Horizontal cylinderical fuel tank gauging

page 76 of Exxon’s “Tables of Useful Information” has a nice table for figuring the contents of horizontal cylindrical fuel tanks. I’d be glad to send you an adobe copy. whitgallman at a0l dor c0m.

its easy to do without math.

Somewhere on the fuel tank there’s probably a label that says how many gallons it holds. Divide that number by four.

The next time he’s filling up his tank, carefully fill it until it’s at the half way mark on his dowel. Next, put in the number of gallons from the 1/4 calculation. Mark that spot on his dowel - thats the 3/4 mark, then mark the mirror of that for the 1/4 mark.

You can do it if you don’t know the number of gallons the tank holds as well. It’s pretty straight forward, but I don’t feel like typing it out right now.

Thanks
J. Webster

Using math, it’s not trivial. Getting an exact solution is quite tricky. But thank goodness for numerical integration! Assuming we’re talking about the 18 wheeler from the show with tank diameter 20 inches and the fuel tank is an exact cylinder, the the 1/4 tank mark should be made at 14.0586 inches, and the 3/4 tank mark should be made at 5.9414 inches.

In the caller’s example, the 1/4 and 3/4 marks should be a approximately 6 inches and 14 inches. For any diameter tank, the 1/4 mark should be at 30% of the diameter, and the 3/4 mark at 60% of the diameter.

I came up with exactly the same idea for the 3/4 and 1/4 marks.

To determine the capacity of the tank, fill to half using the dowel. Then fill to capacity, making sure to note the number of gallons required to get from half to full. Multiply by two.

30% and 70% you mean

I think that should actually be more like 14.03974 inches, and 5.96026 inches, if you want to be accurate.

No fair to just give him a fish. Teach him how to fish! He’s gotta be able to figure it out himself. Here’s an easy way for the driver can do this, with acceptable accuracy, in the comfort of his own kitchen, without getting diesel oil on anything or wasting time draining tanks or cutting up circles. Grab a clean cylindrical glass jar, with lid, from the recycling bin. Fill it to the brim with water and measure its volume by pouring the contents into a measuring cup. Pour exactly 1/4 of the water back into the jar and screw on the lid. Turn jar onto its side and measure the “height” of the water, in millimeters ideally. That’s the 1/4 -full mark on the dipstick! – scaled up, of course, to the dimensions of the diesel tank using the ratios of the diameters. It doesn?t matter if the jar tapers a bit towards the lid-end as long as all of the cross-sections are circles.

Hold on there. Has anyone adjusted for the fact that the ends of the tanks are not flat, but curved out?

Agreed that the problem is not trivial.

Don’t you want to swap your numbers?

100% - the top - 20 inches
75% - three fourths - 14 inches
50% - one half - 10 inches
25% - one fourth - 5.9 inches
0% - none - zero inches

That’s how I’m envisioning the problem.

If anyone is interested in reading further, Wolfram Mathworld has a compact discussion of this topic, in their article “Circular Segment.” http://mathworld.wolfram.com/CircularSegment.html

jeez. If that was the case, I would favor one of the test-and-check methods over computing the answer.

He didn’t say they were, and not all fuel tanks are curved out; some have flat ends.

The tanks that do have curved ends aren’t curved very much, I don’t think it makes an appreciable difference (for the purposes of estimating fuel). They’re not like LP tanks that you see in rural areas that have caps on the end with the same radius as the tank.

If you’re really interesting in playing around with it -
http://www.arachnoid.com/TankCalc/

it doesn’t matter if they are curved, unless its some wacko assymentrical curve. End curves of circular cross-section don’t change the volume metrics.

Here is my answer to Richard’s question about when his gas tank is 1/4 full. We are saying his gas tank is a cylinder on its side, and the cylinder has a diameter of 20’’. So it has a radius r of 10’’.

We can answer the question without calculus, but we will need geometry, trig and algebra. I will type * for times and ^2 for squared.

Suppose the tank is filled to a depth d, which is less than 10’’. So the center of the tank is a height h = r-d above the gas. Draw the tank as a circle of radius r. Draw a horizontal line a distance h below the center of the circle.

