the obvious gain

[ovyyus]

I just dont feel like explaining it.
it is a very simple design and should not need an explanation.

Looks like the brainiac's lose another round to Archurian logic - lol

[jim_mich]

Attached is a screen capture of a WM2D simulation of arthur’s weight on a wheel. Assuming that arthur and maybe others aren't familiar with WM2D I will explain it in more detail than might normally be needed.

Shown is a 1.5 meter wheel. On the wheel is a 0.2 meter blue square weighing 1 kilo. It was initially 1.2 meter out to the right from the center at the start. The green double line is a track that the weight can slide on, much like arthur's tube proposal. Note that it is square pinned to the track so that it can't turn on the track. To the right of the weight is a green spring. Is has an initial tension of zero. It has a force of 5 Kilo/meter which gives it a force of 1 kilo at 0.2 meter. The 'S' shaped green line is a rope used only as a quick and dirty means of measuring the distance of the weight to the wheel center, which is shown on the upper graph as "Length of Rope 17". In the middle of the wheel is a motor that turns at a slow 1 RPM speed so as to cause very little centrifugal force on the weight. I ran the simulation twice. The first time I pinned the weight at 1 meter from the center as can be seen on the graph as a straight line. The second time the weight was free to fall as the spring was stretched, which shows as the curved line.

The bottom graph is the important one. It shows the torque force that the motor produces in order to rotate the wheel. The lower line of the lower graph starts at 1 kilo/meter as would be expected when lifting a 1 kilo weight at a radius of 1 meter and drops to zero as the weight reaches the top. The upper line of the lower graph shows the increased torque force starting at 1.2 kilo/meter needed to lift the i kilo weigh at a radius of 1.2 meters

If what arthur says was to be true and the stretching of the spring is free then the required motor torques would be the same. But they are not the same. More torque is required to lift the weight while stretching the spring than when not stretching the spring, thus proving arthur wrong. Stretching the spring comes at a price of extra torque being needed to lift the weight, contrary to Arthurian logic.

[KAS]

Although I understand Arthur‘s logic in that the spring is being charged for free, it is of little use.
Indeed, at first glance, and in the absence of friction there appears to be a freely available source of potential energy.

There is a “but� however. The payback is that the weight finds itself in a lower position once the charge of the spring is complete. This height cannot be reclaimed no matter what it does for the rest of the rotation.

Height is everything in an OU system and the best the charged spring can do (in theory) is reposition the weight to the very same spot.
But all this is just conjecture as sadly, "friction free" is not available to us.

Kas

[bluesgtr44]

I have tried this in simulation....you are wrong! Your fixed point of 45 degrees with the cord/rope attached will be more of a brake than a governor......it has no "give". When the arm is at 10:30, the cord pulls on the spring while at 4:30 the cord is hanging loose...although the weight is applying its own downward force on the spring and thus pushing outward. So, the springs are going to be a constant tension and when/if this thing were to actually accelerate, these springs will not be as effective as they once were.

I made just one beam so I had two sides 180 degrees apart and gave it a good push start....it actually had enough force to make it through a revolution after that and it did have good launch when it enters and leaves that 10:30 position....but, as is this will never maintain PM......


Steve

[Quartz]

Kas

You said (I understand Arthur‘s logic in that the spring is being charged for free.)

I don’t understand why you would say that. The spring is not charged for free, when the weight is at the 12:00 position before it is allowed to vertically drop charging the spring, it has X amount of starting torque, when that weight dropped vertically charging the spring you lost that original starting torque. What would the power curve look like if you were driving a load off the axel? Though the exchange of energy between the spring and the weight moving from inward to out ward is satisfied, that power curve looks quite deferent.
Or maybe I’ am looking at this wrong.

Ken

[arthur]

I have tried this in simulation....you are wrong! Your fixed point of 45 degrees with the cord/rope attached will be more of a brake than a governor......it has no "give". When the arm is at 10:30, the cord pulls on the spring while at 4:30 the cord is hanging loose...although the weight is applying its own downward force on the spring and thus pushing outward. So, the springs are going to be a constant tension and when/if this thing were to actually accelerate, these springs will not be as effective as they once were.

not sure what youre getting at here.

more "arthurian logic" .....
lol

-----------------------
the spring is always pulling on the rope.
the idea is to set the spring tight enought so that there is never any slack in the rope.

from 10:30 to 4:30, the spring's pull adds positive torque to the rotation.
+
from 4:30 to 10:30, the spring's pull adds negative torque to the rotation.

however,
during most of the rotation from 10:30 to 4:30 the weight hangs from the rope.
+
during most of the rotation from 4:30 to 10:30 the weight hangs from the spring.

thus,
from 4:30 to 10:30, the spring does not pull on the rope as hard as it does from 10:30 to 4:30.

therefore,
from 4:30 to 10:30, the spring adds less negative torque.


now you know what time it is.?

[arthur]

I like this picture better
it is more like my original picture as it shows the weight never falling.

http://www.besslerwheel.com/forum/download.php?id=5662

noone will believe this... but,
Jim, your torque calculations are irrelevant.

the fact remains that 1 kilo falling 1 meter has the force needed to lift the arm.

If 1 kilo fell 1 vertical meter down the same (but opposite) curve, with an arm attached, this arm would have the same torque.

[F.Nepure]

I think what Arthur is suggesting is that the extension spring or elastic is callibrated so that at each point of the lift up to 12 O'clock, the weight extends it the exact distance to allow the weight never to rise.
If the mechanism is ratcheted at this point, then when it reaches the horizontal again, the energy stored in the spring/elastic can be released.
If you think it'll work Arthur then build it! Don't hang around waiting for the approval of the brainiacs. Did Bessler show his designs to Newton for approval before he built his wheel??

