Optimizing Gutter Area
Welbilt Inc.
7439 Allen Highway
Vida, MT 59274
Dear Mr. Cal Q. Luss:
As we were approaching the problem of how to design a gutter which would carry the most water from the roof with a 12 inch wide piece of material, we came up with a thesis; the length of the roof can be any number according to the roof sections, the maximum volume of the gutter will then depend on the maximum area out of the shapes' side (flank) with respect to the 12in-wide-material piece. Thus the areas of four different shapes were found. By process of elimination, the optimal area for each shape was found by deriving the equations of area as well as basic geometry. The shapes used for this project were a semi-circle, triangle, square, and rectangle. For all four shapes we maximized the area rather than the volume because the length of the gutter system does not play an important factor. For the semi-circle, it was easy to assume that the circumference was 24 inches, considering that half of the circumference needed to be equal to 12 inches, the width of the manufactured piece of material. The equation for circumference is C=2πr. By plugging in 24 for C, we were able to find the radius, r, to be 12/π. Since finding the optimum area is needed, we could plug in the radius into the equation A=πr², the area of a circle. By substitution, we found the area to be 144/π or 45.8 in². However, this is the area of the full circle, so half of this number is equal to the semi-circle's area. Thus, 22.9 in² is the maximum amount of water a semi-circle can hold. If we assume the shape of the gutter to be an isosceles triangle, then the angle between the two isosceles sides results in the maximum value of the area of the triangle. It makes sense for both sides, AB and AC, to be of equal length, so each side is 6 in. The diagram of the triangle is drawn below.
AB + AC = 12 (12in-wide-piece)
AB = AC (isosceles triangle)
AB = 12/2 = 6in, AC = 6in also.
The
7439 Allen Highway
Vida, MT 59274
Dear Mr. Cal Q. Luss:
As we were approaching the problem of how to design a gutter which would carry the most water from the roof with a 12 inch wide piece of material, we came up with a thesis; the length of the roof can be any number according to the roof sections, the maximum volume of the gutter will then depend on the maximum area out of the shapes' side (flank) with respect to the 12in-wide-material piece. Thus the areas of four different shapes were found. By process of elimination, the optimal area for each shape was found by deriving the equations of area as well as basic geometry. The shapes used for this project were a semi-circle, triangle, square, and rectangle. For all four shapes we maximized the area rather than the volume because the length of the gutter system does not play an important factor. For the semi-circle, it was easy to assume that the circumference was 24 inches, considering that half of the circumference needed to be equal to 12 inches, the width of the manufactured piece of material. The equation for circumference is C=2πr. By plugging in 24 for C, we were able to find the radius, r, to be 12/π. Since finding the optimum area is needed, we could plug in the radius into the equation A=πr², the area of a circle. By substitution, we found the area to be 144/π or 45.8 in². However, this is the area of the full circle, so half of this number is equal to the semi-circle's area. Thus, 22.9 in² is the maximum amount of water a semi-circle can hold. If we assume the shape of the gutter to be an isosceles triangle, then the angle between the two isosceles sides results in the maximum value of the area of the triangle. It makes sense for both sides, AB and AC, to be of equal length, so each side is 6 in. The diagram of the triangle is drawn below.
AB + AC = 12 (12in-wide-piece)
AB = AC (isosceles triangle)
AB = 12/2 = 6in, AC = 6in also.
The