- quently found in steam 
ad 
PISTON VELOCITY AND ACCELERATION DIAGRAMS 301 
The diagram to the right in Fig. 467 shows the velocity of the 
piston, during one stroke, plotted on a crank angle, or time base. 
If the connecting-rod is of infinite length, AB becomes parallel to the 
line of stroke, BD is perpendicular to CD, and v is equal to V sin @. If V 
is constant the piston has simple harmonic motion, and the velocity-space 
and the velocity-time diagrams for the piston are the same as those shown 
in Fig. 461, p. 297. The polar velocity diagram becomes a circle 
described on the crank as diameter when the latter is perpendicular to the 
line of stroke. 
The arrangement shown in Fig. 468 is the equivalent of an infinite 
connecting-rod. This 
arrangement is _ fre- 
pumps, one rod M being ; 
the steam piston-rod, .@Y) ir 
and the other N the —- 
pump plunger. The : 
crank in this case must 
be an overhung one, or 
the crank shaft must Fia. 468. Fre, 469. 
be divided to allow the slotted piece KL to pass. If the slotted piece 
have two slots at right angles, as shown in Fig. 469, the crank may be 
placed anywhere on the shaft without altering the shaft in any way. 
261. Piston or Slider Acceleration.—Since CD (Figs. 467 and 
470) represents the piston velocity if CB represents the crank pin 
velocity, it follows that since acceleration is rate of increase of velocity, 
the velocity of the point D along CD will be the rate of increase of CD, 
and will therefore be the piston acceleration. 
Consider the point D as a point in the connecting-rod produced, then 
D must be moving at the 
instant in a direction per- © er E 
pendicular to OD with a 
OD 
OB 7 ' 
This velocity may be found z oy 
by construction as follows. “Ss wag 
On OD make OB’=OB. a 
Draw B’C’ perpendicular to Z 
OD and equalto BC. Join A | C 
OC’ and produce it to meet SY 
DE, a perpendicular to OD, N 
at E, then DE=V «OD. 
If DE be resolved into components DF along AB, and DH along 
CD, then DH is the velocity of D along CD, and therefore represents 
the piston acceleration. Draw CK parallel to OD to meet AB at K, and 
draw KL perpendicular to AB to meet AC at L, then it will be shown 
that CL=DH. The triangles CKB and OBD are similar, and 
CK_OD_OD _ DE _DE 
BC OB OB’ BC BC’ 
velocity equal to’ V - 
Fia. 470. 
