Solution sheet ξi ξ < ξ i+1 0 otherwise ξ ξ i N i,p 1 (ξ) + where 0 0

Transcription

1 Advanced Finite Elements MA WS7/8 Solution sheet This exercise sheets deals with B-slines and NURBS, which are the basis of isogeometric analysis as they will later relace the olynomial ansatz-functions from standard finite elements. In the first exercise we will ractice how to construct B-sline functions (of higher order) and how the relevant quantities like number of degrees of freedom, knots, and sline order are linked. In the second exercise we rove some elementary roerties of B-sline functions, that will be useful later on. Exercise three then introduces B-sline curves, how they are formed and how they can be maniulated. In exercise four we finally see how a circle can be reresented exactly within the isogeometric framework using a NURBS-arametrization. Exercise 3 (Constructing B-sline basisfunctions) Given a knot vector Ξ = (ξ, ξ,..., ξ n++ ) R n++ with ξ i ξ j for i j the B-sline basis functions of degree N are defined for i =,..., n as follows: B-sline basis functions: N i, (ξ) := N i, (ξ) := { ξi ξ < ξ i+ otherwise ξ ξ i N i, (ξ) + ξ i++ ξ N i+, (ξ) () ξ i+ ξ i ξ i++ ξ i+ where := We only consider oen knot vectors Ξ, i.e. the first + and last + knots coincide ξ =... = ξ + and ξ n+ =... = ξ n++. a) Give a formula for the number of basis functions of degree that are created by such an oen knot vector with Ξ = K knots in total. b) Usually the first + reeated knots in the knot vector are set to, the last + reeated ones to. How many interior knots are necessary (e.g. equally distributed in the interval (, )) to get a total of n basis functions of degree? c) For the concrete knot vector Ξ = (,,, 3, 3,,, ) sketch the basis functions of degree =,,.

2 Solution 3 a) If K is the total number of knots in the knot vector, we comare with the definition of knot vectors and set K = n + +, where n is the number of basis functions and their degree. Hence n = K basis functions of degree can be created from the knot vector. b) If k denotes the number of interior knots, we can add these together with the first + and last + reeated ones to have a total of K = k + ( + ) knots. Now using the formula from a) we get k + ( + ) = K = n + + and hence k = n. c) The basis-functions of degree zero are constructed trivially. Here it is only imortant not to forget also the ones that are identically zero, like e.g. N, (ξ) which would only be not zero for ξ [ξ, ξ ) but this interval is emty due to ξ = ξ, hence N, (ξ). The basis functions of degree one again start with a constant zero function, since N, is a combination of N, and N, which are both zero. From there on one always combines the ξ ξ i ξ i+ ξ i iecewise constant functions of degree zero with the terms of () giving iecewise linear hat-functions known from classical finite element theory. For the quadratic basis-functions one now starts from the linear hat-functions and again multilies them with terms, yielding iecewise quadratic functions. ξ ξ i ξ i+ ξ i Fig. : B-sline basis of left: degree, middle: degree, right: degree, each for the node vector (,,,,,,, ). 3 3 We give the analytical exressions for the first basis-functions of each degree: degree degree degree N, (ξ) N, (ξ) N, (ξ) = ( 3ξ) χ [,/3) (ξ) N, (ξ) N 3, (ξ) = χ [,/3) (ξ) N, (ξ) = ( 3ξ) χ [,/3) (ξ) N, (ξ) = 3ξ( 3ξ)χ [,/3) (ξ)+ N 3, (ξ) = 3ξχ [,/3) (ξ)+ ( 3ξ)3ξχ [,/3) (ξ)+ N 4, (ξ) = χ [/3,/3) (ξ) ( 3ξ)χ [/3,/3) (ξ) ( 3ξ) χ [/3,/3) (ξ)... If you want to lay around with different knot vectors and degrees, you can find the MATLAB-file draw_b_sline_basis.m on the lecture homeage. It is a function that accets a knot vector and a degree as an inut and lots the B-sline basis-functions of

