Comments on: Pi and Pie: Enjoying some time with trigonometry

By: Juilan Viereck

Juilan Viereck — Sun, 07 Jun 2009 19:55:54 +0000

oh yes, I had to solve this on my own some years ago :D

But to put a little bit more math in there: you should calculate the distance between the center of the pie and the mouse-click-point. If this distance > radius => user clicked outside of the pie => we don’t have to care about this ;)

That’s where you bring in the nice math equation: a^2 + b^2 = c^2.

And who wants to miss this nice equation ;)

By: dion

dion — Fri, 05 Jun 2009 02:42:02 +0000

Awesome comments!

@Joe Ah come on mate, there is always room for a new slice of pie ;)

@Mike: Great point for this solution!

@zznq: Very true. atan2 is so simple though that it didn’t seem like more work really.

By: Ryan Rampersad

Ryan Rampersad — Fri, 05 Jun 2009 01:21:40 +0000

This is neat if only because I’m finishing my school year learning trigonometric identities. I’ve always wanted to make a circle menu.

By: Joe Walker

Joe Walker — Thu, 04 Jun 2009 22:06:11 +0000

Dion's general solution works for an arbitrary number of slices. If you want small, incomprehensible and quadrants only, you could also do: return ["bottom","left","right","top"][(x>y?2:0)+(500-x>y?1:0)]; Needless to say, I'll leave Dion's version in Bespin, if only to prevent him from changing the number of slices. ;-)

By: Will Riley

Will Riley — Thu, 04 Jun 2009 20:50:51 +0000

I did something similar a while ago, it’s in dojoc. I used dojox.gfx to draw the pie, and trig to figure out the angle. There’s also some code there so that it can support any number of menu items.

You can grab the code here: http://svn.dojotoolkit.org/dojoc/sandbox/piemenu/

By: Mike

Mike — Thu, 04 Jun 2009 20:43:22 +0000

Trig is cool but occasionally people resort to it prematurely.
High school linear algebra applies.
You have two lines that each divide the plane in two. One line is y=x and the other line is y=-x. You can compute a boolean for each line that says which side you’re on.
var a = y>-x; var b = !a;
var c = y>x; var d = !c;
If area one is the area that contains (1,0) and you number counter-clockwise,
var area;
if (d & a) area=1;
if (c & a) area = 2;
if (c & b) area = 3;
if (d & b) area = 4;
Optimization left to the reader.
Now suppose you want the wedges that contain (1,0) and (-1,0) to be thinner than the wedges that contain (0,1) and (0,-1). This simply changes the equations of your lines and the resulting definitions of the predicates a, b, c and d.
Suppose the slopes of these lines are .5 and -.5:
y = x *.5;
y = -x *.5;
Change two statements of the code:
var a = y > -x*.5;
var c = y > x*.5;
The rest of the above logic would be the same.

By: zznq

zznq — Thu, 04 Jun 2009 19:10:16 +0000

Using atan2 to solve this seems to be a bit like overkill; a coordinate change (with your easy-ville solution) would probably function better.