energy producing experiments

[jim_mich]

The computer program WM2D is programmed from the rules in your books (such as using moment of inertia). If the books are wrong the program is wrong.

I intend to make the equal lever arms out of graphite arrows I will get the materials ordered, but mean will look at this.

Take a ten kilogram mass and place an eye-bolt and multiple pulleys at the top. Above it on the ceiling place an eye-bolt and multiple pulleys. In this why a cord can be tied to the 10 kg mass or ties to the ceiling. The cord can go through the pulley or pulleys at the top and/or the pulley or pulleys on the 10 kilogram mass.

Thread the cord through a top pulley and then tie the cord to the 10 kg mass; you will have to place a 10 kilogram mass on (what we will call) the right side to balance the 10 kilogram mass on the left side. This pulley arrangement will give you no mechanical advantage.

If you then place an extra 20 kilogram mass on the left side, the system will accelerate as if 20 kilograms is accelerating 40 kilograms; for an acceleration of (20/40 * 9.81) 4.905 m/sec². After a drop of 1 meter, for the 20 kilogram mass, you will have a final velocity of 3.312 m/sec. This is (40 * 3.132) 125.3 units of momentum and (.5 * 40 * 3.132 * 3.132) 196.2 joules of energy. Note that the center of mass of the two 10 kilogram masses remains stationary.

Thread the cord through a top pulley and then tread it through a pulley on the 10 kg mass then tie it to the ceiling; you will have to place a 5 kilogram mass on the right side to balance the 10 kilogram mass on the left side. This will give you a mechanical advantage of 2.

If you then place an extra 20 kilogram mass on the left side the system will accelerate as if 20 kilograms is accelerating 40 kilograms; for an acceleration of (20/40 * 9.81) 4.905 m/sec². After a drop of 1 meter, for the 20 kilogram mass, you will have a final velocity of 3.312 m/sec for the 20 kilograms and the 10 kilograms but the 5 kilograms is moving 6.624 m/sec. This is (30 * 3.132 + 5 * 6.264) 125.3 units of momentum and (.5 * 30 * 3.132 * 3.132 + .5 * 5 *6.264 * 6.264) 245.24 joules of energy. Note that the center of mass of the 10 kilogram and the 5 kilogram masses remains stationary.

Thread the cord so as to make a mechanical advantage of 5 then you will have to place 2 kilograms on the right side to balance the 10 kilograms on the left.

If you then place an extra 20 kilogram mass on the left side the system will accelerate as if 20 kilograms is accelerating 40 kilograms; for an acceleration of (20/40 * 9.81) 4.905 m/sec². After a drop of 1 meter, for the 20 kilogram mass, you will have a final velocity of 3.312 m/sec for the 20 kilograms and the 10 kilograms but the 2 kilograms is moving 15.66 m/sec. This is (30 * 3.132 + 2 * 15.66) 125.3 units of momentum and (.5 * 30 * 3.132 * 3.132 + .5 * 2 *15.66 * 15.66) 392.37 joules of energy. Note that the center of mass of the 10 kilogram and the 2 kilogram mass remains stationary.

Thread the cord so as to make a mechanical advantage of 10 then you will have to place 1 kilogram on the right side to balance the 10 kilograms on the left.

If you then place an extra 20 kilogram mass on the left side the system will accelerate as if 20 kilograms is accelerating 40 kilogram; for an acceleration of (20/40 * 9.81) 4.905 m/sec². After a drop of 1 meter, for the 20 kilogram mass, you will have a final velocity of 3.312 m/sec for the 20 kilograms and the 10 kilograms but the 1 kilogram is moving 31.32 m/sec. This is (30 * 3.132 + 1 * 31.32) 125.3 units of momentum and (.5 * 30 * 3.132 * 3.132 + .5 * 2 *31.32 * 31.32) 637.6 joules of energy. Note that the center of mass of the 10 kilogram and the 1 kilogram mass remains stationary.

The final velocity of free fall for one meter is 4.429 m/sec: so the initial energy in all of the situations was (.5 * 20 kg * 4.429 m/sec * 4.429 m/sec) 196.2 joules; and the experiments produced 196.2, 245.24, 392.37, and 637.6 joules of energy.

I believe that the values in red are a typographical error. (I've made such mistakes in the past.)

Maybe I can sum this up in a few words? (I hope pequaide doesn't mind.)

Briefly, pequaide presents four setups involving block and tackle type arrangements. On the left side there is always a 10 kg weight supported by various number of cords. On the right side are various weights of 10 kg, 5 kg, 2 kg, and 1 kg supported by a single cord thus producing ratios of 1:1, 1:2, 1:5, and 1:10. The 10 kg weight on the left side always balances a single weight on the right side because of the pulley ratios.

