MECH 576 Geometry in Mechanics September 16, 2009 Using Line Geometry
1 Deriving Equations in Line Coordinates
Four exercises in deriving fundamental geometric equations with line coordinates will be conducted. These results may be widely used to formulate constraint equations to solve many useful geometric problems.
• First the derivation of the summations, introduced earlier,
3 3 X X p = Pa ∩ P : Pi = Pijpj,P = Pr ∩ p : pi = pijPj (1) j=0 j=0
that define the plane p on a given axial line Pa and point P are derived using the plane equation Grassmannian. This is followed by inductive reasoning to express the dual relation, i.e., the point P on given radial line Pr and plane p. • Next the set of all lines tangent to a given sphere is derived. This is a simple second order complex; a three parameter set of lines.
• Finally the cylinder and cone of revolution, using a given tangent cylinder and an absolute point on the axis and a Euclidean point on the apex, respectively, are expressed as a one parameter set of ruling line coordinates. These are converted to point form with any two constraint equations, selected from the first set of four expressed by Eq. 1, and setting all Pi = 0.
2 Spanning Plane and Piercing Point
Consider a plane, given by its four homogeneous coordinates,
p{P0 : P1 : P2 : P3} and a line, given by its six homogeneous radial Pl¨ucker coordinates,
Pr{p01 : p02 : p03 : p23 : p31 : p12}
Recall the identical axial line is related as
Pa{P01 : P02 : P03 : P23 : P31 : P12} = λ{p23 : p31 : p12 : p01 : p02 : p03}, λ 6= 0
1 The given plane and line, if linearly independent, intersect on a point. The purpose here is to illustrate how to structure the 4×4 Grassmannian determinant form of the standard point equation so as to produce the well known sum of products piercing point equation
3 X pi = pijPj j=0 to yield the four homogeneous coordinates of the point
P {pi}, i = 0,..., 3,P = Pr ∩ p
Recall that the conventions, pij = 0 if i = j and pij = −pji if i 6= j, apply in this summation. Imagine that Pr is expressed by some plane pair
q{Q0 : Q1 : Q2 : Q3}, r{R0 : R1 : R2 : R3} whose coordinates may be expanded to axial Pl¨ucker coordinates.
Pr ≡ Pa{Q0R1 − R0Q1 : Q0R2 − R0Q2 : Q0R3 − R0Q3 :
Q2R3 − R2Q3 : Q3R1 − R3Q1 : Q1R2 − R1Q2}
So the point equation determinant is written
X0 X1 X2 X3
P P P P 0 1 2 3 = 0 Q0 Q1 Q2 Q3
R0 R1 R2 R3
Expanding on top row minors computes the piercing point homogeneous coordinates as coeffi- cients. But in this case each coefficient is computed not by directly evaluating the 3 × 3 cofactor determinant but by expanding on its top row minors. These are the remaining second row entries of the original 4 × 4 after crossing out the top row minor’s row and column. The first top row minor and cofactor are
P P P 1 2 3
+ Q1 Q2 Q3 X0 ⇒
R1 R2 R3 which expands as
[(Q2R3 − R2Q3)P1 − (Q1R3 − R1Q3)P2 + (Q1R2 − R1Q2)P3]X0 ⇒ which is the same as
[(Q2R3 − R2Q3)P1 + (Q3R1 − R3Q1)P2 + (Q1R2 − R1Q2)P3]X0 ⇒ and this becomes, via the substitution of the appropriate axial Pl¨ucker coordinates
[P23P1 + P31P2 + P12P3]X0 ⇒
2 or via the substitution of the appropriate radial Pl¨ucker coordinates
[p01P1 + p02P2 + p03P3]X0 ⇒
Finally, this compares identically to the first row of the summation where i = 0 and p00 = 0 and p0, the first coefficient of the point equation, is obtained.
p0 = p00P0 + p01P1 + p02P2 + p03P3
The line-point-plane relation is clearly understood by seeing through to the end the computation of the next three point equation coefficients. The second top row minor and cofactor are
P P P 0 2 3
− Q0 Q2 Q3 X1 ⇒
R0 R2 R3 which expands as
−[(Q2R3 − R2Q3)P0 − (Q0R3 − R0Q3)P2 + (Q0R2 − R0Q2)P3]X1 ⇒ and this becomes, via the substitution of the appropriate axial Pl¨ucker coordinates
−[P23P0 − P03P2 + P02P3]X1 ⇒ or via the substitution of the appropriate radial Pl¨ucker coordinates
−[p01P0 − p12P2 + p31P3]X1 ⇒
Finally, this compares identically to the first row of the summation where i = 1, pij = −pji and p11 = 0 and p1, the second coefficient of the point equation, is obtained.
