Planar interpolation with a pair of rational spirals T N T Goodman and D S Meek Abstract Spirals are curves of one-signed monotone increasing or decreasing curvature Spiral segments are fair curves with the advantage that the minimum and maximum curvatures are at their endpoints Two situations where the planar two-point G Hermite interpolation problem can be solved with a pair of rational spiral segments are outlined here Keywords: two-point G Hermite interpolation; pair of spirals; rational spiral Introduction Spirals are curves of one-signed monotone increasing or decreasing curvature Spiral segments are fair curves because they have no internal curvature extrema and no cusps They have the advantage that the minimum and maximum curvatures are at their endpoints The spiral segments used here are restricted to having zero curvature at one end The restriction allows the use of a simple rational spiral that was identified in [] In designing a planar curve the points of zero curvature (they include inflection points) give crucial information on the behaviour of the curve and in many situations these points together with their corresponding tangents may be stipulated by the problem or the designer One situation which commonly occurs is the need to interpolate between two points of zero curvature with given tangents Section 3 gives a construction for a curve segment which joins two consecutive points of zero curvature with the minimum number of spiral segments The curve segment is a C-shaped curve formed by joining two spiral segments Another situation which commonly occurs is the need to interpolate between two points with given tangents and given curvatures of opposite sign Section 4 gives a construction for such a curve segment which again has the minimum number of spiral segments The curve segment is an S-shaped curve formed by joining two spiral segments Department of Mathematics University of Dundee Dundee Scotland DD 4HN e-mail: tgoodman@mathsdundeeacuk Department of Computer Science University of Manitoba Winnipeg Manitoba Canada R3T N e-mail: Dereck_Meek@UManitobaCa
It should be noted that these interpolation problems cannot always be solved with a pair of spirals; tests are given for the solvability The user has some choice that allows the shaping of the curve Once the shape parameters are chosen the spirals are given by explicit formulas Meek and Thomas [5] outlined the sets of two-point G Hermite data that can be matched by a pair of spirals without the restriction that each spiral segment have zero curvature at one end That work covers the sets of G Hermite interpolation used here and its results are briefly summarized in the next section The idea of joining pairs of spirals each of which has zero curvature at one end has been used previously A pair of clothoids is used in [3] Clothoids have the disadvantage that they are transcendental curves and are not NURBS curves Polynomial spirals are used in [6] [7] and [8] Polynomials have the disadvantage that they cannot satisfy full range of G Hermite data that a general spiral can satisfy Preliminaries This section establishes notation for two-point G Hermite interpolation and gives formulas for the spiral that is used in subsequent sections Two-point G Hermite data in the plane are a pair of distinct points A and B a pair of unit tangent vectors T A and T B at A and B and a pair of signed curvatures k A and at A and B The curvature is positive if the centre of curvature is on the left as one moves from A to B The G Hermite interpolation problem is to find a curve that matches those data Spiral segments with one end point having zero curvature are considered here A standard form and labelling of such a spiral segment follows The endpoint with zero curvature will be labelled A (curvature k A = 0) while the other endpoint will be labelled B (curvature > 0) The angle