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### Fuel Left in an Ellipsoidal Tank

```Date: 10/19/2002 at 01:33:03
From: Chris Wells
Subject: How much fuel is left in a tank at any depth, given that the
tank is ellipsoidal?

While working for an oil company 25 years ago, I was encamped in the
desert in western Argentina. It was about 6 hrs round-trip to the
nearest gas station. We had a tanker truck that we used to supply fuel
for our equipment. The tank was shaped like a flattened cigar with
abruptly curved, symmetric ends. Early in the project we realized that
we didn't have an accurate method of estimating volume as fuel was
drawn, so we calibrated a gauging stick when first we refueled. That
worked okay, but I always thought there must be a more elegant
solution.

I hope that you can provide me with a general solution, but I am also
somewhat hopeful that you can guide me to a solution. On the other
hand, this question hasn't exactly been burning in my soul. It just
sort of sits there as an irritating, unfinished, unresolved mystery.

- Chris
```

```
Date: 10/19/2002 at 07:52:23
From: Doctor Jeremiah
Subject: Re: How much fuel is left in a tank at any depth, given that
the tank is ellipsoidal?

Hi Chris,

This can definitely be done, but the really easy way to solve it
involves calculus, and you don't clearly say what level of mathematics
you want to deal with here. Are you interested in a calculus-based

If we assume that the ends are not curved at all but actually
vertical, we can find the partial area of an ellipse and then multiply
it by the length. We certainly could calculate the answer with curved
ends but it shouldn't be necessary for a first approximation.

To calculate the partial area for an ellipse, first you need to
imagine making the shape with sheets of paper. We stack a bunch of
pieces of paper up and we get a 3d shape even though they are very
thin. If the width of the paper gets wider and then narrower again as
we stack them up, we will get something that looks like a elliptical
tank made out of paper sheets.

That's the way to calculate the partial area of an ellipse.

The ellipse's equation can be used to figure out the width of the
sheets as we stack them up, which means we need to have the equation
for the ellipse. Since we don't know the truck's measurements, we
should use a general equation:

y^2/a^2 + x^2/b^2 = 1

The y value is the height of the liquid, but the tank's bottom is not
located at zero: the center of the tank is located at zero. So the
tank is 2a high, where y ranges from -a to a.

We need to find the width (x) of the thin sheets based on the current
height (y),  so we need to solve that equation for x:

y^2/a^2 + x^2/b^2 = 1
x^2/b^2 = 1 - y^2/a^2
x^2 = b^2 - b^2 y^2/a^2
x = sqrt(b^2 - b^2 y^2/a^2)

The area of the tank is the sum of all the thin sheets. Each sheet has
a width of 2x = 2sqrt(b^2 - b^2 y^2/a^2). (They are 2x wide because x
is only the distance from the center of the tank to the edge).  And
the y value runs from y=-a to y=+a, so the whole integral looks like
this:

total_volume = tank_length times the sum from y=-a to y=a of
(current_width times infinitely_thin_thickness)

Or, written mathematically:

L = tank length   2a = max tank height   2b = max tank width

y=a
/
total_volume = L | 2x dy
/
y=-a

which is:

y=a
/
total_volume = L | 2 sqrt(b^2-b^2 y^2/a^2) dy
/
y=-a

and that means the partial volume can be calculated like this:

If the tank is 2a in height then when the liquid inside is "d" deep
(where empty=0 and full=2a) d translates to a y value of:

y=d-a

So the partial volume of the tank is:

y=d-a
/
partial_volume = L | 2 sqrt(b^2-b^2 y^2/a^2) dy
/
y=-a

If you solve that integral and then substitute in the y values that
are the boundaries (-a and d-a) you will get the equation that will
work for any ellipse filled to some arbitrary depth.

If you want help solving the integral or if you have any other

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/21/2002 at 07:43:39
From: Chris Wells
Subject: How much fuel is left in a tank at any depth, given that the
tank is ellipsoidal?

For mearly two decades I tried to solve it algebraically and after
considerable struggle, I realized that I needed a way of summing an
infinite number of infinitely thin intervals. My math background was
starved at that point and would not budge, having never been fed that
particular meal.

Your explanation was very good, indeed. The path to the problem's
solution lies before me, but like Moses' followers, entrance to Canaan
is forbidden me because of my own shortcomings. I fear that I would
have to take a calculus course to understand, else, yours would be a
voice crying in the wilderness.

I would love for you to take the time to explain this to me, but I
would also hate for you to waste your time if a calculus background is
required.

Thanks for your time and effort.

---chris
```

```
Date: 10/21/2002 at 18:54:07
From: Doctor Jeremiah
Subject: Re: How much fuel is left in a tank at any depth, given that
the tank is ellipsoidal?

Hi Chris,

Its really not as complicated as you are thinking. I am going to try
to explain, but keep in mind that there are reasons that entire books
bogged down, just scroll to the end to get the final equation.

The sum of the thin sheets makes up a cross section of the object.
Each sheet has a width and a infinitely small thickness. So the total
cross sectional area is the sum of all the infinitely thin areas.
This sum of an infinite number of infinitely thin areas is called an
integral.

There are ways to do these integrals completely from scratch but
everybody either uses a set of rules to solve integrals or a book with
integrals and their solutions.

