# Volume of a lozenge shaped pond

1. Apr 7, 2006

### TheRobster

Hi all,

First post here so 1) hello everyone, and 2) I have a problem that I am having trouble solving and was hoping to get some help from the folks on here.

I am trying to work out the formula for calculating the volume of a lozenge-shaped pond. I think this is basically a problem requiring some geometry and integration but it’s been a long time since I’ve done this sort of maths and so am a little rusty.

Basically the lozenge-shaped pond consists of an upper and lower surface that each have perfect semi-circular ends of radii r1 and r2 and each also have a middle section that is either square or rectangular. I’ve uploaded sketches of the pond to my website (see links) which should help clarify what I mean.

http://www.sudsolutions.com/misc/pond1.JPG

http://www.sudsolutions.com/misc/pond2.JPG

Both the upper and lower sections of the lozenge have the same shape but the lower section is smaller and would have sloping sides going down from the top. The information available would be the internal and external length of the pond as well as the internal and external width (which is also equal to the radii r1 and r2 respectively). The depth of the pond would also be known and has being denoted ‘d’ in the diagrams.

So I need to come up with a formula for working out the volume of any lozenge-shaped pond based on the information above. I think this involves:

1) Coming up with two formulas: one for the area of the lower surface and one for the area of the upper
2) Taking these as limits and then integrating to come up with an equation that gives the volume

For the lower surface I have:

And for the upper surface I have:

Now in order to come up with a single equation that gives the total volume of the shape I am guessing that I need to integrate these two equations, taking the lower surface area as the lower limit and the upper surface area as the upper limit, but also need to somehow work the variable “depth, d” into the equation as well.

Problem is I am stumped as how to do this, so any help here would be greatly appreciated!

Many thanks
-Rob

*edit* apologies for putting all the images as links but the

Last edited: Apr 7, 2006
2. Apr 7, 2006

### Curious3141

If I've read your problem correctly, it sounds like a truncated pyramid with a base shaped like a "lozenge" (as you describe it).

You can assume that the volume of a pyramid with base area A (the base can be *any* shape) and height h = 1/3*A*h.

Then treat the figure as a large pyramid minus a smaller one to give the truncated shape seen. The heights, etc. can be worked out with similar triangles, like so :

$$\frac{H}{r_2}=\frac{h}{r_1}$$

$$H = h + d$$

where H is the height of the imaginary big pyramid that begins the surface of the pond and tapers downwards to a point, and h is the height of the imaginary smaller pyramid that begins at the bottom of the pond and tapers downwards to a point.

Just do the algebra, work out H and h, then calculate the volumes of the two pyramids and subtract to find the volume of the pond.

This is an easier approach than calculus.

Last edited: Apr 7, 2006
3. Apr 7, 2006

### Curious3141

If my algebra is correct, the final expression I get is :

$$V = \frac{d}{3}(\pi + \frac{2L}{r_1} - 4)(r_1^2 + r_1r_2 + r_2^2)$$

where L is the "internal length" you alluded to. You don't need to specify external length because they are related by the ratio of the radii r2/r1.

4. Apr 8, 2006

### TheRobster

Hi,

Thanks for the reply although I am not sure if that approach works since I don't think the shape can be treated at a truncated pyramid. In order for the truncated pyramid formula to work, do the sides of the pyramid have all meet at a single point (apex?) at the top if they were continued past the upper surface? The pond does not do this - the side slopes were designed on safety grounds and are approximately 1 in 4. If they are continued until they meet then they do not meet at a point, rather they form a sort of ridge.

Cheers
-Rob

Last edited: Apr 8, 2006
5. Apr 8, 2006

### Curious3141

No, you don't understand, this is a truncated inverted pyramid. The upper (higher) surface (the "base") has a larger area than the lower (deeper) one. I meant that if you took the sloping edges that connect these two cross-sectional surfaces and extended them deeper, they will meet at a point. This is the apex, and it is below the depths of the pond.

Correct ?

6. May 9, 2007

### TVG

Dr

Call the area of the upper surface - A and the lower surface B

The variation in surface is linear with respect to the depth

The average area is 1/2 [A+B}

The volueme is d times the average area = d/2 [A+B]

Let me rename r(1) as r
r(2) R to
R squared as Rsq, r squared as rsq

Note that L(1) = L(2) + 2(R -r), call it L

Upper area A(u) = (pi) [ 2rL - 3r sqared]
Lower area A(l) = (pi) [Rsq +2LR - 4R + 2 rR]

The answer is, d/2times [Au) + A(l)]

TVG

7. May 9, 2007

### Integral

Staff Emeritus
Yet another possibility, take your pick!

Break the pond in to its primitive simple shapes. I see a rectangular solid bounded by the frustum of a cone and 2 triangular cross section beams (? for lack of better word).

The volume of the solid rectangle is:

$$V = (IL-2r) * 2r *h$$

the volume of the frustum is

$$\frac 1 3 \pi h (R^2 +r^2 + Rr)$$

The volume of one of the triangular pieces is:

$$\frac h 2 (R-r) (IL -2r)$$

The total volume is the sum of those pieces.