of rotation W from the unit tangent vector T A to the unit tangent vector T B is positive The rational spiral used here imposes the further restriction that W < If the above two-point G Hermite data have A at the origin and T A along the X-axis then the conditions that a spiral segment is possible can be expressed as a region for B [] [4] (see Figure ) where the matrix for rotation about the origin is R ot () = cos sin () sin cos and the point V as V = A + sinw T A + ( cosw)r ot T A () B can be written in polar coordinates (r ) r = B A so the conditions for a spiral segment are
< W ( cosw) < B A sin (3) B A W k A = 0 V T A Figure The allowable region for B with A at the origin T A along the X-axis The rational spiral segment S(t) given in [] and quoted below matches the two-point G Hermite data: A T A k A = 0 B in the region in Figure T B with 0 < W < and > 0 Let C B be the centre of the circle of curvature at B C B = B + R ot T B and let be the angle C B R ot T A makes with T A C B T A k tan = B C B T A Define > 0 and with sin = tan tan W 3 tantan W tan W cos = tan tan W tan The length of a vector Q is Q and is given by Q sin ( 3+ cos)= C B T A 3 With the length of Q now determined define the three vectors P Q and R as 3
P = Q R ot T A Q = R ot ()P R = R ot ()Q It is convenient to define the following value K = + cos Q 3 The spiral segment that matches the two-point G Hermite data is S(t) = A + t cos + ( 6Kcos )t 4 t + 3Kt 3 P 3 + ( cos)t + t 4 + t 3 + t 3 3 6 3Kt + t 4 + ( cos)t + t Q 4 4 3Kt ( 4 cos)t R 0 t (4) + ( cos)t + 4 t 3 A C-shaped pair of spiral segments A C-shaped pair of spiral segments is formed when the two spirals meet with G continuity at their endpoints that have non-zero curvature Without loss of generality the whole C-shaped curve has non-negative curvature Let the joint of the two spiral segments be B and the curvature at the joint be Given an adjacent pair of points A i where the curvature is required to be zero and unit tangent vectors T i at those points i = 0 the interpolation problem is to find a G Hermite interpolating C-shaped curve of maximum curvature comprising two spiral segments one from A 0 to B and the other from B to A The region in which B must be chosen is identified After B is chosen there is a range in which the angle of the tangent at the point of maximum curvature may be chosen Let the non-zero angles from T 0 to A A 0 and A A 0 to T be 0 and and let the ray starting at A 0 in directiont 0 be T 0 and the ray starting at A and in direction T be T (see Figure ) Without loss of generality assume 0 < 0 Since each of the spiral segments in (4) has a tangent vector rotation of less than the total rotation of the tangent vector from T 0 to T 0 is restricted to be less than It will be shown later that if 0 (and 0 ) there is a lower bound on > cos( 0 ) A A 0 sin 0 (3) 4
T k = 0 A α θ W B T α 0 θ0 W 0 A 0 k 0 = 0 T 0 Figure G Hermite data and a pair of spiral segments ( 0 < ) Let the angles of rotation of the tangent vector throughout the spiral segments be W 0 and W 0 < W 0 W < (3) The continuity of the tangents at B requires that W 0 + W = 0 (33) Although of these two angles only W 0 is actually needed W will be retained in some expressions A valid W 0 is one for which it and the corresponding W from (33) satisfy (3) The position of the point B can be expressed in terms of A 0 A and two angles 0 Let the angle from T 0 to B A 0 be 0 0 < 0 < 0 and let the angle from B A to T be 0 < < (see Figure ) The restrictions on B for a single spiral segment are given in Section Here the restrictions for both spiral segments must be considered simultaneously Working from A 0 the conditions corresponding to (3) are 0 < W 0 (34) ( cosw 0 )< B A 0 sin 0 (35) while working from A the conditions are < W (36) T 0 5
