Question

Arc length of an oval that is 6ft by 5ft. i just want 1/4 of the oval. i want to find out how long i would need a wire to go from the ground to the middle of the "dome".

the picture explains it better.

in the end i want to have a wire frame around the garden and than take some netting and use that to keep out the big bugs and birds. i will of course use zip ties to attach the screen/netting to the frame. unless there is a easier and cheaper way to do this.​

Answer 1

You can model the ellipse with the equation

(x/6)² + (y/5)² = 1

If you're comfortable with calculus, read on.

Let x = 6 cos(t ) and y = 5 sin(t ), with 0 ≤ t ≤ π/2. Then the curve C parameterized by (x(t), y(t)) traces out one quarter of the ellipse. The arc length of this curve is given by the integral,

`\displaystyle\int_C\mathrm ds = \int_0^(\pi/2)√(x'(t)^2+y'(t)^2)\,\mathrm dt \\\\ = \int_0^(\pi/2)√(36\sin^2(t)+25\cos^2(t))\,\mathrm dt \\\\ = \int_0^(\pi/2)√(25+11\sin^2(t))\,\mathrm dt \\\\ = 5\int_0^(\pi/2)\sqrt{1+(11)/(25)\sin^2(t)}\,\mathrm dt`

Unfortunately, you cannot find the exact value of this integral. (Well, you can, but it involves introducing a special kind of function; I'll link it in a comment.)

But you can approximate it, and the integral above comes out to about 8.657.

I'm not sure if there's a simpler, purely geometric way to estimate the arc length. But since the semiaxes (5 ft and 6 ft) are fairly close, you could try a circle with a radius somewhere in between. The arc length would range between 7.854 ft and 9.425 ft. If the material isn't too expensive, you can always overshoot and trim off the excess or repurpose it.