|
Theorem 2.1:
Proof: Consider the algebraic curve f(x, y) =0. We want to make a linkage such that if a movable point Bn is moved along a straight line then the point P will describe the curve f(x, y)=0 (1) Consider the parallelogram of fig.1 has side m and n which make angle α, β with the x-axis. The vertex P is a point of the curve, so the coordinates are then
Now the sine of any angle
can be written as the cosine of its complement. Furthermore, the product
and the powers of cosine can be expressed as the sum of cosine. Thus ,
if we substitute equation (2) in (1),we shall have a sum of terms of the
sort:
As the fig.2 show, it is composed of similar crossed-parallelograms. By mean of such mechanisms, we may obtain integral multiples of any angle, for instant, aα or bβ. In fact, it is a Multiplicator. Joining one multiplicator to another will produce the combination aα±bβ. As the fig.3 show, the plate BOK with angle k is connected rigidly to the end bar. Thus we will build up a linkage to produce ∠BOX＝aα±bβ±k. If, in fig.3, OB is taken equal to A, then the x-coordinate of the point B is A.cos(aα±bβ±k). As the fig.4 show, the mechanism is composed of two parallelogram with OB pivoted at O. Within the limits of mechanism, the bars O´B´ can be moved freely in the plane, remaining always parallel to OB. In fact, it is a translator. By combining the linkages of fig.1, 2, 3 and 4, we may erect a chain of links OB, BB1, B1B2, B2B3, B3B4…, with the end point Bn (as shown.). The x-coordinate of the point Bn:
But if P is moved along the given curve, then it’s coordinates x, y satisfy: f(x, y) = 0. Accordingly, the locus of the end point, Bn, of the chain is X+C=0, a straight line parallel to the Y-axis. Conversely, if Bn is moved along this line (with the help of Peaucellier cell, for instance), the point P will generate the curve f(x, y) =0.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Reference: Robert C. Yates 1949 Geometrical Tools P.190-191 |
|