( cosw )< B A sin (37) The inequalities (34) - (37) define a region in which B can lie so that a valid W 0 can be chosen; that region will be detailed in Theorems 3 3 and 33 For any value of W 0 the region described by (34) - (37) is the intersection of two wedges (a similar region R(W 0 ) will be used later in the S-shaped case in Section 4) When W 0 varies circular arcs that form parts of the boundary of the region for B are obtained by intersecting certain pairs of (34) - (37) with equality (see Figure 3) Circular arc A is derived from the intersection of (34) and (36) This circular arc lies to the right of the directed chord A 0 A and subtends an angle of + 0 with that chord (see Figures 4 and 5) Circular arc B is derived from the intersection of (34) and (35) This circular arc is tangent to T 0 at A 0 lies to the right of the directed line segment A 0 A and has curvature Circular arc C is derived from the intersection of (36) and (37) is tangent to T at A lies to the right of the directed line segment A 0 A and has curvature If 0 circular arc D is derived from the intersection of (35) and (37) This circular arc is tangent to both T 0 and T sweeps through an angle of 0 and has curvature It can be shown that circular arc B is superfluous as a boundary when 0 because it always lies "inside" or "to the left of" circular arc A T A circular arc C circular arc A α W (33) circular arc B α 0 W 0 B (34) A 0 T 0 (35) circular arc D Figure 3 The inequalities specifying the position for B (36) 6
A lower bound on If 0 < 0 the allowable region for B (see Figure 4) disappears when circular arc C equals circular arc D If the radius of circular arc C is large enough that circle is tangent to the side T 0 which makes it the same as circular arc D This radius of circular arc C is A A 0 sin 0 tan sin( 0 ) + 0 = A A sin 0 0 sin 0 which gives the lower bound on in (3) If 0 = there is a lower bound on (also given by (3)) In this case W 0 + W = so the inequality (37) becomes ( + cosw 0 )< B A sin The sum of this inequality and (35) gives < B A 0 sin 0 + B A sin = A A 0 sin 0 from which (3) (with 0 = ) follows T A circular arc C circular arc A circular arc B B circular arc D A 0 T 0 Figure 4 The allowable region for B 0 < 7
T circular arc C A B circular arc A T circular arc B circular arc D A 0 T 0 T 0 Figure 5 The allowable region for B 0 > Theorem 3 If 0 < a C-shaped pair of spirals matching the Hermite data can be found if and only if B is in region R defined as follows Region R is the set of points outside circular arcs A and C between the rays T 0 and T and "inside" (or to the left) of circular arc D (see Figure 4) Proof The first part of the proof is to show that if a point B and an angle W 0 satisfy (3) - (37) then B is in R The boundaries that were obtained from (34) - (37) with equality give the circular arcs and it can be seen that with inequality point B must be outside circular arcs A B and C and "inside" circular arc D B is between T 0 and T because W 0 and W are positive Since not all of (3) - (37) were considered simultaneously in the previous discussion it is possible that they describe a smaller region than R Thus the second part of the proof is to show that for any point B in R a valid W 0 can be chosen so that together they satisfy the inequalities (34) - (37) Consider the family of circular arcs each of which sweeps out an angle of 0 and is tangent to both T 0 and T (see Figure 6) An individual arc is identified by its radius r As r 8
increases from the arc moves through the entire allowable region for B Further for any point B in R there is a unique circular arc of the above family that passes through it Thus B can be expressed in polar coordinates (r ) based on the arc that it lies on The conditions that B is in R (outside arcs A B C and to the left of arc D) are: 0 + < 0 (38a) [ cos 0 ]< B A 0 sin 0 (38b) [ cos ]< B A sin (38c) r > (38d) T r r [ - cos(α 0 + α - θ)] r θ T 0 B r [ - cos θ] α 0 + α Figure 6 Expressing B by an arc that is tangent to T 0 and T The lengths of the perpendiculars from B to the lines T 0 and T are shown For a given B and recalling that (34) must be satisfied define W 0min = 0 ; (39) referring to (35) define = if B A 0 sin 0 k W B 0max satisfies [ cosw 0max ]= B A 0 sin 0 otherwise (30a) (30b) 9
