I have a cylindrical fuel tank. of total capacity. = V. if i insert a yard stick,I measure X inches depth?

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wow.. hmmm.. ok.. lemme try to think... fuel tank.. laying on side.. so basically.. starting with a full fuel tank... you have an expanding bubble forming at the top of the tank as it goes down...

Ignoring anything less than half a tank of fuel to start with.. to keep some of the math simple:

Ok.. first.. in geometry, the empty bubble shape at the top is called a segment of the circle... the top of the fuel would be called a chord (a line that meets the circle in two points).

if you draw radius lines from the two points where the top of the fuel meets the circle to the center of the circle.. then the area of the circle inside these lines is called a sector of the circle. (it includes the segment).

The circumferance of your circle is... 2pi(r) or 2pi(12) or 24pi.
The area of the circle is pi*r^2 or pi*(12^2) or 144pi.

The amount of fuel in the tank is represented by the ratio of the area of the segment to the area of the circle... so if we let A = Area of the Segment, and F = Area of Full circle, then 90[(F-A)/F] is how much fuel is in the tank since 90 gallons is the full circle when A=0

your yardstick makes a diameter line when you measure the depth of the fuel.. and this diameter line meets the chord at a right angle.

Imagine a triangle formed by:
1. a line from the center of the circle to the point where your yardstick meets the top of fuel... of length (d-12) if you have more than half a tank of fuel.
2. a radius line from center to the right side where the top of the fuel meets circle... of length 12
3. the right half of the chord (line formed by right side of the top of the fuel).. of.. hmmm.. since this is a right triangle, it's length... sqrt[144 - (d-12)^2]... OR... since you have a circle... it's length can also be described as the sin{arccos[(d-12)/12]} ... cos = adjacent/hypotenuse... sin = opposite/hypotenuse

Now.. the area of this triagle is 1/2{(d-12)*sqrt[144-(d-12)^2]} and with two of them (one on left and one on right) we get a total area for the triangles under the segment as {(d-12)*sqrt[144-(d-12)^2]}

Ok.. remember the trigonometry I mentioned?... arccos[(d-12)/12].... this is HALF of the angle formed by the arc at the top.. so the angle of our segment and therefore of our sector is:

2arccos[(d-12)/12]

and the portion of the full circle is {2arccos[(d-12)/12]}/360

so it has an area of 144pi*{2arccos[(d-12)/12]}/360

but.. we only want the area of the segment.. not the full sector

so we need to subtract the area of the triangles under the segment.. {(d-12)*sqrt[144-(d-12)^2]}

and we end up with:
A = {144pi*{2arccos[(d-12)/12]}/360} - {(d-12)*sqrt[144-(d-12)^2]}
as the area of the segment..

now to get the amount of fuel.. we plug this long value into the equation for the amount of fuel.. namely 90[(F-A)/F] where F is the area of the full circle.. or 144pi

remember.. this only works for fuel levels OVER 1/2 tank.. I'll leave the calculation of those under half up to you....Show more

divide X by the height of the tank.

you can get height as (90*231)/(144*pi)

144=(24/2)^2
231 = cubic inches per gallon...Show more