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Author Topic: re: energy producing experiments  (Read 43944 times)

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #180 on: January 20, 2020, 04:37:31 PM »
In an Atwood's 8.75 grams can accelerate a balanced mass of 1310 grams to a velocity of .09128 m/sec after the 8.75 grams has dropped .064 meters. This occurs when the accelerated mass of  1310 grams is at the same radius as the 8.75 g drive (the mass that provides the force) mass. In this experiment the drive mass is at a 10 mm radius; and the 1310 grams of balanced mass is at the same location.

In an Atwood's 8.75 grams can accelerate a balanced mass of 135 grams to a velocity of .885 m/sec after the 8.75 grams has dropped .064 meters. This occurs when the accelerated mass of  135 grams is at a radius 97 mm while the 8.75 g drive mass remains at 10 mm. The 8.75 grams is still moving .09128 m/sec and the 135 g is moving 9.7 times faster.

The 1310 grams has a momentum of 1.310 kg * .09128 m/sec = .11958 units

The 135 grams has a momentum of .135 kg * .885 m/sec = .11948 units

The 8.75 grams has dropped .064 m; To restore its original position you would have to throw it upwards .064 m. A velocity of 1.12 m/sec is needed to throw a mass upwards .064 m. So .00875 kg * 1.12 m/sec is the needed momentum to restore the original position of the drive mass.

The momentum needed to restore the original position of the 8.75 grams is .00875 kg * 1.1206 m/sec =  .0098005 units

The quantity of momentum available in the 1310 grams or the 135 grams is 12.19 times that which is necessary to recycle the system.  .11952  / .009805

This is possible because the time over which the force acts in a free fall of .064 m is .114 sec.; and the time over which the force acts to produce the motion of the 1310 grams or the 135 grams is 1.4023 seconds.    1.4023 sec  / .114 sec = 12.33 

The 1310 is at a radius of 10 mm. And 1/9.7ths of that mass (135 g) is at 9.7 times that radial distance (at 97 mm).

What would prevent us from placing 1/12.28ths the mass (106.67 g) at 12.28 times (122.8 mm) the distance?

The 106.67 grams would be moving 12.28 times faster than the 8.75 grams moving at .09128 m/sec.  You would have 106.67 grams moving 1.12 m/sec.

106.67 grams moving (.09128 m/sec * 12.28) 1.12 m/sec would have .10667 * 1.12 m/sec = .11947 units of momentum.

The 1.12 m/sec velocity is sufficient to reload the system and you have 106.67 grams instead of 8.75 grams.

Or how about 26.2 grams at a radius of 500 mm?    50 * .09128 m/sec = 4.564 m/sec 

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Re: re: energy producing experiments
« Reply #180 on: January 20, 2020, 04:37:31 PM »

Offline Kator01

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Re: re: energy producing experiments
« Reply #181 on: January 25, 2020, 01:41:32 AM »
Delburt,
wow, 12 years have passed by...you remember ?
For those here to better understand how the dynamics of your system works we go back 12 years where you posted the pictures of the evolving motion in series including description: go to reply#116, January 18th

https://overunity.com/1995/free-energy-from-gravitation-using-newtonian-physic/105/
I am still scratching my head...to find a technical solution
Mike

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #182 on: January 25, 2020, 03:45:20 AM »
A pulley with two radius lengths can be used to make energy. As in a double Atwoods. When the ratio of the length of the long radius to the length of the short drive radius is greater than the square root of the mass ratio between the accelerated mass over the drive mass. The speed of the accelerated mass throws it beyond the distance dropped.

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Re: re: energy producing experiments
« Reply #182 on: January 25, 2020, 03:45:20 AM »
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Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #183 on: January 25, 2020, 04:32:25 AM »
The ratio of accelerated mass to drive mass would be; as in the above experiment 1318.75 g / 8.75 g.  And then you place a 1 / 12.28 mass at 12.28 times the distance.

Offline Kator01

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Re: re: energy producing experiments
« Reply #184 on: January 26, 2020, 09:09:57 PM »
Delburt,
found this simulation-applet, however you can not vary the radius. I myself can not run the simulation , my Linux-System misses the Java-code
https://demonstrations.wolfram.com/DoubleAtwoodMachine/
Now concerning this guy here:
https://www.asc.ohio-state.edu/durkin.2/treb2.htm
Lets do some energy calculations:
Mass of the Brick 1.5 kg, g 0.98 kg*m/sec exp2,
Mass of golfball 0.046 Kg, speed 16 m/s
From the video I only can guess the hight of the falling brick is about 0.8 m
Potential energy of the brick : m * g + h = 1.5 x 9.8 x 0.8 = 11.76 JEnergy of the golfball : 1/2x 0.046 x 16 exp2 = 5.88 J
So this system is not showing energy-increase. I assume it misses the right point in time to release the ball.He mentioned that - accoding to his observation - the brick almost stood still at times while falling.
I would really find more about this Mr. Gordon here : http://garydgordon.com/Despin/
especially how he connected the pucs at the rim so they could be catapulted tangentially.
Good find anyway

Mike



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Re: re: energy producing experiments
« Reply #184 on: January 26, 2020, 09:09:57 PM »
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Offline Kator01

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Re: re: energy producing experiments
« Reply #185 on: January 26, 2020, 10:34:39 PM »

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #186 on: January 28, 2020, 01:11:35 AM »
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%

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Re: re: energy producing experiments
« Reply #186 on: January 28, 2020, 01:11:35 AM »
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Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #187 on: January 29, 2020, 01:27:16 AM »
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. 

