Proof: minimum volume of a notch cut at equal angle of cutting surfaces with horizontal plane

Question

A notch is cut in a cylindrical vertical tree trunk. The edge of the cut reaches the axis of the cylinder and the cut is between two half-circle planes. Each half-circle is bounded by a horizontal line passing through the axis of the cylinder. The angle between the two half-circle planes is θ. Prove that the volume of the notch is minimized (for given tree and θ) by taking the half-circle planes at equal angles with the horizontal plane.

I'm having trouble even visualizing the problem. I think it would be useful to find the total volume of the notch in terms of the angles that the bounding planes form with the horizontal plane and then differentiate to find when the minimum is achieved. Can someone provide a solution?

Answer 1

Imagine four halfplanes $H_1,H_2,H_3,H_4$ which are all bounded by a horizontal line L which intersects the axis at the point $C$. The angle between $H_1$ and $H_2$ is $\theta$, the angle between $H_3$ and $H_4$ is $\theta$ and $H_1$ and $H_2$ are at equal angles to the horizontal but $H_3$ and $H_4$ are at unequal angles to the horizontal. We wish to prove that the volume of cylinder bounded by $H_3,H_4$ is less than the volume of cylinder bounded by $H_2,H_1$.

First, consider the case when the two volumes do not overlap. Choose a point $P$ on the surface of the cylinder which lies between $H_1$ and $H_2$, then rotate the volume bounded by $H_1,H_2$ around the line $L$ until $H_1$ coincides with $H_3$ and $H_2$ coincides with $H_4$. Now the point $P$ is at a greater tilt from the horizontal plane through $L$ than it was before. $P$ must therefore lie inside the cylinder, since the length of a line from $C$ to the surface of the cylinder increases as the line becomes more tilted from the horizontal. Since this is true for any point $P$, the volume bounded by $H_1,H_2$ must fit inside the volume bounded by $H_3,H_4$ after this rotation.

Now suppose that the volumes overlap. Denote by $V_1$ the region which is contained in the volume bounded by $H_1,H_2$ but not contained in the volume bounded by $H_3,H_4$. Define $V_2$ similarly. Now reflect $V_1$ about the horizontal plane containing $L$, then rotate $V_1$ about $L$ until the planes bounding $V_1$ and $V_2$ are on top of each other. Again, any point $P$ on the surface on the portion of the cylinder within $V_1$ gets moved to a point with a higher tilt and this operation therefore fits $V_1$ inside $V_2$.

Answer 2

I would like to solve this by using cylindrical co-ordinate system integrals:-

Let A be the area above horizontal plane and B be the area below horizontal plane

Now lets Calculate A for which we use $\alpha$ as the angle between the horizontal plane and the upper half plane.

$A$ = area of upper half cut cylinder

$\alpha$ = angle above the horizontal reference plane

$\rho , \phi, z$ = coordinate of cylindrical system

$r$ = radius of the cylindrical tree

So, $z= tan(\alpha).\rho. sin(\phi)$ equation of line with slope

$\theta =$ the angle between the two half planes

Now $ A = \int_0^r \int_0^\pi .\rho.tan(\alpha).sin(\phi).\rho.d\phi.d\rho$

$$ = tan(\alpha) \int_0^r \int_0^\pi .\rho^2.sin(\phi).d\phi.d\rho $$ $$ = tan(\alpha) \int_0^r \rho^2.[-cos(\phi)]_0^\pi.d\rho $$ $$ = tan(\alpha) \int_0^r \rho^2.[+1+1].d\rho $$ $$ = 2.tan(\alpha) {[\rho^3]_0^r \over 3} $$ $$ = { 2.tan(\alpha).r^3 \over 3} $$

Now, $$ B = { 2.tan(\theta-\alpha).r^3 \over 3} $$

Total Volume $V = A + B = (2/3).r^3.[tan(\theta-\alpha)+tan(\alpha)] $

For minimum vloume $ dV/d\alpha = (2/3).r^3.[-sec^2(\theta-\alpha)+sec^2(\alpha)] = 0$

$$ => sec^2(\theta-\alpha)=sec^2(\alpha) $$ since $ 0 \leq \alpha \leq {\pi \over 2} $, $$ => \theta-\alpha= \alpha $$ $$ \alpha = \theta/2 $$

So that the volume of the notch is minimized (for given tree and θ) by taking the bounding planes at equal angles to the horizontal plane.

