Posted by MarkB_MI on September 07, 2019 at 16:05:09 from (174.230.7.60):
In Reply to: math posted by ohiojim on September 07, 2019 at 13:21:14:
OK, I'm going to assume the tank is laying on its side, so the circular ends are vertical. I could look up the formula, or maybe use calculus (if I could remember any of it), but this is a case where you can reasonably estimate without a lot of fancy math.
So the circular cross section is πr2 = 3.14 x 192 = 1134 in2
But it's only filled to 15 inches, which is fairly close to the 19 inch radius. If it was half full, the cross section would be 1/2 of 1134, or 567 square inches.
But since it's less than half full, we have to subtract the cross sectional area between the fuel level and the half-full level. This is where we cheat: Since the curvature of the tank is almost vertical at this point, we can pretend that area is a rectangle 38 inches wide by 4 inches high. (It's actually a bit less than that.) 38 x 4 = 152 in2
So the cross sectional area up to the fuel level is 567 - 152 = 415 square inches.
The total volume is the cross sectional area times the length of the cylinder: 415 x 60 = 24,900 in3
Converting to gallons: 24,900 in3 x .00433 gal/in3 = 108 gallons
About 110 gallons, I'd say. Note that this method only works when the tank is close to half full.
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