re: energy producing experiments

[Kator01]

now here another puzzle comes along
https://sciencedemonstrations.fas.harvard.edu/presentations/double-atwoods-machine
Mike

[Delburt Phend]

I need to specify what I mean by a double Atwood's: It is a pulley with two radii, one for the accelerated mass and one for the drive mass. Maybe I should say two radii Atwood's.

I added 245.6 grams at the larger circumference (instead of 135 g) in the two radii Atwood's. I had to add 17.5 grams at the drive mass location of the shaft to get the gate time back to .0291 seconds. This was off of F = ma by 10% instead of the 5% for the 135 grams, and 8.75 grams.  Still: to conserve energy I would have had to added 134.7 grams at the shaft location. One or two grams off is a lot closer than 134 grams.

The energy of the 17.5 grams dropped .064 meters is .010987 joules

The energy of 245.6 grams moving .83 m/sec is .08459 joules   

.08459 J / .010987 J =  7.699   770%

[Delburt Phend]

I set up the double radii Atwood's again; because I intend to extend the legs, so it will be out of service for a while.

In this experiment I placed 200 grams at the drive mass location on the shaft. The gate was tripping at .0299 sec with no mass placed on the circumference.
 
After I placed 245.6 g of balanced (122.8 grams both sides) mass on the circumference the gate trip time increased to .0319 seconds.

To get the gate trip time back to .0299 seconds I had to add 14.58 grams at the drive mass location on the shaft.

So 14.58 grams can accelerate 245.6 grams to .8217 m/sec after the 14.58 grams has dropped .064 m.

The 245.6  grams at 97 mm radius has the same momentum (245.6 g * .027 m / .0299 sec * (97 mm / gate radius = .8127 m/sec)) as 2382.3 ( 2382.3 g * .8127 m/sec * 10 mm / 97 mm) grams moving at the shaft. Previous experiments showed that the two masses (at their locations) would cause the same rate of rotation. Even though the 245.6 grams is at 97 mm it is the same as 14.58 grams accelerating 2396.88 (2382.3 + 14.58 g) grams at the shaft.

So a perfect F = ma at the shaft would have a velocity of: the square root of (.064 m * 2 * 14.58 g / 2396.88 g * 9.81 m/sec) = .08739 m/sec which would give you: .08739 * 9.7 =  .848 m/sec at the circumference.

About 4% error:   .848 m/sec / .8127 m/sec

I think this proves that: no matter what the radius of the mass; you will get the same amount of Newtonian  momentum produced per unit period of time in a double radius Atwood's.  And the rate of rotation will remain the same as long as the quantity of mass is inversely proportional the the length of the radius.

And what is the practical application?       To throw this momentum into a smaller mass.  Or make the second radius large enough that you do not have to throw at all. That is why the next step is to increase the length of the Atwood's mounts. 

[Delburt Phend]

I made a mistake. The real speeds (that were determined by the photo gate trip times) are off because I had the Photo gate in the wrong mode. I thought the time started at one photo gate and ended at the other photo gate. This is true in some modes but not in the simple (gate mode) that I had selected.

The gate was recording the time over which the single gate was started when one side of the flag blocked the gate and then it stopped when the other side of the flag passed. But the flag is a 3/8 inch dowel pin.

So in any situation were I used .027 mm for the gate distance will be incorrect. I have to reevaluate everything I have done for the last few weeks. The ratios concerning radius to mass will remain the same; because the relationship of the gate times will remain the same. But raw speed  and F = ma %s will be inaccurate.  Sorry: I had totally forgotten how the different modes worked.

[Delburt Phend]

The newest data confirms that the concepts are correct: That Atwood's make momentum. And that that momentum is controlled by the 'mass to inverse radius length' concept of the wheel or pulley. For example: The driving force will accelerate 1 kilogram at a radius of 100 just as easily as it will accelerate 100 kg at a radius of one

Now lets place the current experiment on a commercial scale by multiplying the mass by around a million; and the linear dimensions by 20. The throwing radius would then be 32.6 m and the drive radius would be 200 mm. This shaft diameter of 400 mm does not need to support the mass of the system; it needs only to accelerate it. The accelerated or thrown mass is now 1 metric tons. But what is the drive mass?

Lets use an evenly spaced stack of 100 one ton masses stacked one hundred meters high.

To obtain a velocity of 44.29 m/sec at the 32.6 meter radius (of the throwing wheel) we will need a shaft surface speed of .2717 m/sec. This 44.29 m/sec will allow the bottom one ton mass to be thrown back up to the top of the stack of one hundred.

We can now employ the stacking concept to obtain the drive mass. The machine is capably of throwing the accelerated mass of 1 metric tons up 100 meters. The stack of one hundred 1 ton masses is reconfigured when you throw one mass from the bottom to the top. The drive mass would then be (1 tons * 100) 100 metric tons, in an evenly placed stack 100 meters high.

When we employ a drive mass of 100 metric ton we then need to have a wheel with significant rotational inertia. It could have a rotational inertia of 1900 metric tons. We then have 100 ton accelerating 2000 ton (100 ton + 1900 ton) for a acceleration rate of .4905 m/sec/sec.

With an acceleration of .4905 m/sec/sec a drop distance of .07525 m gives you a final velocity of .2717 m/sec at the shaft; and 44.287 m/sec at the 32.6 m circumference. At 44.29 m/sec the one ton mass will rise 100 meters.  This is a total momentum of (2000 ton *.2717 m/sec) 543,400 units. And the one ton rise 100 meters costs 44,290 units.

So the system can throw one ton from the bottom of the stack back to the top of the stack and it has only dropped 7.525 cm. The drop distance of .07525 m goes into 1 m 13.29 times. You can drop it 13 times and you only need one drop of this distance.

The one ton masses are one meter apart so you have .92475 meters of free energy. Further: after the 1 ton is thrown from the 32.6 m circumference the wheel and stack are still moving .2717 m/sec at the shaft. This is 499,113 units of momentum. As the wheel feeds another one ton mass out to the circumference the speed will drop. But to bring the wheel and stack back up to speed the stack only needs to add 44,290 units of momentum. The stack does not need to restart the wheel every time it throws.

When the second (to be thrown) mass has reaches the circumference at 32.6 meters; the wheel will have slowed to .24955 m/sec at the shaft (before any force has been added). The dropping stack needs only to accelerate the system back to .2717 m/sec at the shaft. This will be achieve with less than 7cm of drop; I would guess around 2cm. You will need giant electric generators to hold back this acceleration.

I began collecting data in a different manner. The flag is a 9.3 mm dowel: the distance between the gates is 27 mm: I place the electronics timer in pulse mode. I am now placing the flag between the gates. The flag comes out from between the gates: the first gate starts the timer as the flag interrupts the gate. After nearly one full rotation: the flag interrupts the second gate and the timer stops.

Now the times are in full seconds: 3.73 seconds is common. The times are the same for a balanced mass of 245.6 grams on the circumference at the 97 mm (radius): and for 2383 grams of balanced mass at the shaft of 10 mm. You can also move the driving force: and that will remain equal if you acknowledge the (one tenth mass at 10 times the radius: or five times the mass at one fifth the radius) radius to mass concept. One ton at 32.6 m is like 163 ton being accelerated at .2 meters.