It Would Be More Noble To Clear J. Bessler's Name, than trying to solely profit

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rks1878
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by rks1878 »

Hello Scott,

My avatar and signature have not always been showing up after posting.

What's up?

Regards,

Robert
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by graham »

Hi Robert, I had the same question about the sig not showing up.
It doesn't show up on every post you make, it appears just once on the first post on a new page. I guess that your avatar does the same.

Sounds like you are having some fun with your project, good luck and keep us posted

Graham
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by Wheeler »

Robert
I have had results with weights in real wheel tests, but not totally done yet.
I also have been trying to work with WM2D , but I am not up to speed yet.
Looks like I will have it in a few days.
It is about to happen.
Ken This is similar to what I have.
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by rks1878 »

Wheeler,

That's great!
Something that might be of use to you is a means to make sure all your weights are equal. I bought a 10lb digital postal scale at Staples. They sell 5, 10, 20, and 150lb versions. The 150 is only $129.

One of my other ideas I mentioned before was a crossbar deal that went around about 250-270 degrees, but once the wheel stopped it wanted to slide back around in the opposite direction and to go any farther forward required a huge input of force. So, that seemed very un-pmlike to me and I quit that one.

This new idea....when the wheel stops after 40 degrees or so, there is no "wall" to scale and no reverse travel. To go the other 3-4-5 inches or so radially requires hardly anything because when the activated weight finishes its job, everything is back in place and balanced.

It is not a swinging pendulum, but then the weights don't shift radially or move closer to the axis either.

If you are successful, are you going to inform John Collins first? A PM is a very important discovery. Everything should be done with some class, I think.

We should give God, Herr Bessler, and John Collins due credit and respect on the arrival of a true PM Machine.

Robert
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by rks1878 »

I was wondering if anyone would know how to figure at what RPM a wheel with a diameter of 45 9/16" would have to be for a weight to become "weightless" at the apex of travel?
Weight is 2255g., whose center is 1.5" in from the perimeter. Thanks.
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by Jonathan »

I think RPM=30*sqrt(g/R)/π, in this case RPM=40.66.
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by Wheeler »

Robert
Thanks for the tip on the scale, it is something I have needed.
I am still trying to do this with wm2d, because I am not able to build
it yet for lack of time.
I wish we lived closer and could join together on this.
My dream would be several of the people involved would finish it as a group.
I am new to this, so I do not have a history like others, but it came to me clearly when I found the forum, and read what Bessler had for clues.
By using my own ideas of gravity, I was able to find the theory over the past several weeks.
I hope I am correct
I hope you are correct
I think several other have it also.
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by Wheeler »

no "wall" to scale
sounds like something JB would of wrote.
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by rks1878 »

Thanks Jonathan for figuring that out. I am not an engineer. Have to feel things out like a trombone player finds notes on the slide.

Your solution explains why Mr. Bessler's wheels were said to go "25 or 26 revolutions per minute". And I take it, the larger the diameter, the greater the centrifugal force would be if rpm remains constant.

Anyway, it feels like on my wheel, that about 1lb of radial force is applied in the 1-2 o'clock position, then the weight resumes its position on the perimeter and everything is balanced again in time for the next weight to "fire". With better weight bearings, maybe I can get away with only 8 weights on one side as momentum carries into the next position. The problem is going to be getting some sort of catch made, releasing at the right time and then locking until it's time again. Hmmmmmmmm.

It is like if you turn your bicycle over and with your thumb, stroke the front wheel in one place 8 times each rev.

Will be a couple of weeks, as I have a lot in the shop to do to get done and pay bills with.

Wheeler, I'm afraid that theory, etc., will not be enough. At some point you will have to invest in some floor space and tools and put your hand to the weights too........
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by Wheeler »

Hi Robert
I wrote
I have had results with weights in real wheel tests
You must have a theory in order to start building.
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by rks1878 »

A MORE COMPLEX PHYSICS PROBLEM:

I held my postal scale to the wheel at the point where the weight activates and on through where it is re-placed on the rim.

Figure each of the 8 weights producing this same output at a single location on the rim each revolution.

There is 370g of radial force descending to zero through about 40 degrees.
8 weights, each 2255g, wheel dia 45 9/16. Momentum created with this force, I see now, carries a single activated weight and the 7 others through a little more than 180 degrees then stops with no backtracking.....

Is it possible to predict the terminal rpm, and how long it would take to reach that rpm?

Robert
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by Jonathan »

Regarding the centrifugal force:
http://www.besslerwheel.com/forum/viewt ... 8788&#8788
As for the problem, I can't tell what you're asking, but if I'm correctly getting the gist of it, then no.
Also, a few posts back (this taken with this), and in the one just previous, you've used the word radial in a way that I'm not sure if you're using it correctly. Recently someone else used it incorrectly and I figured it was a typo and didn't say anything, but the use you seem to be using is consistent with that misstatement. So please don't (anyone) take offsense, I just want to be sure we're on the same page: there is angular, radial, and axial movement, each corresponding to θ, r, and z, respectively, of a cylindrical coordinate system. The misstatement refered to angular movement as axial, IIRC. Which makes some sense (movement about an axle), but in fact, they are called by whatever the movement is parallel to.
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by rks1878 »

Hello Jonathan:
Thanks for trying to understand.
My meaning would be #4 here from dictionary.com...

As in: "To go the other 3-4-5 inches or so radially... "
I can't find angular "movement" anywhere, but ther is "angular momentum".


ra·di·al ( P ) Pronunciation Key (rd-l)
adj.

1.
a.Of, relating to, or arranged like rays or radii.
b.Radiating from or converging to a common center.
c.Having or characterized by parts so arranged or so radiating.
2. Moving or directed along a radius.
3. Anatomy. Of, relating to, or near the radius or forearm.
4. Developing symmetrically about a central point.

n.
1. A radial part, such as a ray, spoke, or radius.
2. A radial tire.


Thanks again,
Robert
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by ken_behrendt »

Here's some definitions I regularly use and should be understandable to most people and engineers:

radial = moving toward or away from a center of rotation

angular = moving along a circumference of circle whose center is located at the center of rotation

tangential = moving in a straight line that is perpendicular to another line passing through a center of rotation

centrifugal force = a force that throws things out away from a center of rotation

centripetal force = a force that pushs things inward toward a center of rotation

Hope this can be of use...

ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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re: It Would Be More Noble To Clear J. Bessler's Name, than

Post by rks1878 »

Hello Ken:

I guess I should begin using angular then instead of radial. I was thinking of radial gas engines like on crop dusters....

MY WHEEL

Is going to be tough to complete because the catch and release device is going to really be touchy to make work right.
What I have now as far as the mechanism is sort of rickety, something just to test the angular force.
When I hold the weight with my hand and let it go, everything's fine. The weight moves the wheel, and all the other weights, far enough to the next trip point, but when I let it drop suddenly, like an opened catch would do, there is a lot of shock and the weight's energy is absorbed by all the ricketyness and it won't work.
I need to use all steel parts and tight bearings in the joints, so will see...

Ken, I was reading in one of your other posts about your mercury ailment. What rotten luck. You were only trying to do the right thing to stay healthy. Like I've heard before, "The worst things happen to the nicest people".
Anyway, hope you have strength to adapt and endure....

Robert
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There's never time to do it right the first time, but there's always time to do it over again.
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