[arthur]

I think what Arthur is suggesting is that the extension spring or elastic is callibrated so that at each point of the lift up to 12 O'clock, the weight extends it the exact distance to allow the weight never to rise.
If the mechanism is ratcheted at this point, then when it reaches the horizontal again, the energy stored in the spring/elastic can be released.

exactly.
what you have just described is one way to use the energy stored in the spring.

Kas,
as F.Nepure just pointed out, the energy can be used.

I'm not waiting for approval however I wont be building this thing any time soon.
I just figured Id share the idea.

[Fletcher]

I think Jim & Oystein explained it pretty conclusively, IMO.

Perhaps I can attempt a more user friendly visual description of whats likely to happen in a complete rotation or at least in half a rotation ?

First off - we only really need to consider what happens in the top two quadrants between 3 o'cl & 9 o'cl to see the contribution of positive torque & negative torque from the spring set-up.

As the lever is rotated upwards towards 12 o'cl the weight will slide down in its slot tensioning the spring [already discussed].

To get it to 12 o'cl however, you have to provide more energy into the rotation force - that's because you get a negative torque [back-torque] resisting the upward movement.

i.e. as the spring tensions more & more as the angle of the weight increases [from say 3 o'cl towards 12 o'cl] then the spring will want to release this tensional force being stored in it - in other words, systems that have Potential Energy want to release that energy & like water will opt to take the course or path of least resistance - the spring will want to discharge its Pe & it is easier for it to move back to 3 o'cl [against the direction of rotation & back to where it came from] than go forward to 12 o'cl & further increase the tension & Pe - that is precisely why, IMO, you have a back torque when charging a spring & why as Jim showed you need to input extra energy to get the spring loaded weight & lever to 12 o'cl.

IMO, with spring, weight & gravity systems the old mantra rings loud - What you put in you can get back, less ordinary system losses, but no more.

The total energy [PE] in the system, with or without the spring is the same [less losses] than if no spring were employed, zero sum game, IMO.

[arthur]

I want to remind everyone that in "the obvious gain" scenario,

the spring cannot push or pull the arm, because it's part of the arm, and its not connected to anything else!!!

therefore,
the only force we are dealing with is the force needed to move the weight up the frictionless, curved path,

as if there was no arm, or spring, just a weight and a curved path.

the force of the spring is only used to suspend the weight.

[Fletcher]

Arthur, although the spring force is inline & along the radial [from rim to axle] there is a small leverage factor applied at the axle bearing to axle shaft, which is the coupling & couter-torque, IMO.

Perhaps I have completely misunderstood your concept but a small inexpensive build by you would quickly give some reliable observations & data.

Edit :

Ok, let's use Kas's & Steve's approach - the red line is the approximate flattened arc the weight travels thru the top quadrants - the spring has most tension at 12 o'cl [blue] - on both sides of the center vertical line down thru the axle, at any horizontal distance from that line, the spring tension is equal [symmetrical forces] - the area above the red curve represents the Potential Energy stored in the spring.

How do you propose to use this stored energy to create sustained overbalance ?

If it's a gravity wheel, presumably you want to use the spring force to bring a weight closer to the axle [a closer radius] at some stage, but then it must be moved out again to a greater radius [lifted upwards component] to give the torque you require on the descending side of the wheel ?!

[Oystein]

"the only force we are dealing with is the force needed to move the weight up the frictionless, curved path"
as if there was no arm, or spring, just a weight and a curved path."

Again Arthur this is NOT true !

Don`t you see why ?

Because the curved path you suggest/describe, would be with/of the same path as the spring-connected weight would take, right?

This meas that the curved incline would a bit flatter than any fixed point on the arm, right ? (end up a bit lower than a marked starting-point on the arm)

This "flattening" means that some more of the weight is resting on the incline (versus direction of lifting force) and the force (or more correctly torque if you use an rotating arm) would be reduced, becase the path is not perfectly circular, but a bit flatter !! (Time to push the "scientific button" on the Windows Calculator, sin/cos ;-)

I can draw and calculate a comparison if you like, but your conclution of the comparison (spring vs. curved path) is WRONG ! Sorry..
It will take less work (force * distance ) to roll or slide the weight up a curved fritionless path, than making an arm raise a spring-connected weight, both to the same height !!


Best
Oystein

[rlortie]

Oystein,

It will take less work (force * distance ) to roll or slide the weight up a curved frictionless path, than making an arm raise a spring-connected weight, both to the same height !!

Following this thread I have wondered why the simple inclined plane has not been mentioned. I somewhat disagree with your statement that it will take less work force. True it will take less time and more force for a vertical lift, but more time with less force for the radial version. In either case your force times distance will amount to the same amount of work accomplished.

I have somewhat ignored this thread as I see no reality related to a gravity wheel. All that I perceive has been related to either a quarter or half circle. To move a weight upward from and to a given hight will require the same amount of work no matter what the variables of time * distance may equate. If I am going to lift 2.5 cubic feet of dirt in a wheel barrow, I would prefer doing so on an inclined plane rather than lifting it straight up. Once achieved the only work difference is the distance of the inclined plane measured horizontally. Which in this scenario would be considered "angular motion' as described above by Fletcher.

Ralph

[ovyyus]

To move a weight upward from and to a given hight will require the same amount of work no matter what the variables of time * distance may equate...

Ralph, work = force x distance, not time x distance. I'm sure you know that already. Arthur claims the spring is compressed for free during the lifting. That is clearly incorrect. IMO, arthur should learn physics and/or build experiments rather than simply insist that he's right and everyone else is wrong. Physics was built on experimental observation, not Archurian logic.