Water Depth in an Ellipsoidal Tank

Let's take the center of the ellipsoid to be the origin of the reference frame $Oxyz$ , and assume that the water level is at $z = z_0$

that is, the equation of water surface (a plane) is $z = z_0$

the equation of a general ellipsoid of arbitrary orientation is given by $r^T R D R^T r = 1$

where $D$ is a diagonal matrix whose diagonal entries are the reciprocals of the squares of the semi-axes of the ellipsoid, and R is a rotation matrix.

We will now transform this ellipsoid into a sphere of unit radius by applying the following transformation

$r = T r' = R D^{-1/2} r'$ , i.e. $r' = D^{1/2} R^T r$

Now the original equation of the water surface is

$k^T r = z_0$

where $k = [0, 0, 1]^T$ is the unit vector along the z-axis.

Applying the transformation, we get, in terms of the new coordinate r',

$k^T R D^{-1/2} r' = z_0$

Now, the signed distance of this plane to the center of the unit sphere is

$d = z_0 / \sqrt{k^T R D^{-1} R^T k } = z_0 / h$

One can show that $\sqrt{k^T R D^{-1} R^T k}$ is the maximum elevation of the ellipsoid above the center. To show this, consider a horizontal plane touching the ellipsoid from above,

then the gradient of ellipsoid at the touching point $r_1$ will be parallel to the unit vector $k$ ,

i.e. $R D R^T r_1 = \alpha k$ with $\alpha > 0$

therefore, $r_1 = \alpha R D^{-1} R^T k$

plugging this into the equation of the ellipsoid , we obtain

$r_1^T R D R^T r_1 = 1$ , so $\alpha^2 k^T R D^{-1} R^T k = 1$

hence, $\alpha = \dfrac{1}{\sqrt{ k^T R D^{-1} R^T k } }$

The elevation of $r_1$ is given by $k^T r_1 = \alpha k^T R D^{-1} R^T k = \dfrac{ k^T R D^{-1} R^T k } {\sqrt{ k^T R D^{-1} R^T k }} = \sqrt{ k^T R D^{-1} R^T k }$

which is what we wanted to show.

now the volume of the transformed space occupied by water in the unit sphere depends only on d. It is given by

$V = \pi ( 2/3 + d - d^3 / 3 )$

Therefore, from transformation theory, the volume of water in the original ellipsoid is

$V = | T | \pi (2/3 + d - d^3 / 3 ) = | D^{-1} | \pi (2/3 + d - d^3 / 3 )$

Since the volumes in both cases are equal, then d must be equal because taking the derivative of V with respect to d, we obtain,

$V'(d) = |D^{-1} | \pi ( 1 - d^2 )$

since $d < 1$ , the function is strictly increasing with d.

Hence $V_1 = V_2$ implies $d_1 = d_2$ .

thus, $\dfrac{z_1}{ h_1} = \dfrac{z_2}{h_2}$ .

and therefore, by adding 1 to both sides, we obtain, $\dfrac{(z_1 + h_1)}{ h_1} = \dfrac{(z_2 + h_2)}{ h_2}$

That is, $\dfrac{(z_1 - (-h_1) )}{ h_1} = \dfrac{(z_2 - (-h_2) )}{ h_2}$ , which translates also, into, $\dfrac{(z_1 - (-h_1) )}{2 h_1} = \dfrac{(z_2 - (-h_2) )}{2 h_2}$

but $(z_1 - (-h_1) )$ is the depth of water in the first orientation of the ellipsoid, and $(z_2 - (-h_2) )$ is the depth of water of the second orientation of the ellipsoid, hence

$(z_2 - (-h_2) ) = \dfrac{2h_2}{2h_1} (z_1 - (-h_1) ) = \dfrac{4}{6} (144) = 96$ cm

'

Let $y_0$ be the height of the highest point above the lowest point; $y$ the height to which the ellipsoid is filled; $V_0$ the volume of the ellipsoid; and $V$ the volume of the fluid. Then $\frac{V}{V_0} = f\left(\frac{y}{y_0}\right),$ where $f(\cdot)$ is a function independent of the shape and orientation of the ellipsoid. ( $\star$ )

Thus, if $V'/V'_0 = V/V_0$ , then $y'/y'_0 = y/y_0$ and $y' = y\cdot\frac{y'_0}{y_0} = \SI{144}{cm}\cdot \frac{\SI{4}{m}}{\SI{6}{m}} = \boxed{\SI{96}{cm}}.$

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Proof of claim ( $\star$ ) above.