3 that knot-vector u to the given degree. The ictures above can be recreated using: N = draw_b_sline_basis ([,,,/3,/3,,,], ); The function is nothing else but a direct imlementation of formula (), where secial attention has to be aid to the / := -cases. Exercise 3 (Proerties of B-Sline basis functions) Proof the following roerties of the B-sline basis functions with oen knot vectors. a) Their suort always covers + knot-sans, i.e. su(n i, ) = [ξ i, ξ i++ ]. b) They form a artition of unity: N i, (ξ) on the interval [ξ +, ξ n+ ]. i= c ) Where defined, for the derivatives of the B-sline basis functions there holds: B-sline derivative formula: d dξ N i,(ξ) = ξ i+ ξ i N i, (ξ) ξ i++ ξ i+ N i+, (ξ) () d ) At knot ξ i the basis functions are m i times continuously differentiable, where m i states how often ξ i aears in the knot vector (knot multilicity). e ) If m i for some knot ξ i only the basis function N i, (take i to be the number of the last reetition of the reeated knot, for examle if ξ 3 = ξ 4 = ξ 5 take i = 5) is taking the value at ξ i while all the others are there (interolatory roerty). Solution 3 Since the B-sline basis functions are defined in a recursive way, induction over is the method of choice to rove those roerties: a) For = the claim holds by definition. Assuming it holds for one sees by induction: N i, (ξ) := ξ ξ i ξ i+ ξ i }{{} in (ξ i,ξ +i ) N i, (ξ) }{{} has su =[ξ i,ξ +i ] + ξ i++ ξ ξ i++ ξ i+ }{{} in (ξ i+,ξ i++ ) N i+, (ξ) }{{} has su =[ξ i+,ξ i++ ] Hence su(n i, ) = [ξ i, ξ i++ ]. Here we have excluded the degenerated case of the suort being the emty set, e.g. for the first degree basis function in an oen not vector. b) Obviously the claim holds for =, hence we assume it to hold for and do the induction ste ( ). Using the recursive definition of the basis functions and 3

4 reordering the terms within the sum by an index shift gives: ξ ξ i N i, (ξ) = N i, (ξ) + ξ i= i= i+ ξ i = ξ ξ ξ + ξ N, (ξ) + = ξ ξ + ξ + ξ i= ξ i++ ξ ξ i++ ξ i+ N i+, (ξ) ξ ξ i ξ i+ ξ i N i, (ξ) + ξ i+ ξ ξ i+ ξ i N i, (ξ) i= i= + ξ n++ ξ N n+, (ξ) ξ n++ ξ n+ = ξ ξ N, (ξ) + N i, (ξ) + ξ n++ ξ N n+, (ξ) ξ + ξ ξ i= n++ ξ n+ }{{} = N, N n+, ( ) ( ) ξ ξ ξn++ ξ = + N, (ξ) + N n+, (ξ) ξ + ξ ξ n++ ξ n+ }{{}}{{} = ξ n+ ξ ξ n++ ξ n+ = + (ξ ξ ) N, (ξ) ξ + ξ }{{} = +(ξ n+ ξ) N n+, (ξ) ξ n++ ξ n+ }{{} = The last two underbraces are zero due to the definition of := and the fact that in an oen knot vector ξ = ξ +, ξ n+ = ξ n++ and also N, N n+,. If the knot vector is not oen, the artition of unity roerty only holds on the interval [ξ +, ξ n+ ], where the suort of N, has already ended (due to a)) and the suort of N n+, has not yet started (also due to a)) and hence the terms are again zero. c) Again we will use induction over the degree from ( ). For = the basis functions are iecewise constant and hence their derivative (where defined) is constant zero, which is also the case in the claimed formula when setting =. To do the induction ste from ( ) to we roceed from both sides and in the end find the equality sign in between..) First we differentiate () by using the roduct rule: d dξ N i,(ξ) = ξ i+ ξ i N i, (ξ) + ξ ξ i d ξ i+ ξ i dξ N i, (ξ) = N i+, (ξ) + ξ i++ ξ d ξ i++ ξ i+ ξ i++ ξ i+ dξ N i+, (ξ) By the induction remise for degree we know that: d dξ N i, (ξ) = N i, (ξ) N i+, (ξ) ξ i+ ξ i ξ i+ ξ i+ d dξ N i+, (ξ) = ξ i+ ξ i+ N i+, (ξ) ξ i++ ξ i+ N i+, (ξ) 4

