10.3 ASEA IRB-1000

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10.4 UNIMATION PUMA (360, 550, 560 SERIES)

•In general,

-an articulated arm with 3 dof for positioning, and 3 dof for orientation

-left/right arm configurations are possible

-uses DC servo motors for drive

-uses 110-130 VAC, 50-60Hz, 1.5KW

-weight 120 lb

-repeatability 0.004in

-RS-232C port for dumb terminal

-32 parallel I/O lines

-memory 16K

-programming language is VAL

•joint 1 (Waist)

10.5 PRACTICE PROBLEMS

2. Write a short program to direct a robot to pick up and put down a block. Assume the points

have already been programmed with the teach pendants.

a)Write program for the IBM 7535.

b)Write program for the Seiko RT-3000.

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ans. a) NEWPROG:BLOCK; RELEASE; -- open the gripper

DELAY(5); -- delay 1/2 second to allow the gripper to open PMOVE(OVER); -- move to the point over the pickup point called ‘OVER’ DOWN; -- move the arm down

DELAY(2); -- wait for the motion to complete and settle GRASP; -- close the gripper

DELAY(2); -- wait for the gripper to close UP; -- raise the block

DELAY(20); -- wait for a couple of seconds

DOWN; -- drop the block back to the surface of the table OPEN; -- open the gripper

UP; move the arm away from the block END; - terminate the program

10. You have been asked to write a program for a Seiko RT-3000. The program is to pick up

a part at point T1, move to point T2, and then load the part into a pallet. The robot should then

return to point A to pick up then next part. This should continue until the pallet is full.

T1 = (300, 300, 20)

T2 = (-300, 300, 0)

Pallet has 6 rows and 7 columns

Pallet origin T3 = (300, 0, 0)

Pallet end of row T4 = (350, 0, 0)

Pallet end of column T5 = (300, 60, 0)

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10 T1 = 300. 300. 20. 0. ans. 20 T2 = -300. 300. 0. 0.

30 T3 = 300. 0. 0. 0.

40 T4 = 350. 0. 0. 0.

50 T5 = 300. 60. 0. 0.

60 R = 6

70 C = 7

80 OUTPUT +OG3

90 DEF PA2(R, C) T3 T4 T5

100 FOR I = 0 TO R-1

110 FOR J = 0 TO C-1

120 MOVE T1

130 OUTPUT -OG3 200

140 MOVE T2

150 MOVE PA2(J, I)

160 OUTPUT +OG3 200

170 NEXT J

180 NEXT I

190 STOP

11. An IBM 7535 industrial robot is to be used to unload small 1 lb. cardboard boxes (5” by

4” by 1”) from a conveyor, and stack them in a large cardboard box (20” by 8” and 2” deep). After

the large box is loaded, it will be removed automatically and replaced with an empty one. The

conveyor will be controlled by a robot output, and it will be stopped when an optical sensor

detects a small box. When the box is full the conveyor will be stopped and a light turned on until

an unload button is pushed. The entire system uses a start and stop button combination. The stop

button is not an e-stop, but it will stop the cycle after the small box is placed in the large box.

a)Layout the position of the conveyor, sensor, large box and robot so that all positions can be reached. Indicate critical points of objects.

b)Design a robot gripper to pick up the boxes.

c)Develop a flow chart for the robot operations.

d)Write an AML program for the flowchart.

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ans. a)

First, we need to convert the given dimensions to mm. small boxes = 127x101.6x25.4mm

large boxes = 508x203.2x76.2mm

Next, we need to overlay these on the robot workspace. In this case there is abundant space and can be done by inspection.

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ans. b)

For this application, vacuum grippers should work effectively because the mass is light, and the boxes should have clean cardboard faces. Because the application has been designed to lift the boxes in the centers, we should be able to use a single suction cup, but a large factor of safety will be used to compensate (>= 3). We will assume that we are using a venturi valve to generate the suction, so a pressure differential of 3psi is reasonable.

( W) FS = PAmin

dmin = 1.13in

Based on this calculation I would select a suction cup that is 1.25” or 1.5” dia.

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ans. c)

no

reset button?

yes

12. Repeat the previous problem for the Seiko RT-3000 robot.

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ans. a)

First, we need to convert the given dimensions to mm. small boxes = 127x101.6x25.4mm

large boxes = 508x203.2x76.2mm

Next, we need to overlay these on the robot workspace. In this case there is abundant space and can be done by inspection.

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ans. b)

For this application, vacuum grippers should work effectively because the mass is light, and the boxes should have clean cardboard faces. Because the application has been designed to lift the boxes in the centers, we should be able to use a single suction cup, but a large factor of safety will be used to compensate (>= 3). We will assume that we are using a venturi valve to generate the suction, so a pressure differential of 3psi is reasonable.

( W) FS = PAmin

dmin = 1.13in

Based on this calculation I would select a suction cup that is 1.25” or 1.5” dia.

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ans. c)

no

reset button?

yes

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ans.

10 R = 3: C = 4: H = 0 ‘ define rows and column variables 20 SPEED 100 ‘ set the robot speed

30 T1 = 0. 449.2 0. 0. ‘ set point A

40 T2 = -350. 449.2 -190.5 0. ‘ set point B 50 T3 = -248.4 449.2 -190.5 0. ‘ set point C 60 T4 = -350. 449.2 190.5 0. ‘ set point D

70 T5 = 0. 0. -50. 0. ‘ a displacement to the conveyor height

80 T6 = 0. 0. -100.4 0. ‘ a displacement to the bottom layer of the large box 90 T7 = 0. 0. -75. 0. ‘ a displacement to the top layer of the large box

100 DEF PA2(4,2) T1 T2 T3 ‘ define pallet

110 WAIT +IE1 ‘ wait for external input #1 to go on, this is the start button 120 FOR H = 0 TO 1 ‘ set box layers

130 FOR I = 0 TO R-1 ‘ loop for rows 140 FOR J = 0 TO C-1 ‘ loop for columns

150 OUTPUT +OE1 ‘ turn on external output #1, this is the conveyor 160 MOVE T1 ‘ move to the conveyor pickup point

170 WAIT +IE2 ‘ wait for the input from the optical sensor to go on 180 OUTPUT -OE1 ‘ turn off the conveyor

190 MOVE T1 + T5 ‘ move to pick up box

200 OUTPUT +OG1 ‘ turn on suction cup on gripper 210 MOVE T1 ‘ pick up the box

220 MOVE PA2(I, J) ‘ move to the pallet position in the large box 230 IF H = 1 THEN GOTO 260 ‘ jump if on the top layer

240 MOVE PA2(I, J) + T6 ‘ move to the bottom layer of the box 250 GOTO 270

260 MOVE PA2(I, J) + T7 ‘ move to the bottom layer of the box 270 OUTPUT -OG1 ‘ turn off the suction cup

280 MOVE PA2(I, J) ‘ move out of box

290 IF NOT IE3 THEN GOTO 310

300 WAIT +IE1 ‘ wait for the start button

310 NEXT J: NEXT I: NEXT H ‘ end of the loops 320 OUTPUT +OE2 ‘ turn on box full light

330 WAIT +IE4 ‘ wait for the reset button

340 GOTO 110 ‘ go back to start anew

14. The IBM 7535 robot arm moves its TCP to point (-450, 250)mm at speeds programmed

by ‘payload(5)’ and decelerates from the resultant speed to zero in 0.5 seconds. The tool has a