What is the approximate weight of the air inside the tire in English Engineering Units (tire outside diameter = 49", rim diameter-22", tire width-19" see: tire side view. ndf?

1.265 Pounds

Explanation:

Data provided:

Tire outside diameter = 49"

Rim diameter = 22"

Tire width = 19"

Now,

1" = 0.0254 m

thus,

Tire outside radius, r₁ = 49"/2 = 24.5" = 24.5 * 0.0254 = 0.6223 m

Rim radius, r₂ = 22" / 2 = 11" = 0.2794 m

Tire width, d = 19" = 0.4826 m

Now,

Volume of the tire = π (r₁² - r₂²) * d

on substituting the values, we get

Volume of air in the tire = π (0.6223² - 0.2794²) * 0.4826 = 0.46877 m³

Also,

Density of air = 1.225 kg/m³

thus,

weight of the air in the tire = Density of air * Volume air in the tire

or

weight of the air in the tire = 1.225 * 0.46877 = 0.5742 kg

also,

1 kg = 2.204 pounds

Hence,

0.5742 kg = 0.5742 * 2.204 = 1.265 Pounds