Then in each example pequaide adds a 20 kg weight to the 10 kg weight on the left side causing all three weights to accelerate. In all four cases the two weights on the left side accelerate downward for 1 meter attaining a velocity of of 3.132 m/second, attaining momentum of 93.96, and attaining 147.14 Joules of kinetic energy.

What happens on the ascending right side is significant. I'll present this as a chart:

Code: Select all

Ratio Weight Velocity  Momentum Joules
 1:1  10 kg  3.132 m/s  31.32   49.04 J
 2:1   5 kg  6.264 m/s  31.32  125.28 J
 5:1   2 kg  15.66 m/s  31.32  245.24 J
10:1   1 kg  31.32 m/s  31.32  490.47 J

The numbers speak for themselves!

[pequaide]

Thank you Jim; it is amazing how you can scan over a number and not see that it is incorrect. I shall try to be more careful.

[Fletcher]

I really have to go now - was just closing down.

See pic & sim below - put together quickly so no time to double check it yet.

Perhaps you'd be so kind ?

The upshot is that although the dropping masses don't travel 1 m for the sim the lifted 2 kg mass does [same relationships].

N.B. FYI the CoG does drop in this sim & this is reflected in the Netted out KE plus PE = Zero Joules.

Unless, in my haste I've got this setup wrong too ?

[pequaide]

WM2D gets several things correct. It predicts that the total potential energy of the 2 kg and 10 kg masses is zero. The center of mass of the 2 and 10 kilogram masses does not move, therefore it has not dropped. You need a drop to have gravitational potential energy.

WM2D says that at a velocity of .9757 m/sec the (2 kg, 10 kg, 20 kg) system has a Kinetic energy of 38.08, which is correct. But where does the velocity come from? The velocity has to come from a drop, and the acceleration of that drop is 4.905 m/sec/sec.

So let’s find out how far the 20 kg was dropped (d = ½ * v²/a). Therefore the 20 kg was dropped a distance of (.5 * .9757 m/sec * .9757 m/sec / 4.905 m/sec/sec) .09704 meters.

Waite a minute: A joule is a newton dropped a meter. Our 20 kilogram exerts a force of (9.81 N/kg * 20 kg) 196.2 newtons. To get 38.08 joules in free fall the 20 kilograms will have to be dropped 38.08/196.2 = .19408 m. But we only dropped it .09704 m in the block and tackle to get the same 38.08 joules.

If we dropped the 20 kilograms .19408 meters in the block and tackle the 30 kilograms would have a velocity of 1.378 m/sec for a kinetic energy of (1/2 * 30 kg * 1.378 m/sec * 1.378 m/sec) + ( ½ * 2 kg * 6.89 m/sec * 6.89 m/sec) = 75.95 joules. This is 75.95/38.08 = 1.9945 times greater than the potential energy of the 20 kilogram at .19408 m.

My predicted energy increase (for the 2 kg block and tackle) in the first post was 392.37 J / 196.2 J = 1.999. So if you use the same drop distance WM2D agrees with my calculations. An energy increase of 1.9945 is equal to 1.999 when rounded to significant figures.

The real problem here is that WM2D just picked drop distances out of the blue sky to get the answers it wanted. It apparently is programmed to do that. The program did not have any problem that the drop distances it picked out of the blue were not even the same.

[broli]

I'm starting to lose hope for this peq. You don't need the concept of moment of inertia to see it fall apart.

Let's consider a torque of 10Nm acting on two setups:

1kg at 10m
10kg at 1m

You say these are equivalent.

Doing some basic math you will find that both masses will undergo the same linear acceleration of 1 m/s/s.

The problem is that in the first this is at 1m radius and 10m in the second.

If you calculate the acceleration of the second from 1m it would give 0.1m/s/s.

So any hanging weight that's causing the torque will accelerate 10 times slower.

This all is using simple linear concepts.

[pequaide]

Make these two lever arms oppose each other in a balanced system.

Set up a quantity of force that would rotate the 10 kg to 1 m/sec, that same quantity of force applied for the same time at the same distance from the pivot point would rotate the 1 kilogram around the circle a 10 m/sec. This would be the same rpm and the same momentum, but the energy is not the same.

Make a balanced rod with a double 10 kg at 1 and then make a double 1 kg at 10 and rotate them at the same rpm. It will take the same amount of force but they will not have the same amount of energy.

How can it take more force to rotate one side of a balanced system than the other side?

