# Aunt and Uncle's fuel oil tank dip stick problem

This problem first came to me in high school, and a couple times since, and I even assigned it for extra credit in one of my calculus classes after I became a teacher. So I know the solution. What I am looking for is other WAYS to obtain the solution. I've been told there exists a solution using only arithmetic, but have never figured it out. Other solutions using ordinary calculus, trigonometry, algebra of conic sections, and so on are also possible.

The problem is usually stated in the form of a letter from an Aunt and Uncle:

Dear niece/nephew, How are things going for you and your folks? We hear you are doing quite well it school. Keep it up! Given this success, we were hoping you could help us figure out a little dilemma. As you know, our home is heated by fuel oil, and we have a big tank buried in the side yard. The tank is a cylinder, 20 feet long and 10 feet in diameter, lying on its side five feet deep, with a narrow tube coming to a fill cap at ground level. Your uncle has a 15 foot length of old pipe that we'd like to utilize as a dip stick in order to know when we are getting close to needing a fill-up. We know that 0 feet is empty, 5 feet is half full, and 10 feet is completely full. Trouble is, we don't know how to mark any other points. We are pretty sure they will not be uniformly spaced. What we really want is to know, within the nearest 0.01 foot, where to mark the dip stick for every multiple of 10% from 0% to 100%. Can you figure this out for us? Of course, we will want to see details of your solution and check it ourselves, and it would especially help if you could draw us a scale model of the dip stick. Love, Auntie Flo and Uncle Jim

That last sentence shows the teacher influence on the problem. So, my challenge to this community is not to find any old solution, but to find the solution at the lowest possible grade level, so to speak.

Thanks.

UPDATE: To those who are focusing in on the .01 feet accuracy, I apologize. The intent was merely to state, it is acceptable to estimate. If the exact answer is sqrt(2)*pi/2 or some other silly thing, go ahead and just write 2.22 feet, for example.

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Empty tank. Add 10% capacity of oil, dip pipe, mark pipe just above the black stuff. Repeat. How's that for "lowest possible grade level"? You don't even have to clean off the pipe in between steps. :) (Offered as comment, since it doesn't satisfy the requirements of the question.) –  Blue Sep 1 '10 at 1:01
I think my car must have been designed by these people because the digital odometer is more accurate than the analog fuel gauge. –  Dan Brumleve Sep 1 '10 at 1:13
This is deceptively difficult. According to mathworld.wolfram.com/CircularSegment.html there is no analytic solution for the locations of the pipe markings. Does this also mean that it can't be solved on the ground with stakes and ropes? Day's empirical method has the advantage that it works for any tank with a convex (to avoid bubbles) cross-section. –  Dan Brumleve Sep 1 '10 at 2:50
I've updated my answer to the +/-.01 feet accuracy range. –  Américo Tavares Sep 2 '10 at 19:09
All: We have a few excellent responses so far, and I have up voted them. Please allow me to leave this one open for a while before accepting an answer. I really am curious what variety of response this will generate. –  cobaltduck Sep 6 '10 at 21:54

Upper figure: cylinder oil tank cross-section perpendicular to its horizontal axe. The vertical coordinate is the oil level in percentage.

Lower figure: graph of oil volume/max. volume (in %) versus oil level $l$ (in feet). The horizontal straight lines represent the area/volume ratio $A(l)/A(10)=V(l)/V(10)$ (in %) for every multiple of 10% from 0% to 100%.

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Since the tank radius is $5$, the oil level with respect to the bottom of the tank is given by $l=5-5\cos \frac{\theta }{2}$, where $\theta$ is the central angle as shown in the figure. The area of the tank cross section filled with oil is

$A(\theta )=\frac{25}{2}\theta -\frac{25}{2}\sin \theta$

or

$A(l)=25\arccos (\frac{5-l}{5})-\frac{25}{2}\sin (2\arccos (\frac{5-l}{5}))$

The area ratio $A(l)/A(10)=V(l)/V(10)$ where $V(l)$ is the oil volume.

Let $f(l)$ denote this area ratio in percentage:

$f(l)=\frac{100}{\pi }\arccos \left( 1-\frac{1}{5}l\right) -\frac{50}{\pi }% \sin \left( 2\arccos \left( 1-\frac{1}{5}l\right) \right)$

Here are the the sequence of $f(l)$ for $l=0,1,2,\ldots ,10$. The graph of $% f(l)$ is shown above.