5 which we substitute into the resective terms from above to arrive at: d dξ N i,(ξ) = ξ ξ ( ) i N i, (ξ) ξ i+ ξ i ξ i+ ξ i ξ ξ ( ) i N i+, (ξ) ξ i+ ξ i ξ i+ ξ i+ + ξ ( ) i++ ξ N i+, (ξ) ξ i++ ξ i+ ξ i+ ξ i+ ξ ( ) i++ ξ N i+, (ξ) ξ i++ ξ i+ ξ i++ ξ i+ + N i, (ξ) N i+, (ξ) ξ i+ ξ i ξ i++ ξ i+ (3a) (3b) (3c) (3d) (3e).) From the other side we start with the claimed formula () and substitute the recursion formula () at the resective ositions, to obtain: d dξ N i,(ξ) = N i, (ξ) + N i, (ξ) ξ i+ ξ i ξ i+ ξ i N i+, (ξ) N i+, (ξ) ξ i++ ξ i+ ξ i++ ξ i+ = N i, (ξ) N i+, (ξ) ξ i+ ξ i ξ i++ ξ i+ + ( ) ξ ξi N i, (ξ) ξ i+ ξ i ξ i+ ξ i + ( ) ξi+ ξ N i+, (ξ) ξ i+ ξ i ξ i+ ξ i+ ( ) ξ ξi+ N i+, (ξ) ξ i++ ξ i+ ξ i+ ξ i+ ( ) ξi++ ξ N i+, (ξ) ξ i++ ξ i+ ξ i++ ξ i+ (4a) (4b) (4c) (4d) (4e) 3.) To finish the roof, it remains to show, that equations (3a-3e) and (4a-4e) are indeed the same. This is obvious for (3e) and (4a), for (3a) and (4b) and for (3d) and (4e). The equality is not evident for (3b) and (4c) as well as for (3c) and (4d), hence we will maniulate (3b) and (3c) to match with their counterarts. Both of those terms have the common factor ( )N i+, (ξ), which we will exclude in the following exressions. Excet from this factor (3b)+(3c) reads: ξ ξ i (ξ i+ ξ i )(ξ i+ ξ i+ ) + ξ i++ ξ (ξ i++ ξ i+ )(ξ i+ ξ i+ ) Finding a common denominator one can also write this as: (ξ i ξ)(ξ i++ ξ i+ ) + (ξ i++ ξ)(ξ i+ ξ i ) (ξ i+ ξ i )(ξ i++ ξ i+ )(ξ i+ ξ i+ ) 5