In case (30a) (using (38a)) 0 + < 0 < so 0 < = W 0max In case (30b) (using (38b)) [ cos 0 ]< [ cosw 0max ] 0 < W 0max Recalling (39) both cases of (30) give W 0min < W 0max Similarly for a given B define W min = (3) and = if B A sin (3a) k W B max satisfies [ cosw max ]= B A sin otherwise k (3b) B The definitions (39) and (3) along with (38a) give W 0min + W min = 0 + < 0 (33) Looking at the angle in the polar coordinates of B in case (30a) < 0 < so < W 0max In case (30b) r [ cos]= B A 0 sin 0 = [ cosw 0max ] and with (38d) it again follows that < W 0max Similarly both cases of (3) give 0 < W max Adding these two results + ( 0 ) = 0 < W 0max + W max (34) The results (33) and (34) show that there are values W 0 and W W 0min < W 0 < W 0max W min < W < W max such that W 0 + W = 0 It is easy to see that (39) - (3) imply (34) - (37) Theorem 3 If 0 = a C-shaped pair of spirals matching the Hermite data can be found if and only if B is in region R defined as follows Region R is the set of points outside circular arcs A and C and between rays T 0 and T The proof of this theorem is fairly simple and so is omitted Theorem 33 If 0 > a C-shaped pair of spirals matching the Hermite data can be found if and only if B is in region R defined as follows Region R is the set of points outside circular arcs A 0
C and D between rays that are a distance of [ cos( 0 ) ] inside T 0 and T right of a perpendicular to T 0 at A 0 and left of a perpendicular to T at A (see Figure 5) Proof Since W 0 + W = 0 > it is necessary to use the restrictions W 0 W < in the proof of this theorem It was not necessary to consider those upper limits in the proof of Theorem 3 The first part of the proof is to show that if a point B and an angle W 0 satisfy (3) - (37) then B is in R Following the argument used previously (34) - (37) imply that point B is outside circular arcs A B C and D The straight line boundaries perpendicular to T i and a specified distance from T i arise from W i = i = 0 in (34) - (37) It can be seen that with W i < B must be in region R The second part of the proof is to show that any point B in R a valid W 0 can be chosen so that together they satisfy the inequalities (34) - (37) Again consider the family of circular arcs each of which sweeps out an angle of 0 and is tangent to both T 0 and T As the radius of an arc r increases from the arc moves through the entire allowable region for B Notice that here the region of interest is outside arc D and the restriction (38d) expresses that condition Four more restrictions (resulting from W 0 W < ) are added to those in (38) 0 < (35a) < (35b) [ cos( 0 ) ]< B A 0 sin 0 (35c) [ cos( 0 ) ]< B A sin (35d) The lower and upper bounds of W 0 and W can be defined as in (39) - (3) Restrictions (35a) and (35b) are needed to show that W 0min < W 0max in (30) and (3) Inequality (33) is still valid Cases (30a) and (3a) give 0 < = W 0max + W max cases (30a) and (3b) using (35d) give 0 = + ( 0 ) < W 0max + W max cases (30b) and (3a) using (35c) give
