6 Replies
      Latest reply on Jun 3, 2020 11:32 PM by Claude31
      RJStover Level 1 Level 1 (0 points)

        I would like to draw text along a complex bezier curve. I've seen several examples on the web using CGPathApply

        to do this. But those examples seem to be wrong. CGPathApply does not supply you with a series of points along

        the curve. I just get back the points, and control points, I just used to specify the Bezier curve. I want actual

        pixel coordinates of points along the curve. Is there some way to do this? Do I have to write my own

        function to reproduce the Bezier curve points given the end points and control points? Has Apple documented

        exactly the algorithm they use for doing this? Or is there some CG function I have overlooked that

        does the evaluation for me? Or is there a CG function that takes a string and a path as arguments and

        does the job for me?

         

        Thanks.

        • Re: Text along a curve
          Claude31 Level 8 Level 8 (9,215 points)

          You have visibly searched a lot, so I fear I will not bring any new information.

           

          However, this post seems interesting

          https://stackoverflow.com/questions/26857850/get-points-from-a-uibezierpath/26872803

           

          Hope it will at least hint in a good direction.

           

          Good luck.

            • Re: Text along a curve
              Claude31 Level 8 Level 8 (9,215 points)

              Here is how to compute directly what you need.

               

              Has Apple documented exactly the algorithm they use for doing this?

                   It is not Apple, it is Mr Bezier, an automotive engineer in the 80's.

               

              There is a simple parametric form of a Bezier curve with n control points (Pi).

              Detailed here:

              h ttps://en.wikipedia.org/wiki/Bézier_curve

               

              if t is a parameter between 0 and 1 (from start to end),

              The formula can be expressed explicitly as follows (cannot print the forumla with nice typo on the forum, so I did my best to edit):

               

              B(t) = Sigma   [ C(n,i) (1-t)^(n-i) * (t ^ i) ] Pi

                         (0...n)

               

              where C(n,i) are the binomial coefficients and Pi the control points ((x,y) coordinates).

              For example, for n = 5:

               

                   B(t) = (1-t)^5 P0 + 5 t (1-t)^4 P1 + 10 t^2 (1-t)^3 P2 + 10 t^3 (1-t)^2 P3 + 5 t^4 (1-t) P4 + t^5 P5                0 ≤ t ≤1

               

              With this formula, you can compute the exact coordinates of any point on the curve and find also the orientation of the tangent (so that letter "sits" on the curve by tilting with that tangent angle).

              If P is (x,y), coordinates of the tangent vector at t are given by derivative (I split in 3 lines to make it more readable):

              - C(n,0) n (1-t)^(n-i-1)  P0

              +

              Sigma   C(n,i)  [ - (n-i) (1-t)^(n- i -1) * (t ^ i) + (1-t)^(n-i) * i (t ^ (i-1) ) ] Pi

                         (1…n-1)

              +

              C(n,n) t^(n-1) Pn

               

              For t = 0

              -C(n,0) n P0 + C(n,1) P1 = n (P1- P0): the direction is from P0 to P1 as well known.

               

              Note: How many control points have you for your Bezier Path ?

            • Re: Text along a curve
              janabanana Level 1 Level 1 (20 points)

              There is sample code from Apple called textLayoutDemo that shows how to layout text in a circle.  I'm not sure if this will help but it might be worth exploring.  It dates back to about 2007 (I think).

              • Re: Text along a curve
                Claude31 Level 8 Level 8 (9,215 points)

                Did you try the proposed solution posted May 23, 2020 7:10 AM ?

                 

                Does it work as expected ?