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### Horizontal Gas Tank Content Formula

```
Date: 01/17/99 at 10:21:46
From: Frank Austrino
Subject: Horizontal Tank Content Indicator

I'm trying to find a chart that can be used by our drivers to check the
content of a fuel tank by inserting a sort of yardstick into the tank
and checking the inches of liquid in the fuel tank. The tanks are 97-
gallon horizontal tanks. I have a horizontal tank content indicator
chart but it is for larger tanks. I need something that a driver, given
the length and height of the tank, can use to calculate the remaining
fuel by measuring the height of the remaining fuel. Our objective is to
reduce the amount of fuel purchased on the road and more accurately
calculate the amount of fuel left on board. The smallest tank size on
the chart I have is 300 gal. Thanks in advance for your help.

Frank Austrino
Leaseway Auto Carriers
Lordstown, Ohio
```

```
Date: 01/17/99 at 12:00:25
From: Doctor Anthony
Subject: Re: Horizontal Tank Content Indicator

The problem is not difficult. We simply need to calculate the area of
a segment of a circle, representing the cross-section of the tank that
contains the fuel. If this cross-sectional area is multiplied by L, the
horizontal length of the tank, we have the volume.

Suppose r = the radius of the tank and h = the height of the fuel
measured from the lowest point to the surface of the fuel (the dipstick

If theta is the angle, in radians, between the vertical through the
centre of the circle and a radius drawn from the centre to the point of
contact of the fuel and the side of the tank, then

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

So theta = cos^(-1)[(r-h)/r)]

Note that:

** area of sector of circle with angle 2 theta  =  r^2 theta
(To believe this, first calculate the area of a circle with radius
r, then calculate the area of a half circle with radius r, and
compare the two.)

** area of triangle to be subtracted = (1/2)r^2 sin(2 theta)
Formulas section of the Dr. Math FAQ, at:

Thus:

Cross-section of fuel = r^2 theta - (1/2)r^2 sin(2 theta)

= r^2[theta - (1/2)sin(2 theta)]

= r^2[theta - sin(theta) cos(theta)]

And so:

Volume of fuel  =  r^2 L[theta - sin(theta) cos(theta)]

It is probably better to leave the formula in this form and use the
other formula

to complete the calculation. It becomes rather untidy if you express
the whole thing in one big formula in terms of h. The sin(theta) term
is the problem.

sqrt[r^2 - (r-h)^2]     sqrt(2rh - h^2)
sin(theta) =  -------------------  =  ---------------
r                     r

and you end up with

Volume = r^2 L[cos^(-1)[(r-h)/r] - sqrt(2rh-h^2)/r (r-h)/r]

= r^2 L[cos^(-1)[(r-h)/r] - (r-h) sqrt(2rh-h^2)/r^2]

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry
High School Trigonometry

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