COVERAGE FOR DIFFERENT SHAPES

COVERAGE FOR DIFFERENT SHAPES

TRIANGLE	To find the number of sq.ft. in any shape triangle or 3 sided surface, multiply the height by the width and divide the total by 2. 40’ height x 50’ width 1,000 sq.ft. 50’ 2,000 sq.ft. 2 2,000
SQUARE	Multiply the base measurement in feet times the height in feet. 40’ 40’ x 40’ = 1.600 sq.ft. 40’
RECTANGLE	Multiply the base measurement in feet times the height in feet. 20’ 40’ x 40’ = 800 sq.ft. 40’
CYLINDER	When circumference (distance around cylinder) is known, mul- tiply height by circumference. 157’ circumference x 100’ height 15,700 sq.ft. 100’ 157’ When diameter (distance across) is known, multiply diameter by 3.1416. This gives circum- ference. Then multiply by height. 3.1416 157’ circumference x 50 diameter x 100’ height 157.0800 feet 15,700 sq.ft. NOTE: Figures do not include end area. See circle.
SPHERE	To find the number of sq.ft. of a sphere or ball, multiply the diameter (distance across) by itself and then multiply this total by 3.1416. If you haven’t the diameter, you can find it by measuring the circumstance and multiplying it by .31831. 50’ diameter 2,5000 x50’ diameter x 3.1416 2,500 7,854.0000 sq.ft.
CIRCLE	To find the number of sq. ft. In a circle, multiply the diameter (distance across) by itself and then multiply this total by .7854 50’ 50’ diameter 2,5000 x50’ diameter x .7854 2,500 1,969 sq.ft.

CHAIN LINK FENCES

Figure the square foot area as a solid and double it to allow for both sides. For spray, this will be fairly accurate since overspray must be considered. For roller, the coverage indicated in the product data sheets should be increased.

WALLS -- When figuring the square foot area, openings of less than 100 square feet should not be deducted.

OPEN WEB STEEL JOISTS

Original equipment manufacturers and fabricators generally dip these joists, as a first or shop coat. On all repaint work, by spray, these manufacturers recommend the paint be estimated by thinking of the joist as a solid rather than open web. Double for both sides.

STACKS

To compute the square foot area of a stack multiply height (B) by the average diameter (A) and multiply that total by 3.

EXAMPLE: Diameter of stack at the top – 5 feet.

Diameter of stack at the bottom – 15 feet.

( 2 /( 5 + 15)

Average diameter = 10 feet.

Height 60 feet.

60 x 10 = 600.

600 x 3 = 1800 square feet of surface area.

DOWNSPOUTS & GUTTERS – Measure the circumference in inches. To figure square feet, divide the circumference by 12 and multiply by total length in feet. Double this figure if the inside is to be painted.

PICKET FENCE – Multiply the height by the length and multiply the result by four.

Citation of chapter in book: Guy E. Weismantel, Chemical Engineering, "Job Preparation" in

Paint Handbook, ed. Harold B. Crawford and Beatrice E. Eckes (The Kingsport Press, 1981)

pp.18-9 & 18-10.

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