An ellipsoid (centered at the origin) is obtained by linear transformation of the unit sphere. Rotating the ellipsoid is also a linear transformation. So any ellipsoid can be written in the form $\left(\begin{array}{c} x \\ y \\ z \end{array}\right) = T\:\left(\begin{array}{c} x_0 \\ y_0 \\ z_0 \end{array}\right),$ where $T$ is a $3\times 3$ -matrix and $x_0,y_0,z_0$ are points on the unit sphere. In fact, using the rotational freedom of the unit sphere, we can write $T$ in the form $T = \left(\begin{array}{ccc} a & b & u \\ c & d & v \\ 0 & 0 & h \end{array}\right).$ Since $z = hz_0$ , it is obvious that the highest and lowest points of the ellipsoid are $\pm h$ . Also, its volume is $V_0 = \frac{4\pi}3 \text{det}\ T = \frac{4\pi}3 h (ad - bc).$ Intersecting the unit sphere at height $z_0$ results in a circle $C_0(z_0)$ with area $A_0(z_0)$ ; likewise, intersecting the transformed unit sphere at height $z = hz_0$ results in a linearly transformed circle (i.e. ellipse) $C(z)$ defined by $\left(\begin{array}{c} x \\ y \end{array}\right) = \left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \left(\begin{array}{c} x_0 \\ y_0 \end{array}\right) + \left(\begin{array}{c} u \\ v \end{array}\right);$ its area is $A(z) = A_0(z_0) \text{det}\ \left(\begin{array}{cc} a & b \\ c & d \end{array}\right) = A_0(z/h)\:(ad - bc)$ .

If the unit sphere is filled from the lowest level $z_0 = -1$ to a given level $z'_0$ , the volume of the liquid is $\int_{-1}^{z'_0} A_0(z_0)\:dz_0 =: v(z'_0).$ Likewise, when the ellipsoid is filled to level $z' = hz'_0$ , $V = \int_{-h}^{z'} A(z)\:dz = \int_{-h}^{z'} A_0(z/h)\:(ad - bc) = h \int_{-1}^{z'/h} A_0(z_0)\:(ad - bc)\:dz_0 = h(ad - bc)\:v(z'/h),$ and $\frac{V}{V_0} = \frac{h(ad - bc)\:v(z'/h)}{\tfrac{4\pi}3\:h(ad - bc} = \frac{v(z'/h)}{4\pi/3}.$ The latter expression is equivalent to $f(y/y_0)$ in the discussion above.

This is a very interesting problem. The solutions provided by Hosam Hajjir and Arjen Vrengdenhil are excellent and in-depth.

First, check out Afffine Transforms , which is really at the heart of this problem.

This will be an effort at a brief, if incomplete, proof (or explanation). Consider three ellipsoids, all of them having the same height of $400 cm$ from floor to top, and same volume. The first has a vertical axis and circular cross sections perpendicular to that axis, the radius of the midsection circle being $50\sqrt{3}$ . The second similarly has a vertical axis and elliptical cross sections perpendicular to that axis, the major and minor axes of the midsection ellipse being $100\sqrt{\dfrac{2}{3}}$ and $50\sqrt{\dfrac{2}{3}}$ . The last is the tilted ellipsoidal tank. The second is the original upright tank that has been "squeezed" into a shorter and wider tank of the same volume, using an affine transform.

At any given height on the vertical axes, the area of the cross section of both the $1st$ and $2nd$ tanks are the same. If it can be shown that the $3rd$ tank can be obtained from the $1st$ by an affine transform limited to the planes perpendicular to the axes, then the area of the cross section of both the $1st$ and $3rd$ tanks are the same. Hence, the area of the cross sections of both the $2nd$ and $3rd$ tanks are the same as well. Hence, a given volume water in the $3rd$ tank will be of the same depth as the same volume of water in the $2nd$ tank. From this, we can deduce that if a volume of water in the original upright tank has a depth of $144$ , then the same volume of water in the $3rd$ tank will have a depth of $\dfrac{2}{3}\cdot144=96$

Suppose an afffine transform (limited to the planes perpendicular to the axes) obtains the $3rd$ ellipsoid from the $1st$ . Then all of the elliptical cross section areas of the $3rd$ ellipsoid corresponding to the circular cross sections of the $1st$ ellipsoid will have the same eccentricity, and thus the the areas will differ by a factor that is a constant. However, for both the $3rd$ and $1st$ ellipsoids to have the same volume, this constant has to be $1$ . Hence the corresponding cross section areas are the same.

This not a substitution for a rigorous proof as how Hosam or Arjen has done it, but hopefully it would serve as a visual explanation of how this works. Here, two separate affine transforms were used, instead of a single general 3D affine transform.

A great variety of mathematical problems can successfully analyzed using affine transforms, a very useful tool to have.