algorithm - How to compute the intersection between a quadratic bezier curve and a horizontal line? -


what trying x coordinates @ bezier curve crosses horizontal line (a y coordinate). moment, have code:

function self.getx(y)     if y > maxy or y < miny         return     end     local = y1 - y     if == 0         return     end     local b = 2*(y2 - y1)     local c = (y3 - 2*y2 + y1)      local discriminant = (b^2 - 4*a*c )      if discriminant < 0         return     else         local abytwo = 2*a         if discriminant == 0             local index1 = -b/abytwo             if 0 < index1 , index1 < 1                 return (1-index1)^2*x1+2*(1-index1)*index1*x2+index1^2*x3             end         else             local thesqrt = math.sqrt(discriminant)             local index1, index2 = (-b -thesqrt)/abytwo, (-b +thesqrt)/abytwo             if 0 < index1 , index1 < 1                 if 0 < index2 , index2 < 1                     return (1-index1)^2*x1+2*(1-index1)*index1*x2+index1^2*x3, (1-index2)^2*x1+2*(1-index2)*index2*x2+index2^2*x3                 else                     return (1-index1)^2*x1+2*(1-index1)*index1*x2+index1^2*x3                 end             elseif 0 < index2 , index2 < 1                 return (1-index2)^2*x1+2*(1-index2)*index2*x2+index2^2*x3             end         end     end      end

a few specifications:

  • this lua code.
  • local means variable local chunk of code, not affect code's functionality.
  • y1, y2, , y3 y coordinate of 3 points. same applies x1, x2, x3.
  • y y coordinate of horizontal line computing.
  • maxy biggest of 3 y's.
  • miny smallest.

for moment code gives me this:

enter image description here

  • there 8 bezier curves
  • the green ones generated using normal method: (1-t)^2*x1+2*(1-t)*t*x2+t^2*x3
  • the red dots control points.
  • the white lines generated using method described code above.
  • the straight lines lines, ignore them.
  • there should 8 curves, 4 rendered.

thanks in advance,

creator!

the bézier curve has

y(t)=(1-t)^2*y1+2(1-t)*t*y2+t^2*y

which expands to

(y1-2*y2+y3)*t^2+2(y2-y1)*t+y1

you have swapped a , c in quadratic equation a*t^2+b*t+c=0 required solving y(t)=y.


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