Question

Compute the **x and y coordinate** of the Point
P(t=0.2)P(t=0.2) on the cubic B-Spline

P(t)=∑nk=1PkBk,4(t)P(t)=∑k=1nPkBk,4(t)

Control Points: P1P1 = ( 3|5), P2P2 = ( 6|6), P3P3 = ( 7|10), P4P4 = ( 20|11)

Use the following blending functions

B0,4=16(1−3t+3t2−t3)B0,4=16(1−3t+3t2−t3)

B1,4=16(4−6t2+3t3)B1,4=16(4−6t2+3t3)

B2,4=16(1+3t+3t2−3t3)B2,4=16(1+3t+3t2−3t3)

B3,4=16(t3)B3,4=16(t3)

Px =

PY=

Show in steps complete calculation

Answer #1

Here is a matlab code just to solve in one click,

B = zeros (4,4)

%t = [0:0.1:1]

t = 0.2;

B(1,4) = 16*(1-3*t+3.*t.^2-t.^3) %B0,4 In matlab index starts from
1 not from 0

B(2,4) = 16*(4-6.*t.^2+ 3.*t.^3) %B1,4

B(3,4) = 16*(1+3*t+3.*t.^2-3*t.^3) %B2,4

B(4,4) = 16*t.^3 %B3,4

xt = [3, 6,7,10]

yt = [5,6,10,11]

Px = 0;

Py = 0;

for i = 1 : 4

Px = Px +xt(i).*B(i,4) ;

Py = Py + yt(i).*B(i,4);

end

Px

Py

The answers for Px and Py at t = 0.2, t = 0.2 are,

Px(t=0.2) = 579.0720

Py(t=0.2) = 676.9920

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