Truck tank dipstick level problem without calculus (for dummies)



Draw a circle on a piece of cardboard and cut it into two equal semicircles. (You will probably need several semicircles as this is a iterative process.)

Draw a line parallel to the straight edge of the semicircle where you think the 25% level may be. Make sure that the line and edge are parallel. Make a primitive balance with a rod or stick and,say, thread or whatever. There are a million ways to do this - if you have a precision micro-balance you are better off than most!

Balance the two pieces of the semicircle and keep cutting semicircles until the pieces balance as accurately as you have the patience. Remember the cut line dividing the semicircle must always be parallel to the straight edge of the semicircle.

Do some elementary math to proportion the heights of the pieces to establish the place to mark the truck dipstick. This method should work even if the dipstick is not vertical.

I first heard of this type of procedure to try to find the area of a very irregular surface. You cut the shape out and accurately measure the weight. If you cut out a known area and weigh it, you have the weight per unit area used to find the unknown area.



As a preface, this is how I calibrated the dowel for the Diesel tank in the keel of my Sailboat. It is more non-linear than a cylinder, since the walls of the tank take many turns.

Let’s say the capacity of the full tank is 200 gallons. Completely empty one of the diesel tanks. We know that when the tank is 1/4 full, it should have 50 gallons in it, since 200/4= 50. So, drive the truck to the fuel pump and pump exactly 50 gallons of diesel into the tank. Then insert the dowel into the tank. Put a notch on the dowel where it changes from wet to dry. Then pump 100 more gallons into the tank. The tank is now 3/4 full (150 gallons), insert the dowel and notch the appropriate spot. There, you’re done.

The only way this would be incorrect, is if the pump is calibrated incorrectly. But that’s a case for the Dept. of Weights and Measures to investigate.



Your way should work perfectly, naturally. However, think of the fun you are missing by not making the semicircles, the balance, etc. This begs the eternal question: Can one have too much common sense?



Glad to see someone else subscribes to the “Edison’s light bulb method” for solving volume problems. I’ll just add that one can get to the solution without completely draining the tank first.

Since the diameter of the tank is 20" and the driver already has a dipstick marked at 10" = half a tank, all he needs to do is wait till the tank is exactly half full or slighty under half full - in which case he slowly adds fuel till it’s at the half mark. Once at the half mark, he now adds exactly 1/4 of the rated gal. capacity for the tank. He dips again and makes a 3/4 full mark. Measuring distance of the 3/4 mark from the half mark, he then marks the 1/4 mark at the same distance from the other side half full mark.


Although my mathematics degree was issued in 1964, I still maintain a fascination with numbers and problem solving. After hearing about this “dipstick” issue this morning, I remembered 2 problem solving techniques-simplify the problem and create a model. I choose the second by using an empty pasta jar, filling it, then after emptying it, filling it with 1/2 of the water and taking some measurements then using 1/4 of the water and taking some more measurements. Using some simple ratios, it appears that 1/4 of the volume is nearly identical to 1/4 the diameter of the cylinder.Thus the dipstick could be marked at 1/4 of the diameter or 5" to measure the 1/4 capacity of the fuel in the tank.

I may have over-silplified the problem but since I remember little of my Calculus, I gave it a try.
I wait with bated breath for your analsis next week on your show!
An avid listener from Pittsburgh, Clare


Diesel tank in the keel Interesting. I don’t recall ever seeing that setup.

In any case I like that method. BTW my degree is in Math and calculus was part of my professional career.


A (slightly) simpler method than ‘henway1’ proposed (doesn’t require draining the tanks, but is only applicable to tanks where the top half is a mirror image of the bottom half; ie: a cylinder):
Calculate total volume of tanks = (10" Radius Squared x pi x length) / 231 cubic inches per gallon.
Pull into a gas station when the tank is less than 1/2 full (per dipstick with a mark at 10").
Fill tank until dipstick indicates exactly 1/2 full (10"). Add 1/4 of total tank volume.
Mark stick at new reading = (3/4 full).
Mark stick at same distance below 1/2 full as above.


