Mechanism for consideration

[Trevor Lyn Whatford]

Hi, the speed needed to reset the rods per degree of wheel tern required would need a lot of torque and be equal to the out of balance torque, this should be looked in to and worked out as you would gearing systems! take a look at free energy news gravity wheel section and you will see the rod lift motor, electric solonoids would be the best reset, it would make a good build but I do not see any gain of free energy!

[Mangyhyena]

My thinking on this, in terms of a self runner, might well be off. Physisists disagree with the way I see this, for sure.

Look at an individual lift, rather than all the lifts required per revolution/cycle. Two pounds will be lifted 1 foot. The one foot lift will create an overbalance of one foot. The distance of that fall will depend on the length of the rod. Use a six foot rod and you'll get more than six feet of fall for your overbalanced weight even though it was lifted one foot. Longer rod will fall farther, and a shorter rod will fall a shorter distance. The distance of the fall or the height of a weight at its highest has absolutely nothing to do with how high the lifter lifts.

All right, stop right there. How many falling pounds, overbalanced by one foot, would it take to supply the energy for the lift during the time it takes to complete the lift? (read that last part carefully. During the time it takes to perform the lift, not how many weights need to fall from top to bottom to obtain the energy needed for a single lift) In other words, if a lift takes one second to complete from start to finish, and it requires x amount of power, we need the other falling weights to supply x amount of power within one second. The machine must supply x amount of energy every second it operates. If it operates for one minute, it needs to supply x amount of energy times sixty during that minute.

Let's say that to perform that lift, you must have ten times the weight falling just to break even. (for the sake of simplicity, it takes ten times the falling weight in equal weights falling at a one foot overbalance.) Since its a two pound lift, it would take twenty weights falling in order to supply the power to perform the lift during the second it takes to complete the lift. Twenty-one rods on an axel would give you that ratio, in this example, one rod being lifted, twenty falling weights to power that lift.

So, the twenty-one rod machine completes the lift. The rod that was just lifted is now actively falling, becoming one of the twenty falling weights. At the same time, the next rod is above the lifter. Since all weights are equal, this rod, too, requires ten times its weight to supply enough power to complete the lift in the allotted one second. Because the reset rod just went back to falling as another rod went out of play above the lifter, the ratio of falling weights to the one in need of a lift remains exactly the same. Twenty rods are falling to supply power for your next lift, though the one previously lifted may be in a different cycle than the rest.

A maintained ratio means there are always the same number of weights falling, no matter which weight is in need of a lift or which cycle it happens to be in. If the actual ratio to obtain the required power for a lift, during the second the lifter is lifting, is higher than 20 to 1, than so be it. Put 101 rods on the axel. The equipment will have to be beefed up in terms of strength and size to allow more room to fit them all and to allow longer rods, but it is very possible to fit that many rods on one axel, and then some. Now we have a 100 to 1 ratio. What changed with the addition of more weights? Not the distance of the lift or the number of lifts to be done at the same time. (at the same time). What changed is the number of falling weights to supply power and the number of times per revolution a single lift must be performed. The weight of the lift may get slightly heavier due to the longer rod, it's true, but not by much and certainly not by the same percentage the total falling weight on the machine changed. We're still performing the lift of one rod and using the power from all the falling weights to power that lift-during the second it takes the lifter to complete its lift.

Now, add another rod to the minimum number of falling weights it takes to power the lift of one rod for the duration of the lift. At this point the machine should be generating more power than it takes to operate itself. Double the number of rods and it should generate twice as much energy, second to second, as the lifter requires.

Most physisists look only at one cycle to determine if the machine will produce more than it takes to run. Lift all the weights at one time and then drop them to see if it required more energy to lift them than they generated when they fell. But, how does this method apply to a machine that gets energy from different weights in 2 different cycles at the same time? in other words, this machine isn't picking up all the weights on it at the same time, then dropping them all at the same time. It maintains a constant ratio that greatly outweighs the one weight being lifted. More than that, once a weight is lifted it immediately goes back to work, joining the other falling weights that have not yet made it down to the lifter.

All I'm looking at is the energy required for one lift, the duration of the lift, how many falling weights it will take to deliver that energy during the time it takes for the lifter to complete a lift, and how to maintain that ratio from one lift to the next. Get enough weights falling to supply that power, maintain that ratio at all phases of the revolution, then add more weights to gain excess energy for useful work.

Physisists either don't see this concept because they were not trained to look at it this way, or they know it can't work this way from experiments they've done, or they've been taught to ignore any measurement other than single cycle operations.

The machine may have 20 or more weights on it, but it does not pick them up and drop them, then run out of energy because there are no longer enough weights falling to power another lift.

