“…1, with the initial parameters (the control point vector and the knot vector) and a combination with (3), it is able to generate the B-spline basis functions, note that the variable in (3) now becomes in (7) with = 0, … ,10 with respect to the example. Since it has a total of seven control points, it requires seven B-spline basis functions of degree 3 which are 0 3 ( ), … , 6 3 ( ) to be able to calculate the trajectory using (1). After calculating all of the B-spline basis functions, note that in every knot span [ , +1 ), there are at most + 1 basis functions are not null, namely − , … , , for example there are 4 basis functions 0 3 , 1 3 , 2 3 , 3 3 are not null in the interval [ 3 , 4 ) as shown in Fig.…”