p1 = p10P0 + p11P1 + p12P2 + p13P3
The third top row minor and cofactor are
P P P 0 1 3
+ Q0 Q1 Q3 X2 ⇒
R0 R1 R3 which expands as
[(Q1R3 − R1Q3)P0 − (Q0R3 − R0Q3)P1 + (Q0R1 − R0Q1)P3]X2 ⇒ and this becomes, via the substitution of the appropriate axial Pl¨ucker coordinates
[−P31P0 − P03P1 + P01P3]X2 ⇒ or via the substitution of the appropriate radial Pl¨ucker coordinates
[−p02P0 − p12P1 + p23P3]X2 ⇒
3 Finally, this compares identically to the first row of the summation where i = 2, pij = −pji and p22 = 0 and p2, the third coefficient of the point equation, is obtained.
p2 = p20P0 − p21P1 + p22P2 + p23P3 The fourth top row minor and cofactor are
P P P 0 1 2
− Q0 Q1 Q2 X3 ⇒
R0 R1 R2 which expands as
−[(Q1R2 − R1Q2)P0 − (Q0R2 − R0Q2)P1 + (Q0R1 − R0Q1)P1]X3 ⇒ and this becomes, via the substitution of the appropriate axial Pl¨ucker coordinates
−[P12P0 − P02P1 + P01P2]X3 ⇒ or via the substitution of the appropriate radial Pl¨ucker coordinates
−[p03P0 − p31P1 + p23P2]X3 ⇒
Finally, this compares identically to the first row of the summation where i = 3, pij = −pji and p33 = 0 and p3, the fourth coefficient of the point equation, is obtained.
p3 = p30P0 + p31P1 + p32P2 + p33P3 This completes the elaboration of the intersection of a projective line and plane. It is not necessary to do the same for the intersection of a projective line and point to get the projective plane so defined. This is because one may invoke the principle of duality and exchange point with plane coordinates and radial line coordinates with axial ones. Consider now the degenerate cases, i.e., if the given elements are linearly dependent. This means that the piercing point does not exist because the line is on the plane or the spanning plane does not exist because the point is on the line. All four coordinates vanish, pi = 0 or Pi = 0. This is a very useful way to express that P ∈ P or p ∈ P and will be used to derive implicit equations of ruled surfaces which are easily formulated using line geometric concepts.
3 The Spherical Tangent Line Complex
The set of all lines T{t01 : t02 : t03 : t23 : t31 : t12} tangent to a given sphere, radius R, centred on point M{m0 : m1 : m2 : m3} can be deduced by equating the square of the moment magnitude of a typical tangent line’s direction numbers, about the point M, obtained in two different ways. First, the square of the difference between the typical tangent line’s moment t23 : t31 : t12 and the moment of a line on M which has the direction numbers t01 : t02 : t03 of the typical tangent line. This can be expressed as
2 1 2 t23 − (m2t03 − m3t02) t23 m1 t01 m0 1 1 t − m × t = t31 − (m3t01 − m1t03) 31 2 02 m0 m0 1 t12 m3 t03 t12 − (m1t02 − m2t01) m0
4 Then the square of the moment of the typical line’s direction numbers about M is formed directly as 2 t01 R t02 t03 2 Therefore the desired equation of the line complex is the difference multiplied by m0.
2 2 [m0t23 − (m2t03 − m3t02)] + [m0t31 − (m3t01 − m1t03)] 2 2 2 2 2 +[m0t12 − (m1t02 − m2t01)] − (m0R) (t01 + t02 + t03) = 0
The Spherical Tangent Line Complex (a 3-parameter set of lines described by a 2nd order equation in Pleucker coordinates)
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Figure 1: Complex Geometry
4 Cylinder of Revolution
Beginning with the equation of the tangent line complex on the sphere
2 2 [m0t23 − (m2t03 − m3t02)] + [m0t31 − (m3t01 − m1t03)] 2 2 2 2 2 +[m0t12 − (m1t02 − m2t01)] − (m0R) (t01 + t02 + t03) = 0
5 consider a cylinder of revolution of radius R and whose axis is line
A{a01 : a02 : a03 : a23 : a31 : a12}
Note that if the axial direction is given as on M, the sphere centre, and A{0 : a1 : a2 : a3}, an absolute point then A is immediately available as A = A ∩ M. If your sensibilities are offended by the use of ∩, intersection rather than union, consider that in the space of dual, plane coordinates the these two points assume plane-like form. First the line complex is converted to the surface equation in terms of typical ruling lines
P{p01 : p02 : p03 : p23 : p31 : p12} by noting that t0j = a0j and moments mit0j − mjt0i = aij and making these substitutions to yield
2 2 2 2 2 2 2 (p23 − a23) + (p31 − a31) + (p12 − a12) − R (a01 + a02 + a03) = 0 This becomes a one parameter set of lines when combined with the Pl¨ucker condition, in this case a linear equation. a01p23 + a02p31 + a03p12 = 0 The point equation is produced by using two elements of the expansion P ∈ P which gives the degenerate plane defined by a line and a point on it.
