Horizontal Gas Tank Content FormulaDate: 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/ |
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