6 factoring out the enumerator one realizes that ξ i ξ i++ cancels with its own negative and when adding a zero, i.e. in the enumerator adding +ξ i+ ξ i+ ξ i+ ξ i+ the whole fraction transforms to: (ξ i++ ξ i+ )(ξ i+ ξ) + (ξ i+ ξ i )(ξ i+ ξ) (ξ i+ ξ i )(ξ i++ ξ i+ )(ξ i+ ξ i+ ) Searating this again into two fractions: ξ i+ ξ (ξ i+ ξ i )(ξ i+ ξ i+ ) ξ ξ i+ (ξ i+ ξ i+ )(ξ i++ ξ i+ ) and again adding the common factor ( )N i+, (ξ) that was droed, one arrives at: ( )(ξ i+ ξ) (ξ i+ ξ i )(ξ i++ ξ i+ ) N ( )(ξ ξ i+ ) i+, (ξ) (ξ i+ ξ i+ )(ξ i++ ξ i+ ) N i+, (ξ) which are exactly (4c)+(4d). d) Again by induction over : = : The degree zero basis functions are discontinuous by definition, so indeendently of the knot multilicity m i = m i is less than and hence the claim holds. = : For j {,..., n}: ξ ξ j ξ j+ ξ lim N j, (ξ) = lim χ ξ ξ i ξ ξi ξ j+ ξ [ξj,ξ j+ )(ξ) + lim χ j ξ ξi ξ j+ ξ [ξj+,ξ j+ )(ξ) j+ } {{}} {{}, ξ i < ξ j < ξ j+, ξ i < ξ j+ < ξ j+, ξ i = ξ j < ξ j+, ξ j < ξ j+ = ξ i =, ξ j < ξ j+ < ξ i, ξ i < ξ j = ξ j+, ξ i = ξ j = ξ j+, ξ j = ξ j+ < ξ i, ξ i = ξ j+ < ξ j+, ξ j+ < ξ j+ = ξ i =, ξ j+ < ξ j+ < ξ i, ξ i < ξ j+ = ξ j+, ξ i = ξ j+ = ξ j+, ξ j+ = ξ j+ < ξ i ξ ξ j ξ j+ ξ lim N j, (ξ) = lim χ ξ ξ i ξ ξi ξ j+ ξ [ξj,ξ j+ )(ξ) + lim χ j ξ ξi ξ j+ ξ [ξj+,ξ j+ )(ξ) j+ } {{}} {{}, ξ i < ξ j < ξ j+, ξ i < ξ j+ < ξ j+, ξ i = ξ j < ξ j+, ξ j < ξ j+ = ξ i =, ξ j < ξ j+ < ξ i, ξ i < ξ j = ξ j+, ξ i = ξ j+ < ξ j+, ξ j+ < ξ j+ = ξ i =, ξ j+ < ξ j+ < ξ i, ξ i < ξ j+ = ξ j+, ξ i = ξ j = ξ j+, ξ j = ξ j+ < ξ i, ξ i = ξ j+ = ξ j+, ξ j+ = ξ j+ < ξ i There is only one case that might lead to a discontinuity: ξ i = ξ j+ > ξ j (= ξ i ) the limit from below is, which can only hold for the limit from above for ξ i = ξ j+ < ξ j+ (= ξ i+ ), 6

7 i.e. ξ i < ξ i < ξ i+ and hence ξ i must have multilicity to avoid discontinuity. The other way round, if ξ i has multilicity or more, we assume w.l.o.g. that (... = ξ i = )ξ i = ξ i ξ i+ and choose j = i, then ξ i = ξ j+ < ξ j+, hence the limit from above is, but the limit from below can not be since ξ i = ξ i contradicts ξ i = ξ j+ > ξ j = ξ i. ( ) : By means of the derivative formula () the continuity of the derivatives of a degree basis function is layed back to the continuity of the degree ( ) basis functions, i.e. the k-th derivative of N i, is a linear combination of {N j, k } j, which have regularity k m i, hence N i, has regularity m i. e) One more time a roof by induction: = : Assuming ξ i is resent in the knot vector at least once, we have for all j =,..., n by (): N j, (ξ i ) = ξ i ξ j χ ξ j+ ξ [ξj,ξ j+ )(ξ i ) + ξ { j+ ξ i, i = j + χ j ξ j+ ξ [ξj+,ξ j+ )(ξ i ) = j+, else Hence N i, (ξ) = N i, (ξ) is the searched for basis-function. ( ) : We assume that ξ i is resent times in the knot vector, i.e. ξ i = ξ i =... = ξ i +. Again by () we have for all j =,..., n. N j, (ξ i ) := ξ i ξ j N j, (ξ i ) + ξ j++ ξ i N j+, (ξ i ) ξ j+ ξ j ξ j++ ξ j+ N j, and N j+, are both of degree with ξ i having multilicity (one larger than degree). By the induction remise, we have that only one secial N k, takes the value one at ξ i, yielding: N j, (ξ i ) = ξ i ξ k δ j,k + ξ k+ ξ i δ j+,k ξ k+ ξ j ξ k+ ξ k Also by induction remise the k is chosen as k = i ( degree of the ( ) function) = i ( ) = i +. Plugging this in yields: N j, (ξ i ) = ξ { i ξ i + ξ i+ ξ i, i = j + δ j,i + + δ j+,i + = ξ i+ ξ j ξ i+ ξ i +, else }{{}}{{} = since ξ i =ξ i + = since ξ i =ξ i + Hence again N i, (ξ) is the searched for function. 7

