Question 6 of 6: Geneva Mechanism & Dial-Indexing Production Line
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2016 — 04-Bio-B7, 3 hours, closed book (one aid sheet allowed,
written on both sides; approved Casio/Sharp calculator only). Six questions are printed; the
first five as they appear in the answer book constitute a complete exam paper (all six are
answered here as a complete study resource). Each question is of equal value; some require an
essay-format answer.
This solution follows the paper's true subject and cites robotics/manufacturing references
accordingly.
Reference texts: M. P. Groover, R. Weiss, R. N. Nagel & N. G. Odrey,
Industrial Robotics: Technology, Programming, and Applications (2nd ed. — robot
configurations, end-effectors/grippers, machine vision, sensors and transducers); M. P. Groover,
Automation, Production Systems, and Computer-Integrated Manufacturing (5th ed. —
Geneva mechanisms/dial indexing, PLC ladder logic, production-rate and line-efficiency
analysis).
Question 6: Geneva Mechanism & Dial-Indexing Production Line (20 marks)
Check — stated assumptions (per the exam's own Note 1):
(1) the dwell portion of the cycle is set equal to the slowest station's required processing time
$T_p=4.3\ \text{s}$ — the minimum synchronized cycle at which no station is ever left
waiting; (2) “rotational speed of the driven member” follows Groover's usage for this problem
type, in which the driven member is the continuously turning Geneva drive (crank) that the
motor turns once per cycle, $N=1/T_c$. The star wheel's own average speed ($N/n$) is reported
alongside it for completeness. (3) Line efficiency and the proportion of downtime use Groover's
definitions, $E=T_c/(T_c+FD)$ and $FD/(T_c+FD)$. The ideal cycle $T_c$ includes the indexing time,
because the machine is up (not broken down) while it indexes.
Given.
Quantity
Symbol
Value
Number of stations (slots)
$n$
8
Slowest station processing time
$T_p$
4.3 s
Frequency of line stops
$F$
0.075 stops/cycle
Average downtime per stop
$D$
5 min = 300 s
Find. (a) indexing (handling) time $T_h$; (b) rotational speed of the driven
member (Geneva drive); (c) production rate $R_p$; (d) line efficiency $E$; (e) proportion of downtime.
Fig. 6 — continuously-rotating driver pin engages the 8-slot star
wheel; smooth entry/exit fixes the driver's indexing rotation to $180^{\circ}-360^{\circ}/n$.
Approach. Use the standard Geneva-wheel geometry to split the driver's
rotation into an indexing (motion) angle and a dwell angle, set the dwell time equal to the
station bottleneck $T_p$ to get the ideal cycle time, then layer in stoppages using the standard
production-line relations $T_c'=T_c+FD$, $R_p=1/T_c'$, $E=T_c/T_c'$.
Geneva indexing/dwell angles. Smooth (impact-free) pin entry and exit requires
the crank-to-pin line to be perpendicular to the slot at engagement, which fixes the driver's
rotation angle during indexing to $$\theta_{\text{index}}=180^{\circ}-\frac{360^{\circ}}{n}=180-\frac{360}{8}=135^{\circ},$$
leaving a dwell angle of $360-135=225^{\circ}$ (fraction of cycle dwelling
$=225/360=0.625$).
Part (a) — ideal cycle and indexing time. The dwell portion must equal
the bottleneck station time: $$T_c=\frac{T_p}{0.625}=\frac{4.3}{0.625}=6.88\ \text{s},\qquad T_h=T_c-T_p=6.88-4.3=\boxed{2.58\ \text{s}}.$$
Part (b) — rotational speed. The Geneva drive (the member the motor turns)
makes one revolution per cycle, and its dwell arc must last the 4.3 s station time:
$$N=\frac{\theta_{\text{dwell}}}{360\,T_p}=\frac{225}{360(4.3)}\ \text{rev/s}=\frac{60}{6.88}\ \text{rpm}=\boxed{8.72\ \text{rpm}}.$$
The star wheel indexes only $1/8$ revolution per cycle, so its own average speed is
$N/n=8.72/8=1.09$ rpm (for reference).
Part (c) — average cycle time with downtime. $$T_c'=T_p+T_h+FD=4.3+2.58+0.075(300)=4.3+2.58+22.5=29.38\ \text{s}.$$ Production rate: $$R_p=\frac{3600}{T_c'}=\frac{3600}{29.38}=\boxed{122.5\ \text{pieces/hr}}.$$
Part (d) — line efficiency. Uptime is the whole ideal cycle $T_c$ (processing plus indexing), so
$$E=\frac{T_c}{T_c+FD}=\frac{6.88}{29.38}=\boxed{0.2342\ (23.4\%)}.$$
Part (e) — proportion of downtime. $$\frac{FD}{T_c'}=\frac{22.5}{29.38}=\boxed{0.7658\ (76.6\%)}.$$ Check: $E+D=0.2342+0.7658=1.000$, so uptime and downtime together make up the whole average cycle.