CODEWITHSUNDEEP
actionscript-3curvebezier
Tim J
Tim J

Reputation: 245

AS3 Calculating evenly distributed points on a line of points

I have been tasked with trying to create a drawing tool that draws dotted lines as you drag the mouse across the stage. I can easily capture the points on MouseEvent.MOUSE_MOVE and store them in a vector and then draw the points as dots:

The problem is that I need to calculate evenly distributed points on an ever growing Vector of points so I can only draw the line between say every 5th point (say using modulus). I have been battling away with Bezier curve equations both Quadratic and Cubic but still can't quite figure out how to convert my Vector of points into an evenly distributed Vector of Points without sucking the life from the CPU.

Anyone help me? I see that George Profenza has come close to this one here on stack overflow...George?

Upvotes: 1

Views: 832

Answers (1)

Flion
Flion

Reputation: 10932

hmm, I would try it like this: go over the points, calculate the distance between one and the next, keep track of the total distance, keep track of how many dots you placed already. Then, for each next point in the vector, see how many evenly distributed dots you would need to put between the new one and the last one and draw them on Bezier to make fancy, but straight should already be nice even.

example: 3 points in vector, total distance 22. distance per evenly distributed dot: 5. Hence, dots drawn on screen so far: 4. New point has distance 7 to last one, makes total distance 29. You need 5 points now (Math.floor(29/5)=5), you have 4, so you need to draw 1. Rest distance = 22 - 4*5 = 2. So then distance to do = 5-2 = 3. And 3 / distance between new and last point (9) = 0.333 -> so place this point on 1/3rd of the line between the new point in your vector and the last point. As in dot.x = seccondLastPoint.x + ((lastPoint.x - seccondLastPoint.x) * 0.333.

I'm pretty sure that will give you the desired result. Do you think you can build the code from this description?

Upvotes: 1

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