8 Exercise 33 (B-sline curves, sensitivity and tangents) For a given (oen) knot vector Ξ R n++ with associated B-sline basis (N i, (ξ)) n i= and a set of control oints B i R d the d-dimensional B-sline curve of degree is defined as: B-sline curve: γ : [ξ, ξ n++ ] R d, γ(ξ) = N i, (ξ) B i (5) i= a) Prove that a B-sline curve of degree with an oen knot vector interolates all its control oints. b) Exlain why this is not always the case for higher degrees than (e.g. for the quadratic curve in fig. ) c) The following icture shows a B-sline curve of degree with control oints. Exlain why the curve changes only locally when a single control oint is moved, i.e. exlain why the right art of the curve does not change at all, if a control oints at the left end is moved Fig. : Quadratic B-sline curves to the knot vector Ξ = (,,,,, 3, 4, 5, 6, 7, 8, 9,,, ). The control oints (red dots) are both the same excet for the fourth one counted from left, that was moved from x = 3 to x = 4 for the right curve. The small red x show the hysical ositions of the ξ i. -) U to which oint exactly is the curve influenced by the moved control oint? -) How would the situation change when we use a quintic curve (i.e. degree = 5) or even an octic ( = 8) with the same control oints? d) Let Ξ be an oen knot-vector where the first + and last + knots are res. and let γ(ξ) be a B-sline curve of degree that is built via the basis-functions associated to Ξ. Prove that the tangent of γ at ξ = always oints towards the second control oint, even if the curve itself does not interolate that oint. Mathematically: Show that γ () = c (B B ), where c is some scaling factor. 8

9 Solution 33 a) In the knot vector each knot is resent at least = times. By exercise,d) this means that for every ξ i, i =,..., n there is exactly one basis-function being at ξ i while all other basis-functions are zero there, furthermore this basis function is always given by N i, (ξ). By definition of the B-sline curve, we can evaluate it at ξ j+ to get: γ(ξ j+ ) = N i, (ξ j+ ) B i i= n+ = N i, (ξ j+ ) B }{{} i i= =δ i,j+ = B j Hence the curve certainly goes through the oint B j, where j {,..., n} was arbitrarily chosen. b) In contrast to the first order basis-functions, higher order bases are in general not interolatory, i.e. for a certain ξ [ξ, ξ n++ ] there are always more than just one basis functions with N i, (ξ). Even though they do form a artition of unity the weight they contribute to the unity might be distributed between them and not lie on just one of them. In other words: In the linear ( = ) case we have seen in exercise art a) that at ξ j+ only the basis function associated to B j is active. For > there are also some other basis functions each associated to an other control oint such that γ(ξ j+ ) becomes a linear combination of those control oints lying somewhere in between them. c) To answer the question recisely lets have a look at the B-sline basis functions associated to the given knot vector that are also used to create the curves via definition (5) Fig. 3: Quadratic B-sline basis functions to the knot vector Ξ = (,,,,, 3, 4, 5, 6, 7, 8, 9,,, ). The basis function associated with the fourth control oint (the one that was moved) is also the fourth one from the left, i.e. the urle one. By exercise,a) we knot that 9