This problem in math is known as the area of a segment of a circle. And after a little algebra and trig, I find that the 1/4 point in the tank will be about 35% from the bottom which for the 20" tank give a distance of close to 7".

For the math inclined, the solution involves subtracting the triangular area within a section of a circle. Having the resulting area equal 1/4 and solving the implicit equation for the angle, and then take the sine of the angle. Assuming for simplicity a diameter of 1 and then multiplying the result of the sine by the tank’s diameter.


I guess there are two solutions. The mathematical solution and the empirical solution. Both approaches seem to get you to the correct place. I think one of the advantages of the empirical method, is that it can be used for a tank of any shape, not just a cylinder.

I think if I were a truck driver I would mark a couple of dowels. In the event that one got lost or broken, I wouldn’t have to go through the whole excercise again.

I find that the 1/4 point in the tank will be about 35% from the bottom which for the 20" tank give a distance of close to 7".

I think you’re on the right track, but I have to disagree with your math. On one of the other threads about this, several of us have calculated it and have come up with answers just a little shy of 6 inches.

You can use this link to simulate different tank sizes as well, and see the results of different fuel levels. On this simulator, 6 inches gives 25.25% of a full 20 inches, on a tank with radius 10 inches.


20" = F
14" = 3/4
10" = 1/2
6" = 1/4
3 5/8" = 1/8
Here is a URL to help you calc the values.
This will give you the other formulae to calc ‘h’ accurately.


I worked this out using a straight algebraic solution with no trigonometry involved and also came up with about 7 inches.
As mentioned on the show, assuming this is a symmetrical tank and the dipstick is inserted top center we can work with a 2-dimensional area.
r = radius of circle
ra= distance from bottom of the circle to the fill line (what we’re solving for)
A = area below fill line.

Since A is 1/4 the area of the full circle,
And since A is also 1/2 the area of a circle with radius ra,

So, we solve for ra:
(pir^2)/4 = (pira^2)/2
ra=r/sqrt(2) or approximately 0.7*r

And in this case the radius of the tank is 10 inches, so that gives us approximately 7 inches.


Along the same lines (and assuming I’m not an idiot here), one could calculate how full the tank is based on the number of inches the fill line is from the bottom (ra).

f = (ra^2)/(2*r^2)

?And in this case the radius of the tank is 10 inches, so that gives us approximately 7 inches...
Yeah, but that?s wrong. You?re disagreeing with the several people (myself included) who have calculated this using trigonometry, the people who have calculated it using the ?too much for me to digest with the sinus headache I have? formula found here -, and the tank volume calculation program found here - I?m just having trouble buying that you?re correct.


I think the issue may be that my calculation assumes that the ends of the tank are flat making the tank a cylinder. Realistically, I guess that’s probably not likely.


That’s not the problem, because that’s the same assumption that everyone else used as well - all that was needed (as pointed out on the show) was to figure out the area on a two dimensional circle. That’s all that’s being done in the formula that was linked to, and a tank can be simulated on the tank program with flat ends. I’m not sure where the formulas you’re using came from, but it seems they are not correct, or not being correctly applied.


Well, in that case I don’t know. My formulas didn’t come from anywhere. I’m just using the formula for area of a circle.
I agree that I’m likely wrong since everyone else is coming up with 6", but I don’t see where my logic is wrong that 1/4 the area of the full circle is equal to the area of the filled semicircle (which is half the area of a circle with that radius).


Scratch that… yes I do see what’s wrong with that. It’s not a semicircle. It’s a partial elipse.

I’m going to go crawl into a hole now.


Here’s the calculus for a 53 GMC short school bus with an OVAL tank. 1. Drive into a level gas station on fumes as usual. 2. Fill tank with exactly five gallons. 3. Mark stick. 4. Repeat steps 2 & 3 until tank is full. You are now the proud owner of a fuel gauge more accurate than the original. Super persnickety people could even mark their stick at each gallon if they had all the time in the world or were waiting for their wife to get out of the restroom.


I love math as much as the next guy and love mathematical solutions to problems.

However, I will have to say that the mathematical solution will be infinitely more difficult in a tank that is an unusual shape. Some fiberglass tanks are very unusually shaped. The empirical method will work on these complex tanks.