The Fuel Tank Problem?

Question

Imagine you have cylindrical tank 4 units in diamiter and and 10 units long. When the installation company put the tank in the ground they laid it on its side. Now you must devise a measuring rod to put in the take in order to show the volume of fuel contianed within. What formula do you use to know how to make the measuring stick.

Answer 1

You know, from the pythagorean theorem, that (w/2)²+(r-h)²=r², where r is the radius of the tank, h is the height of the fuel from the bottom, and w is the width of the tank at the surface of the fuel. It follows that w=2√(r²-(r-h)²). Since the diameter of the tank is 4, it follows that the radius is 2, so this is w=2√(4-(2-h)²). The area of the part of the circular cross section below height h is then given by [0, h]∫2√(4-(2-x)²) dx. Simplifying this:
[0, h]∫2√(4-(2-x)²) dx

We need to find an antiderivative of 2√(4-(2-x)²). let u=x-2. Then:
∫2√(4-(-u)²) du
∫2√(4-u²) du
Let 2 sin v = u
∫2√(4 - 4 sin² v) * 2 cos v dv
∫4√(4(1-sin² v)) cos v dv
∫4√(4 cos² v) cos v dv
∫8 cos² v dv
∫4(2 cos² v - 1 + 1) dv
∫4(2 cos² v - cos ² v - sin² v + 1) dv
∫4(cos² v - sin² v + 1) dv
∫4(cos (2v) + 1) dv
4∫cos (2v) dv + 4∫1 dv
2 sin (2v) + 4v + C
4 sin v cos v + 4v + C
4 sin v √(1-sin² v) + 4v + C
Now, originally we had 2 sin v = u. Therefore:
2u√(1-(u/2)²) + 4 arcsin (u/2) + C
2u√(1-u²/4) + 4 arcsin (u/2) + C
And since u=(x-2)
2(x-2)√(1-(x-2)²/4) + 4 arcsin ((x-2)/2) + C
2(x-2)√(1-(x²-4x+4)/4) + 4 arcsin (x/2-1) + C
2(x-2)√(x-x²/4) + 4 arcsin (x/2-1) + C

Evaluating this big ugly integral from 0 to h:

2(h-2)√(h-h²/4) + 4 arcsin (h/2-1) + 2π

Note that the radius of the tank is 2, so this formula should say that when the tank is full (i.e. h=4), the cross-sectional area occupied by the fuel is 4π, that the area is 2π when the the tank is half-full (h=2), and the area is 0 when the tank is empty (h=0). You will note that this formula does in fact give these values. Of course it should, since it was derived from a valid method, but with the manipulations as complex as they are, it helps to do a sanity check.

Finally, since the tank is 10 units long, multiply the cross-sectional area occupied by fuel h units high, and multiply by 10 to get the total volume. Therefore, the volume of the fuel at h units high is:

20(h-2)√(h-h²/4) + 40 arcsin (h/2-1) + 20π

Phew.

Answer 2

Hi. If you just want the level it is, of course, simple. But you want to know what the volume is at, say 1/3 full or 1/2 full or 1/6 full, right? The stick needs to be at least 4 units long in it's measurement area. The number is 0 when empty, pi r^2 h when full (62.83 units) and half that value at half tank.

Answer 3

Basically the volume is the length times the area of a part of the circle defined by the height of its chord (the fuel level). This is the formula I wrote for the semicylinder part of my oil tank. It's in Basic; ATN is arctangent, SQR is square root, H is fuel depth, R is tank radius. Obviously if the tank is more than half full you input the distance of the fuel level from the top to find the airspace area and derive fuel area from that.
T1 = SQR(R ^ 2 - (R - H) ^ 2)
'T1 is halfwidth of fuel surface
T2 = ATN(T1 / (R - H)) * R ^ 2
'T2 is 2 * area of sector defined by 2 radii: one vertical, one to the edge of the filled portion
Area = T2 - (R - H) * T1
'Area is 2 * sector area minus 2 triangular airspace portions in T2
EDIT: The above code and Pascal's calculus-based approach generate the same results.

Answer 4

If it is 4 units in diameter, the rod should be 4 units long (+ the depth of the fuel tank position). The tank can maximally contain 4*4*Pi*10=160Pi units cubed of fuel.
So when you use the rod and it is X units "wet" you can calculate the actual amount of fuel this way:
X:4=(amount of fuel in units cubed):160Pi
So, amount of fuel in units cubed is 160Pi*X/4.