Wagner Wheel Math

[Jonathan]

I've recently read John's reproduction of "Das Triumphirende...", and have done an analysis on Wagner's Bratenwender (turnspit). The attachments show my conclusions if anyone's interested.

[Jonathan]

I'm surprised there have been no replies. Does anyone see any errors (beside the mispelling of "circumferential" in the drawing)?
I notice that the views of the thread and dowloads of the document don't match, so maybe I should clarify, that in the document I prove that the device as Wagner presented it couldn't have functioned the way that Bessler's did (I do note in it that if there is a huge pulley reduction, then it could work, but the rate at which it lifts the load would be ridiculus and nothing like Bessler's).

[ovyyus]

Hi Jonathan,

Good analysis. It seems that Wagner was more interested in his own position in the World than in the facts surrounding Bessler's demonstrations. It also seems pretty clear that Wagner could offer no supporting evidence that his turnspit mechanism was anything more than another speculative attempt to discredit Bessler.

As you've shown, Wagner's theoretical turnspit arguement was based on flawed reasoning and could not replicate Bessler's wheel demonstrations. He must have been a furious and very frustrated man after the completion of Bessler's long duration test!

[Jonathan]

Thanks Bill.
It occured to me that the proof relies on the engraving being an accurate representation of Wagner's device, and since there are a few subtle mistakes, I've checked it's consisentcy with the dimensions of the Merseburg wheel further. The wheel thickness in the engraving is 21.5 mm, which in the real world translates via the scale as 21.5mm*(170.7cm/121mm)=30.33cm=11.94in, which is very close to the 11.15in posted on your website. The iron axle radius I used in the math is .71cm=.28in, whereas the same of the Merseburg wheel was reportedly .375in. This is a big difference since the error margin approaches the size of the numbers themselves, but since a smaller version of this radius is beneficial to Wagner's case, it's conservative to leave it that way. Your website also says that the wooden axle radius was about 3in, whereas the axle I used in the math is 10.7cm=4.2in. Because this radius is greater, this was less conservative. I should have checked this before and altered my math. Here are the alterations using the more conservative wooden axle radius of 2.8in (=5.6in dia./2):
T_w=6.16831Nm
S_t=1.265987*10^11 Pa
Which is still much greater than 5 to 20*10^8 Pa.

[Jonathan]

I found an even bigger error recently, so I've altered the paper accordingly. And while I was at it, I swiched to using the stamps as a load instead of friction and a box of bricks.

[John Collins]

Nice work Jonathan. I wonder if I might use your analysis in my forthcoming book - subject to your approval and any refining you might wish to add?

John Collins

[Jonathan]

Well yes, but as you can see, at any moment I might look back at one of my papers and notice a glaring error. The problem has been fixed where I write
T = d/(2(θf/2π)) = dπ/θf
instead of
T = d/(2θf).
The idea was to take the axial length of the fusee, divide by twice the number of turns of chain around it, and you'll have the thickness of one chain (the former equation). The problem before was that I divided by twice the number of radians subtended by chain (the latter equation), and that deflated the thickness of the chain by 2π. That in turn deflated the cross-sectional area of the chain by 4π², which inflated the tensile stress by the same. This means that in the original paper, the tensile stress should have been about 4.7*10^9. That's better than the figure I posted in the recent figure, but there is too much uncertainty about the pulley reductions, and the fictional load is ugly to calculate.
Back to your question, when it comes time, I'll probably want to rephrase it a bit.