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.They constitute intrinsic properties of the curve and togetherthey form the moving trihedron of the curve, which can be considered a local coordinatesystem for the curve.The three vectors constitute the local coordinate axes and thethree planes divide the space around point P(i) into eight octants.The curve passesthrough the normal plane and is tangent to both the osculating and rectifying planes.NormalplanehijTNBFigure 1.14: The Moving Trihedron.1.6.6 CurvatureThe curvature of a curve is a useful entity, so it deserves to be rigorously defined.Intuitively, the curvature should be a number that measures how much the curve deviatesOsculatingplaneRectifyingplane 1.6 Curvature and Torsion 31from a straight line at any point.It should be large in areas where the curve wiggles,oscillates, or makes a sudden direction change; it should be small in areas where the curveis close to a straight line.It is also useful to associate a direction with the curvature,i.e., to make it a vector.Given a parametric curve P(t) and a point P(i) on it, we calculate the first twoderivatives Pt(i) and Ptt(i) of the curve at the point.We then construct a circle thathas these same first and second derivatives and move it so it grazes the point.This iscalled the osculating circle of the curve at the point.The curvature is now defined asthe vector º(i) whose direction is from point P(i) to the center of this circle and whosemagnitude is the reciprocal of the radius of the circle.Using differential geometry, it can be shown that the vectorPt(t) × Ptt(t)|Pt(t)|3has the right magnitude.However, this vector is perpendicular to both Pt(t) andPtt(t),so it is perpendicular to the osculating plane.To bring it into the plane, we need tocross-product it with Pt(t)/|Pt(t)|, so the result isPt(t) × Ptt(t) × Pt(t)º(t) =.(1.19)|Pt(t)|4Figure 1.13 shows that the curvature (vector b) is in the direction of the binormal N(t),so it can be expressed as º(t) =Á(t)N(t) where Á(t) is the radius of curvature at pointP(t).Given a curve P(t) with an arc length s(t), we assume that dP/ds is a unit tangentvector:dP(t) dP(t) ds(t) Pt(t)= =.(1.20)ds dt dt st(t)Equation (1.20) shows the following:1.dP(t)/ds and Pt(t) point in the same direction.Therefore, since dP(t)/ds is aunit vector, we getdP(t) Pt(t)=.ds |Pt(t)|2.st(t) =|Pt(t)|.We now derive the expression for curvature from a different point of view.Thecurvature k is defined by d2P(t)/ds2 = kN, where N is the unit principal normal vector(Section 1.6.2).The problem is to express k in terms of the curve P(t) and its derivatives,not involving the (normally unknown) function s(t).We start withd Pt(t)d2P(t) d Pt(t) dt |Pt(t)|= =ds2 ds |Pt(t)| st(t)Ptt(t) Pt(t) d|Pt(t)|- ·|Pt(t)| |Pt(t)|2 dt=.(1.21)|Pt(t)| 32 1.Basic TheoryThe identity A " A = |A|2 is true for any vector A(t) and it impliesd|A(t)|A(t) " At(t) =|A(t)|.dtWhen we apply this to the vector Pt(t), we getd2P(t) Ptt(t) Pt(t) " Ptt(t)= - (1.22)2 Pt(t),ds2 Pt(t) " Pt(t)Pt(t) " Pt(t)which can also be writtenPt(t) × Ptt(t) × Pt(t)d2P(t)kN = = (1.23)2.ds2Pt(t) " Pt(t)1.6.7 TorsionTorsion is a measure of how much a given curve deviates from a plane curve.The torsionÄ(i) of a curve at a point P(i) is defined by means of the following two quantities:1.Imagine a point h close to i.The curve has rectifying planes at points h and i(Figure 1.15).Denote the angle between them by ¸.hiTiRectifyingplane iBiRectifyingplane h¸Figure 1.15: Torsion.2.Denote by s the arc length from point h to point i.The torsion of the curve at point i is defined as the limit of the ratio ¸/s whenh approaches i.Figure 1.15 shows how the rectifying plane rotates about the tangentas we move on the curve from h to i.The torsion can be expressed by means of thederivatives of the curve and by means of the curvature|Pt(t) Ptt(t) Pttt(t)| |Pt(t) Ptt(t) Pttt(t)|Ä(t) = = Á(t)2.|Pt(t) × Pt(t)|2 |Pt(t)|6BhAngle betweenplanes 1.6 Curvature and Torsion 33(The numerator is a determinant and the denominator is an absolute value.This ex-pression is meaningful only when Á(t) False,ViewPoint->{-0.846,-1.464,3.997},DefaultFont->{"cmr10",10}];Figure 2.8: A Bilinear Surface.The first is a vector in the xz plane, while the second is a vector in the y = 1 plane.The"P(u,1) "P(1,w)following two tangent values are especially simple: =(0.5, 0, 0) and ="u "w(0, 1, 0).The first is a vector in the x direction and the second is a vector in the ydirection.Finally, we compute the normal vector to the surface.This vector is normal to thesurface at any point, so it is perpendicular to the two tangent vectors "P(u, w)/"u and"P(u, w)/"w and is therefore the cross-product [Equation (1.5)] of these vectors.Thecalculation is straightforward:"P "PN(u, w) = × =(1 - w, 0.5(1 - u), 1 - 0.5w).(2.13)"u "wThere are two ways of satisfying ourselves that Equation (2.13) is the correct expressionfor the normal:1.It is easy to prove, by directly calculating the dot products, that the normalvector of Equation (2.13) is perpendicular to both tangents of Equation (2.12).2.A closer look at the coordinates of our points shows that three of them have a zcoordinate of zero and only P00 has z = 1.This means that the surface approaches aflat xy surface as one moves away frompoint P00.It also means that the normal shouldapproach the z direction when u and w move away from zero, and it should move awayfrom that direction when u and w approach zero.It is, in fact, easy to confirm thefollowing limits:lim N(u, w) =(0, 0, 0.5), lim N(u, w) =(1, 0.5, 1).u,w’!1 u,w’!0Exercise 2.12: (1) Calculate the bilinear surface for the points (0, 0, 0), (1, 0, 0), (0, 1, 0),and (1, 1, 1).(2) Guess the explicit representation z = F(x, y) of this surface.(3) Whatcurve results from the intersection of this surface with the plane z = k (parallel to the 2.3 Bilinear Surfaces 63xy plane)
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