3 X Pi = Pijpj = 0 j=0 Substituting radial for axial line coordinates, the cylinder is given by
2 2 (a23p0 + a02p3 − a03p2) + (a31p0 + a03p1 − a01p3) 2 2 2 2 2 +(a12p0 + a01p2 − a02p1) − R (a01 + a02 + a03) = 0 The obvious advantage of this formulation is that it gives one the option to write the implicit form of this quadric surface directly from its geometric definition, without resort to coordinate transformation or even trigonometric functions.
5 Cone of Revolution
Consider the cone of revolution, apex on point X{x0 : x1 : x2 : x3}, with all generators tangent to a sphere centred on point M{m0 : m1 : m2 : m3} and of radius R. Note that R is not the radius RM of the circular section on M.
Rh 2 2 RM = √ , h = (x − m) h2 − R2 where x and m are the position vectors of points X and M. Again, the formulation starts with the tangent line complex on the sphere of radius R centred on M.
2 2 [m0t23 − (m2t03 − m3t02)] + [m0t31 − (m3t01 − m1t03)] 2 2 2 2 2 +[m0t12 − (m1t02 − m2t01)] − (m0R) (t01 + t02 + t03) = 0
6 However this time the set of generators is not on the point that closes A = M ∩ X but upon the apex X. This condition is expressed by the degenerate plane
3 X Xi = Tijxj = 0 j=0 Switching to radial coordinates produces
0 + x1t23 + x2t31 + x3t12 = 0
−x0t23 + 0 + x2t03 − x3t02 = 0
−x0t31 − x1t03 + 0 + x3t01 = 0
−x0t12 + x1t02 − x2t01 + 0 = 0
2 2 2 Since the cone is ruled by Euclidean lines and t01 + t02 + t03 6= 0, the second and third equations can be augmented with the Pl¨ucker condition to eliminate t23, t31, t12. x0 0 0 t23 x2t03 − x3t02 0 x0 0 t31 = x3t01 − x1t03 t01 t02 t03 t12 0 which is solved for the tij vector.
x0t23 = x2t03 − x3t02
x0t31 = x3t01 − x1t02
x0t03t12 = −(x2t03 − x3t02)t01 − (x3t01 − x1t03)t02 Substituting this into the complex yields
2 2 [m0(x2t03 − x3t02) − x0(m2t03 − m3t02)] t03 2 2 +[m0(x3t01 − x1t03) − x0(m3t01 − m1t03)] t03 2 +[m0(x1t02 − x2t01)t03 − x0(m1t02 − m2t01)t03] 2 2 2 2 2 −(m0R) (t01 + t02 + t03)t03 = 0 2 Setting m0 = x0 = 1 and eliminating t03 produces the line equation of the cone. 2 2 [(m3 − x3)t02 − (m2 − x2)t03] + [(m1 − x1)t03 − (m3 − x3)t01] 2 2 2 2 2 +[(m2 − x2)t01 − (m1 − x1)t02] − R (t01 + t02 + t03) = 0 This is converted to a point equation by invoking
3 X P ∈ T ,Pi = Tijpj = 0 j=0 and the second and third equations of
0 + t23p1 + t31p2 + t12p3 = 0
−t23p0 + 0 + t03p2 − t02p3 = 0
−t31p0 − t03p1 + 0 + t01p3 = 0
−t12p0 + t02p1 − t01p2 + 0 = 0
7 are used to eliminate t02 and t03.
" (x p − x p )(x p − x p ) #" t # " 0 # 3 0 0 3 2 0 0 2 02 = 0 (x1p0 − x0p1) t03 (x3p0 − x0p3)t01 which is solved for the t0j vector.
(x1p0 − x0p1)t02 = (x2p0 − x0p2)t01
(x1p0 − x0p1)t03 = (x3p0 − x0p3)t01
Substituting these and simplifying
2 [(m3 − x3)(x2p0 − x0p2) − (m2 − x2)(x3p0 − x0p3)] 2 +[(m1 − x1)(x3p0 − x0p3) − (m3 − x3)(x1p0 − x0p1)] 2 +[(m2 − x2)(x1p0 − x0p1) − (m1 − x1)(x2p0 − x0p2)] 2 2 2 2 −R [(x1p0 − x0p1) + (x2p0 − x0p2) + (x3p0 − x0p3) ] = 0
The surface is expressed in ordinary Cartesian coordinates of P and X by setting p0 = x0 = 1.
