# Homework Help: How big is this cylinder?

1. Nov 18, 2013

### curtmorehouse

1. The problem statement, all variables and given/known data
I have a cylinder laid on it's side, a.k.a. a storage tank. I need to track the volume of liquid stored in said tank. I have a 'tank chart' that shows the volume of liquid in the tank for a specific 'depth' measurement. I think the measurements of the tank are wrong (and therefore the tank chart is wrong), but can not see the tank because it is in the ground.

I need to know the tank dimensions given a couple sets of known measurements for increases and decreases. The tank is said to be 80" diameter by 204" long.

For an increase of 6001 gallons of liquid, the depth changed from 12.5" to 80.375"
For a decrease of 207.2 gallons of liquid, the depth changed from 80.375" to 78.25"

I can provide more measurements like these if needed.

2. Relevant equations
$h =$tank_depth;
$d =$tank_diameter;
$r =$d / 2;
$l =$tank_length;

$x = pi() *$r * $r / 2;$y = $r *$r * asin(1-$h /$r);
$z = ($r-$h)*sqrt($h*(2*$r-$h));
$area =$x - $y -$z;
$volume =$area * $l;$volume = \$volume / 231;

3. The attempt at a solution
MY attempts at a solution so far result in finding that increasing the size of the tank's length or diameter can make one set of depth/volume measurements work, but make other measurements fail.

2. Nov 18, 2013

### Simon Bridge

You don't have enough information [edit: we-ell, maybe...] ...
The reason is that the change-in-height when adding a volume of liquid will depend on how much liquid was already in the tank. [... and it's easier on the math to have a controlled experiment.]

[Chestermillar (below) has valid points about the measurements.]

I'd start with: $V=LA$ ... where L is the overall length of the tank, and A is the area of the circular segment at the end - which depends on the height h of the liquid.

Putting axis with the origin at the bottom of the tank...
The equation of the end-circle of the cylinder is: $x^2+(y-R)^2=R^2$
The area of the cord below a given height y=h is the area of the triangle subtracted from the area of the segment you get when you sketch it out.

If we define $\alpha$ to be the half-angle at the center, then we can write: $$A= \pi R^2 \frac{2\alpha}{2\pi}-x(R-h)$$... deliberately left unfinished.

Off this you can get an equation for volume as a function of height with L and R as unknown parameters.

Now do an experiment: measure height, add fixed volume, measure again, repeat for as many measurements as you can stand. Ideally you want a bunch of measurements close to when the tank is half-full, and close to when it is completely full or completely empty (either), and a few in between. Say 20-30 measurements in all.

Then you graph your data - and try to adjust L and R to fit the resulting curve.

There's usually an easier way - depending on how accurate you need to be.
i.e. you could use a length of rebar as a probe, or a metal detector, to determine the dimensions of the tank much more easily... but I'm guessing the tank is plastic and it is below a slab of concrete or asphalt or something.

If you find the math tricky then you may just have to make lots of measurements, filling the tank in small increments, from empty to full, and just using that list. If it is worth money to you - consider hiring someone.

Last edited: Nov 18, 2013
3. Nov 18, 2013

### Staff: Mentor

Something definitely seems wrong. For the dimensions that you gave, I calculate a total volume for the tank of about 4400 gallons. That's less than the 6000 gallon change in volume that you measured when you increased the depth from 12.5" to 80.375". Incidentally, if the tank diameter is only 80", how can you have a depth of 80.375?

4. Nov 18, 2013

### Simon Bridge

I still think the problem is being handled by a more difficult way than strictly needed.

80.375in sounds like "overfull" - 0.375" sounds like the lip on the top opening and/or a slight angle on the measuring rod. Which suggests that we know the diameter!

i.e. The "height" of the liquid can be measured, then the diameter of the tank can be determined directly. i.e. a weighted line can be lowered through an opening until the bottom is struck, or a long rod can be lowered vertically into the tank etc.

The length of the tank is then obtained, knowing the diameter, from the maximum volume it can hold.

So if you've got the physical tank right there - why not just measure what you need.
It sounds like the sort of thing that would be fun to set as an exercize for students though.

5. Nov 18, 2013

### haruspex

If the diameter and depths are taken to be correct, the two volumes are in pretty much the right ratio. So it remains to calculate the length that will give you those volumes. Can you figure out the area in the liquid for a vertical cross-section at one of those depths?

6. Nov 19, 2013

### curtmorehouse

Correction: The tank is said to be 108" diameter by 204" long.

7. Nov 19, 2013

### Staff: Mentor

With these new numbers, what does your tank chart predict for the changes in volume in the two cases you considered, and what does your computer program predict?

Chet

8. Nov 19, 2013

### Staff: Mentor

It seems more and more like that 204" is a typo, and the actual length is 240". That would make the tank 9 ft in diameter and 20 ft long.

Chet