Furcurequs (aka Dwayne) questions Jim_Mich

[Tarsier79]

Peq. Moving the drive mass does not change the inertia of the wheel. Still you do not understand the difference between the factors of torque and inertia.

[Wubbly]

It is unfortunate peq has not been able to move forward in the last two years.

Free energy devices do exist and the energy does not come from an Atwoods, or peq's "mr" hypothesis. The best free energy devices are electromagnetic in nature. People who try to reveal working devices to the world are quickly silenced by the trillionaire thugs who are interested in making money with the petroleum based economy and running the planet into the ground.

William Colby, a former CIA director, was going to give Steven Greer's group a working free energy device, along with 50 million in startup money to get it out to the world. The week before he was to meet with Greer's board, he was found dead in the Potomac river.

peq should have no worries though, since his experiments don't produce excess energy.

[Furcurequs]

This is jim_mich's original mistaken answer which he obtained from his own calculations. It is off by about 524%, so not only is it not in the ballpark, it is not even in the same state as the ballpark (or in the same country as the ballpark if the ballpark is in Europe).

Time = 133.48 seconds to accelerate the mechanism up to a speed of 50 RPM.

http://www.besslerwheel.com/forum/viewt ... 402#104402

Here is my correct answer which I apparently posted within 24 hours of jim_mich's mistaken one. I got my answer from equations I derived myself and then checked the answer using an energy balance.

So, plugging in the distance and velocity of our 500 gram mass we can determine how long it took to come up to speed...

t = 2 * 5.6 meters / 0.523 or about 21.4 seconds

http://www.besslerwheel.com/forum/viewt ... 426#104426

Later that same night I plugged the relevant numbers into Wubbly's equations and got my same correct answer once again.

I've since checked my math by plugging the numbers into Wubbly's formula found here:

http://www.besslerwheel.com/forum/download.php?id=11164

...and I get the same rates of acceleration using both methods.

http://www.besslerwheel.com/forum/viewt ... 437#104437

Here is the result of jim_mich's Working Model 2D simulation of the problem from January of this year confirming that MY answer has been correct all along.

As a double check I just now went back and simulated it using WM2D, which came up with an answer of 21.358 seconds.

http://www.besslerwheel.com/forum/viewt ... 976#118976


Here is the result of Fletcher's Working Model 2D simulation of the problem from January of this year confirming that MY answer has been correct all along.

21.375 seconds to achieve 50 rpm.

http://www.besslerwheel.com/forum/viewt ... 992#118992

Here is the result of Wubbly's spreadsheet calculation confirming that MY answer has been correct all along.

The spreadsheet got a number of 21.4 seconds for the fall time. This equates with Jim's WM2D simulation that got 21.358 seconds.

http://www.besslerwheel.com/forum/viewt ... 615#119615

Here is jim_mich's most recent mistaken answer which he obtained from his supposedly corrected equations. It is off by 19%, so though it is not in the ballpark either, it may actually be in the same city as the ballpark this time at least.

The bottom line results should be a time of 25.5367 seconds to accelerate the mechanism up to a speed of 50 RPM.

http://www.besslerwheel.com/forum/viewt ... 976#118976

I owe jim_mich no apology with this stuff. So, I'm certainly not gone.

Sorry, though, for my little absence.

Dwayne

[jim_mich]

Thanks for the detailed reply Jim. Lets assume for theory that there is no friction, the rod, rope and pulley are weightless and the following dimensions:
Weights on rod = 1kg each
Rod weights are 1m from the center.
Pulley has a radius of 10cm
Falling mass = 500g

Is there a formula that will tell me how much time and how far the weight would need to fall to reach 50RPM?

Answered by Furcurequs = 21.4 seconds
Answered by Jim_Mich = 133.48 seconds (With a disclaimer: Hopefully no mistakes & no guarantee.)
Answered by WM2D = 21.358 seconds
Answered by Fletcher = 21.375 seconds
Answered by Wubbly = 21.4 seconds
Answered by Jim_Mich = 25.5367 seconds (Corrected answer, I had made one math error.)

