Online calculator: Cutting a circle

Below you can find two calculators which calculate how to cut a circle into equal parts - traditional and non-traditional way. By traditional way I assume cutting a circle into sectors, just like you usually cut a pie or pizza. And by non-traditional way I assume cutting a circle into equal slices with parallel lines or with parallel chords, if you like. Both calculators present a drawing which illustrates the result. And you can find all formulas and math in the article below the calculators.

Cutting a Circle into equal sectors

Circle Radius

Number of Sectors

Calculation precision

Digits after the decimal point: 2

Angle of a Sector

Length of an Arc of a Sector

Length of a Chord of a Sector

Cutting a Circle into equal parts with Parallel Cuts

Circle Radius

Number of Parts

Calculation precision

Digits after the decimal point: 2

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Cutting a Circle into Sectors

Ok, you need to cut a circle into number of sectors (even non-even number). To do this, you need to find parameters of a sector. It is a very simple task:

Find the angle of a sector in radians by dividing 2π (which represents 360 degress in radians) by number of sectors.

$\alpha=\frac{2\pi}{N}$

Find the length of an arc of a sectos by multiplying a radius by an angle of a sector in radians.

$a=\alpha R$

Find the length of a chord of a sector by using Law of cosines (a chord is the base of the isosceles triangle, with two radiuses as legs and sector angle as apex angle).

$c=R^2+R^2-2RR \cos \alpha$

This completely defines all N equal sectors.

Cutting a Circle with Parallel Cuts

This way is more interesting. For simplicity I will consider half of a circle, since it is symmetrical.

Cutting a Circle into Three Parts with Two Parallel Lines

Let's cut it with vertical cuts. In this case we need to find the x-coordinates of parallel chords, which should split our circle to equal-area parts. (see points x1 and x2 on the picture above). Let's derive the general formula for an area of a left slice.

Our half-circle can be thought of as a function y=f(x), where x - is the coordinate along abscissa axis, and y is the function equal to value of corresponding half-circle point.

Using pythagorean theorem the y function is

$y=\sqrt{R^2 - x^2}}$

To find an area of a left slice, you need to integrate this function from -R to x. The antiderivative of our function is :

$F(x)=\frac{1}{2} (x\sqrt{R^2-x^2} + R^2 \arctan(\frac{x}{\sqrt{R^2-x^2}}))+C$

We need to find value of constant. Obviously, at the point where x equals -R area should be zero. If we plug -R instead of x into the formula above we get

$F(-R)=-\frac{\pi R^2}{4}+C=0$ , hence

$C=\frac{\pi R^2}{4}$

Our final integral is

$F(x)=\frac{1}{2} (x\sqrt{R^2-x^2} + R^2 \arctan(\frac{x}{\sqrt{R^2-x^2}}))+\frac{\pi R^2}{4}$

Now how do we find x of the first cut? We know we area we should get - Nth part of the total area (note the half-circle)

$S=\frac{\frac{\pi R^2}{2}}{N}=\frac{\pi R^2}{2N}$

Thus we can equate

$\frac{1}{2} (x\sqrt{R^2-x^2} + R^2 \arctan(\frac{x}{\sqrt{R^2-x^2}}))+\frac{\pi R^2}{4}=\frac{\pi R^2}{2N}$

Which gives us

$\frac{1}{2} (x\sqrt{R^2-x^2} + R^2 \arctan(\frac{x}{\sqrt{R^2-x^2}}))+\frac{\pi R^2}{4}-\frac{\pi R^2}{2N}=0$

This is transcendental equation, and we need to use numerical methods to solve it, for example, Bisection method or Newton's method. Here I used Newton's method.

Next points of cut can be found with the same approach. We need to cut two times more for second point $S_2=2\frac{\pi R^2}{2N}$ , three times more for third point $S_3=3\frac{\pi R^2}{2N}$ and so on.