jim_mich wrote:Dunesbury,
Your estimate seems good. I would expect four or six iron hoops rather than just your two.
You used red oak flooring as a reference. Red oak is native to North America. I doubt Bessler would have used red oak from North America. It's weight is usually about 0.027 lbs/cu/inch.
European white oak is heavier than red oak. It is usually about 0.029 lb/cu/inch. Using white oak rather than red oak would add maybe 4 or 5 lbs more to the weight of the axle, which is insignificant compared to all that must be guessed at.
![Image](http://my.voyager.net/~jrrandall/Jim_Mich.gif)
Stamper material on power output page listed as red oak. I used that as reference for calculation. Is hammer material white oak, or is it insignificant?
I estimate axle boards milled lengthwise for joining strength. Maybe V shaped profile, 2 hoops at end of plugs. I guess he did not dismantle axles but reused them unless they were part of prime mover.
Depending on how weight levers connected to wheel, material for these and crossbars could have been even smaller dimensions. If connected through axle and to 12 foot wide octagon shaped periphery,(for 8 weight configuration) to make math easy say 1 x 1 inch x 12 foot for crossbars. If needed, additional pieces could be attached midway of crossbars, forming 6 foot octagon.
These pieces for crossbars with additional 6 foot octagon for strength, and periphery, would be about 103 feet of 1 inch square (4.592 feet for each outer periphery piece, 2.296 feet for each inner octagon piece).
103 x 12 x .029 ~ 36 lbs.
Framework for linen or veneer was round.
Did he steam bend strips of wood for this? If so, these strips might have been 1/2 x1/2 inch. To make math easy, say 1x1 inch.
Circumference 37.7 feet.
2 sides = 75.4 feet
Width 18 inches
8 pieces to connect frame together ( and also connect to lever weight crossbars so all goes round together), 12 feet.
75.4 + 12 = 87.4 feet
87.4 x .029 x 12
~ 30.4 lbs.
8 boards weights landed gently on, unknown size, say 8 lbs total.
8 levers unknown length. Max length ~ 2.5 feet, 20 feet of 1x1 inch ~ 7 lbs.
Linen covering unknown weight.
Axle,levers, crossbars, periphery and linen frame total weight could have been ~
75 + 7 + 36 +30.4 + 8 = 156.4 lbs. plus linen.
If two way wheels had 8 four lb. weights , total weight ~ 188 lbs. plus linen, if my math is correct, using these dimensions for construction.
From page 3 of this thread
furcurques wrote:So, that means that just the 32 pounds of weights alone moving on the rim of the 12 foot diameter wheel rotating at 26 RPM would have enough kinetic energy to lift your 200 pound person about 8 inches into the air.
So, although the 32 pounds of mass on the rim would have enough kinetic energy to lift the 200 pound person about 8 inches into the air, the person would only be lifted about 1 inch by suddenly grabbing hold of the rim.
This, of course, was neglecting the mass of the wheel itself, which most likely would dominate when it comes to our person launching.
I'll spare you the details on this one, but using the moment of inertia equation for a solid disk, one in which there is a uniform mass distribution throughout the volume, I calculated the numbers for a 300 pound wheel of the same dimensions and speed as before.
A 300 pound, 12 foot diameter solid disk rotating at 26 RPM should, then, have enough kinetic energy to lift the 200 pound person 3 feet 1 inch into the air, but for the person grabbing onto the rim where we have to consider the conservation of momentum, the person would only be lifted 1 foot 1 inch.
These calculations, of course, don't take into account any extra drive force, but only the stored kinetic energy in the flywheel.