Little Math Help Needed

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Little Bill

Little Bill

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I have to give a quote on a pool and the site is not available yet. I need to figure how much bond wire to include. I usually measure for this but can't as of now.
If I have a rectangular shape pool with dimensions of 16' x 42'. Now add the required 18" - 24" from the outside wall of the pool. How would I get the amount needed from these dimensions pretty close? I know I can add the two sides and two ends plus the added 2' but was thinking there might be a area/radius formula to use to double check myself.
 
If I’m reading this right you got it.
18x44, or 124’. Then add what’s needed to get up to panel

area is 18 X 44 or 792ft^2
 
Is the bond wire a simple rectangle, 18" - 24" from the outside wall? If so, and we use 24", that means the bond wire will be a rectangle that is 20' x 46'. So the perimeter is 20' + 46' + 20' + 46' = 132'.

Or if you want 24" radius corners, it will be a little less. You can just take the perimeter of the pool and add the perimeter of a 24" circle (because you need (4) quarter circles, or one full circle). So that's 16' + 42' + 16' + 42' + pi * 4' = 129'.

Cheers, Wayne
 
Just subtract the pool area from the overall area. Large rectangle minus small rectangle.

Or look at it like a two-foot-wide sidewalk on each side of the pool plus the corners.
 
Little Bill said:
I have to give a quote on a pool and the site is not available yet. I need to figure how much bond wire to include. I usually measure for this but can't as of now.
If I have a rectangular shape pool with dimensions of 16' x 42'. Now add the required 18" - 24" from the outside wall of the pool. How would I get the amount needed from these dimensions pretty close? I know I can add the two sides and two ends plus the added 2' but was thinking there might be a area/radius formula to use to double check myself.


Bill is this for the equipotential bonding around the perimeter. If so, then you just need to add the sides-- you don't need anything else.
Click to expand...
 
ptonsparky said:
Wouldn’t the extra 24” make it a 20x46?
Click to expand...
Yes, that's how I see it. Linear dimensions so why bring in area? Anyway, (20+46)*2 or 132 ft. Making the corners a quarter circle arc would reduce that by about 12 ft.
But more joints. And how easy is it to make an accurate radius arc? Is it worth doing?
 
wwhitney said:
It would replace a 4x4 square (16 feet perimeter) with a 4' diameter circle (12.56 feet perimeter), so it would reduce it by about 3'.

Cheers, Wayne
Click to expand...
You'd also lose 2ft of each end of each of the straight sections.
 
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