$f(0)=0$, $f(1)=5.2044$, $f(2)=14.238$, $f(3)=25.232$, $f(4)=37.353$, $f(5)=50$,

$f(6)=62.647$, $f(7)=74.768$, $f(8)=85.762$, $f(9)=94.796$, $f(10)=100$

Edit: One still needs to solve the nonlinear equation $f(l)-10k=0$ for $k=1,2,3,4,6,7,8,9$, e. g. by the Secant Method.

Edit 2: The problem of solving graphically, as shown in the secong figure, is that it would be very difficult, or impossible, to get the required accuracy of 0.01 (feet).

Update: The oil level marks (in feet) should be placed at

$0,1.57,2.54,3.40,4.21,$

$5,5.79,6.60,7.46,8.44,10$

corresponding to the oil volume percentage of

$0,10,20,30,40,$

$50,60,70,80,90,100$.

This calculation was based on the following $f$ function values:

$f(0)=0.0$, $f(1.5648)=10.0$, $f(2.5407)=20.0$, $f(3.40155)=30.0$, $f(4.21135)=40.0$

$f(5)=50.0$, $f(5.7887)=60.0$, $f(6.59845)=70.000$, $f(7.4593)=80.000$,

$f(8.4352)=90.000$, $f(10)=100.0$

Update 2 Figure of marks:

[Rearranged to show the sequence of editions and updates.]

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The graph looks nicely linear in the middle; maybe interpolation might more expediently give intermediate values than having to solve a nonlinear equation. –  Ｊ. Ｍ. Sep 1 '10 at 12:44
J. M. Yes, more expediently, but the question asks for an accuracy of 0.01 (feet). –  Américo Tavares Sep 1 '10 at 13:30
+1: For finding f(i). –  Aryabhata Sep 1 '10 at 16:24
@Moron: Thanks! I have added the "scale model of the dip stick". –  Américo Tavares Sep 16 '10 at 21:20
Correction: "Here is the sequence of" instead of "Here are the the sequence of" –  Américo Tavares Sep 17 '10 at 20:32

It is sufficient to consider the circular cross-section of the tank and volumes below 50% (marks for those above 50% are the reflection image of those below 50% across the 50% mark). Consider the radius of the tank to be 1 unit. For some amount of oil in the tank, consider the central angle formed by the points on the circular cross-section at the top of the oil and the center of the circle--call this $\alpha$, measured in radians.

The area of the cross-section of the oil is the area of the sector determined by $\alpha$ ($\frac{\alpha}{2\pi}\pi r^2=\frac{\alpha}{2}$) minus the area of the triangle that is part of the sector but not part of the cross-section of oil ($\frac{1}{2}ab\sin C=\frac{\sin\alpha}{2}$), $\frac{1}{2}(\alpha-\sin\alpha)$. The portion of the circular cross-section (and hence the portion of the volume) corresponding to this angle is $\frac{\alpha-\sin\alpha}{2\pi}$. Setting this expression equal to 0.1, 0.2, 0.3, and 0.4 and solving for $\alpha$ will give the values of $\alpha$ corresponding to 10%, 20%, 30%, and 40% full--solving here is done numerically/graphically, as there is no algebraic method to solve these equations. For each value of $\alpha$, the distance from the center of the circle to the oil level is $\cos\frac{\alpha}{2}$, so the depth of the oil is $1-\cos\frac{\alpha}{2}$. (Note that these are for a radius of 1 unit and need to be rescaled for the original problem's specific numbers.)

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@Isaac: I think you have wanted to write $\frac{\alpha}{2}r^2$, in your 1st formula (anyway this doesn't matter since you are using the unitary radius). –  Américo Tavares Sep 1 '10 at 10:58
@Américo Tavares: My intent was that I was substituting in 1 for r across that = in the first formula, hence no more $r^2$ term. –  Isaac Sep 1 '10 at 11:25
@Isaac: I was able to see it, but only in your note at the bottom line. –  Américo Tavares Sep 1 '10 at 15:06
+1 for your elegant solution. –  Américo Tavares Sep 1 '10 at 19:36
Flo and Jim say: This answer is really getting to the heart of the problem (i.e. reducing volume in the cylinder to area under a chord of a circle, reducing to the simplest case then going back and substituting actual values, etc.) But can you tell us how your angle alpha relates to h, the height of the shaded region along a vertical radius? That is how the dipstick will be marked. –  cobaltduck Sep 1 '10 at 23:45

I will give the calculus-based solution myself just for the sake of argument, taking note that I am still hoping to obtain a wide variety of other solutions, if possible.

For this, I am going to mentally rotate the tank (or the oil within it) $90^\circ$ clockwise, cut it in half, and center it at $(0,0)$, so that the upper half of the tank is represented by $y=\sqrt{25-x^2}$ and the volume of the oil by $\int_{-5}^h \sqrt{25-x^2}\mathrm{d}x$.

