How wide is needed to make a corrugated panel 52 inches wide 4 inches thick?
A manufacturer of metal roofing corrugated wants to produce panels that are 52 inches wide and 4 inches thick in the treatment of flat sheets of metal in the form indicated above The profile of the roof takes the form of a sine wave. If x = 0, y = 0 is the left end of the profile shown above and x = 52, y = 0 is the right end, what is the equation y = f (x) profile? It was part A and the answer was y = 2 * sin ((2 * pi * X) / 13) Pat B asks: What is the width (in inches) of a flat sheet is necessary to make a 52 inch wide corrugated panel 4 inches thick? You can bring the response within 0.01 inches. Could someone please explain in Part B for me. Thank you
You must calculate length of the curve. Let (x1, y1) and (x2, y2) are two points of the curve very close to each other. The distance between them is L = √ (x2-x1) ² + (Y2-y1) ² = | x2-x1 | √ [1 + ((y2-y1) / (x2-x1)) ²] When x2 and x1 are infinitely close to each other, their distance is dl = dx √ (1 + y ²) The length of the curve is L = ∫ (x from a to b) √ (1 + y '²) dx In your case, y = 2 sin (2π x / 13), y '= (4π / 13) cos (2π x / 13). Then L = ∫ (x 0 to 52) √ [1 + ((4π / 13) cos (2π x / 13)) ²] dx appears to be a part elliptical. I calculated the length numerically, and obtained L = 61.2658
METAL ROOFING – Chimney Counter Flashing