Remember that this is the integral:

y=d-a
/
partial_volume = L | 2b sqrt( 1-(y/a)^2 ) dy
/
y=-a

Where L is the length, a and b are the axis of the ellipse and d is
the depth of the fluid.

Its really messy looking! If we want to make it easier we can
substitute in something for y/a. It turns out that a trigonometric
solution works well.

Let's choose  y/a = sin(z)  which is the same as  y = a sin(z)

The dy isn't just there for show. It is an infinitely thin bit of y,
so if we change variables we must change the dy as well. dy is
actually the differential value of y (the slope of the graph of y) so
if y = a sin(z) then dy = a cos(z) dz

If we substitute  y/a = sin(z)  and  dy = a cos(z) dz  into the
integral, we get:

/
partial_volume = L | 2b sqrt( 1-sin(z)^2 ) a cos(z) dz
/

But remember that sin(z)^2+cos(z)^2 = 1, which means that 1-sin(z)^2 =
cos(z)^2 and that makes our integral:

/
partial_volume = 2Lab | cos(z)^2 dz
/

But the double angle formula  cos(z)^2 = (1 + cos(2z))/2 fits right
into things. If we do that we get:

/
partial_volume = 2Lab | (1 + cos(2z))/2 dz
/

And if you distribute everything through the parenthesis:

/            /
partial_volume = Lab | 1 dz + Lab | cos(2z) dz
/            /

Applying the "rules" of solving integrals at this point gives us this:

partial_volume = Lab z + Lab sin(2z)/2

We can't solve this with an sin(2z) in the answer, but the double
angle formula  sin(2z) = 2sin(z)cos(z) can be used to make things
work:

partial_volume = Lab z + Lab 2sin(z)cos(z)/2

= Lab z + Lab sin(z)cos(z)

And remember that sin(z)^2+cos(z)^2 = 1 which means:

partial_volume = Lab z + Lab sin(z)sqrt(1 - sin(z)^2)

Now remember that y/a = sin(z), which means that z = arcsin(y/a)

partial_volume = Lab arcsin(y/a) + Lab (y/a)sqrt(1 - (y/a)^2)

We are back to something with y in it. The value for y ranges from
-a to d-a, so to find the volume we must subtract the partial volume
value at the bottom (-a) from the partial volume value at the top
(d-a)

The value at the top is:

partial_volume @ (d-a) = Lab arcsin((d-a)/a)
Lab ((d-a)/a)sqrt(1 - ((d-a)/a)^2)

partial_volume @ (d-a) = Lab arcsin(d/a - 1)
Lab (d/a - 1)sqrt(1 - (d/a - 1)^2)

partial_volume @ (d-a) = Lab arcsin(d/a - 1)
Lab (d/a - 1)sqrt(1 - (d^2/a^2-2d/a+1))

partial_volume @ (d-a) = Lab arcsin(d/a - 1)
Lab (d/a - 1)sqrt(2d/a-d^2/a^2)

The value at the bottom is:

partial_volume @ (-a)  = Lab arcsin((-a)/a)
Lab ((-a)/a)sqrt(1 - ((-a)/a)^2)

partial_volume @ (-a)  = Lab arcsin(-1)
Lab (-1)sqrt(1 - (-1)^2)

partial_volume @ (-a)  = Lab arcsin(-1)

partial_volume @ (-a)  = Lab (-Pi/2)

partial_volume @ (-a)  = -Lab Pi/2

So the complete final formula is:

partial_volume = partial_volume @ (d-a) - partial_volume @ (-a)

partial_volume = Lab arcsin(d/a - 1)
+ Lab (d/a - 1)sqrt(2d/a-d^2/a^2)
- -Lab Pi/2

Or this:

partial_volume = Lab arcsin(d/a - 1)
+ Lab (d/a - 1)sqrt(2d/a-d^2/a^2)
+ Lab Pi/2

Lets say the tank is half full. Then d=a because a is the distance
from the center to the bottom. That means:

partial_volume @ 50% full = Lab arcsin(a/a - 1)
+ Lab (a/a - 1)sqrt(2a/a-a^2/a^2)
+ Lab Pi/2

partial_volume @ 50% full = Lab arcsin(0) + Lab Pi/2

partial_volume @ 50% full = Lab Pi/2

Lets say the tank is completely full.  Then d=2a.  That means:

partial_volume @ 100% = Lab arcsin(2a/a - 1)
+ Lab (2a/a - 1)sqrt(2(2a)/a-(2a)^2/a^2)
+ Lab Pi/2

partial_volume @ 100% = Lab arcsin(1) + Lab Pi/2

partial_volume @ 100% = Lab Pi/2 + Lab Pi/2

partial_volume @ 100% = Lab Pi

Don't worry if you didn't follow all of it.  Like anything else, to
understand it completely would mean writing an answer the size of a
book. However, if you are interested in knowing more or if you have

In the end, the formula you seek is this:

partial_volume = Lab arcsin(d/a - 1)
+ Lab (d/a - 1)sqrt(2d/a-d^2/a^2)
+ Lab Pi/2

Make sure that your arcsin does not return degrees; it must return
radians (values from -Pi to Pi).

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/

```
Associated Topics:
College Calculus
High School Calculus

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