163 
of Edinburgh, Session 1875-76. 
QF : FP : : Lm : MK : Lq : Kp 
: : BQ (1 - cos 0) : AP (1 - cos <p) 
: : BQ • 2 sin 2 ? : AP ■ 2 sin 2 £ 
A A 
: : BQ • O 2 ■ AP • f . . . . (1J) 
: : BQ • AP 2 : AP ■ BQ 2 ... (2) 
: : AP : BQ 
0 0 
(] ) For small angles, sin ^ =-^ nearly. 
(2) Ultimately, BQ sin 0 = AP sin <p, or, BQ : AP : : sin : sin 0 : : <p : 0. 
In connection with Watt’s system we are led to consider the 
motion of the pantagraph, which has been a means of extending 
the parallel motion. Thus, while to the point F is attached the 
air-pump rod of an engine; by means of the pantagraph, whose 
property is that it describes similar curves on a smaller or larger 
scale, another point will trace another parallel line, and to it there- 
fore may be fixed the end of the piston-rod. By superadding the 
parallelogram of bars to the Watt parallel motion, and making 
E, — the angle of the parallelogram — the connecting point of the 
piston rod, so as to concentrate the force ; we have the point E 
describing a curve similar to F, when 
QF : BQ : : PQ or HE : BH. (Fig. 3.) 
H 
E 
Fig. 3. 
The problem of the pantagraph may be proved as follows. Let 
the initial position be at an angle of 90°, the arms lying along the 
axes of x and y. (Fig. 4.) 
Given the fulcrum A and the ratio AB : AC, to find B 1? the posi- 
tion which B the pencil assumes when the tracer C moves to C r 
When the position of C n is given, the angle CjDjA is known, for 
ADj and CDj are constants. Join BBj and CCj ; AB X and ACj 
(which latter are not assumed to be coincident). 
In the triangles CjDjA and BjFjA, the angle BjFjA is equal 
to the angle CjDjA; and the sides about these angles are propor- 
