Straight Line motion with rigid sets

Transcription

1 Straight ine motion with rigid sets arxiv: v [math.mg] 9 Jan 04 Robert Connelly and uis Montejano January 7, 08 Abstract If one is given a rigid triangle in the plane or space, we show that the only motion possible, where each vertex of the triangle moves along a straight line, is given by a hypocycloid line drawer in the plane, and a natural extension in three-space. Each point lies on a circle which rolls around, without slipping, inside a larger circle of twice its diameter. Introduction Consider three straight lines in the plane or in three-space. When can you continuously move a point on each line such that all the pairwise distances between them stay constant? One case is when the three lines are parallel. Figure shows another way. Are these the only possibilities? We answer the uestion affirmatively with Theorem in the plane, and Theorem in higher dimensions. Figure

2 The example of Figure is known as a hypocycloid straight-line mechanism, and Figure shows a model from the Cornell Reuleaux collection. []. Figure It is easy to see from Figure that as the inner circle rolls around the central point with a fixed radius, the triangle, formed from the other points of intersection with the three fixed lines form a triangle with fixed internal angles, and therefore the triangle moves rigidly. In dimension three, one can form a cylinder rotating inside a larger cylinder of twice the diameter. But in the plane and three-space, the three lines all have a fourth line that is perpendicular and incident to all three lines. The planar case We first describe the motion of a segment of fixed length d sliding between two fixed non-parallel lines, and, first in the plane. We assume, without loss of generality, that and intersect at the origin and are determined by two unit vectors v, v, where v v = cos α = c, α being the angle between v and v, and < c <. et t, t be the oriented distance from the origin of the points p, p, so that p = t v, and p = t v. Then

3 treating the suare of vector as the dot product with itself, d = (p p ) = (t v t v ) = t + t c t t. () Thus in t, t space the configurations of the line segment form an ellipse centered at the origin whose major and minor axes are at 45 from the t, t axes. Thus there are constants a = d / ( c ), b = d / ( + c ) such that t = a cos θ b sin θ, t = a cos θ + b sin θ () describes the full range of motion of the line segment for 0 θ π. It is also clear from Figure 3 that the length of the image of each p i on i, i =, is an interval of length d / sin α centered about the intersection of and. We can now state the situation for the plane. p α d p Figure 3 Theorem If a triangle, with fixed edge lengths, continuously moves with each vertex p i on a line i, i =,, 3 in the plane, then either all the lines are parallel or they all intersect at a point forming a hypocycloid straight line drawer as above, with the range of each point an interval of the same length on each line, while the midpoint of each range is the intersection of the two lines. Proof. Choose any two of the the three lines, say and. The parametrization discussed above shows that any position of the d segment can described 3

4 by the euations in (). These define the positions of p and p as a function of θ, for all 0 θ π. Then the position of p 3 is determined also as a function of θ, p 3 (θ), since it is carried along rigidly. The image of p 3 is a continuous curve in the plane and it is symmetric about the intersection of and. (Actually it is generally an ellipse, also, but that is not needed for this part of the argument.) If that image is to be in a straight line 3, then 3 should intersect the intersection of and. So if p 3 (θ) satisfies the euations corresponding to () for 3 and 3 replacing, for an interval of values of θ, then it must satisfy those euations for all θ and the image of each p i in i is the same length for i =,. But applying this argument to another pair such as 3, shows that all the images are the same length. 3 The higher dimensional case In higher dimensions, each pair of non-parallel lines i, j, i, j =,, 3 have two points ij i, for i j, such that ij ji is perpendicular to both i and j. The shortest distance between i and j is ij ji = D ij = D ji. We construct six variables t ij, i, j =,, 3, i j, where p i = ij + t ij v i. Then we can project orthogonally into a plane spanned by the vectors v i, v j to get the analog of Euation (), which is the following representing the three euations for each of the three edge lengths d ij = d ji of the triangle: d ij D ij = (p i p j ) = (t ij v i t ji v j ) = t ij + t ji c ij t ij t ji. (3) Similar to Euation (), we define constants a ij = d ij D ij / ( c ij ), b ij = d ij D ij / ( + c ij ) such that a ij = a ji, b ij = b ji and t ij = a ij cos θ ij b ij sin θ ij, t ji = a ij cos θ ij + b ij sin θ ij (4) and θ ij = θ ji is the parameter as before. This gives three separate parameterizations for each line segment between each pair of lines. Furthermore the euations of (3) give a complete description of position of each p i, two ways for each line, in terms of t ij and t ik, where t ij t ik = ij ik, a constant. So t ij = t ik + e ijk, where e ijk = ± ij ik is a constant. See Figure 4 for a view of these coordinates. 4

5 v v v 3 Figure 4 Theorem If a triangle, with fixed edge lengths, continuously moves with each vertex p i on a line i, i =,, 3 in a Euclidean space, then either all the lines are parallel or they all intersect a line perpendicular all of them, with the range of each point an interval of the same length on each line, while the midpoint of each range is the intersection of the two lines. Proof. Start with Euation (3) for the, lines defining the variable θ = θ which, in turn defines the positions for p (θ ) on and p (θ ) on. Note that t = t 3 + e 3 and t = t 3 + e 3. So we can regard t 3 and t 3 as linear functions of t and t. Similarly, t 3 = t 3 + e 3, so we can regard t 3, as a function of t 3. So if we subtract the 3 and 3 euations for (3), we are only left with linear terms in t 3. The suared terms have cancelled. Thus either t 3 has no term in the difference of the two euations, or t 3 is a non-zero rational function of t and t and thus θ. If the linear t 3 terms have disappeared, then the and 3 euations imply the 3 euation, which means that the edge of the triangle makes a full 360 turn and the image of p in is symmetric about. If the t 3 term is a non-zero rational function of t and t and thus θ, and again the image of p in is symmetric about. 5

6 Applying the above argument to each euation for (3) for ij, we see that each image of p i in i is symmetric about i. If all the lines lie in a three-dimensional Euclidean space, and no two lines i are parallel, then the midpoints of the images of each p i must lie on a line perpendicular to all of the i, and as claimed. So these lines are just a three-dimensional lift of the two-dimensional case. If the lines span a higher dimensional space, then there is a non-zero vector perpendicular to each v i, for i =,, 3, and the lines i can be projected orthogonally into a three-dimensional space, and the projection of the points on the projected lines will be a mechanism, one dimension lower. Then similarly the three-dimensional mechanism comes from a two-dimensional mechanism, from the argument above. The following are some immediate corollaries. Corollary If a polygon in the plane is continuously moves as a rigid body so that each vertex stays on a straight line, then those vertices all lie on a circle or the three lines are parallel. Corollary Given three lines,, 3 on a Euclidean space such that they are are not all perpendicular to a fourth line, then there are at most 8 configurations of a triangle p, p, p 3 with fixed edge lengths, where p i is on i, i =,, 3. This is because each edge length is determined by a degree two euation, and Bézout s theorem [] implies that there are at most 3 = 8 individual solutions. References [] Keith Kendig. A guide to plane algebraic curves, volume 46 of The Dolciani Mathematical Expositions. Mathematical Association of America, Washington, DC, 0. MAA Guides, 7. [] John Saylor. Kinematic Models for Digital Design ibrary (KMODD). url=\ 6