If my way of looking at the possibility of looping static forces with a maintained ratio turns out to be correct, any static force could be looped to provide energy with this principle. Gravity motors, magnet motors, buoyancy motors---even springs or rubber bands could be used, if the timing could be worked out to provide a favorable, maintained ratio. If I'm wrong, it wouldn't be the first time.

[Tarsier79]

...The one foot lift will create an overbalance of one foot. The distance of that fall will depend on the length of the rod. Use a six foot rod and you'll get more than six feet of fall for your overbalanced weight even though it was lifted one foot. Longer rod will fall farther, and a shorter rod will fall a shorter distance. The distance of the fall or the height of a weight at its highest has absolutely nothing to do with how high the lifter lifts.

Perhaps you should rethink your first assumption. If you have a six foot rod and your lifter only lifts it an extra foot, the power you will get out, minus the power it takes to reset that weight will be the same as the input into lifting that one foot. Eventually, that one weight has to be lifted back to where it started, right?

[Mangyhyena]

Absolutely, that weight will have to be reset with another lift. The power released when that one pound weight falls will not be the same as one pound falling. If the overbalanced weight falls down on the right side, the opposite weight at the other end will be going up, which requires power, which it gets from its falling counterpart, which means the falling pound can't fall with the force of one pound. No free energy using this method of lift alone. If momentum or speed were involved, there would be more factors to examine, but I would expect this machine to turn slowly.

What a one foot lift on two pounds with a rod between them does do is set a weight many times that height much faster than lifting one weight the entire distance of the fall. This allows for a maintained ratio.

If you want 5 weights to fall out of phase, you need the distance of the fall (height of your machine) to be high enough to accomodate those 5 weights. If you want more weights, you will have to build a taller machine, which will require a higher lift, which will require more energy, a lot more energy if you have to increase the speed of the lift as well.

With the rods, if you need more room to accomodate more rods, you increase the length of the rod. The height/speed of the lift remains the same, which should mean the energy required would remain roughly the same.

If there is a possibility of free energy in this machine, it'll be in its maintained ratio. Putting the rods between two weights allows the machine to maintain a constant ratio of falling weights to the one lrod being lifted without having to increase the energy required per lift if you want to upscale it/increase the ratio of falling weights to the one being lifted.

If I'm wrong about a maintained ratio, the machine is a dud and shouldn't be bothered with.

[Tarsier79]

"you increase the length of the rod"

See last post. If there is a weight each end of the rod, increasing the length of the rod makes the fall slower, but does not produce any extra energy.

[Trevor Lyn Whatford]

Hi, Ok lets look at the torque for and against, you have out of balance weight in a 8 rod device, that is 8 weights on both sides of the wheel but on one side they are away from the axle to give a unbalance in the wheel, at any time in the cycle you would have to direct lift the total true weight of 1 rod and 2 weight very quickly indeed, if not then you need to lift 2 rods and 4 weights in a direct lift, but you are only lifting with out of balance, the greater the out of balance the greater the true weight lift!

Edit, think of rope and pulleys, how much weight would you need to lift 1 weight at 8 x the speed?
Edit, what true weight do you have to do this with?

[jim_mich]

Look at an individual lift, rather than all the lifts required per revolution/cycle. Two pounds will be lifted 1 foot. The one foot lift will create an overbalance of one foot. The distance of that fall will depend on the length of the rod. Use a six foot rod and you'll get more than six feet of fall for your overbalanced weight even though it was lifted one foot. Longer rod will fall farther, and a shorter rod will fall a shorter distance. The distance of the fall or the height of a weight at its highest has absolutely nothing to do with how high the lifter lifts.

You need to learn about torque.

Your example uses a six foot rod with one pound at each end and one foot of lift. Thus initially you have a first weight at 6 o'clock at 3-1/2 feet below center. And you have a second weight at 12 o'clock at 2-1/2 feet above center. The distance between the two weights is the six foot rod. Am I correct?

You then lift the two weights a distance of one foot. This puts the top weight at 3-1/2 foot above center and the bottom weight at 2-1/2 foot below center.

You seem to claim that this will give more rotation to the wheel. so lets look at how much torque is actually produced. assume the weights have rotated around to 3 & 9 o'clock where maximum torque will be produced.

A one pound weight at 3 o'clock at 3-1/2 foot distance produces 1 × 3.5 = 3.5 foot/lbs clockwise torque.

A one pound weight at 9 o'clock at 2-1/2 foot distance produces 1 × 2.5 = 2.5 foot/lbs anti-clockwise torque.

3.5 - 2.5 = 1 foot/lbs of total rotational torque.

