ME 486 - Exam #2 - Automation Building BlocksFall, 2006
Name______Solution
(30) 1. True/False (circle the correct answer)
a. Binary/blob analysis requires that a threshold value be set to 128 (for 8 bit CCD’s). T F X
b. An absolute encoder is generally more expensive than a relative encoder. T X F
c. Resolution and sensitivity are used to describe the smallest instrument response. T X F
d. WIP can be considered as inventory in the factory. T X F
e. An inductive sensor can measure proximity of a glass bottle. T F X
f. An inductive AC motor requires brushes to deliver a voltage to the rotor windings. T F X
g. Any vibration frequency in a bowl feeder will move a small part on the feeder ramps. T F X
h. The gripper shown is electrically actuated. T X F
i. Toyota has over 50 manufacturing facilities outside of Japan. T X F
j. Lab 506 uses machine vision to estimate robot inaccuracy. T X F
k. Over 90% of machine vision applications is 1-D or 2-D. T X F
l. Absolute encoders used on a robot require a robot home state. T F X
m. A 13 bit encoder has a resolution of up to 0.0439 degrees. T X F
n. Laser encoders are the more expensive encoders. T X F
o. The picture to the right shows a laser encoder. T F X
(30) 2. The average part produced in a certain batch manufacturing plant must be processed through an average six machines. Twenty (20) new batches of parts are launched each week. Average operation time = 6 min; average setup time = 5 hr; average batch size = 25 parts, and average non-operation time per batch = 10 hr/machine. There are 18 machines in the plant. The plant operates an average of 70 production hours per week. Scrap rate is negligible. Determine (a) manufacturing lead time for an average part, (b) plant capacity, (c) plant utilization, and (d) how you would expect the non-operation time to be affected by the plant utilization?
Solution:
(a) MLT = 6(5 + 25(0.1) + 10) = 105 hr eqn (2.22)
(b) TP = (5 + 25 x 0. 1)/25 = 0.30 hr/pc eqn (2.10)
RP = 3.333 pc/hr eqn (2.11)
PC = 70(18)(3.333)/6 = 700 pc/week eqn (2.17) where n = 18, SH = 70, RP = 3.333, no = 6
(c) Parts launched per week = 20 x 25 = 500 pc/week.
Utilization U = 500/700 = 0.7143 = 71.43% eqn (2.18)
(d) As utilization increases towards 100%, we would expect the non-operation time to increase. When the workload in the shop grows, the shop becomes busier, but it usually takes longer to get the jobs out. As utilization decreases, we would expect the non-operation time to decrease.
(20) 3. Expand equation (5.4) in the text specifically for a 10 bit DAC. If the reference voltage is 20 V, what is the value of the output voltage for the binary value 1000011101? What is the resolution of the DAC in terms of output volts? Prove that your answer is correct by performing an ADC (A/D) to get back the correct binary number.
Solution DAC:
Eo = 20 V {0.5 B1 + 0.25 B2 + 0.125 B3 + 0.0625 B4 + 0.03125 B5 + 0.015625 B6 + 0.0078125 B7
+ 0.00390625 B8 + 0.001953125 B9 + 0.000976563 B10 }
Eo = 20 V {0.5 + 0.015625 + 0.0078125 + 0.00390625 + 0.000976563} =
Eo = 10.56641 V Resolution is 0.01953 V
20 V range / Quantitization / Bit / Value10.56641 / 10 / 1 / 10
0.56641 / 5. / 0
0.56641 / 2.5 / 0
0.56641 / 1.25 / 0
0.56641 / 0.625 / 0
0.25391 / 0.3125 / 1 / 0.3125
0.09766 / 0.15625 / 1 / 0.15625
0.0585975 / 0.078125 / 1 / 0.078125
0.019535 / 0.0390625 / 0
0.00976938 / 0.01953125 / 1 / 0.01953125
Total = 10.56641
ADC:
(20) 4. Given that you have one blob in a binary image as shown in the figure, write down the equations that you would use to determine the X centroid and the Y centroid of the blob. Use Cx and Cy as the pixel to mm conversion factors in the X and Y axes. Cx and Cy have been determined previously from calibration methods. Explain each term in your equations and show how they are used to determine the centroid.
Solution: