# A 17'.6" Span straight line - If I do an ark that goes 2 1/2' OR 3' high ( In the middle, How long is the arc ?

Relevance

If the arc is a part of a circle, that's one thing; if it's a catenary or a parabola, that's something else again! I'm going to assume the arc is part of a circle.

Let's say the chord of the circle is 17'6" in length, that is, 17.5 feet, and that the arc goes 2.5 feet high. Then three points on the circle might be represented as (-8.75,0), (0,2.5), and (+8.75,0). These three points must be equidistant from the center of the circle, which I'll call (0,-c). In particular,

sqrt(8.75^2 + c^2) must equal c + 2.5, so you have:

8.75^2 + c^2 = c^2 + 5c + 6.25, or

5c = 8.75^2 - 6.25 = 70.3125, and

c = 14.0625.

The radius of the circle is 16.5625, and the central angle between the two radii that go to (-8.75,0) and (+8.75,0) is 2*arcsin(8.75/16.5625), or 63.78 degrees.

Finally, the arc length is (63.78/360)*(2*pi*16.5625) = about 18.44 feet.

That's the answer for a 2 1/2' rise.

To find the answer for a 3' rise, go back to my calculation and replace the "c + 2.5" by "c+3" and repeat the rest of the calculation.

• We will assume a circular arc.

Let's define some points to make the calculation easier to talk about. Let A and B represent the ends of the span, and X its midpoint. The center of the arc (above X) will be C, and the center of the circle of which the arc is a part will be O.

Assuming your span is 17 feet 6 inches, not 17 feet 0.6 inches, the distance AX is 17.5/2 = 8.75 feet. You want the distance CX to be 2.5 to 3 feet, so the angle XCA will be

.. XCA = arctan(8.75/[2.5, 3]) = [74.05, 71.08]°

The angle COA will be twice the complement of this angle, so will be

.. COA = [31.85°, 37.85°]

The corresponding distance AO will be

.. AO = 8.75/sin(COA) = [16.5625, 14.2604] . . . . feet

The arc length is the product of this radius and twice the angle COA (in radians), hence

.. arc ACB = [18.44, 18.84] . . . . . feet

An arc length of 18' 6" will give a height between 2.5 and 3 ft in the middle.