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galactus · May 9, 2008, 05:14 AM

Here is a toughy. ebaines, whatcha think?

Given a semicircular trough of length L and radius r that is full of water, derive a formula for the amount of water in the trough wrt the angle it is tilted to the horizontal.

I believe this requires more than one integral for the different scenarios.

i.e. the top(end tilted up) is partially covered and the top end is not covered.

ebaines · May 9, 2008, 06:01 AM

Galactus - you're idea of fun is definitely twisted! I'll give it a shot and get back.

galactus · May 9, 2008, 10:29 AM

I know. This problem is a booger. There are various scenarios to consider.

Are the slices parallel to the water surface? Is the bottom partially covered or not covered? On and on. It'll probably require 3 or 4 different triple integrals.

ebaines · May 11, 2008, 01:14 PM

I don't think it's all that complicated. Take slices parallel to the end cap, integrating the incremental volume of water for each slice depending on the depth at each point along the length of the trough. The depth is equal to R-x* tan(alpha), up to the point where either x = L or x = R/tan(alpha).

I'm traveling this week, but hope to get you an answer in the next day or so.

galactus · May 14, 2008, 10:42 AM

Here's something I came up with.

$\int_{-r}^{r}\int_{-sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}}\int_{0}^{\frac{y+r}{tan{\theta}}}dzdydx \;\ = \;\ \frac{{\pi}r^{3}}{tan{\theta}}$

galactus · May 15, 2008, 06:18 PM

Oops, mistake in the above. That's for a circular, not semi-circular trough. There is no way to edit or delete. Wonder why?

$\int_{-r}^{r} \;\ \int_{0}^{\sqrt{r^{2}-x^{2}}} \;\ \int_{0}^{ycot{\theta}}dzdydx=\frac{2r^{3}}{3}cot{ \theta}$

There are different ways to approach this. And depending on the level of the water, determines our integral.

The equation for the trough would be $y=\sqrt{r^{2}-x^{2}}$

The equation of the water surface is a plane $z=ky$

k is determined by the angle the trough is tilted.

We could slice the water perpendicular to the y-axis and get rectangular cross sections.

We could slice the water perpendicular to the z-axis and get sections of circles and use polar coordinates.

ebaines · May 16, 2008, 06:25 AM

I don't understand what your integral equation is - are you suggesting the volume of water is not dependent on the length of the trough? Also, for $\theta = 0$ you get $\infty$ , surely that's not right?

Here's what I've got - I used a coordinate system with x along the length of the trough and y parallel to the end cap. The angle of the trough to the ground is $\theta$ . The depth of the water at any point along the length of the trough is then:

$ h=R-x \cdot tan\theta \text { for } x \le R \cdot tan \theta$ , and $0$ for $x \ge R \cdot tan(\theta)$ .

The cross-section area of the water of depth h is found from:

$ A=\frac {\pi R^2} 2 - 2 \displaystyle \int _0 ^h \sqrt {R^2 - y^2} dy \\ =\ \frac {\pi R^2} 2 - (R-h) \sqrt {2Rh-h^2} + R^2 sin ^{-1} ( \frac {R-h} R)\\ $

Subbing in $h = R-x tan \theta$ :

$ A = \frac {\pi R^2} 2 - x tan \theta \sqrt {R^2 - x^2 tan^2 \theta } - R^2 sin ^-1 (\frac {x tan \theta} R ) $

The volume is then:

$ V= \displaystyle \int _0 ^L A dx$ ,

where
$L = min(L, \frac R {tan \theta})$ .

From this I arrived at:
For $\theta =0:\ \ V = \frac {\pi R^2} 2 L $

and for $\theta >0:$

$ V = \frac {\pi R^2} 2L + \frac { tan^2 \theta} 3 [( \frac {R^2} {tan ^2 \theta} - L^2)^{3/2} - (\frac {R^2} {tan^2 \theta} )^ {3/2} ] - R^2 L sin^ {-1} ( \frac {tan \theta} R L) - \frac {R^3} {tan \theta} [ \sqrt { 1 - \frac {tan^2 \theta} {R^2} L^2} -1 ] $
where $L = min (L, \frac R tan \theta} ). $

This is admittedly a bit ugly - perhaps there's an elegant way to simplify this expresion which I'm not seeing.

galactus · May 16, 2008, 08:39 AM

Hey ebaines:

That one I posted is for a particular scenario. When the water does not touch the other end.

There are obviously two integrals needed when the water does touch it and that is what you have. That's similar to my workings.

Very cool, ebaines. You're a 'smart feller'.

I will post my other solution this afternoon.

galactus · May 16, 2008, 11:38 AM

Here's what I came up with when the bottom is covered and the top is only partly covered.

$\int_{-r}^{Ltan({\theta})-r} \;\ \int_{0}^{\sqrt{r^{2}-x^{2}}} \;\ \int_{0}^{\frac{y}{tan{\theta}}}dzdydx+\int_{Ltan{ \theta}-r}^{r} \;\ \int_{0}^{\sqrt{r^{2}-x^{2}}} \;\ \int_{0}^{L}dzdydx$

Barring any typos.:)