I know from simple $A=\pi r^2$ that the total cross-section area of the tank is $25\pi$ and thus for the upper half from -5 to +5 is $12.5\pi$. I will set the result of the integral to the various 10% proportions of this value, knowing that 10% of the upper half will occur at the same position as 10% of the entire circle etc.

The integral is $\frac12\left(x\sqrt{25-x^2} + 25\arcsin\left(\frac{x}{5}\right)\right)$ evaluated from -5 to h, so the equation we need to solve is: $\frac12\left(h\sqrt{25-h^2}+25\arcsin(\frac{h}{5})\right)+\frac{25\pi}{4}=12.5\pi P$

Substituting values of .1, .2, .3, .4, and .5 successively for $P$ and using various tools to estimate $h$, my results are -3.43424, -2.45931, -1.59846, -0,788681, 0. Adding 5 to account for the central displacement, rounding off, then reflecting these values for the right (upper) values, confirms the values given earlier by Americo.

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I fixed your TeX. 1: You forgot to enclose 'em in dollar signs. 2: Here's where you can practice syntax (or you can use as a crutch if you're like me who can't be bothered to remember): codecogs.com/latex/eqneditor.php –  Ｊ. Ｍ. Sep 7 '10 at 0:30
+1 for a completely different solution. –  Américo Tavares Sep 12 '10 at 19:00

Looking @Isaac's answer, the appearance of "$\alpha - \sin\alpha$" and "$1-\cos\frac{\alpha}{2}$" made me think that power series would come in handy, with their respective initial terms vanishing. They didn't turn out to be as handy as I might have hoped.

Write $w$ for the ratio of oil to the tank's capacity, and $d$ for the ratio of the oil's depth to the radius of the tank. Then, by Isaac's work ...

$$2\pi w = \alpha - \sin\alpha = \alpha-\sum_{m=0}^\infty\frac{(-1)^m}{(2m+1)!}\alpha^{2m+1}= \sum_{m=1}^\infty\frac{(-1)^{m+1}}{(2m+1)!}\alpha^{2m+1}$$ $$d = 1 - \cos\frac{\alpha}{2}=1-\sum_{m=0}^\infty\frac{(-1)^m}{(2m)!}\left(\frac{\alpha}{2}\right)^{2m}=\sum_{m=1}^\infty\frac{(-1)^{m+1}}{(2m)!}\left(\frac{\alpha}{2}\right)^{2m}$$

Write $w_n$ and $d_n$ for the values associated with lopping off the power series at the $n$-th term (that is, from $m=1$ to $m=n$). In particular, we have

$$2\pi w_1 = \frac{\alpha^3}{3!}$$ $$d_1 = \frac{\alpha^2}{2^2 2!}$$

Eliminating $\alpha$ yields the relation

$$32 d_1^3 = 9 w_1^2 \pi^2$$

This relation, though, isn't very accurate (as shown by the $n=1$ curve in the diagram).

If we take $n=2$ ...

$$2\pi w_2 = \frac{\alpha^3}{3!}-\frac{\alpha^5}{5!}$$ $$d_2 = \frac{\alpha^2}{2^2 2!}-\frac{a^4}{2^4 4!}$$

... then eliminating $\alpha$ isn't quite so easy unless you do something like invoke Mathematica's Resultant function, which yields

$$4718592 d_2^5 - 13762560 d_2^4 + 10035200 d_2^3 - 768000 d_2^2 w_2^2 \pi^2 + 3024000 d_2 w_2^2 \pi^2 - 2822400 w_2^2 \pi^2 + 1875 w_2^4 \pi^4 = 0$$

Also not terribly accurate at we get close to $w=0.5$ (which should have $d=1.0$). With $n=3$ we're very close, and with $n=4$ and above, we've pretty much converged on our target (at least according to the graph). I won't give the $n=4$ equation (the coefficient on $d^9$ is $2^{71} 5^4 7^2$), but here are some data points:

$$(0.0, 0.000000)$$ $$(0.1, 0.312953)$$ $$(0.2, 0.508164)$$ $$(0.3, 0.680452)$$ $$(0.4, 0.842809)$$ $$(0.5, 1.001730)$$

Of course, power series and resultants aren't exactly "lowest possible grade level", but this is the best I can do at this point.

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I guess this rigorously confirms Américo's note that approximating by a linear function mid-range won't give the required accuracy. –  Ｊ. Ｍ. Sep 1 '10 at 21:29