Call the center of the tank C, and the places where the horizontal line hits the circle, points A and B. Call the angle ACB, in degrees, “t”. Then
(the cross-sectional area of the gasoline) = (area of a pie-slice of the circle with angle t) - (area of triangle ACB).

(area of a pie-slice of the circle with angle t) = (area of circle) (t / 360) = pi * r^2 * t / 360.

Area of triangle BCA = .5 * base * height = .5 * 2rsin(t/2) cos(t/2) = .5r^2 * sin(t).

Cross-sectional area of gas is:
pi * r^2 * t / 360 - .5r^2* sin(t)
= r^2[ pi * t / 360 - .5* sin(t)].

If the tank is 1/4 full, we want
(cross-sectional area of gas) = (1/4) pi* r^2 = pi* r^2/4.

So (cross-sectional area of gas) = pi* r^2/4(Area of circle) = pi * r^2.

So we want:
pi* r^2/4 = r^2[ pi * t / 360 - .5* sin(t)].

Cancelling the r^2:
pi/4 = pi * t / 360 - .5* sin(t).

This is a transcendental equation, so I solve it by the very advanced method known as “trial and error”.

I get t is about 132.3 degrees.

So h = r* cos(t/2) = 10 cos(132.3/2) or about 4.04’’

And the depth of the gas is about 10’’ - 4.04 ‘’ or 5.96’’. This is about 5 and 15/16 inches, but Richard might as well mark his stick at 6’’ from its end.

Chris

According to the tank calculation program I linked to above, it does. If you create a tank with spherical end caps with a 20 inch radius (a very shallow cap), and calculate fuel level at 5 inches, it’s approximately 19.07 percent of a full tank (20 inches). However, if you create a tank with end caps of radius 10 inches (looks like a scaled-down LP tank), 5 inches is only 18.5 percent. That’s not to mention that the fuel tanks that do have curved ends probably have more of an elliptical shape to them, further complicating things.

Even without the benefit of this tank calculator, I just don’t see how curved ends could not affect the answer. The change may be small, but it certainly seems that there is indeed a change.

I took the same approach to solving this. It makes more sense to me to break it up into smaller pieces - the triangles and pie-shaped area - and solve from there. Rather than use trial and error, or work the problem backwards, I just put all the pieces into formulas in an Excel sheet and used the “goal seek” function. It’s easier than working the problem in reverse, or writing one really long equation and solving with a graphing calculator.

This is a transcendental equation, so I solve it by the very advanced method known as “trial and error”.

Not sure what your “trial and error” method entailed, but there is a systematic approach known as “finding the fixed point of a function” which entails rearranging the equation to be in the form “x=f(x)” and noting that for any x supplied, the result f(x) will be closer to the solution. In other words, f(f(x)) will be closer to the solution than f(x). All that is left is to keep applying f() until the amount of change is acceptable.

Re-arranging your formula to be in the form “x=f(x)”:

t = (pi/2 + sin(t)) * (180 / pi)

Each time the formula is iterated – using the previously computed result as the next ‘t’ – the result becomes closer to the fixed point solution.

   f( 90.0) = 147.2
  f(147.2) = 120.9
  f(120.9) = 139.1
  f(139.1) = 127.4
  f(127.4) = 135.4
  f(135.4) = 130.1
  f(130.1) = 133.7
  f(133.7) = 131.3
  f(131.3) = 132.9
  f(132.9) = 131.9
  f(131.9) = 132.6
  f(132.6) = 132.1
  f(132.1) = 132.4
  f(132.4) = 132.2
  f(132.2) = 132.4
  f(132.4) = 132.3
  f(132.3) = 132.3

This technique is easily implemented in a computer program – as described in Section 1.3.3 of the Structure and Interpretation of Computer Programs (http://mitpress.mit.edu/sicp).

My brother who is smarter than me whiteboarded it for me.

The problem is that fuel lines in semis don’t go all the way to the bottom of the tank, especially on older trucks. This allows for dirt and gunk to settle in the bottom instead of going into the engine. New highly refined diesel fuel makes this less necessary, but this is probably an older truck we are talking about. [i]Like so many of the questions in this forum, this one can’t be answered without knowing the year, make, and model.[/i] If each tank holds 150 gallons, I am willing to bet the unusable fuel for each tank is at least five gallons, but we won’t know for sure unless we know more about the truck.