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 
reading):

   

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)
   (For more information, see the Parallelogram in the Geometric  
   Formulas section of the Dr. Math FAQ, at: 
   http://mathforum.org/dr.math/faq/formulas/faq.quad.html   )

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

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

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/