0 = ( 0 ) + < W 0max + W max and cases (30b) and (3b) work as before so that all four cases give (34) Again (33) and (34) show that W 0 and W W 0min < W 0 < W 0max W min < W < W max can be chosen such that W 0 + W = 0 As in Theorem 3 (39) - (3) imply (34) - (37) Summary Choose (maximum curvature subject to (3) if 0 ) and choose B (this will define 0 and ) from the regions in Figures 4 and 5 Choose W 0 (its bounds are given in (39) and (30)) and calculate W from (33) Finally apply (4) to find the two spiral segments 4 An S-shaped pair of spiral segments An S-shaped pair of spiral segments is formed when the two spirals of opposite signed curvature meet with G continuity The interpolation problem to be solved in this Section is to find an S-shaped curve S comprising two spiral segments that joins the point B 0 on a circle C 0 to the point B on a circle C while matching tangent vectors T i and signed curvatures k i (k 0 > 0 k < 0) at B i i = 0 The two spirals in S join at the inflection point A and the circles C i are the circles of curvature of S at B i i = 0 The region in which the point A must be chosen is identified After A is chosen there is a range in which the angle of the inflection line T A the tangent at A may be chosen The radii of circles C 0 and C are r 0 = k 0 and r = k Assume the distance between the centres of the circles C i is greater than r 0 + r so that the circles are non-overlapping Let the line joining the centres of the circles C i meet the circles at D i and the common tangent T touch the circles at E i i = 0 (see the Appendix) as illustrated in Figure 0 It is assumed that B i lies on the lesser arc joining D i to E i i = 0 Let i be the angle from T i to B B 0 i = 0 and without loss of generality assume 0
Circle C 0 D 0 Circle C r 0 T E B B 0 β 0 β E 0 r T 0 D T Figure 0 Notation for the circles C 0 and C Let W 0 and W 0 < W 0 W < denote the unsigned angles turned through by a tangent vector to the first and second spirals of S respectively Let 0 < W 0 be the angle from T 0 to A B 0 and let < W be the angle from T to B A (see Figure ) Since the signed rotation of the tangent vector to S from B 0 to B is 0 0 W W 0 = 0 0 Hence W is not actually needed in the following however it will be retained in some expressions T A W β 0 θ B B 0 θ0 β T 0 A T W 0 Figure Notation for the inflection point A 3
The conditions for the spiral segments to exist give conditions on the position of their join point A For the first spiral segment from A to B 0 the conditions are W 0 0 < W 0 or W 0 < (4) 0 r 0 ( cosw 0 )< B 0 A sin(w 0 0 ); (4) and for the second spiral segment from A to B the conditions are W < W or W < (43) r ( cosw )< B A sin(w ) (44) Define the open region R(W 0 ) as illustrated in Figure The lines L 0 M 0 L and M are obtained from equality in (4) - (44) Define F i and G i as the intersection points of the pairs of lines L i M i and L i M -i respectively i = 0 Note that M i is tangent to C i at F i i = 0 and M 0 and M are parallel Consider the lines L 0 M 0 L and M as directed lines with the directions G 0 F 0 G F 0 F G and F G 0 respectively R(W 0 ) is the set of points strictly left of L 0 right of M 0 right of L and left of M and any point A in R(W 0 ) satisfies (4) - (44) If W 0 is increased F 0 G 0 and F 0 G move to their left sides in a non-self-intersecting manner while G 0 F and G F move to their right sides in a non-self-intersecting manner For a suitable choice of W 0 = W 0min M 0 and M will coincide with the common tangent of the two circles T as in Figure0; F i = G i = E i i = 0 and R(W 0 ) becomes empty The angle that F 0 B 0 makes with T 0 is W 0 while the angle B F makes with T is W W 0 0 Since W 0 W W 0 the directed line L 0 is at a greater (or equal) angle to T 0 than the directed line L For a suitable choice of W 0 = W 0max L will intersect C 0 at F 0 = G and again R(W 0 ) will become empty 4
