Hi,
The function that I gave you (A*x +x^2) was an arbitrary example.
Looks like you'd need to use what ever bezier math is used by the Flash
player to represent the line. Not the easiest thing to figure out, but
here's a link that might help:
http://www.moshplant.com/direct-or/bezier/math.html
There might be tricks that we're not thinking of. For instance - take a
bitmap snapshot of it in memory. Then use getPixel(x, y) to walk which ever
column you're in until you reach a black pixel (or non-alpha if you use
getPixel32). That's your y.
Another trick is to position the clips on the line where they need to be
when the line is at rest. Then displace them vertically based on the
amplitude of the center displacement.
Good luck.
Jobe Makar
http://www.electrotank.com
http://www.electro-server.com
phone: 252-627-8026
mobile: 919-609-0408
fax: 919-882-1121
----- Original Message -----
From: "leolea" <[EMAIL PROTECTED]>
To: <flashcoders@chattyfig.figleaf.com>
Sent: Wednesday, April 25, 2007 4:28 PM
Subject: Re: [Flashcoders] Curves question for math gurus
Hi, thanks for your reply!
My curve isn't exactly a circle. Here's what my animated curve would look
like:
http://pages.videotron.com/poubou/flash/cannes01.html
The curve is drawn using the drawing API:
example:
mc.moveTo(0,0);
mc.curveTo(400,900,0,800);
So, I know the 3 bezier points that define my curve:
startpoint = 0,0
middlepoint = 400,900
endpoint = 800,0
With those values in hand, how can I apply them to your function:
f(x) = A*x + x^2
Do I make any sense?
On 4/25/07 3:50 PM, "Jobe Makar" <[EMAIL PROTECTED]> wrote:
Hi,
Your typical funciton looks something like this in math books:
f(x) = A*x + x^2 //just an example
Where f(x) is essentially 'y'. So, you just need the equation that
defines
your curve. The curve in your jpg appears to be a circle.
y = sqrt(x^2 + r^2) //where r is the radius
That actually yields + or - and you just pick what fits your situation
best.
So, you pump in an x and get you 2 y's. Pick the best y and use it.
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