Furcurequs, I explained why my results were different from other people's results.

The discrepancy is that with my calculations the weight masses out on the ends of the rod are fixed to the rods and thus are rotated, whereas the weights of WM2D were allowed to free-wheel and thus not rotate. Fixing the weights to the ends of the rods rather than letting them free-wheel results in it taking a slightly longer time to accelerate up to speed. In this case about 4.17 seconds longer.

But I'll try to explain it in more detail here so that maybe you will understand. In all the other people’s equations including yours, it was assumed that the weights out on the ends of the rod were not themselves rotated. When the Atwood method was used, the weights simply dropped and thus the weights themselves did not spin. When Fletcher made his WM2D model, he attached the weights using a rotational pin rather than a fixed pin. This again allowed the weights themselves to not spin. When you look at Wubbly's setup, you will see that the weights did not themselves spin. They drop straight downward.

Now then, I used a formula that took into account the spin of the mass. The original problem did not define the size and shape of the weights and it did not say if the weights on the ends of the weightless rod were fixed or rotatable. If they were rotatable, then the weights will act like point-mass objects. If they were fixed to the ends of the rod, then the mass of the weight is rotated in two ways. It rotates around the center of its mass and it rotates around the center of the rod. But since the size of the weight was unknown and since a point-mass weight will result in zero rotational effect for rotation around the weight's center of mass, I assumed the two weights were each a point-mass. And since they were equal mass and equal radial distance, I could assume that the two weights on the rods were one single point-mass.

Now then, when the non-rotating falling weight is added into the equation, you end up with another point-mass type weight. The assembly as a whole is no longer a single point-mass. You now have two point-mass objects, and now must take into consideration what happens during their rotation. Both objects are rotated at the same rotational speed. Because there are two point-mass objects which rotate around the pivot center, you have the equivalent of a solid object that rotates both around the rod center and also rotates around the venter of mass of the two point-mass objects.

Thus, to but this in the most simple words, gravity acting upon the falling weight must accelerate both weights in a linear fashion and must also accelerate the two weights in a rotational manner as if they were a two-molecule solid. Except in this case they are two individual point-mass objects. This rotational component must still be included into the equation when dealing with real-life situations.

There are demonstrations on the internet where two cylinders of equal mass are rolled down an incline and one rolls faster than the other, due to one cylinder having most of its mass near the rim and while the other having uniform mass. It illustrates how radius of gyration can affect the results.

Said another way, all other peoples calculations that were posted only looked at the linear acceleration and ignored the rotational acceleration. If the weights on the ends of the rods are attached in such a manner as to allow them to spin freely around their attachment point, then the acceleration will take about 21.4 seconds. But if the weights are attached in a fixed rigid manner to the ends of the rods (which was my assumption) then it takes a little more energy to spin the weights themselves in addition to spinning the weights around the rod center. This will result in a slightly longer acceleration time.

This additional time and energy will vary depending upon the size (i.e., the diameter, etc.) and the shape (i.e. square, disc, sphere) of the weights. Even assuming the weights to be point-mass objects (thus eliminating their size and shape), when you combine weights rotating at different radii, then you no longer have a single point-mass situation, and thus in real life situations, additional energy (beyond simple acceleration) is needed to spin the weight-mass. Thus in real-life situations, the time needed is closer to 25.5 seconds rather than 21.4 seconds.

But of course there are no real-life point-mass objects, the weights have a size and shape bigger than a point. In real life there are no weightless rods, ropes, or pulleys. If someone wants real-life calculation then the calculations must include real-life formulas that take into account real-life shapes and sizes and include all components. So this whole discussion is pointless since it ignores a substantial portion of the real-life situation.

And by now I've lost everyone’s attention by including all the blasted bothersome detail.

And I've wasted hours of fruitless typing while writing this post about something that should have been dropped and forgotten a very long time ago. Good grief, my original math error was made 491 days ago. I'm sorry if Furcurequs doesn't understand why my results, where slightly (19%) different from the other members results. I explained the difference at the time.

