How much does a 10 foot diameter by 3 foot high pool weigh when filled with water?

Answer 1

Assuming the pool is the shape of a perfect circle, the shape of the pool would be a cylindrical solid. The volume of a cylindrical solid is determined by multiplying the area of the circle at one end by the length of the cyllinder. In this case: (pi x r x r) x h 3.14 x 5 x 5 x 3 = 235.61944875 cubic feet The weight of water is 62.42796 pounds per cubic foot. Therefore, not including the weight of the actual pool, the weight of the filled pool would be 14706 pounds, 3.8 ounces, or 7 tons, 706 pounds, 3.8 ounces.

Answer 2

Here's a formula for round pools:

distance across*distance across*depth*5.9

for rectangles:

length*width*depth*7.5

or for a deep end:

length*width*((shallow end depth + deep end depth)/2)*7.5

so in your case, 10*10*2.5*5.9=1500 gallons

Answer 3

The area is pi times diameter squared over 4 or

3.14x10*10/4 = 78.5 sq ft

the volume is height times area or

3 x 78.5 = 235.5 cu ft

there are 7.48 gallons in a cu ft so there are

7.48 x 235.5 = 1761 gallons of water

Answer 4

The pool has a volume of 196.35 cu ft and a maximum capacity of 1,468.9 gallons.

Answer 5

To fill it the pool to a depth of three feet would require about 1,770 gallons of water.

Answer 6

7.48 gallons per cubic foot * 20 feet * 10 feet * 3.5 feet = 5236 gallons