[greendoor]

I'm starting to lose hope for this peq. You don't need the concept of moment of inertia to see it fall apart.

Let's consider a torque of 10Nm acting on two setups:

1kg at 10m
10kg at 1m

You say these are equivalent.

Doing some basic math you will find that both masses will undergo the same linear acceleration of 1 m/s/s.

The problem is that in the first this is at 1m radius and 10m in the second.

If you calculate the acceleration of the second from 1m it would give 0.1m/s/s.

So any hanging weight that's causing the torque will accelerate 10 times slower.

This all is using simple linear concepts.

The way I see it this is the whole point. Slowing down the rate of descent of the falling mass. Why? Because the total Force acting on the falling mass does not change - therefore the Force x Time Impulse is multiplied the longer we prolong the fall. This equates to more Momentum being obtained.

As long as momentum isn't being lost through friction or other, it is conserved and becomes available for use at the end of the Acceleration period.

Previously - we were looking at increasing the mass of a balanced system to slow down the fall. That required a momentum conversion to turn heavy/slow into light/fast.

This way we use leverage on a light mass instead - so the energy conversion is just being done at the same time.

You are absolutely correct that the acceleration is slowed right down on the descent side - i'm just saying that this is being done for a reason. But on the ascent side, the leverage multiplies the velocity of the small weight.

[Fletcher]

WM2D gets several things correct. It predicts that the total potential energy of the 2 kg and 10 kg masses is zero. The center of mass of the 2 and 10 kilogram masses does not move, therefore it has not dropped. You need a drop to have gravitational potential energy.

WM2D says that at a velocity of .9757 m/sec the (2 kg, 10 kg, 20 kg) system has a Kinetic energy of 38.08, which is correct. But where does the velocity come from? The velocity has to come from a drop, and the acceleration of that drop is 4.905 m/sec/sec.

So let’s find out how far the 20 kg was dropped (d = ½ * v²/a). Therefore the 20 kg was dropped a distance of (.5 * .9757 m/sec * .9757 m/sec / 4.905 m/sec/sec) .09704 meters.

Wait a minute: A joule is a newton dropped a meter. Our 20 kilogram exerts a force of (9.81 N/kg * 20 kg) 196.2 newtons. To get 38.08 joules in free fall the 20 kilograms will have to be dropped 38.08/196.2 = .19408 m. But we only dropped it .09704 m in the block and tackle to get the same 38.08 joules.

If we dropped the 20 kilograms .19408 meters in the block and tackle the 30 kilograms would have a velocity of 1.378 m/sec for a kinetic energy of (1/2 * 30 kg * 1.378 m/sec * 1.378 m/sec) + ( ½ * 2 kg * 6.89 m/sec * 6.89 m/sec) = 75.95 joules. This is 75.95/38.08 = 1.9945 times greater than the potential energy of the 20 kilogram at .19408 m.

My predicted energy increase (for the 2 kg block and tackle) in the first post was 392.37 J / 196.2 J = 1.999. So if you use the same drop distance WM2D agrees with my calculations. An energy increase of 1.9945 is equal to 1.999 when rounded to significant figures.

The real problem here is that WM2D just picked drop distances out of the blue sky to get the answers it wanted. It apparently is programmed to do that. The program did not have any problem that the drop distances it picked out of the blue were not even the same.

Perhaps that's why I asked you for a drawing Pequaide, before I attempted to interpret your math & model ? - Clearly the CoG does drop & I don't need a sim to see that !

Your assumptions about what the WM sim is showing are incorrect - the 10 kg masses each side will not move without a force - that force is gravity applied to the 20 kg mass attached to the left side 10 kg's - the TOTAL CoG does move downwards contrary to what you say - this results in a difference in Ke's & a difference in in Pe's which zero out.

WM doesn't pick out of thin air distances etc - I quickly put together the sim, as I explained, & had the lifted side travel approx 1 m - the Pe's were zeroed out in the meter box so that a net figure could be arrived at for comparison to the summed Ke's of movement.

So, whilst there is a difference in Ke's each side this is equated by an exact same loss in Pe's [i.e. netted out].

I don't believe you can treat the 10 kg mass system separate from the acceleration mass & force which is provided by the 20 kg mass & gravity - take away the 20 kg mass & nothing moves.

No energy created here !

[pequaide]

I stated many times that the 20 kg is dropped, but the center of mass of the remainder of the system does not move. The drop of the 20 kilograms is the potential energy used up and the kinetic energy of the system is the energy given out. There is a large increase in energy.

