Aunt and Uncle's fuel oil tank dip stick problem

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 is the sequence of $f(l)$ values 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.]

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.)

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.