Volume of a Cyldrical tank on it’s side by measuring its content’s height.
Example:
2000 gallon fuel tank which is on it side. (like a soda can tipped over)
Determine gallons by openingin a cap on the top, drop in a graduated measuring stick. Read the inches of fuel on the stick and determine gallons in tank.
I need a formula ladies and gentlemen.
Here is what we have for numbers (in inches):
A=(Length of tank)Mine is 144
B=(Radius)Mine is 30
C=(Inches of fuel)
D=(Gallons of fues)

Nope, this is a real life application…
It is a real fuel tank my father has. I remember sitting in Highschool saying “This algebra crap is NEVER going to help me in the real world” Ah to be young and dumb.

D = (K/pi) x (H x sqrt(1 - H²) + arcsin(H)) + (K/2)

where K is the capacity (2000 gallons) and H is computed as above from your B and C. H should always be a value from -1 to +1, as a check.

Note that the tank’s length (your “A”) is not needed, assuming you were honest in reporting the tank’s volume of fuel when full. I didn’t verify this though.

Note also that this problem is beyond the reach of algebra. It’s really a calculus problem.

You measure the height h. The radius of the tank is r. The distance from the center to the liquid is (r - h). You can compute the angle between the perpendicular (of length r - h) and the radius that just meets the surface of the liquid (of length r), using trig:

theta = acos ((r-h)/r)

(theta is in radians). Now, we can compute the area of the smallest wedge from the center of the circle that includes all the liquid, it is:

A * (2theta)/(2pi)

where A is the area of the whole tank, or 2pir^2. Simplifying:

2 * theta * r^2

To find the area of the cross section of the liquid, you have to subtract a couple of the triangle forming the point of the wedge. It has height (r-h), and using the Pythagorean theorem, we see that it has a (half) base equal to

sqrt(r^2 - (r-h)^2)

Simplifying

sqrt(2rh - h^2)

So the (whole) area to remove is:

(r - h) * sqrt(2rh - h^2)

Or, subtracting:

(2 * theta * r^2) - ((r - h) * sqrt(2rh - h^2)

Now, if the tank is more than half full, let h be the height above the liquid, and subtract the area above from the total area of the tank.

Of course, that’s a cross sectional area. Don’t forget to multiply by the length of the tank (and don’t tell me that it has hemispherical endcaps, please).

Draw a circle, radius R, and draw a horizontal line (below centre) to represent fuel. Draw a vertical line through the centre from top to bottom to represent your rod. Now from the centre draw two lines to where your “fuel” meets the circle. You should now have two right angle triangles. Label their common line H and considering them as one triangle label the top angle (theta). Okay here goes:

Area of a segment (your fuel) = area of sector - area of triangles

Area of sector = (pi)R^2(theta) / 360

Using trig: cos(theta/2) = H/R. Since H + C (your inches in fuel) = R => H = R - C

So: cos(theta/2) = (R - C) / R = 1 - (C/R)
(theta/2) = inv(cos)(1-(C/R))
(theta) = 2 x inv(cos)(1-(C/R))

Area of sector = (pi)R^2(2 x inv(cos)(1-(C/R))) / 360
= (pi)R^2(inv(cos)(1-(C/R)) / 180

Still with me?

Triangle with hypotenuse R and height H has base equal to:

While that’s true, you don’t actually need any calculus to solve it. You might think you need to integrate to find the area under a curve, buy you don’t. You can simply leverage the fact that someone else has already done that integration (when they derived the formula for the area of a circle).

Fair enough; I overstated the case. There are probably several non-calculus approaches to solving this particular problem. (I found an integral to be the easiest, but that’s just me.)

I can’t seem to create an easy formula. I have come up with several difficult ones, though. Here’s one not posted yet. Got a good calculator? It is calculating the area of the circle that’s tough. After that it is just multiplying by the length of the cylinder.

However, that said…

Suppose you measure the height of the liquid and find it to be h. Let r be the radius of the cylinder’s circle (half the diameter). First, calculate an intermediate variable t which will be an angle measured in degrees by
t = arccos(1 - h/r)
arccos might be cos[sup]-1[/sup] on your calculator in which case
t = cos[sup]-1[/sup](1 - r/h)
t should always be between -1 and 1.

The area in question is then
(t/180)(pir[sup]2[/sup]) - 0.5*r[sup]2[/sup]*sin(2t)
Multiply this by the length of the cylinder and you’re done.

This will work no matter what the height is. No flip-flopping formulas.

Not quite. There are standard algebra formulae for calculating these volumes (as well as wetted and surface areas) of cylinder problems like this in the backs of engineering reference manuals. My PE study manual even has a simple algebra solution. I have a spreadsheet that will do this, and take into account the volume (and area) of hemispherical, dished, toriconical, and other heads on the ends of the cylinder too.

SORRY!! 1 - h/r should always be between -1 and +1. t should be between 0 and 180. Also note that in the second version I flipped it. It should be h/r in both of them. Sheesh.

and letting theta be in radians (so you can cancel pi with 180):

theta * r^2 - 0.5r^2sin(2*theta)

Notice that you only differ from the other formulas:

theta * r^2 - (r - h) * sqrt(2rh - h^2)

by that second factor, which is identical (one using trig, the other using the Pythagorean theorem).

My first smartass answers were to stand the tank on its end, and measure the volume of the cylinder of fluid; or else to weigh the tank and use the density of the fluid and the (estimated) weight of the tank; or else to fill it with a measured amount of water until it overflowed.

I doubt that the sides of this tank are flat, so you will need to use a formula that will account for the volume in the heads. See page 3 of the link below.

Ah, not quite. Without using that identity and sticking with mine, it doesn’t matter what the height is, though, since sin(2theta) will change signs when theta>90.

I didn’t notice we did the same thing, though. It seems like the best way to do it.