Hi Farhad.

Using those two measurements, you can determine the arc length by first calculating (in your head) the radius r.

Then, as Harley described in http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/dale2.html, you can use that to put the result into the form A arctan B, where A and B are values you've calculated in your head.

However, that uses arctangent. Arctan can be approximated using the Taylor expansion:

tan^-1(x) = x - x³/3 + x⁵/5 - x⁷/7 ... and on to infinity ...

or until you figure you are precise enough for your purposes.

So given all this, let's say I have a circular arch 6 meters along the ground and 1 meter tall in the center. How long is the arc of the arch?

You can try this in your head, I'll write it out.

First the radius:

r² = (c / 2)² + (r - h)²
r² = (6 / 2)² + (r - 1)²
r² = 9 + r² - 2r +1
2r = 10
r = 5.

[Yes, I am making the numbers "work out" nicely. In your head, with real world measurements, this could be more challenging....]

Now the formula Harley worked out for the arc length (I call it s):

s = 2r tan^-1[ (c / 2) / (r - h)]
s = 2(5) tan^-1[(6 / 2)/(5 - 1)]
s = 10 tan^-1(0.75)

And now the Taylor series to calculate an approximation of the tan^-1 value:

s ≈ 10 [ 0.75 - (0.75³) / 3 + (0.75⁵) / 5 ] = 6.57 m

if we take another term, we get

s ≈ 10 [ 0.75 - (0.75³) / 3 + (0.75⁵) / 5 - (0.75⁷) / 7 ] = 6.38 m

That's a substantial difference. A fifth term makes it 6.46m. A sixth term makes it 6.42m. A seventh term makes it 6.44m which is getting close to what my calculator says is the more precise arctangent of 0.75 radians: 6.435m.

So you see, we needed 7 terms just to get the accuracy to a centimeter rounded off. That means calculating, among all the rest, 0.75 to the thirteenth power, in your head.

I'm impressed if you can do that, even with a number as round as 0.75.

Cheers,
Stephen La Rocque.