M 0 circle C 0 M W B 0 G L F L R(W 0 0 ) F 0 W 0 G 0 B T 0 Figure Allowable region R(W 0 ) for A circle C T The region in which A can be chosen is R = U{ R(W 0 ):W 0min < W 0 < W 0max } Let P be a point in the closure of R(W 0 ) which lies on the boundary of R The point P must be the intersection of at least two of L 0 M 0 L and M For the situation illustrated in Figure this implies that P is F 0 G 0 F or G For 0 < and large enough W 0 the region R(W 0 ) will be bounded by the three lines L 0 M 0 and L By decreasing W 0 it is seen that the intersection of L 0 and L cannot lie on the boundary of R and thus when W 0 is large enough P must be F 0 or G Theorem 4 The allowable region R for A is the bounded open region outside circles C 0 and C with boundary comprising an arc of C 0 an arc of C a segment from E 0 to C of the curve 0 given by ( X + Y )( Y + r ) XY[ B B 0 cos 0 r sin( 0 )] ( ) B B 0 sin 0 r cos( 0 ) + X Y [ ]= 0 (45) X = B 0 A cos 0 Y= B 0 A sin 0 (46) and a segment from E to C 0 of the curve given by ( x + y )( y + r 0 ) xy[ B B 0 cos r 0 sin( 0 )] ( ) B B 0 sin r 0 cos( 0 ) + x y [ ]= 0 (47) x = B A cos y= B A sin (48) 5
The region R in the statement of Theorem 3 is illustrated in Figure 3 The values (X Y) in (46) are Cartesian coordinates of a system with the origin at B 0 the X-axis along T 0 and the Y-axis along R ot T 0 The centre of circle C in this coordinate system is C X = B B cos r sin( ) 0 0 0 C Y B B 0 sin 0 r cos( 0 ) and 0 in (45) can be written in the simple form ( X + Y )( Y + r ) XYC X + ( X Y )C Y = 0 (49) Note that 0 is independent of r 0 and the position of B on C Similarly (x y) in (48) are Cartesian coordinates of a system with origin at B the positive X-axis along T and the positive Y-axis along R ot T If (C 0x C 0y ) T denotes the centre of circle C 0 in this coordinate system then (47) can be written ( x + y )( y + r 0 ) xyc 0x + ( x y )C 0y = 0 (40) which is independent of r and the position of B 0 on C 0 Proof of Theorem 4: Since G i lies on circle C i it remains to show that G i (see Figure ) lies on i i = 0 The case i = 0 is treated below; the case i = is similar and is omitted Now A = G 0 if and only if A lies on L 0 and M that is (4) and (44) are satisfied with equality Since W = W 0 0 this gives r [ cos( 0 0 )]= B A sin( 0 0 ) (4) Note from Figure B 0 A cos( 0 0 ) + B A cos( ) = B B 0 (4) B A sin( ) = B 0 A sin( 0 0 ) (43) Using (46) (4) and (43) may be rewritten as B A cos( ) = B B 0 Xcos 0 Ysin 0 (44) B A sin( ) = Xsin 0 Ycos 0 (45) Employing (46) (44) (45) and some simplification (4) becomes (45) The boundary points H 0 and H in Figure 3 are now identified 6
Γ 0 circle C 0 Γ E 0 H 0 B 0 B Γ 0 R T 0 T H E Γ circle C Figure 3 Allowable region R for A Theorem 4 The curve 0 given by (49) touches the circle C at the point H given by X = C X r sin Y= C Y + r cos tan = r + C Y (46) C X and given by (40) touches C 0 at H 0 given by x = C 0x r 0 sin y= C 0y + r 0 cos tan = r 0 + C 0y C 0x Proof: The formula for H will be derived below; the derivation for H 0 is similar and is omitted Any point on circle C is of the form X = C X + r cos Y= C Y + r sin (47) for some Substituting (47) into (49) gives after some simplification r ( + C Y )C X cos + ( r + C Y ) [ C X ]sin + ( r + C Y ) + C X = 0 With satisfying (46) this gives sincoscossin += 0 or sin() = Thus = + (mod) and so from (47) there is a point of intersection of 0 and C which is H given by (46) The intersection is unique so 0 is tangent to C at H 7
circle C 0 A B 0 B circle C Figure 4 A sector containing the direction of T A A summary of the procedure for constructing the curve S follows The inflection point A can be chosen anywhere in the region R (see Figure 3) There are two ways the choosing of A could be done A geometric way would be to draw the region R on the screen for interactive picking The curve 0 can be drawn in the B 0 system as follows The points E 0 and H 0 give the range of Y-coordinates of 0 The X corresponding to any Y can be found from (49) which is a quadratic equation in X An algebraic way to verify that A is chosen in region R is to let the user pick a tentative point for A (X Y) in the B 0 system which is equivalent to (x y) in the B system and then test whether or not X + (Y r 0 ) > r 0 x + (x r ) > r and both (49) and (40) hold with = replaced by < If all four inequalities are true the point A is in R The choice of A determines the angles 0 and (see