Math CentralQuandaries & Queries


Question from farhad, a parent:

i need to measure length of arc by having only two measurements, first the length of the chord and
the height of the curve .
if i have a simple calculator that haven't sin cos tang (i want to calculate it in my mind)

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, 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.

Stephen La Rocque.

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