I have read the thread to make this happen for 4 points but only in 2D space here .

I have implemented the answer for 3D but only for 3 control points here

I have read this post but don’t understand the sudo code or the math

Can anyone simplify in java? I don’t want to draw the curve as 2 segments of 3 points

Formula for cubic Bezier curve component (say X):

X(t) = P0.X*(1-t)^3 + 3*P1.X*(1-t)^2*t + 3*P2.X*(1-t)*t^2 + P3.X*t^3

where `P0`

and `P3`

are end points, and `P1`

an `P2`

are control points.

We have four points for curve (`SrcPt`

array in that answer), end points coincide with Bezier endpoints, and two internal points on curve should define two control points `P1`

an `P2`

of Bezier curve. To calculate, we **must** know – what `t`

parameters correspond to `SrcPt[1]`

and `SrcPt[2]`

. Let these parameters are 1/3 and 2/3 (possible issues are in linked answers).

So substitute `t=1/3`

and `t=2/3`

into the formula above:

SrcPt[1].X = SrcPt[0].X*(1-1/3)^3 + 3*P1.X*(1-1/3)^2*1/3 + 3*P2.X*(1-1/3)*1/3^2 + SrcPt[3].X*1/3^3 SrcPt[2].X = SrcPt[0].X*(1-2/3)^3 + 3*P1.X*(1-2/3)^2*2/3 + 3*P2.X*(1-2/3)*(2/3)^2 + SrcPt[3].X*2/3)^3

and solve this system for unknown `P1.X`

and `P2.X`

. So we will have all needed points to describe Bezier curve. Linked answer implements the solution.

Example – how changing `t`

values for the same internal points influences on curve: