Linkage System: more powerful than you think

In the machine era, as a theoretical tool for mechanical structures, the linkage system once became the most popular topic in the field of mathematics. The so-called connecting rod system is a mechanical device formed by connecting some rigid small rods with a rotating shaft at the endpoint. After the positions of some vertices are fixed, other dynamic points can draw Interesting tracks. For example, if one of the endpoints of a fixed line segment AB is a vertex A, vertex B depicts a circle around the point.

The most exciting thing about the linkage system is that some simple linkage devices can plot very complex curves. For example, the right graph above is a connecting rod consisting of five lines of the same length. After the endpoints A and B are fixed, c and d clearly depict the arc, but the trajectory of E is hard to imagine. In fact, the trajectory of E point is quite strange and must be described in complex algebra.

 

 
In the connection system field, there is a problem that has plagued humans for nearly a hundred years-Can a straight line be drawn using the connection system? At that time, it seemed that this problem was so difficult that it was widely speculated that even attempts to prove that a straight line connecting rod could not exist. In 1864, a French Naval Officer Charles-Nicolas peaucellier invented the first connecting rod system capable of drawing a straight line, which caused a great sensation. The principle of peaucellier connecting rod is not difficult to understand. It is sufficient to prove the correctness of peaucellier connecting rod by using Junior High School ry knowledge.

Peaucellier is composed of seven connecting rods. AC = AD = A, BC = Ce = ED = DB = B, and Ob is of any length. Fix the location of A and O points so that the distance between OA is exactly equal to OB, then E points will depict a straight line perpendicular to AO. It is easy to see that a, B, and E are on the same straight line. We first show that AB AE is a constant. Through point C for ch AE, the vertical foot is H. So AB · AE = (AH + He) · (AH-HB) = ah2-BH2 = (ac2-CH2)-(BC2-CH2) = a2-b2 = constant.

Why can we ensure that the trajectory of point E is a straight line When AB · AE is a constant? The diameter am of the circle o is generated after the point, and a point of N is found on the ray am so that AM. An is equal to this constant. Because am an = AB AE, we can immediately know △abm △△ane, so login ane = 127abm = 90 °, that is, en and an are always vertical. This proves that the trajectory of point E is indeed a straight line perpendicular to AO.

 
 
After resolving the problem of linear connections, mathematicians are obviously not satisfied. There are many indications that the linkage system is more powerful than we thought, and it does not seem to be a problem to draw some strange figures.
There is a very simple structure that almost instantly enhances the functions of the linkage system, making people more confident in the possibility of building a complex linkage. Although the linkage system requires that the rod and the rod must be connected at the endpoint, we can use the triangle stability to directly link one end of a rod to the middle of the other. For example, although AB and BC are two connecting rods that can respectively rotate around B, but simply use a triangle to fix them, AB and BC will become a line segment AC. Using this basic structure, we can directly connect the end of the connecting rod to the center of another connecting rod.

 
This basic structure inspires us a lot. Why don't we use the most basic structure to construct a more practical basic structure, just like studying the ruler plot, how can I build a building with a link? In 1877, the British mathematician Alfred Kempe followed this idea and finally came to an astonishing conclusion: the connecting rod system can not only draw a straight line and a circle, but also draw a hyperbolic, parabolic, and elliptical, even a complex curve, such as a semi-cubic parabolic curve and dual-link line. In fact, any algebraic curve F (x, y) = Σ (I = 1 .. n) Σ (j = 1 .. n) CI, J · Xi · YJ = 0 can be drawn using the linkage system!

 

The basic idea of this proof is as follows. First, construct two diamond shapes with the O as the endpoint. Using the peaucellier linkage, we can keep the X and Y points moving along two vertical straight lines. After the O point is fixed, a plane Cartesian coordinate system is established. Next, we need to rotate point y around the origin clockwise for 90 degrees. Assuming that the side length of the diamond ocyd is l, the connecting rod OC '= C' y' = y' d' = d' o = L, CC' = dd' = √ 2 L is constructed, in this way, the oy length is transferred to the X axis. Next, we will use a series of connecting rods to construct a point t so that t is always at the position (f (x, y), 0) in the coordinate system. Then we construct a vertex s so that S is always in the coordinate system (x, y. Finally, we fix the T point at (0, 0), then the s point will depict the f (x, y) = 0 image.
To obtain T (f (x, y), 0), we only need to perform the following four operations on the points on the X axis:

(1) Add the coordinates of a point to a constant C
(2) Multiply the coordinates of a certain point by a constant C
(3) Add the coordinates of the two points
(4) Multiply the coordinates of two points

 

The first two operations are not difficult. For a point P on the X axis, to obtain the point Z = P + C, you only need to fix two points A and B with a distance of C and construct a series of parallelogram. To obtain the point Z = C · P, we only need to construct a group of similar triangles oap and obz so that Ob = C · OA and Bz = C · AP. Add a connecting rod PC so that the Quadrilateral ABCP is a parallelogram to ensure that the two triangles are similar. Note: In the multiplier structure, we use the basic structure mentioned above, that is, the link is directly connected to another link.

 

Adding two variables is easier than you think. In fact, we can not only add two dynamic points on the X axis, but also directly implement a stronger basic operation-adding two vectors on the plane. It is easy to prove that the vector Oz is the sum of the vector op and OQ by constructing a series of Parallelogram.

 

However, it is troublesome to multiply two variables on the X axis. Note that since P · q = (p + q) 2-(p-q) 2)/4, as long as we can implement square operations, there is a way to implement multiplication. Because 1/P-1)-1/(p + 1) = 2/(P2-1), as long as we can implement the reciprocal operation, there is a method to implement the square. When proving the correctness of the peaucellier connecting rod, we have proved that there is z · P = a2-B2 in the connecting rod, and we can use it to realize z = (A2-B2) /P. Take A and B as appropriate values, and we can get the reciprocal of P.

Since X and y' are already on the X axis, we can get T (f (x, y), 0 by using these basic operations above ). In addition, we can obtain the vectors and S = (x, y) of ox and Oy by Using Vector processors ). Place the T point in the O position of the origin, and the S trajectory is the F (x, y) = 0 image.
The most surprising conclusion of Kempe is that, since various curves can be roughly described using algebraic curves, the linkage system can almost be considered omnipotent. Therefore, if you are patient enough, you can even construct a linkage system that can check out your name!