Looking at is a little differently. Start with the two weight at equal distance from the center. Then move them 1/2 foot upward. You end up with a total weight of 2 pounds moved 1/2 foot off center. Thus you get 2_lb × 1/2_ft = 1 foot/lbs of torque. Which is the same results as my first calculation.

You could use a 100 foot rod. If you lift the weights one foot (1/2 foot off center) then you will get the same torque.

Sure, a weight falling 50.5 feet will take a lot longer, but do not forget that it is also lifting the other weight 49.5 feet.

The bottom line is weights always give back exactly the same energy when falling as was given to them when lifted up. There is no magical arrangement that will break this rule of nature and allow a PM wheel based upon gravity.

[Trevor Lyn Whatford]

Hi Jim,

With all due repects, I hope in time to prove you wrong with your last statement!

The bottom line is weights always give back exactly the same energy when falling as was given to them when lifted up. There is no magical arrangement that will break this rule of nature and allow a PM wheel based upon gravity.

Regards Trevor

Edit, I do however agree it is a good rule of thumb for weight shift designs, out of balance wheel type designs!

[daanopperman]

Mangyhyena ,
180 deg of rotation later than your 1 ft lift , you will need to lift 2 ft to produce ob , 1 ft will just put you back in balance .

[jim_mich]

Trevor,

I'm confused by you contradictory statements.

First you say you hope to prove me wrong with my last statement.

Then you quote my two last statements.

So I must assume that of the two statements, the part that you hope to prove me wrong is, "There is no magical arrangement that will break this rule of nature and allow a PM wheel based upon gravity."

Then you turn around and agree with me by saying you, "do however agree it is a good rule of thumb for weight shift designs, out of balance wheel type designs!"

This I take to mean that you agree with my first statement, "The bottom line is weights always give back exactly the same energy when falling as was given to them when lifted up."

So my confusion is: You agree with my first statement that rising and falling weights give and take the same energy. Then why do you fail to agree with me that there can be no working PM wheel based upon gaining energy from gravity?

I'm left scratching my head.

[getterdone]

Jim, I agree with Trevor, although I had to learn the hard way that there is no meccanical advantage from simply shifting weights, we do have an axle that is fixed to the ground.
We dont have to fallow the same path, I dont think we can rule out the possibility that a wheel can turn with just gravity and common sense

[Trevor Lyn Whatford]

Hi Jim,
there is no contradiction, I believe an out of balance wheel to be impossible, but I believe a gravity wheel to be possible, one direction force is used to drive a wind turbine, and the wind turbine is rotated to the direction of that force, ( angle of attack) with gravity you know where the angle of attack is so it is easy to use. It is a force that can be converted just like the wind, first you have to understand this and then design to it, you do not have to lift a weight directly against gravity to reset its torque values when done right!
Regards Trevor

[AB Hammer]

IMHO the biggest problem with the over balanced wheel is that people keep trying the same things over and over again. But also IMHO I can't see how anybody can exclude gravity in the actions of Bessler's wheel.

Completely new ideas are needed.

[Tarsier79]

I don't think anyone doubts Besslers use of gravity in his wheels. I am of the opinion that Bessler built a gravity-environment wheel, because that is what was the easiest way to sell it, while hiding the internals.

If someone discovers the principle, I am sure it will be able to power a grounded axle wheel, and cyclic non-gravity based mechanisms, or horizontal inertia type setups. Even so, it is however quite probable that the prime mover does have some sort of gravity component, which would account for the timing of the weights being lifted at the correct time into the OOB position.

[Mangyhyena]

First off, thank you for all the responses. I did title this thread as a machine for consideration. No claims of success can be made on any gravity motor unless there is a working prototype, which there is not in this case. I posted because it seemed like a different approach and I don't have the training/education to evaluate it myself with enough accuracy.

You all explained the problems with this setup well. I think I understand why it can't work now. Thank you again for spending the time to explain.

I was going for an efficient juggling of the weights, hoping the multiple falling weights could power a single lift and get that dead weight back to a falling position before the next weight reached the 6 o'clock position, maintaining the ratio of falling weights to the single lift.

Because increasing the rod length will only increase the distance of the fall, but not the total power released during the fall, the longer you make the rod, the less power per inch or foot of fall you will get on the weight's way back down. I may be able to get 100 weights falling and apply their power to a single lift, but because all those weights combined will release far less energy than 3 pounds worth of energy per inch of fall, they will never release enough energy to power a 2 pound lift, no matter how many rods I add to the mix. The more rods I add, the longer the rods need to be, the less power per inch they'll release. I could even increase the distance between the drilled holes from 1 inch apart to one foot, but there will still not be enough power from all the other weights to power the lifter with its 2 pounds during that foot of rotation/fall.

That sound about right?

Considering the impact free energy would have around the world, I had to find out if this setup were viable.