Parabolic Arch Math

Archimedes formula for parabolic arches says that the area under the arch is 2 3 the base times the height.

Parabolic arch math. Answered by harley weston. The equation of a parabola which opens down is y y v a x x v 2 where x v y v is the vertex in your case this is 0 25 and a is a constant affecting the curvature. Unlike a catenary arch the parabolic arch employs the principle that when weight is uniformly applied above the internal compression see line of thrust resulting from that weight will follow a parabolic curve. What is the distance between the two walls of the arch.

The second arch is created by the transfomation of the vertex 7 0. Then you have a suitable equation. One parabola is f x x 2 3x 1 and hyperbolic cosine is cosh x e x e x 2. A hotel entrance makes a parabolic arch that can be represented by the quadratic function y x 2 8x 24 where y is the height of the arch and x is the distance from wall to wall in the feet.

Thus your equation is just y ax2 25. B graph the line that is symmetric to the parabola and its transformation. Of all arch types the parabolic arch produces the most thrust at the base. The height is 9 units so using archimedes formula the area under the arch is 2 3 6 9 36 square units.

A using graph paper graph the equations of the two arches on the same set of axes. This parabola intersects the x axis ay x 3 and hence the length of the base is 2 3 6 units. An equation of the first arch is given to be y x 2 9 with a range of 0 y 9. Use one of the end points of the arch such as 60 0 to find the value of a.