Answer 3

Here’s an intuitive “proof” (not mathematically rigorous but still a solid line of reasoning):

Consider a constant angle $θ$ (angle between the planes), and an angle $α$ (angle between the plane bisecting the upper and lower planes described in the problem and the horizontal plane).

Imagine positioning the notch so that the upper plane is vertical. The volume of the notch would obviously be infinite. The current value of $α$ is the maximum possible value of $α$.

Now imagine slowly changing $α$ so that the notch tilts downwards. At any instant in time, the change in volume is equal to the infinitesimal amount of $($volume gained$)$ $-$ $($volume lost$)$. These volumes can be approximated with the area of the two moving boundaries.*

As you can see, when $α$ is greater than $0$, the upper plane is always larger in area than the lower plane, as the cross-section of a cylinder at a steeper angle is always larger. This means that the volume lost is greater than the volume gained as $α$ decreases when $α > 0$, meaning the notch is decreasing in volume.

Similarly, when $α < 0$, the bottom plane is larger, meaning that the volume gained is larger. The notch increases in volume as it moves downward.

Eventually, when $α$ reaches its minimum value, the volume of the notch will again be infinite. (Imagine flipping the second diagram upside-down).

Here’s my rough sketch of a volume vs time graph for the notch based on the logic above:

It makes sense that the minimum value is reached when $α$ is $0$ and the ‘middle’ plane is horizontal, meaning that the angles between the horizontal plane (which is also the bisecting plane) and the planes above and below are the same!

It doesn't matter what the tree's radius or $θ$ are (as long as $θ$ is less than 180^\circ rad).

In summary, by showing that the derivative of a volume function $V(α)$ wrt. α is negative when $α > 0$ and positive when $α < 0$ for $|α|<90^\circ-{1\over2}θ$, you can conclude that volume is minimum when $α=0$.

*The infinitesimal changes in volume are not exactly proportional to the areas since the planes are rotating and not just translating. But since the relationship between the area and change in volume is monotonic, the reasoning holds true.

Answer 4

Let $r$ be the trunk radius, and $\alpha$ and $\beta$ the angles formed by the two cut-planes with the horizontal plane, thus $\alpha+\beta=\theta$.

The equation for the trunk cylinder can be written as $x^2+y^2=r^2$, and the two cut-planes as $z_1=y\tan\alpha$ and $z_2=-y\tan\beta$. Then, integrate $z_1-z_2$ over a half-circle in the $xy$-plane to obtain its volume,

$$V = \int_{-r}^{r}\int_0^{\sqrt{r^2-x^2}}[y\tan\alpha -(-y\tan\beta)] \>dydx=\frac23 r^3(\tan\alpha+\tan\beta)$$

Next, with $\beta=\theta-\alpha$, take the derivative with respect to $\alpha$ and set it to zero to minimize $V$,

$$\sec^2\alpha - \sec^2\beta=0$$

which yields $\alpha = \beta$. Hence, the notch volume is minimal at equal angles of the cutting surfaces with the horizontal plane.

Answer 5

Compute upper and lower notch volumes $V_u,V_v$ cut out fron a cylindrical tree trunk

$$V_u = \int_{0}^{a}\int_0^{\sqrt{a^2-x^2}}z \,dy \, dx\,= \int_{0}^{a}\int_0^{\sqrt{a^2-x^2}} x \tan u \, dy\,dx= \int_0^{\sqrt{a^2-x^2}} \tan u \, a\,d{(x^2/2)}= \frac{a^3}3\, \tan\,u $$

Similarly lower notch volume $$V_v=\frac{a^3}3\, \tan\,v $$ Total notch volume $$ V= V_u+V_v = k( \tan\, u+\tan\, v ) \tag1 $$ we are given total notch/wedge angle at vertex fulcrum

$$w= u+v \tag2$$ is constant $w$ as shown. The yellow wedge rotates around $O$ as fulcrum to remove a notch volume along its two arms.

We can now use calculus of variation approach with two variables using Lagrange multiplier $\lambda $ where Lagrangian comprising $(V,w)$ are object and constraints:

$ L= V -\lambda \, w \tag3$

results in

$$\frac{\partial V_u}{\partial V_v}=\frac{\partial w_u}{\partial w_v}= \lambda= 1 \tag4$$

$$\rightarrow u=v = w/2 \tag 5$$

The result says that the arms of total notch volume should be cut out symmetrically about $x=0$ for minimum $V$.