Yup - .polar is working GREAT (and yes, using radians)
now lets say I have a position somewhere and I want to see where an angle
from it intersects a rectangle (assuming the point is within the
rectangle)... is there a similar method to get that x, y?
On Tue, Mar 17, 2009 at 11:36 AM, Piers Cowb
Lots of cool stuff in the Rectangle class too, in case you didn't
know :)
Piers
On 17 Mar 2009, at 15:00, Eric E. Dolecki wrote:
Thanks man, you just changed my world.
On Tue, Mar 17, 2009 at 10:26 AM, Merrill, Jason <
jason.merr...@bankofamerica.com> wrote:
I have the angle (say 0 degre
: [Flashcoders] Intersection on circle based on angle
Jason,
Could you elaborate please?
Point.polar( length, angle );
I have the angle (say 0 degrees). The length would then be pixels from the
center (ie. radius)?
On Tue, Mar 17, 2009 at 9:53 AM, Merrill, Jason <
jason.m
Thanks man, you just changed my world.
On Tue, Mar 17, 2009 at 10:26 AM, Merrill, Jason <
jason.merr...@bankofamerica.com> wrote:
> >> I have the angle (say 0 degrees).
> >>The length would then be pixels from the
> >>center (ie. radius)?
>
> Yes, radius is pixel distance from starting point (in
>> var coord:Point = Point.polar( _radius, angle );
Yeah, did you see my post? :) Point.polar takes radians, not angle. They are
not the same thing. I posted some methods for you which do the heavy lifting
and conversion of angle to radians. Try it out and see if any of those methods
work for
After seeing Jason's mail, I notice I should add that the angle is
supposed to be in radians (i.e., degrees / 180 * Math.PI). However, he
already provided a convenient set of functions, so you probably don't
need that anymore... :)
Mark
On Tue, Mar 17, 2009 at 3:50 PM, Mark Winterhalder wrote:
On Tue, Mar 17, 2009 at 2:17 PM, Eric E. Dolecki wrote:
> I've done this before but can't find my file(s). I am currently googling but
> haven't found the answer yet (I'm sure it's out there, just need to
> rediscover it).
> I have a circle, registration at 0, 0.
> If I have an angle from it's cen
sorry - found my pilot error. this ROCKS - thanks, I never would have found
that. LOTS of cool stuff in the Point class.
On Tue, Mar 17, 2009 at 10:34 AM, Eric E. Dolecki wrote:
> Okay this thing may be the coolest thing I've seen in a while. However it's
> acting funny.
> If I do this:
>
> var c
Hi,
You have the position of the point in polar coordinates (radius,
angle), so you want to get the point in cartesian (x, y) - this function
converts it. You need the angle in radians though.
It's shorthand for:
x = radius * sin(theta); y = radius * cos(theta);
Where theta =
Okay this thing may be the coolest thing I've seen in a while. However it's
acting funny.
If I do this:
var coord:Point = Point.polar( _radius, angle );
It works great. But I am after a point further out than _radius. So
naturally I pad it a little...
var coord:Point = Point.polar( _radius + 10,
>> I have the angle (say 0 degrees).
>>The length would then be pixels from the
>>center (ie. radius)?
Yes, radius is pixel distance from starting point (in your case, the center
point of the circle). Then you just need to convert angle to radians and
you're good to go.
var cartesianPoint:Poi
Jason,
Could you elaborate please?
Point.polar( length, angle );
I have the angle (say 0 degrees). The length would then be pixels from the
center (ie. radius)?
On Tue, Mar 17, 2009 at 9:53 AM, Merrill, Jason <
jason.merr...@bankofamerica.com> wrote:
> In AS3, Point.polar()
>
>
> Jason Merr
In AS3, Point.polar()
Jason Merrill
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From: flash
This Tutorial proved handy for me:
http://www.codylindley.com/Tutorials/trigonometry/
On Tue, Mar 17, 2009 at 9:17 AM, Eric E. Dolecki wrote:
> I've done this before but can't find my file(s). I am currently googling but
> haven't found the answer yet (I'm sure it's out there, just need to
> redi
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