The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Making a Pool Tarp

Date: 9/20/95 at 14:26:7
From: Anonymous
Subject: geometric slope

I am trying to construct a pool frame out of PVC that will be 
placed over a pool.  We already have the pool tarp, but we need 
to build something that will shed water so it has to have a 
slope.  PVC only comes in 90 degree and 45 degree angles and 
straight sections.  How do I figure out the maximum height the 
frame can stick straight up in the air to utilize the tarp. 
The pool tarp is:

        272.5 inches width
        484 inches length

The pool frame (that needs to be constructed out of pvc can 
not be smaller than):

        226 inches
        412 inches

How do figure out the maximum height based on the pool cover 
dimensions so that the cover will fit on the frame width wide.  
I can make up for the length but the width is crucial.... (this 
is difficult to explain a drawing would be better).

How do I figure out if this will even work?  HELP!!!

Date: 9/22/95 at 16:22:5
From: Doctor Andrew
Subject: Re: geometric slope

If I understand your problem correctly, you want to build 
something that looks like this (not as steep though) to cover 
your pool:

    /|\      -
   / | \     |
  z  |  z    h
 /   | A \   |
/    |    \  |
|~~~~~~~~~|  -   
|  pool   |

You want to have the biggest height h possible in order to  
have a steepest slope possible on your pool frame.  Let x be 
half the width of the base of the frame.  Let z be half the 
width of the tarp. It turns out, as might be intuitive, that 
constructing the frame symmetrically across the pool yields the 
largest h.  (You can check this with a piece of paper or 

The Pythagorean Theorem states that for a right triangle (one 
with two perpendicular sides), like the triangle marked "A" in 
the diagram, the sum of the squares of the two smaller sides 
equals the square of the longer side.  So,

z^2 = x^2 + h^2 

Now, we need to solve for h in terms of x and z.  Well, 
subtracting x^2 from both sides of the equation we get

h^2 = z^2 - x^2. 

Taking the square root of each side, we get 

h = square root(z^2 - x^2).

In order to make h as big as possible we want the largest z and 
smallest x we can use.  The largest z is half the width of the 
tarp.  The smallest x is half the width of the smallest base of 
the frame we can use.

So, in your case, h = square root ((272.5 / 2)^2 - (226 / 
2)^2) = 76.1253 in.

You can also check to make sure you have enough length using 
the same technique.  If width is ignored, the optimal h for the 
length of the tarp is

h = square root ((484 / 2)^2 - (412 / 2)^2) = 126.996 in.

So, it looks like width is, as you already said, the limiting 

Before you go cutting any PVC you could cut yourself a 272.5 
inch and a 226 inch piece of string.  If you pin the shorter 
one at each end, the string should be able to fit a 76.1253 
inch stick standing upright between the ground and the longer 

I sure hope I've answered the right question.  Please try the 
string experiment in case I've made a mistake. Good luck!

-Doctor Andrew, The Geometry Forum

Date: 9/27/95 at 13:28:25
Date: Wed, 27 Sep 1995 13:28:54 -0500
Subject:  Re: geometric slope -Reply

I appreciate your math!  I will do what you say and try it 
with string first.... but it does seem to me that 76 inches 
is a HUGE amount of height considering that the overlap 
between the pool tarp width and the pool frame is 46.5 
inches.  I don't see how you could have a height of apx 8 
ft.... something isn't right with the equation.... do you 
have any ideas???

P.S. Your tarp drawing is correct.  I just don't think that 
eight feet high would be the correct answer if the overlap is 
only 46.5 inches in the width.

Date: 9/27/95 at 14:58:38
From: Doctor Andrew
Subject: Re: geometric slope -Reply

It sure seems huge, but I've checked it.  It certainly isn't 
intuitive.  There is another problem where you have a steel 
belt around the equator of a earth and you add a foot to it.  
Suddenly it is off the face of the earth by miles if I 
remember correctly.  Anyway, with regard to this problem, you 
can just use the Pythagorean Theorem to check your answer:

If 113 if half the frame width and 76 is the maximum height:

sqrt(113^2 + 76^2) = 136 

136 is half the tarp, right?

So it seems to work.  When I worked on your problem I didn't 
really stop to think about the physical implications of the 
result, but it's pretty neat.

I have a feeling you're not going to be building a 6 ft pool 
frame, but it sure looks like you have enough tarp.  Still, 
I'd try it with the string. :^)

- Doctor Andrew, The Geometry Forum

Associated Topics:
High School Geometry
High School Practical Geometry
High School Triangles and Other Polygons

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.