[FunWithGravity2]

Peq

With all due respect, lift the 20 KG back to start height or maybe its time to concede that your maths might be flawed. I believe that their is a fundamental error with the linear acceleration you are using to accelerate your flywheels and the translation of that into angular momentum and i think when you try to reverse the procedure you will see a disproportionaly large amount of energy used in the translation back into linear acceleration, every conceivable method that i can see of reacclerating the dropped mass require as much or more energy than you have been able to create.

I think that you are doing a great job, and i don't want to rush you, but maybe a little reality check is sometimes what we need to move on or see past current blockage. I'm sure you have tried to accomplish the lift many times by now without much success otherwise we would be viewing it (your sharing is your strongpoint).

As it seems everyone adds their 2 cents about what you should do to prove or disprove your current designs i will also add my 2 cents, but to offer suggestion for how to overcome your current dilemna. IMHO your biggest issue is the stopping of the system, i know that everyone thinks that this is the best easiest way to work the math, but its not the best for actually getting more energy out of the system. Initial acceleration of the system or falling mass is causing issues. If your sytem/wheel/mass are already moving then things get alot easier. IMHO


Crazy Dave

[pequaide]

The mathematics is over three hundred years old and has been proven to be correct millions of times. Even WM2D got the math right and then Fletcher denied that it did.

I am capable of believing the math; to me Kinetic energy is just as acceptable as height.

The following paragraph is a discussion about a rod with 10 kilograms on one end and one kilogram on the other. It has a bearing at the center of mass and it is rotating.

At some point in a balanced vertical rotation the small mass will be moving up. I contend that you release it there and let it rise. Height can be a stored form of energy. Let the remaining 10 kg side swing (down and then) up to its maximum height and lock it there. At 10 meters per second for the 1 kg mass the 10 kilogram (a third part of the system) drive mass will have only needed to drop .1529 meters; this is (.1529 m * 98.1 N) 15 joules of energy. The 10 kilogram rotating mass will rise .05097 meters for (.05097 m * 98.1 N) 5 joules. The one kilogram mass will rise 5.097 meters for (9.81 N * 5.097 m) 50 joules. 55 joules is greater than 15 joules.

These energy contents are expressed in rise or height: which is Potential Energy.

[Fletcher]

The mathematics is over three hundred years old and has been proven to be correct millions of times. Even WM2D got the math right and then Fletcher denied that it did.

'Woolshirt' Pequaide - I denied no such thing - the Potential Energy of Position was converted into Ke but the sum of both Pe + Ke [at any height from a datum] remained the same, therefore NO Energy Was Created, is actually what I said, confirmed by WM.

What you are suggesting in affect is that by letting a cylinder or ball roll down a ramp you are creating Energy - clearly unless the ball is returned to its starting position then the system is incomplete & no better at explaining OU or Energy Creation than citing a water wheel as another fine example of Creating Energy.

But then that has been the 'close the loop' argument all the way thru this thread !

[pequaide]

WM showed that the same amount of energy was made with half the drop distance.

[Fletcher]

Peq. .. here is the same sim again after a bit of time tidying it up & putting in colours, boxes & arrows to direct you around the pics.

Pic one is start conditions i.e. static situation with zero velocities & datum of -1.2 meters.

Pic two is a snap shot just before finish where the masses are still accelerating - note that the Net gravitational Pe is the sum of the left side & right side from the datum height - the Total Ke is the Ke's of all masses summed.

The blue boxes show the comparison of Net Pe v's Total Ke's & they zero sum at any time.

You can also see the height that the right side 2 kg achieves [0.999386 m] & the displacement of the left side masses [-0.199877 m] - you can also see the velocities of each mass [yellow meters].

N.B. the sim accuracy has been upped to 5000 frames per second to make it easier to compare trends to your math predictions.

[greendoor]

I believe that their is a fundamental error with the linear acceleration you are using to accelerate your flywheels and the translation of that into angular momentum and i think when you try to reverse the procedure you will see a disproportionaly large amount of energy used in the translation back into linear acceleration, every conceivable method that i can see of reacclerating the dropped mass require as much or more energy than you have been able to create.

Pequaide gives as many linear momentum examples and experiments as he does rotary ones. The block & tackle example is pure linear momentum.

Personally, I think he has every reason to believe that the momentum of a rotary mass is directly equivalent to linear momentum. If a chunk of mass falls off a rotating wheel, it continues on in a straight line with the same linear velocity as it previously had as peripheral (angular) velocity.

I don't believe there is any issue with the use of rotary wheels. If the oversimplification of using linear momentum maths worries you, stick to the linear experiments. But I believe Pequaide has enough experimental evidence to justify the use of rotary experiments where they are more practical to build.