Bessler Wheel Math

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Jonathan
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re: Bessler Wheel Math

Post by Jonathan »

Maybe Alden Park has access to some documents unknown to me, and if so I'd like to see copies of them. But I've not seen any copies of documents where Bessler mentions Orffyrean -bearings, or -pendula, especially to the extent that one could come up with a theory as detailed as his. Nor is it mentioned even in passing in any of John's books (I've read three, I think that's all of them so far). It's been pointed out before that the Orffyrean bearings Park talks about bear an uncanny resemblance to Wilkes' Rolamite bearing, shown in Fig. 16 in the patent (granted in '69) at the end of the page.
Certaintly Bessler could have had bearings like that. But since even today such a thing would be rarely seen, you'd think that weird bearings would call for very close scrutiny, and the eyewitnesses would've make note of it in their testimonies and correspondence.
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re: Bessler Wheel Math

Post by jim_mich »

I think Jonathan is right. I've also not seen any evidence of 'Orffyrean -bearings' except from Alden Park.

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re: Bessler Wheel Math

Post by Jonathan »

I've done an analysis of the Archimedian screw. It's not as powerful as I'd hoped, but it was worth a try. I thought I'd post it anyway, its nice to know.
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re: Bessler Wheel Math

Post by scott »

Jonathan, you're a monster! Thanks and keep up the good work.
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re: Bessler Wheel Math

Post by Jonathan »

I'm glad you appreciate it. This is probably the last one, it is the math for the approximate power for the Merseburg wheel's stamps (simplest and shortest I've done).
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re: Bessler Wheel Math

Post by winkle »

does enyone know how many stamps rose and fell in one revolution of the wheel ?
the uneducated

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re: Bessler Wheel Math

Post by ken_behrendt »

winkle...

From studying the woodcuts, I think each of the four stamps was lifted and then dropped twice during each rotation of the wheel. That means each rotation of the wheel would have produced exactly 8 impacts of the stamps against the box below them. I am still convinced that the main purpose of this add on gadget was not to demonstrate the power of the wheel, but, rather, to produce distracting noises that would mask any sounds coming from the wheel's interior. I think Bessler did not want witnesses being able to precisely locate where within the rotating drum the sounds it's weight shifting mechanisms made were originating from.


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: Bessler Wheel Math

Post by jim_mich »

Jonathan calculates the Merseburg Wheel to be 28.7 watts.
See http://www.besslerwheel.com/forum/download.php?id=593 and http://www.besslerwheel.com/forum/files/stamps.doc
I get lost when Jonathan uses Greek letters. So I did my own calculation using his dimensions. I used my old Machinery's Handbook as a guide to formulas and my old HP-48sx scientific calculator to convert units.

Stamp length = 1.955 meters = 76.9685 inches
Stamp thickness = 0.0752 meters = 2.9606 inches (square)
Stamp lift = 0.0752 meters = 2.9606 inches = 0.2467 feet
Number of stamps = 4
Number of lifts per revolution = 2
RPM = 40 (or more) = 0.025 sec./rev.(wrong)
RPM = 40 (or more) = 1.5 sec./rev.

Weight of wood = 0.029 lbs/cu.in. for white oak

Stamp volume = 76.9685 * 2.9606 * 2.9606
Stamp volume = 674.6542 cu.in.

Stamp weight = cu.in. * 0.029
Stamp weight = 19.565 lbs.

Weight lifted each rev. = stamp_weight * number_of_stamps * lifts_per_rev.
Weight lifted each rev. = 19.565 * 4 * 2
Weight lifted each rev. = 156.5198 lbs.

HP = (Lb * Ft) / (550 * Sec.)
HP = (156.5198 * 0.2467) / (550 * 0.025)(wrong)
HP = 2.808(wrong)

HP = (156.5198 * 0.2467) / (550 * 1.5)
HP = 0.0468

Watts = HP * 745.6999
Watts = 2.809 * 745.6999(wrong)
Watts = 2094.11(wrong)

Watts = 0.0468 * 745.6999
Watts = 34.9

This is substantially more than what Jonathan shows so I think one of us is wrong.
Edited:
This is close to what Jonathan shows. It was I that was wrong. Please forgive me for my math error.

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Last edited by jim_mich on Thu Nov 17, 2005 11:16 am, edited 1 time in total.
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re: Bessler Wheel Math

Post by Jonathan »

I think it's you, we can be pretty sure it didn't take 2 kW to lift the stamps! The problem is that 40 rpm≠.025 sec/rev, it is 1.5 sec/rev. Using the rest of your numbers, that is 34.9 W, or .047 HP. I got a lower power because I used .023844 lb/in³.
PS I used only three greek letters, in the standard way: pi, rho, and omega!
Last edited by Jonathan on Thu Nov 17, 2005 6:26 am, edited 1 time in total.
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re: Bessler Wheel Math

Post by rlortie »

Ken,

I recall you having said at one time you calculated this. I am not sure but didn't you come up with 26 watts or there about.

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re: Bessler Wheel Math

Post by ken_behrendt »

Ralph...

You are right. I did perform a previous estimate of the power output of the Kassel wheel and came up with a figure of 25 watts. At first I thought that I had to have made a math error, but, upon checking the calculations several times, they did, indeed, seem to indicate a small power output of only 25 watts...just enough to light one small lightbulb! And that coming from a machine 12 ft in diameter!

If, as I suppose, this wheel was driven in either direction by only one set of 8 weights, then the CG of those weights could not have been displaced more than a few inches at most from the wheel's axle.

In light of the miniscule power output, is it any wonder that Count Karl was in no hurry to offer Bessler 100,000 thalers to buy the rights to the invention?

To make Bessler's inventions practical, it will be necessary to boost their power output up into the thousands of watts. Whether this can be done will, unfortunately, have to await our complete understanding of what the secret mechanism was and how it operated.


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: Bessler Wheel Math

Post by AgingYoung »

Ken,

The stampers in the Bessler wheel might have acted as a governor loading the wheel with a small amount of work to keep it from running away. Is this
25 watts the power that the stampers consumed or is it based on the amount of work it did to lift weight.

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re: Bessler Wheel Math

Post by ovyyus »

Gene, the wheel was also demonstrated running with no loads attached. You should read John's books. You could start with "Perpetual Motion - An Ancient Mystery Solved?", which you can download for just a few bucks. I think it could really clear a few things up for you.

Jim, 2kW? - wish it were true!
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re: Bessler Wheel Math

Post by ken_behrendt »

Gene...

My calculation of 25 watts output power was done years ago and was based on how I thought at the time that the 8 4 lb weights inside of the drum were dropping as they shifted on the descending side of the wheel. The mechanism I then believed was his secret proved not to be, but the calculation seems to be "in the ball park" anyway.

Unfortunately, we can not go from power calculations based on the external work the wheel performed to what its internal mechanism must have looked like. However, I think the low power output does indicate that he used relatively light weights (only 4 lbs each) that were about 6 ft from the wheel's axle.

Yes, I agree that the stampers would definitely prevent the wheel from "running away" and actually reduced its rotational velocity when it was in motion. However, I can not shake the feeling that the stampers main job was to mask any internal sounds issuing from the hollow drum by its shifting weights that might have given the curious critical hints as to how the device operated.


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: Bessler Wheel Math

Post by rks1878 »

Ken and others:

The power output was the same force that 38 lbs exerts on a six foot long arm, in a horizontal position.
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