All 
sents the forces at the point of intersection. Thus the point V 
in figures 1 and 3, p. 412, is represented by the parallelogram 
BODE in figure 2, and the point A in figure 2 is represented 
by the parallelogram PRQS in figures 1 and 3. 
Peaucellier’s linkage consists of the four equal pieces form- 
ing the jointed rhombus PQRS together with two equal arms 
OS and OR. 
When these arms are longer than the sides of the rhombus 
the linkage is said to be positive; when they are shorter the 
linkage is said to be negative. 
When Peaucellier’s linkage is employed as a machine it is 
acted on by three forces, applied respectively at the fulcrum 0, 
and the two tracing poles @ and 8. 
These three forces, if in equilibrium, must meet in some 
point 7. We may therefore suppose them to be stresses in 
three new pieces OT, QT, ST, which will complete the frame. 
Let us suppose that both O and Tare outside the rhombus, 
and that OS intersects PZ’ in the point V, and let us apply 
Bow’s method to construct the diagram of stress reciprocal to 
this frame. 
If we letter the areas as follows, putting — 
A for the rhombus PRQS, 
B for the triangle PSV, 
C for the triangle OTY, 
D for the quadrilateral ORPV, 
E for the quadrilateral QSVT, 
and F' for the space outside the frame, 
then, in the diagram of stress, the stresses of the four sides of 
the rhombus will meet in A, and since the opposite sides of the 
rhombus are parallel, the lines HA and AD will be in one 
straight line, and the lines BA and AF will also be ina straight 
ieee: ) 
Also since in the frame the pieces OR and OS are equal, 
the angles ORP, PSV are equal, and the corresponding angles 