The discrepancy is that with my calculations the weight masses out on the ends of the rod are fixed to the rods and thus are rotated, whereas the weights of WM2D were allowed to free-wheel and thus not rotate.

That 19% difference represents the extra time and gravity needed to spin the weights themselves in addition to spinning them around the rod axis. It represents real-life verse lab equations.

Along these same lines of thought, physics books will give you a formula for calculating the swing-time of a theoretical pendulum. But if you build a real-life pendulum with the center of its weight located at the theoretical distance, it will not swing as calculated. You must include the mass of the rod into the calculation. And even then it will not swing as calculated. You must include calculations (radius of gyration) that account for the shape and size of the pendulum bob. Only then will the calculation match real-life.

Now Furcurequs, get off my back! It is not my problem that you don't understand such things. And as I've stated before, from now on I'm never going to try to help anyone, for if I make a mistake (we are all human) then the trolls here crucify me.

If he can show me that his words are true like I showed him that mine were, then I'm out of here. ...oh, and I'll even add in an apology.

Furcurequs, you owe me an apology for your relentless vicious verbal attacks towards me.

[pequaide]

I predicted 2.14 seconds. Because the 500 gram mass is at the pulley not the edge. And the target speed is only .523 m/sec. I do these experiments and it would not take 21.4 seconds.

These people buy a sim and they think it makes them physicists.

[pequaide]

One half kilogram in free fall for 21.4 second gives you 105 units of momentum. v = at

The 2.14 sec. for .5 kg on the pulley ( or 2.18 sec) gives you about one tenth that much. So Ft = mv checks.

But the other physicists are producing only 10.4 units of momentum while using 21.4 second, instead of 2.14 sec. They are looking at your math when they should be looking at their own.

[jim_mich]

pequaide, I have no idea what you are trying to say or prove.

Let us just drop this whole thread and move on, OK?

[pequaide]

Take a 20 kg block on dry ice and accelerate it with .5 kg draped over a pulley. You will get 10.7 units of momentum in 2.188 seconds. The simsters are pretending that Ft = 1/2mv², but it is Ft = mv. To bad they can not do experiments.

Drop it and move on means; accept their failure. But it is your thread; so OK.

[ruggerodk]

I made one small error...
The line that reads:
It will be 97.804878÷10 = 9.7804878_g of force acting on 2500 g of mass.
Should be:
It will be 500_g × 97.804878÷10 = 51.1222_g of force[/color] acting on 2500 g of mass.
By mistake I failed to multiply by the 500_g.

And of course this error trickles down through the rest of the calculations.
The bottom line results should be a time of 25.5367 seconds to accelerate the mechanism up to a speed of 50 RPM.

500_g × 97.804878÷10 = 51.1222_g.....?????

Please forgive me - but shouldn't that be: 48.902439_g...?

regards
ruggero ;-)

[Furcurequs]

Thanks Ruggero,

Do you know if that accounts for a 19% error in his answer?

Maybe jim_mich would like to redo his calculations and see - rather than clinging to another lame excuse for his errors.

I had told him after looking over his original math that while even allowing for a mistake with his units his equations didn't look like they would yield the correct answer. He said, though, that they would and so that's the reason why I made my little wager - so I could perhaps have a chance to get to see what he was at least trying to do.

He, of course, showed us that he left out THE DRIVE MASS from his original equation - which he called a "multiplier" - thus confirming what I suspected and that there was indeed something wrong there beyond a mere mistake with his units.

...and then, of course, he still got the answer wrong.

I don't know that I would hire someone to work for me as a "real world" engineer who would use just a bunch of excuses to explain away a 19% error. Would you? His more recent excuse is not valid, btw.

Anyway, maybe he'll get the right answer eventually.

Also, he probably shouldn't really be speaking down to me since I'm actually the degreed engineer here who was doing simple problems like this one over 30 years ago when I was still but a teenager.

If the masses on the ends of the rod are fairly concentrated the point mass approximation that I did - which was also the equivalent of Wubbly's equations - would be very very close to the correct answer. Certainly nothing like the 19% error that Jim was wanting to defend.