10 su(n 4, ) = [ξ 4, ξ 7 ] = [, 4 ]. By changing B 4 we change the contribution of N 4, to the curve, which is only not-zero in the given suort interval. Hence the geometry of γ is only changed between ξ = and ξ = 4, which is the region between the first and fourth red x in the icture above, before and after these oints, the curves are exactly the same! The following icture shows an overlay of the both curves to highlight the difference / similarity - regions: Fig. 4: Quadratic B-sline curves: blue: before the change, black: after For the same curve but with order = 5 the knot vector, in order to still have an oen one, is recomuted to: Ξ = (,,,,,, 7, 7, 3 7, 4 7, 5 7, 6 7,,,,,, ). In case of = 8 it even becomes: Ξ = (,,,,,,,,, 4, 4, 3 4,,,,,,,,, ). The resective B-sline bases look as follows: Fig. 5: left: Quintic B-sline basis, right: Octic B-sline basis For the suorts of the fourth basis functions each there holds: su(n 4,5 ) = [ξ 4, ξ ] = [, 4 7 ] and su(n 4,8) = [ξ 4, ξ 3 ] = [, ], i.e. the fourth quintic basis function already influences more than half of the arametric interval, while the octic-basis function influences even the whole curve. The following overlay-lots show the situation:

11 Fig. 6: Left: Overlay lot of quintic B-sline curves with one control oint changed, right: the same for octic sline functions. Esecially in the octic case it is imossible to see that indeed the whole curve changes. This is due to the fact, that N 4,8 is extremely small towards the right end of its suort (around 39 ), however(!) it is not zero there, so really the whole curve is influenced by the moved control oint. Summarizing the revious results we can conclude that with increasing degree the suort of individual basis functions grows and hence each control oints influences a larger art of the B-sline curve. d) Using linearity of the differentiation and () we get: γ (ξ) = N i,(ξ) B i i= = i= ξ i+ ξ i N i, (ξ) B i ξ i++ ξ i+ N i+, (ξ) B i Since the knot-vector is suosed to be oen, ξ = is reeated + times and by exercise,d) is only one basis-function of degree not being zero at ξ = ξ =... = ξ +, i.e. N + ( ), (ξ) = N, (ξ). This lets us reduce the sum to n = and we get: γ () = N, () B ξ + ξ N, () B ξ + ξ + + N, () B ξ + ξ N 3, () B ξ +3 ξ where the first and last term vanish and we remain with: γ () = ξ + ξ (B B )

12 Exercise 34 (NURBS-reresentation of a circle) Consider the quadratic B-sline curve C : [, 4] R 3 with knot vector Ξ = (,,,,,,, 3, 3, 4, 4, 4) and control oints B i = (±,, ), (, ±, ), (±, ±, ) which are ordered clockwise starting with (,, ) (and also again ending with (,, ), i.e. this control oints aears twice and hence the curve is closed), also see figure 7. a) Homogenize the coordinates of the B-sline curve, i.e. roject it onto the z = -lane of R 3 and identify this lane as R to obtain a lane curve P. Using the oint (,, ) as the rojection center, the homogenization rocedure can be summarized in the following formula: P x (ξ) P y (ξ) P z (ξ) = C x(ξ) C z(ξ) C y(ξ) C z(ξ) C z(ξ) C z(ξ) b) Show that the curve P (ξ) is an exact circle of radius. c) How does this relate to the NURBS-setting / definition of NURBS? Solution 34 Since the considered sline-curve C(ξ) is rotational eriodic with eriod π/, it suffices to rove the statement for the first quarter of the curve (that this will be rojected onto a quarter of a circle). The arametrization of the sline-curves is done over the interval [, 4] (knot vector [,,,,,,, 3, 3, 4, 4, 4]), so for the first quarter it is sufficient to restrict the arametrization to [, ]. On that subinterval only the first three B-sline basisfunctions are non-zero and hence the curve is exclusively comosed by them (see figure 7,c)). Exlicitely these three first B-sline basisfunctions can be exressed as follows: N, (ξ) [,] = ( ξ), N, (ξ) [,] = ξ ( ξ) and N 3, (ξ) [,] = ξ. Now emloying the definition of a B-sline-curve and using the secific control oints given in figure 7 yields: C x(ξ) C y(ξ) C z(ξ) [,] N,(ξ) + N,(ξ) + N 3,(ξ) = N,(ξ) + N,(ξ) + N 3,(ξ) N,(ξ) + N,(ξ) + N 3,(ξ) ( )ξ + ξ = ( )ξ + ( )ξ + ( )ξ + ( )ξ + = ξ( ξ) + ξ ( ξ) + ξ( ξ) ( ξ) + ξ( ξ) + ξ (6) a) P x (ξ) P y (ξ) P z (ξ) = [,] C x (ξ) C z (ξ) C y (ξ) C z (ξ) C z (ξ) C z (ξ) = [,] ( )ξ + ξ ( )ξ + ( )ξ + ( )ξ + ( )ξ + ( )ξ + ( )ξ + (7)

