In jewel-bearing movements energy losses due to
friction are influenced by the position of the pivot axis, the
load on the pivot, maximum contact pressure developed and
minimum working torque available. These losses are at a
minimum when the spindle axis is vertical, and in this situ-
ation both pivot and jewel radii are significant factors. In the
case of a movement with the spindle axis horizontal, jewel
radius has no bearing on performance (other than establish-
ing the maximum working pressure between pivot and
jewel).
For conically-ended vertical spindles, frictional errors
related to minimum available torque and move-ment weight
may be determined from Fig. 1. Since losses are a minimum
when working pressure between pivot and jewel is a maxi-
mum, these data are based upon an assumed maximum
value for steel pivots of 500,000 psi.
Graphical solutions are also based on a coefficient of fric-
tion of 0.1. Practical coefficients for steel pivots and sap-
phire jewels may be somewhat higher, such as 0.15 for good
average practice, or as high as 0.2 in some cases. Graph
solutions readily may be corrected to any practical coeffi-
cient by multiplying by l0µ, where µ is the practical coeffi-
cient of friction in question.
Example 1 (Fig.1)
To find the frictional error with a movement weighing 10
grams and having a minimum operating torque of 150 dyne-
cms —
Trace the torque ordinate to intersect the 10 gram diagonal
and read off a frictional error of 0.14 percent.
Example 2 (Fig.1)
To find the minimum operating torque required to operate a
4 gram movement with a maximum error of 0.025 percent:
Trace along the 0.025 abscissa to its intersection with the 4
gram diagonal. This point corresponds to a 210 dyne-cm
operating torque.
Example 3 (Fig.1)
To find the frictional error as in Example 1, but with a coef-
ficient of friction of 0.18:
Factor the original answer by 10 x 0.18
or , 0.14 x 1.8 = 0.252 percent.
The corresponding graph for horizontal spindles (Fig. 2)
is calculated on the basis of I gram of weight with diagonals
corresponding to pivot radius in thousandths of an inch; or,
alternatively, for a pivot radius of 1 / 1000 inches with diag-
onals relating to movement weight in grams. In either case
the answers may be factored to accommodate either a
movement weight or pivot radius greater than unity.
Example 1 (Fig.2)
To find the frictional error with 1,000 dyne-cms minimum
operating torque per gram weight of movement with a pivot
radius of 0.005 inches: Find intersection point between the
0.005 diagonal and 1000 ordinate to obtain frictional error
of 0.125 percent. If the movement in question weighs N
grams, the frictional error in this case would be N x 0.125
percent.
Example 2 (Fig.2)
Find the pivot radius required with a 20 gram move-ment
and 2,000 dyne-cms minimum operating torque to hold a
maximum instrument error of 0.5 percent. Frictional coeffi-
cient is 0.15.
Reading from the graph, a 20 gram movement with a 0.00 1
inch pivot radius gives 0.25 percent error at 2,000 dyne-
cms, for a friction coefficient equal to 0.1. Thus error with
µ = 0.15, is 1.5 x 0.25, or 0.375 percent.
This must be multiplied by 1.8 to equal 0.5 percent. Hence,
required pivot radius is 0.0018 inches maximum.
It is necessary, however, to check the choice of pivot radius
against jewel radius (the latter normally produced to stan-
dard dimensions) to establish that the working pressure
between pivot and jewel does not exceed the maximum safe
working level. (Fig. 3). These data are related by the
function
where r
1
is the radius of the pivot in thousandths of an inch
and r
2
the radius of the jewel in thousandths of an inch.
Example 1 (Fig. 3)
Knowing the value of r1 or r2, establish the required rela-
tionship between them for an 8 gram load:
In the case of a vertically mounted pivot, establishing this
relationship to give a maximum safe value of working pres-
sure gives minimum frictional loss. For a horizontally
mounted pivot, performance of the movement is not affect-
ed and it is only necessary to establish that the relationship
falls on or below the maximum working level for the pivot
material used. From the graph, maximum working pressure
is achieved for an 8 gram load when 8 r1 = 0.58.
r
=
(
r
1
2
r
)
r
1
2
r
1
1
1
r
DESIGN DATA
Calculating Frictional Losses in Jewel Bearing Movements
R. H. Warring,
Design Consultant, London, England
Reprinted from DESIGN NEWS.