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #188 on: February 04, 2020, 03:34:46 AM »
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.

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Re: re: energy producing experiments
« Reply #188 on: February 04, 2020, 03:34:46 AM »
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Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #189 on: February 10, 2020, 07:02:05 PM »
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. 

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #190 on: February 14, 2020, 08:37:09 PM »
By changing the dimensions of this experiment we can get a doable measurable experiment. One hundred meters is too high for an amateur to measure but 10 meters is not. You can tell the difference between 30 feet and 60 feet. I have already throw to the top of our trees, which is 80 to 90 feet.

So lets rearrange the above experiment to throw 10 meters upwards. This is also the height of the stack of drive masses.

And lets let the drive distance drop be .0667 meters. Lets place one kilogram in this drive distance of .0667 meters. One meter can be divided by .0667, which shows that 15 one kilogram masses can be evenly spaced along the distance of one meter.

There are 10 meters for such 15 mass units so the drive mass is 150 kilograms.

The rotational inertia of the wheel should be about 19 times (19 * 150 kg = 2850 kg) this is so that the total mass accelerated is 20 times the quantity of the drive mass. This 20 to one will give us an acceleration of .4905 m/sec².    9.81 m/sec² / 20 = .4905 m/sec²

So the speed (at the shaft) after the drive mass has dropped .0667 meters is going to be about .2557 meter/sec.  The square root of (.0667 m * 2 * .4905 m/sec²) = .2557 m/sec.

Now this is 3000 kilogram moving .2557 m/sec which is 767.1 units of momentum.

One kilogram moving 14.007 will rise 10 meters. This is 14.007 units of momentum. One kilogram rising ten meter will restore the configuration of the stack; which has dropped .0667 meters.

So the system can reconfigure with 14.007 units of momentum and it produces 767.1 units.

Now we need to apply the radius to mass rule for the rotational inertia of spinning objects. If the accelerated mass (3000 kg) is at ten times the radius of the shaft then the wheel only needs to have a mass of 300 kilograms. But this 300 kg is moving 2.557 m/sec. For the same 767.1 units of momentum.

The thrown mass is 1 kilogram: which has come off the bottom of the 10 meter stack.

So: as in the cylinder and spheres experiment we attach two ½ kilograms masses to the outside of the wheel. We spin the wheel and release the spheres on their tethers and one of them will rise up to the tree tops. And the other sphere will thump into the ground; I have done it many times.

But not with a 300 kilogram rim mass wheel. So lets start downsizing.

I use 152 gram one inch spheres. Two of these are 304 grams instead of  the 1000 grams in the above experiment. This means that the 300 kilograms for the one kilogram above becomes 91.2 kg for the .304 grams of thrown mass.      .304 * 300 kg = 91.2 kg.

Using a thrown mass of .304 kg give a drive mass of .304 kg * 150 = 45.6 kilograms. And an accelerated mass of 912 kg and a momentum of 912 kg * .2557 m/sec = 233.19 units.

This means that I can walk into the yard and spin a 91.2 kilogram rim up to 2.557 m/sec and see if one .152 kg mass can rise 10 meters up.

Well maybe some of you are 91 kilogram hombres; but I am more like a 3 kilogram guy. But if I use a 3 kilogram wheel does that mean that I can spin it 30 times faster? 91.2 kilograms moving 2.557 m/sec is 233.1 units of momentum. And 3 kilograms moving 76.71 m/sec is also 233.1 units of momentum.

So this means I can take a 3 kilogram rim that has two 152 grams masses attached, in the cylinder and spheres fashion, and I can spin it as fast as I can. And after I release the spheres when one of them reaches the tops of the trees then it proves that the .0667 meter drop can be repeated. The spheres have more energy (½ * 1 kg * 14.007 m/sec * 14.007 m/sec  + Extra meters) than  (½ * 3000 kg *.2557 m/sec .2557 m/sec) when they rise higher than 10 meters.

Did you notice that the 3 kilogram wheel already has a velocity of over 5 times faster than the spheres would need to rise 10 meters?

Okay lets check some of the relationships to see if I am staying consistent.

The stack of one kilogram masses has 767.1 units of momentum for the one thrown kilogram to work with; and a thrown mass of .304 kg has 233; checks

The one kilogram needs 14.007 units of momentum to cycle and the .304 kg needs 4.26

300 kilogram moving 2.557 m/sec has 767.1 units of momentum: and  3 kilograms moving 77.6 m/sec has 233 units   14.007 / 767.1 and 4.26 / 233 is only  2% to cycle.

Baseballs have a mass of 142 grams and they make great tree ornaments. 

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Re: re: energy producing experiments
« Reply #190 on: February 14, 2020, 08:37:09 PM »
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