2 [(m3 − x3)(x2 − p2) − (m2 − x2)(x3 − p3)] 2 +[(m1 − x1)(x3 − p3) − (m3 − x3)(x1 − p1)] 2 +[(m2 − x2)(x1 − p1) − (m1 − x1)(x2 − p2)] 2 2 2 2 −R [(x1 − p1) + (x2 − p2) + (x3 − p3) ] = 0
The obvious advantage of this formulation is that it gives one the option to write the implicit form of this quadric surface directly from its geometric definition, without resort to coordinate transformation or even trigonometric functions.
6 Elliptical Cone
Examine Fig. 2. Two circular sections are shown. The given circle appears in both principal views while an elliptical section appears in the auxiliary projection. Also shown is the given apex X. Consider the following three sets of point-on-line specifications. The line in question is a typical generator given in axial Pl¨ucker coordinates.
Pa{P01 : P03 : P03 : P23 : P31 : P12}
The last three coordinates are the line direction numbers. the first three are the moment of each direction number about the origin. The first set states that any point P (p1, p2, p3) on the given circle is on the line. P01p1 +P02p2 +P03p3 = 0 −P +P p −P p = 0 01 12 2 31 3 (2) −P02 −P12p1 +P23p3 = 0 −P03 +P31p1 −P23p2 = 0
8 X
x 2
K A
(FD3)ECone7Ch x 1 P
given circular section right truncated base A
x 3 P K
A-A x 1 X P
K X
given circular section
Figure 2: Three Views of an Elliptical Cone
The second set states that X(x1, x2, x3) the given cone vertex is on the line.
P01x1 +P02x2 +P03x3 = 0 −P +P x −P x = 0 01 12 2 31 3 (3) −P02 −P12x1 +P23x3 = 0 −P03 +P31x1 −P23x2 = 0
The third set maps the line to an arbitrary variable point K(k1, k2, k3) on the cone surface.
P01k1 +P02k2 +P03k3 = 0 −P +P k −P k = 0 01 12 2 31 3 (4) −P02 −P12k1 +P23k3 = 0 −P03 +P31k1 −P23k2 = 0 Only two equations in every set are linearly independent so the second and fourth equation in each set will be chosen for solution by eliminating all Pl¨ucker coordinates. First, however, the equations from set Eq. 2 are rewritten to lie on the plane p2 = 0 of the circle and to be at distance R from the circle centre, chosen on the origin. q 2 2 P01 + P31p3 = 0,P03 − P31 R − p3 = 0
9 Eliminate p3 from these and rewrite the remaining five equations thus.
2 2 2 2 P01 + P03 − R P31 = 0
−P01 + P12x2 − P31x3 = 0
−P03 + P31x1 − P23x2 = 0
−P01 + P12k2 − P31k3 = 0
−P03 + P31k1 − P23k2 = 0 (5)
Set P23 = 1 because Pl¨ucker coordinates are homogeneous. No cone generator may be normal to this direction therefore P23 6= 0. Then eliminate the five other Pij. This results in a product of two trivial double solutions and the quadric we seek. In Fig. 2 the problem is defined, without loss of generality, in a frame where x3 = 0 to simplify Eqs. 6 and 7.
2 2 2 2 2 2 2 2 2 2 2 2 2 (k2−x2) (k3−x3) [R x2−2R x2k2−x2k1 +2x1x2k1k2+(R −x1−x3)k2 +2x2x3k2k3−x2k3] = 0 (6) The first four solutions all map to the cone vertex. Eq. 7 presents the quadric in symmetric coefficient matrix form. The cone of revolution is engendered therein. 2 2 2 R x2 0 −R x2 0 1 0 −x2 x x 0 k 2 1 2 1 [1 k1 k2 k3] 2 2 2 2 = 0 (7) −R x2 x1x2 R − x1 − x3 x2x3 k2 2 0 0 x2x3 −x2 k3 7 Conclusion
Other line geometric formulations will be encountered later. These include other, non-degenerate quadrics like the hyperboloid of one sheet and hyperbolic paraboloid as well as the two lines that intersect four given ones. These surfaces play important rˆolesin solving problems in manipulator kinematics. The ambitious student may try to find more formal and authoritative literature on line geometry by searching libraries and related web-sites. The author has obtained little success in this way. When asked, knowledgable central European authorities inevitably reply, “I don’t know where this is written but everyone knows all about it anyway.” In 1908 Felix Klein[1] summed up the situation elegantly. “In geometry we possess no unified textbooks corresponding to the level of the science, . . . ” -note that he considers mathematics to be science with profound engineering application- “ . . . such as exist in algebra and analysis, thanks to the model French Cours. We find, rather, a single page here, another there, of an extensive subject, just as it has been developed by one or another group of investigators.” This situation has persisted and worsened during the intervening century.
References
[1] Klein, F.C. (2004) Elementary Mathematics from an Advanced Standpoint -Geometry-, Dover, ISBN 0-486-43481-8, p.-v-.
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