?
_______________________________
p.s. from the formulas it turns out that for any initial speed V1> 0, it should already be OU
in reality, friction losses should be taken into account, losses due to the fact that the impulse of an external force is transmitted with an efficiency of <100%.
but in practice must be V1min, after which OU > 0.
the formula shows OU = M * V1 * ΔV (ideal conditions), in reality minus losses.
Of course, this is all theory
need experimentation.
p.p.s. it seems that all the OU generators where something massive rotates get OU from MECHANICS
and all kinds of coils, spirals, nozzles, etc., are purely engineering solutions for implementation
conditions for obtaining OU...
____________________________________________________________________

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