Another arithmetic question

From an online tire size calculator:
Tire Height (Diameter) 31.7 in 806 mm
Wheel Size 16 in 406 mm
Sidewall Height 7.9 in 200 mm
Tire Width 9.3 in 235 mm
Tire Circumference 99.7 in 2,532 mm

Just rough calculations using these dimensions and simple area formulas and addition/subtraction, the external surface area of a tire that size is approximately 2100 square inches.

(31.7/2)^2 X pi X 2 {area of the sidewalls of the tire}
-
(16/2)^2 x pi x 2 {area of the sidewalls of the rim}
+
99.7 x 9.3 {area of the tread}
=
~2100 square inches
x
80 pounds per square inch
=
168,000 pounds of force

Based off the inside dimensions of the tire it will be a bit less, but you get the idea.
 
(quoted from post at 07:13:59 06/29/23) At a glance I'd say you want cubic measurement, not square? gm

Air pressure is measured in pounds per SQUARE INCH. That is an area calculation. It only works against the surface of the vessel containing it. The volume of air is not really relevant to the force that is being contained.

Notice the handle, BarnyardENGINEERING. We did pressure vessels in college.
 

In two different posts he asked for square inches as in surface area inside a tire.

Since a tire does not expand like a ballon, the surface area of a tire does not really change much.

A rough measurement can be had by using a flimsy or clothe tape measure from the bead, down the sidewall, across the tread and up to the opposite bead, inside the tire.
Then measure the inside circumference of the tire tread.
Then multiple those two dimensions for square inches.
 
You could get a rough estimate by calculating the volume and surface area a similarly sized torus.

https://www.engineersedge.com/volume_calc/torus.htm
 
OD is 30.73 and ID is 15. Using area exposed to pressure is 3.14*30.73^2/4 minus 3.14*15^2/4 results in 565 square inches. At 80psi that's 45,000 lbs trying to blow out sidewalls. There's also the tread area to consider. The tread width is 9.25 inches. Tread area is 3.14*30.73*9.25 = 892 square inches. At 80PSI that's 71,000 lbs total but of course that's spread across the whole surface. For me, the 45,000 lbs is the reason I cringe whenever adding air to old tires.
 
(quoted from post at 11:49:05 06/29/23) OD is 30.73 and ID is 15. Using area exposed to pressure is 3.14*30.73^2/4 minus 3.14*15^2/4 results in 565 square inches. At 80psi that's 45,000 lbs trying to blow out sidewalls. There's also the tread area to consider. The tread width is 9.25 inches. Tread area is 3.14*30.73*9.25 = 892 square inches. At 80PSI that's 71,000 lbs total but of course that's spread across the whole surface. For me, the 45,000 lbs is the reason I cringe whenever adding air to old tires.

45,000lbs per sidewall. There are two sidewalls, so 90,000lbs total. Then add in 71,000lbs for the tread.

That's 161,000lbs total force against the surface of the tire. Pretty close to my calculation of 168,000.

Air pressure acts in all directions, against every surface at all times. It doesn't pick and choose which surface to act against.

Fortunately, tires don't normally pop instantly like a balloon. They normally split and release the pressure in a localized area, so you only get a fraction of the total destructive power of the tire.
 
No but when they do it is not good blew one hauling
cows yesterday and it lifted the back of my one tone
truck with the trailer hooked on I was empty .
 
It is not the PRESSURE per se that makes a tire so dangerous. It's the ENERGY stored in the compressed air. If you had a tire mostly filled with water and a little compressed air, it wouldn't do much when it blew because water is nearly incompressible and can't store much energy. But there's a lot of energy in compressed air; when a tire blows that energy is instantly released.
 
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