Figure ) The direction of the tangent T A at A is determined by W 0 (see Figure ) where W 0 can be any angle satisfying (4) - (44) with W = W 0 0 Geometrically the conditions (4) and (4) require that the direction of T A lies between that of the tangent to C 0 through A and the tangent to C 0 through the point of intersection of C 0 and B 0 A as illustrated in Figure 4 Similarly conditions (43) and (44) require that the direction of T A lies in the corresponding sector involving B and C Algebraically W 0 can be chosen so that W 0min < W 0 < W 0max (see the Appendix) Having chosen A and T A the two spiral segments comprising the curve S can be determined from formula (4) 5 Forming a curve of many spiral pairs There are two interpolatory situations in which the results of Sections 3 and 4 can be applied to give a G continuous curve with the minimum number of spiral segments 8
Given a set of points I i where the curvature is zero and a set of unit tangent vectors at those points T i i = 0 n one wishes to put a pair of spiral segments forming a C-shaped curve between each adjacent pair of points I i (see Figure 5) This problem can be solved (when it is possible to solve it) using the results in Section 3 No new points with zero curvature will be introduced and the user must specify a maximum curvature in each segment although this maximum curvature can have a lower bound determined by the geometry (if 0 (3) applies) The maximum curvature can be viewed as a shape parameter T I 0 I I T T 0 Figure 5 G Hermite data described by points of zero curvature I i Given a set of circles C i i = 0 n and points of those circles B i one wishes to put a pair of spiral segments forming an S-shaped curve between adjacent pairs of B i points (see Figure 6) In Figure 6 S-shaped curves from B 0 to B and from B to B 3 are required This problem can be solved (when it is possible to solve it) using the results in Section 4 One inflection point will be introduced for each pair B B B 0 B 3 Figure 6 G Hermite data described by points on circles References [] T N T Goodman D S Meek and D J Walton An involute spiral that matches G Hermite data in the plane submitted to Computer Aided Geometric Design Aug 004 [] H W Guggenheimer Differential Geometry McGraw-Hill New York 963 9
[3] D S Meek and D J Walton The use of Cornu spirals in drawing planar curves of controlled curvature Journal of Computational and Applied Mathematics 5 (989) 69-78 [4] D S Meek and D J Walton Clothoid spline transition spirals Mathematics of Computation 59 (99) 7-33 [5] D S Meek and R S D Thomas Hermite interpolation with a pair of spirals Computer Aided Geometric Design 0 (993) 49-507 [6] D J Walton and D S Meek A planar cubic Bézier spiral Journal of Computational and Applied Mathematics 7 (996) 85-00 [7] D J Walton and D S Meek A Pythagorean hodograph quintic spiral Computer-Aided Design 8 (996) 943-950 [8] D J Walton D S Meek and J M Ali Planar G transition curves composed of cubic Bézier spiral segments Journal of Computational and Applied Mathematics 57 (003) 453-476 Appendix: formulas for Section 4 The following points and lines are expressed in the B 0 coordinate system: the origin at B 0 and the X-axis along T 0 Let d = B B 0 and let r() = cos sin dsin( A = B 0 A r( 0 ) = ) sin( 0 + 0 ) r( ) 0 C i are centres of circles C i i = 0 0 C 0 = r 0 C = dr( ) r r 0 ( ) 0 F 0 = C 0 + r 0 r + W 0 = sin W 0 r r W 0 0 F = C + r r + W 0 = dr( 0 )+ r r + W 0 r ( ) 0 G 0 = F 0 + dsin(w ) r cosw 0 0 0( 0) r cosw sin W 0 ( ) r W 0 0
dsin W + r 0 sin W 0 sin 0 G = F 0 sin W line L 0 : F 0 + t r W 0 line L : F + t r W 0 W line M 0 : F 0 + t r( W 0 ) line M : F + t r( W 0 ) W 0min is the smaller root of r( W 0 ) [ dsin 0 + r 0 + r cos( 0 )]cosw 0 + [ dcos 0 r sin( 0 )]sinw 0 = r 0 + r which a quadratic equation in tan W 0 tan W 0max = E 0 = sin W 0min dcos 0 dsin 0 r 0 r W 0min + r 0 sin 0