He could actually try the thing with his simulation that he spoke of - but apparently couldn't be bothered to do - if he doubts this. ...or others could certainly try it. The effects he was speaking of, of course, do depend upon the actual geometry of the masses, though.

I did a back of an envelope calculation to see how spread out the masses on the ends would have to be to give his answer and it appears they would have to be something like 1.2 meter wide disks. ...lol

I really don't think any of us envisioned the initial problem as looking quite like that.

Anyway, thanks again.

Dwayne

[Dunesbury]

Jim-mich wrote:

"Now then, I used a formula that took into account the spin of the mass. The original problem did not define the size and shape of the weights and it did not say if the weights on the ends of the weightless rod were fixed or rotatable. If they were rotatable, then the weights will act like point-mass objects. If they were fixed to the ends of the rod, then the mass of the weight is rotated in two ways. It rotates around the center of its mass and it rotates around the center of the rod. But since the size of the weight was unknown and since a point-mass weight will result in zero rotational effect for rotation around the weight's center of mass, I assumed the two weights were each a point-mass. And since they were equal mass and equal radial distance, I could assume that the two weights on the rods were one single point-mass."

If Bessler's wheel weights were rotatable on the ends of their arms, rather than fix, or fix rather than rotatable, would that change the secret principle?

[Furcurequs]

Dunesbury,

I can't pretend to know what Bessler's secret was (yet?), but I am exploring some ideas of my own. I certainly don't believe, though, that it would be a bad idea to think about such relative motions and where the energy is as viewed from whatever reference frame.

I'll try to explain the sorts of things that are formally taught about your question, then, at least.

If we assume all the mass is concentrated in a point, we can determine the rotational kinetic energy of that mass rotating around the axis of rotation using the formula found here:

http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html#rke

...where the rotational energy equals one half the moment of inertia multiplied by the square of the angular rate of rotation:

KE=1/2*I*w^2 (I'll use w for omega here)

...where in this case the moment of inertia is simply the mass times the square of the radial distance (R) to the mass.

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mix

I=m*R^2

So,

KE=1/2*(m*R^2)*w^2

If we wanted to consider the actual geometry of the mass and calculate its rotational energy about its own center of mass then we can use the same rotational energy formula but where the moment of inertia is now that of the mass around its own center of mass. If we use a cylinder, then the formula for the moment of inertia around its axis is one half the mass times the square of the radius (r) of the cylinder.

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#cmi

I=1/2*m*r^2

So,

KE=1/2*(1/2*m*r^2)*w^2

If, like in the problem we were discussing before, the cylindrical mass is affixed to the end of a rod, it will rotate around its own center of mass at the same rate the mass is rotating around the other end of the rod (where in this case the other end is at the axis of rotation).

Since the formulas, then, for the rotational energy of the mass around the big circle and the rotational energy of the mass around its own center of mass are the same except for the parts involving their moments of inertia, we can see the difference in the relative energies by only considering those.

One is the mass times the square of the radius of the circular path.

I=m*R^2

The other is one half the mass times the square of the radius of the cylinder.

I=1/2*m*r^2

If the radius of the circle (R) is one meter and the radius of the cylinder (r) is 0.1 meter, for example, then we can calculate the ratios of the energies.

(1/2*m*r^2)/(m*R^2)

1/2*r^2/R^2

1/2*(0.1)^2/(1)^2 = 0.005

Since the masses cancel out and the units cancel out, we can see that the energy of rotation of the cylinder about its own center of mass is then only 0.005 times that of the energy of the equivalent point mass moving around the larger circle.

So, if one is using solid cylindrical masses as weights on the end of a rod, they have to be rather large in diameter before their rotational energies about their own centers of mass are more than a fraction of a percent of their kinetic energy about the greater circle.

(...and for those paying attention, this is also why jim_mich's excuse for his 19% error in a prior post is totally unacceptable)

I hope that helps.

Dwayne

[Dunesbury]

Yes that does help. We know Bessler's weights not that big, even if they free to swing at end of arm they wouldn't have added much energy that way. In your example, cylinder is 1/10 of circle. In Bessler wheel, weights were small can size (not 1.2 feet) inside 12 foot.