13 Projection of the B-sline curve Projection of the control olygon.5.5 z z.5.5 y x y x a) b) quadratic B-sline-basisfunctions used to create the rojection curve to-down view at the curves and control olygons Ni;(9) y x c) d) Fig. 7: a) B-sline curve of order with knot vector Ξ = (,,,,,,, 3, 3, 4, 4, 4) (blue) and its rojection (green) onto the z = lane, which is an exact circle with radius. b) The control oints of the B-sline curve (red) and their rojections onto the z = lane (orange). c) The B-sline basisfunctions used to create the B-sline curve in a). d) A to-down view at the entities from a) and b) (same colours) to better see the ositions of the control oints: (±,, ), (, ±, ), (±, ±, ) for (red, (,, ) aears twice, as first and last oint) and (±,, ), (, ±, ) and (±, ±, ) for (orange). b) Identifying the z = lane from R 3 with R finally gives: ( P x (ξ) P y (ξ) ) = P x (ξ) + P y (ξ) = [ ( )ξ + ξ ] [ + ( )ξ + ( )ξ + ] [ ( )ξ + ( )ξ + ] = (4 6 )ξ 4 6ξ 3 + (4 4 )ξ + (8 6)ξ + 4 (4 6 )ξ 4 6ξ 3 + (4 4 )ξ + (8 6)ξ + 4 c) Obviously a rational arametrization of a quarter-circle (again after identifying the z = lane from R 3 with R ) has been found in equation (7), which can be glued together iecewise with other quarter-circles to a full circle (again see figure 7). To further examine the structure of the rational arametrization of the quarter-circle, one again invokes equation (6) (only the first = sign) to write: 3

14 ( N, (ξ) + N, (ξ) + N 3, (ξ) ) ( ) Px (ξ) = P y (ξ) [,] N, (ξ) + N, (ξ) + N 3, (ξ) N, (ξ) + N, (ξ) + N 3, (ξ) Since all the other B-sline basisfunctions N i, (ξ), i = 4,..., 9 are zero on the interval [, ] (see figure 7,c)) one can add them in the enumerator as well as in the denominator to arrive at: ( Px (ξ) P y (ξ) ) [,] = 9 N i, (ξ) B i x i= B iy (8) 9 N i, (ξ) B iz i= For the second quarter-circle (ξ [, ]) now the B-sline basisfunctions N, (ξ) and N, (ξ) as well as N i, (ξ), i = 6,..., 9 are identically zero, which means equation (8) can be derived in the same way also for ξ [, ]. Finally this holds true for any of the four quarter circles so that in the end, the rational arametrization of the full circle also reads: ( Px (ξ) P y (ξ) ) = 9 N i, (ξ) B i x i= B iy (9) 9 N i, (ξ) B iz i= The structure of this arametrization of the unit circle, which is comonentwise a fraction of olynomials, more recisely linear combinations of the B-sline basisfunctions, and hence a rational function is exactly what was given in the definition of NURBS and NURBS-curves, where NURBS stands for Non-Uniform (the knot vector doesn t have to be uniform in general) Rational B-Sline. 4