The conundrum of "The Oil Tank Problem"

[jessi1236w]

In my Calc 162 class, we were recently assigned a problem that goes a little like this:

"American Transport delivers fuel by oil trucks to underground tanks. Upon delivery, the driver lowers a verticle ruler (like a car's dipstick) into the tank to measure the amount of fuel left in the tank." This precaution enables the driver to stop the truck's pump before overfilling the tank. For this purpose however, the driver needs a means to convert the reading on the ruler to an estimate of remaining volume in the fuel tank. A conversion chart would be ideal. Is there a formula to generate such charts by computer? In each case, the tank is in the shape of a horizontal circular cylinder. "

Basically, I don't even know where to start. We have been recently working with more advanced integrals and I believe they are involved. (ie we have covered integration by parts, trigonometric subsitution, and strategies for integration) I tried working with some basic algebra to manipulate the problem and also tried working with the solids of revolution to tease the equation into integral format but didn't really get anywhere. Any ideas?

 

[galactus]

Yes, you can generate a general formula using calc.

The equation of a circle of radius r is given by \(\displaystyle \L\\\sqrt{r^{2}-x^{2}}\)

Center the circle at the origin. If the driver sticks the ruler down into the tank he measures the fuel of depth h.

You can use the integral:

\(\displaystyle \L\\2L\int_{-r}^{-r+h}\sqrt{r^{2}-x^{2}}dx\)

r is the radius of the tank, L is the length of the tank, and h is the depth of fuel measured on the rule. All the info you need to find the volume at a designated depth.
Integrating this gives a formula in terms of r and L and h.

I would use some sort of technology to generate it. I used my TI-92.


You know, there's another thing you can do.

The formula for the depth h, knowing the radius and the central angle is given by:

\(\displaystyle \L\\h=r(1-cos(\frac{{\theta}}{2}))\)

Solve this for theta:

\(\displaystyle \L\\{\theta}=2cos^{-1}(\frac{r-h}{h})\)

Sub this into the formula for the area of a circular segment:

Area of circular segment is: \(\displaystyle \L\\\frac{1}{2}r^{2}({\theta}-sin({\theta}))\)

Then multiply that by the length, L. It'll be in terms of r, h, and L. This will give you the same thing without calc.

Rather cumbersome, but not bad with technology.

 

[soroban]

Hello, jessi1236w!

American Transport delivers fuel by oil trucks to underground tanks.
Upon delivery, the driver lowers a verticle ruler (like a car's dipstick)
into the tank to measure the amount of fuel left in the tank.
This precaution enables the driver to stop the truck's pump before overfilling the tank.
For this purpose, however, the driver needs a means to convert the reading
on the ruler to an estimate of remaining volume in the fuel tank.
A conversion chart would be ideal.

Is there a formula to generate such charts by computer?
In each case, the tank is in the shape of a horizontal circular cylinder.

              * * *
          *           *
        *               *
       *                 *

      *         O         *
      *         *         *
      *       / θ \       *
          R /   |   \ R
       *  /     |C    \  *
       A* - - - * - - - *B
          *     |d    *
              * * *

We want the area of the segment at the bottom of the diagram
. . with depth \(\displaystyle d\).

The area of sector \(\displaystyle OAB\) is: \(\displaystyle \:\frac{1}{2}R^2\theta\)

The area of triangle \(\displaystyle OAB\) is: \(\displaystyle \:\frac{1}{2}R^2\sin\theta\)

Hence, the area of the segment is: \(\displaystyle \:A\:=\:\frac{1}{2}R^2\theta\,-\,\frac{1}{2}R^2\sin\theta\)

. . \(\displaystyle A \:=\:\frac{1}{2}R^2(\theta\,-\,\sin\theta)\)

And the rest is not a simple problem . . .


Suppose we want to know when the tank is \(\displaystyle \frac{1}{4}\) full.

Then we have: \(\displaystyle \:\frac{1}{2}R^2(\theta\,-\,\sin\theta) \:=\:\frac{1}{4}\pi R^2\)

. . and we have: \(\displaystyle \:\theta\,-\,\sin\theta\:=\:\frac{\pi}{5}\)

This equation cannot be solved by any elementary methods.
. . We can only approximate the answer.
(A computer program would be an excellent tool to use.)

Once we find \(\displaystyle \theta\), we can find \(\displaystyle d\) with: \(\displaystyle \:d\:=\:R\left[1\,-\,\cos\left(\frac{\theta}{2}\right)\right]\)