[Furcurequs]

Dunesbury,

That would be my thinking, too.

Now, if the "weights" were themselves on axles and were perhaps rolling against something or in some other way geared to the main rotation (intermittently or not) then they might have more stored rotational energy (being a function of the their rate of rotation SQUARED, of course), but such small cylinders probably wouldn't make for very good flywheels under such conditions, I don't believe.

...and, of course, we don't really know how the "weights" were moving within the wheel, either, or even if or how any unconventional source of energy was truly being tapped into.

Take care.

Dwayne

[jim_mich]

Click here for picture of the setup similar to as originally described...

Two 1 kg large weights at 1 meter radii must be accelerated up to 50 rpm while one 0.5 kg small weight falls pulled by gravity thus the small weight must also accelerate up to the equivalent of 50 rpm but at its smaller radius.

This setup would act the same if the 0.5 kg weight were fastened to the setup and then a rotational force of 0.5 kg were applied to the small weight.

A weight at twice the radial distance of another equal weight will require four times a much rotational energy to accelerate or decelerate as the more inward weight. Stated simply, the energy required to accelerate or decelerate any weight changes according to the square of its radial distance.

When the weight is spread out in any fashion (rather than being a point-mass-object) then some of the weight is more inward and some more outward. In a rotating environment, the portions of the weight-mass that are more inward require less energy to accelerate and decelerate. The portions of the weight-mass that are more outward require more energy to accelerate and decelerate. Because it takes 4 times as much energy to double the speed of an object, the more outward weight-mass requires much more energy be added or subtracted in order to change its rotational speed.

A steel cube bolted to a wheel will cause the wheel to act differently than a sheet of steel having the same weight bolted at the same location. This makes things complicated when trying to calculate how a wheel will act. But men much smarter than myself have devised formula that allow for calculating real-time results.

This is where Radius of Gyration formula come into play. If you move all the more inward mass outward and also move all the more outward mass inward until all of the mass is at a same radial distance from the center of rotation, and you choose a distance that results in the rotation acting exactly the same as before, then this radial distance is called the Radius of Gyration. Different shaped objects such as a steel cube verses a sheet of steel rotating around a center axis will have different Radius of Gyration distance. Click here for picture of the Radius of Gyration formula for the large weight cylinder used in this described setup.

So now back to our original problem, 2 kg of weight is rotating at 1 meter and 0.5 kg of weight is rotating at 0.1 meter. To simplify this problem we need to move the 2 kg weight inward and move the 0.5 kg weight outward until both are located at the same radial distance where the resultant rotating assembly will act exactly the same as the original situation.

Men much smarter than myself have devised the formula for calculating this imaginary radius. This formula has been published in Machinery's Handbook since 1914, a hundred years ago. A pendulum is simply an unbalanced wheel acted upon by gravity. A pendulum having two weights located at different radii will swing the same as combining the total weight into a point-mass at a radius calculated by the Radius of Gyration formula. Click here for picture of how to use such a formula for this setup.

The formula says that when the total weight of 2.5 kg (5.5115 lbs) is located at a radius of 0.9815 meter (38.6414 inch) then the whole setup will act like the original setup.

The original setup applied 0.5 kg of gravity force located at 0.1 meter out from the axis. The law of levers says that a smaller force located at a larger distance will result in the same torque being applied to rotate the lever. Converting kg to lbs, 0.5 kg of gravity force equals 1.1023 lbs of gravity force. So 0.1123084 lbs of force at 38.6414 Radius of Gyration equals the same torque force as 1.1023 lbs (0.5 kg) located at 3.937 inches (0.1 meter).

The rest of the calculations are simple.
Determine the velocity desired at the 38.6414 Radius of Gyration to be 202.326 inch/sec.
Determine the time required to accelerate the 5.5115 lbs up to said velocity to be 25.72 seconds.
Determine the dropping distance of the 0.5 kg weight, if it were located at 0.1 meter radius to be 22.09 feet.

My calculation say the time required (about 25.72 seconds) will be longer than what other members say.

Best regards,
Jim from Michigan