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Let's see if my Math/Physics is accurate OK.
The surface area of a torus is: 4 ? (pi squared) ? R ? r See image for what R and r are. ![]() I don't have a 6" wheel in front of me, but let's say R is 5" and r is 2": 4 x 3.14159 x 3.14159 x 5 x 2 = 394 square inches. So 1 psi would exert 394 lbs of force distributed evenly. Not a tremendous amount of force. |
Well . . . - lets think Bagel . . .
OK, this is off the OP, but lets take the tire assembly and slice in the middle plane like a bagel. Then the projected area of the slice is the area acted on by the pressure.
If the ID is 5" and the OD is 12 " the it is 3.14 X (12^2-5^2)/4 or 93 sq inches. So lets say it is 100 sq-in , so 1 psi would be about 100 lb force. 28 psi is 2800 lb force, half of which is equivalent to acting on the OD perimeter of the tire. So, the rim sees 1400#. Still not a lot for 3 1/4 AN bolts, but the axle nut would retain the halves (of the RV design) if the rim fasteners were to fail. The Cessna in the video posted earlier would NOT be retained by the axle nut, so all wheels are not created equal. Thinking is in order. I had not thought about this before and will certainly give consideration to the potential energy involved. Knowledge helps in flying and working safely. To the OP, I did not see the assembly procedure on your blog, did I just blank out and not see it? |
Thanks for the knowledge that is passed on
I am neither a mechanical or electrical engineer (though I was a fire department Engineer for two years) and have decided that when I change my first tire ever, I'll let the air out first - just don't like to see things flying around the maintenance area at near the speed of sound - just makes sense to me!
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