[CST-2] Graphics, knot vectors

[Kevin Chan]

I think you have your equations all mixed up.

Firstly for the N[i,1] cases they should typically be over a range

e.g.

N[1,1] =     {   1, 0<= t< 0
                        0, o/w

and more importantly

N[3,1] =     {   1, 0<= t< 1
                        0, o/w


Then your other equation should be:

N [2,2] = ((t - t2)/(t3-t2)) * N[2,1] + ((t4 - t)/(t4-t3)) * N[3,1]
                      ^^^                                          ^^^
^^^

This is probably where you are getting confused.  Just check the equation
through carefully and it should make sense.  You have just confused yourself
with subscripts =)  Easily done =)

I'm guessing that this should help you with the rest of the question.  I
haven't really looked at it

Kev




----- Original Message -----
From: "Lucy Lian" <lxl20@hermes.cam.ac.uk>
To: <CST-2@srcf.ucam.org>
Sent: Monday, June 04, 2001 2:59 PM
Subject: [CST-2] Graphics, knot vectors


> Can anyone tell me what do we do with the "division by zero"s when
> finding knot vectors? cos I don't quite understand the whole topic.
>
> E.g. if the knot vector is [0 0 0 1 1 1], and for (0 <= t < 1)
> (using []s for subscripts below):
>
> N [1,1] = 0
> N [2,1] = 0
> N [3,1] = 1
> N [4,1] = 0
> N [1,2] = 0
>
> then (no.s appending t's are subscripts too)
> N [2,2] = ((t - t1)/(t3-t2)) * N[2,1] + ((t4 - t)/(t3-t3)) * N[2,1]
>         = 0 + (1 - t)/0
> where 0 is generated from (t3 - t3).
>
> Working backwards from results given in notes N [2, 2] should be (1-t), so
> should we just pretend that the zero is not there, or have I done
> something wrong and it shouldn't be there?
>
> The rest of the basis functions I got for this example are
> (again for (0 <= t < 1)):
>
> N [3,2] = t
> N [1,3] = (1-t)^2
> N [2,3] = 2t(1-t)
> N [3,3] = t^2
>
> but I'm not sure if they're correct, so it'd be great if someone could
> check it.
>
> Many thanks,
> Lucy
>
>
>
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