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346 CAM DESIGN HANDBOOK

P
h = out . (11.68)
P
in
To make physical sense, the value of h must lie between zero (completely inefficient) and
one (perfectly efficient). The power lost due to dissipation is simply the difference between
the input and output power or
P = P - P = (1 h P ) . (11.69)
-
eff in out in
The power dissipated depends upon the amount of power being transmitted through the
subsystem and so, like Coulomb friction, depends upon both the applied load and the speed
of the system. Gearboxes, lead screws, and other transmission devices are often success-
fully modeled using an overall efficiency to account for system damping. Mechanical effi-
ciency is somewhat different than the viscous or dry friction models presented above since
its inclusion alters the effective mass, stiffness, and damping of the system. Qualitatively,
the loss of efficiency associated with an inefficient transmission makes everything down-
stream of the transmission “feel” heavier, stiffer, and more damped. This effect is illus-
trated in the example in the following section.

11.5.4 Combinations and Equivalent Dampers

As mentioned before, one of the most common ways of adding damping to a system is to
choose an appropriate form for the damping and determine the parameters experimentally.
If, however, damping values are known for individual dampers in the system, these ele-
ments may be combined according to the same rules as springs. If two dampers have
damping coefficients of b 1 and b 2, a single equivalent damper has a damping coefficient of
b = b + b (11.70)
eq 1 2
if the dampers are combined in parallel and

1 1 1
= + (11.71)
b b b
eq 1 2
if the dampers are combined in series. The same relations hold for rotary dampers com-
bined in series or parallel. Furthermore, if dampers are integrated into a system in a way
that incorporates mechanical advantage, this must also be taken into account to determine
an equivalent damping coefficient.

EXAMPLE Consider a modification to the gear example in Sec. 11.3 with viscous damping
added to both the input and output shafts and an overall mechanical efficiency assigned
to the gear pair (Fig. 11.21a). Assume for simplicity that the inertia of the gears are small
enough relative to the inertia of the load that they may be ignored and the goal is to reduce
this system to the equivalent system in Fig. 11.21b. The power into the gearbox is
P = ( T - b w )w
in in rin in in
while the power out of the gearbox is
P = ( b w + J w ˙ )w .
out rout out load out out
Using the definition of efficiency from Eq. (11.67), the input and output power expres-
sions give