[Math] Calculate origin of pie slices

circles

I need to draw a pie graph with slices similar to the following…

enter image description here

The required information I need to draw a single slice is the X & Y coordinate of the origin of the slice and the angle of the slice. Not too difficult to draw a single slice. However to draw slices in the fashion shown above, a new origin is required for each slice which is denoted by red dots in the image.

I can't seem to figure out a generic algorithm or formula to determine the origin of the next slice. Math is not my strongest point, but any guidance would be helpful.

Best Answer

enter image description here Hint: You have 12 slices so the central angle $360^0$ is going to be divided by 12 which will give you the angle of separation for each slice, ie. ($\frac{360^0}{12}=30^0$).

This is literally the unit circle( just use a pie of radius 1), anddelete the angles $45,135,225,315$. Lastly, just write equations for line that go through those points and the origin, Notice you will only need to make 4 equations of lines.

Related Solutions

[Math] How to determine the number of degrees overlap between two circular slices

Let the sectors be $\theta_1\pm\Delta\theta_1$ and $\theta_2\pm\Delta\theta_2$ in your notation. The angle between the sector centers is $\phi = |\mathrm{wrap}(\theta_1-\theta_2)|$, where $$\mathrm{wrap}(\theta) = \begin{cases} \theta & \text{if $-180^\circ\le\theta\le180^\circ$,} \\ \theta+360^\circ & \text{if $\theta < -180^\circ$,} \\ \theta-360^\circ & \text{if $\theta > 180^\circ$} \end{cases}$$ is the angle normalized to lie between $-180^\circ$ and $180^\circ$. Then the angle of overlap is simply $\Delta\theta_1+\Delta\theta_2-\phi$; if this is negative, there is no overlap.

[Math] Finding formula for an arc of a circle that fills a rectangle

In addition to the answer pointed out by @n_b I would like to show how to reach there:

The final result:

$$\LARGE\boxed{R=\frac H2+\frac{W^2}{8H}}$$

float cos_t = (R-H)/R;
P_x=(rect.width/2);
P_y=-R.cos_t;

Process:

enter image description here

Without using calculus(actually it would never be solved with calculus) and only using trigonometry:

The component of R along $\theta$ is $R.\cos\theta$ and perpendicular to it is $R.\sin\theta$

Now remaining component is $R-Rcos\theta=R(1-\cos\theta)$

Now we have: $$ W=2.R.\sin\theta\implies\sin\theta=\frac W{2R}\\ H=R(1-\cos\theta)\implies\cos\theta=1-\frac HR=\frac{R-H}R $$ Using the identity $\sin^2\theta+\cos^2\theta=1$ $$\left(\frac W{2R}\right)^2+\left(\frac{R-H}R\right)^2=1$$ $$\frac{W^2}{4R^2}+\frac{R^2+H^2-2RH}{R^2}=1\\\text{ I assume you are familiar with }(a\pm b)^2=a^2+b^2\pm 2ab$$ $$\frac{W^2}4+R^2+H^2-2RH=R^2$$ $$\frac{W^2}4+H^2=2RH$$ $$R=\frac{W^2}{8H}+\frac H2$$ It is easy to see that center is at the midpoint of width. Also y-coordinate is $-R.\cos\theta$