



 
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  x^{3}/3 + x^{5}/5  x^{7}/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:
[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):
And now the Taylor series to calculate an approximation of the tan^{1} value: s ≈ 10 [ 0.75  (0.75^{3}) / 3 + (0.75^{5}) / 5 ] = 6.57 m if we take another term, we get s ≈ 10 [ 0.75  (0.75^{3}) / 3 + (0.75^{5}) / 5  (0.75^{7}) / 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,  


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