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Algorithm to generate radial gradient

开发者 https://www.devze.com 2023-01-03 09:46 出处:网络
I have this algorithm开发者_如何学运维 here: pc = # the point you are coloring now p0 = # start point

I have this algorithm开发者_如何学运维 here:

pc = # the point you are coloring now
p0 = # start point
p1 = # end point
v = p1 - p0
d = Length(v)
v = Normalize(v) # or Scale(v, 1/d)

v0 = pc - p0

t = Dot(v0, v)
t = Clamp(t/d, 0, 1)

color = (start_color * t) + (end_color * (1 - t))

to generate point to point linear gradients. It works very well for me. I was wondering if there was a similar algorithm to generate radial gradients. By similar, I mean one that solves for color at point P rather than solve for P at a certain color (where P is the coordinate you are painting).

Thanks


//loop through vector
//x and y px position
int x = i%w;
int y = i/w;
float d = distance(center,int2(x,y));
//if within the grad circle
if(d < radius)
{
  //somehow set v[i] alpha to this:
  float a = d/r;
}


Linerise over atan2(dy,dx) where dx is x-center, and dy is y-center.

cx # center x
cy # center y

r1 # ring is defined by two radius
r2 #  r1 < r2

c1 # start color
c2 # stop color

ang # start angle 

px # currect point x,y
py 
if( px^2 + py^2 <= r2^2 AND px^2 + py^2 >= r1^2  )  # lies in ring?
    t= atan2(py-cy,px-cx)+ang
    t= t+ pi # atan2 is from -pi to pi
    if (t > 2* pi) # it might over 2pi becuse of +ang
       t=t-2*pi
    t=t/(2*pi) # normalise t from 0 to 1
    color = (c1 * t) + (c2 * (1 - t))

Problem whit this algorhitm is that ang is actualy wrong and should be rotated by pi and normalized between 0 and 2pi.


Based on the comment, what you want can still be viewed as a linear gradient -- i.e. you have a line from the center to the outside of the circle, and you have a linear gradient along that line. As such, the calculation is virtually identical to what you already had.

Edit: Okay, apparently I misunderstood what you want. To figure a gradient running around a radius, you still basically linearize it -- figure out the circumference at that radius (2*Pi*R), and then do a linear interpolation along a line of that length.

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