Re: FLASH: formule needed

[unique]

>Hello chosen flash ones!

>

>Can someone think with me on this one:

>I need a formule which calculates every x & y cordinates in a circle.

>let say:

>

>radius	=10

>x	=  0

>y	=  0

>

>so first point

>x=10;y=0

>

>So the next coordinates are?


I've read about folks' "square root" solution to this question, but 

there may be something lacking in going the square root route :-) , 

if you're simply calculating y for incremental values of x, or vice 

versa.


If you want *equally spaced* points along the circumference of the 

circle, you'll need to work with trig instead of algebra. You want 

constant/equal degrees of rotation.


Joe has determined that the first point, which we'll call the 0� 

point, is at (10,0).


Let's say that we want 1� increments. If we want our next point at 1� 

of rotation, we need to consider the trigonometry of a right triangle 

with an angle of 1�. I'll draw it in accordance with the way Jos has 

defined the initial 0� point:


(This will look like crap if you don't have a monospaced font in place.)


                                     *

                                        *

                                           *

                                             *  (x,y)

                                          *****

                                     *****    **

                                *****         * *

                           ***** 1�           *  *

                    (0,0) *********************  *

                                          (x,0)


Since we know the length of the hypotenuse, we can use the formulas 

for sine and cosine to retrofit a number for x and y above (x and y 

are the lengths of the sides of the triangle).


So we end up with:


sin 1� = y / 10 and

cos 1� = x / 10


x = 0.175 and

y = 9.998


Now all ya have to do is work some ActionScript finesse to do the 

same thing 360 total times (in this case of 1� increments), and, 

there you go: equal arcs, all the way around.


Sorry if this was long-winded for some, but now it's "newbie friendly", too.


HTH,


Ken Sherwood



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