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-composite filament layup

-spot welding

-arc welding

-material/work manipulation

-collision detection

-deburring

-inspection

-kinematic and dynamic simulation

-controller simulation

•The simulators available for the robots in the lab allow off-line programming and simulations.

39.6 PRACTICE PROBLEMS

1. a) What are some basic functions expected on a robot teach pendant

b) Describe how a computer can help avoid debug robot programs without a robot being used

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.

c)Write program for the Mitsubishi RV-M1.

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

3. We plan to use a pneumatic gripper to pick up a 4 by 8 sheet of glass weighing 40 lbs. Suggest a gripper layout and dimensions of the cups. State any assumptions.

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ans.	For stability we want to set up an array of cups. A set of 3 or 4 would be reason-
	able to help support the sheet. - I will pick 4. Now, the diameter of the cup should
	be determined. We will assume that the vacuum pressure will be 5 psi below
	atmosphere, and we will use a factor of safety of 2.
	FS( L)			= Nπ r2P
	2( 40lb)			= 4π r25psi
	4	= r	2	r > 1.128in
	--
	π

4.A vacuum pump to be used in a robot vacuum gripper application is capable of drawing a negative pressure of 4.0 psi compared to atmospheric. The gripper is to be used for lifting stainless steel plates, each plate having dimensions of 15” by 35”, and weighing 52 lbs. Determine the diameter of the suction cups to be used for the gripper if it is decided to use two cups for greater stability. A factor of safety of 1.5 should be used in the computations.

5.Consider the following gripper design problems.

a)We plan to use a friction gripper to pick up a 50 lb iron plate. Suggest a gripper design and specify the force required.

b)Design an end effector, and describe the path planning approach for a robot unloading satellites from the space shuttle.

6.What is the workspace for each of the robots below, and can the robots reach all positions and orientations in the workspace?

y

y

y

	x
				x					x
y				y

x

x

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7.Suggest a type of robot suitable for the following tasks. Briefly explain your suggestion. a) placing pallets on rack shelving

ans. cartesian - well suited to cartesian layout of shelves. b) electronics assembly

ans. scara - will work on a flat table well.

c) loading and unloading parts from an NC mill

ans. articulated - can easily move around obstructions.

8.Suggest a type of robot suitable for the following tasks. Briefly explain your suggestion. a) a gas pump robot for placing the gas nozzle into the fuel tank.

b) for drilling holes in a printed circuit board. c) to vacuum a hotel.

9.Why are 5 axis enough for some robotic applications (eg. welding) and all NC milling operations?

10.You have been asked to write a program for a robot (you can choose either the Seiko RT-3000 or Mitsubishi RV-M1). 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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	ans.	10	T1	= 300. 300. 20. 0.	10	PD 1, 300, 300, 20, 0, 0
		20	T2	= -300. 300. 0. 0.	20	PD 2, -300, 300, 0, 0, 0
		30	T3	= 300. 0. 0. 0.	30	PD 30, 300, 0, 0, 0, 0
		40	T4	= 350. 0. 0. 0.	40	PD 32, 350, 0, 0, 0, 0
		50	T5	= 300. 60. 0. 0.	50	PD 31, 300, 60, 0, 0, 0
		60	R = 6		60	PA 3, 7, 6
		70 C = 7			70 GO
		80 OUTPUT +OG3			80 SC 31, 0
		90 DEF PA2(R, C) T3 T4 T5			90 RC 7
		100 FOR I = 0 TO R-1			100 SC 32, 0
		110 FOR J = 0 TO C-1			110 RC 6
		120 MOVE T1			120 PT 3
		130 OUTPUT -OG3 200			130 MO 1
		140 MOVE T2			140 CG
		150 MOVE PA2(J, I)			150 MO 2
		160 OUTPUT +OG3 200			160 MO 3
		170 NEXT J			170 OG
		180 NEXT I			180 IC 32
		190 STOP			190 NX
					200 IC 31
					210 NX

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.

		( 0, 650, 0)	y
	A
		photo	x
	127/2mm	sensor
			z
D
( –650, 0, 0)
			( 650, 0, 0)
B	C	A = (0, 650-101.6/2, 0) = (0, 599.2, 0)
		B = (-400, -1.5*127, 0) = (-400, -190.5, 0)
		C = (-400 + 101.6, -1.5*127, 0) = (-298.4, -190.5, 0)
		D = (-400, 1.5*127, 0) = (-400, 190.5, 0)

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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

1lb3		lb
	= 3------Amin
		in2
Amin	= 1in2
		dmin	2
		---------
Amin ≤ π		2	2
2		dmin
1in		---------
	≤ π	2

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)

	Start
	reset pallet values
no	start
	button pushed?
	yes
	pick up small box
	index pallet
	move above box
		no
	no	yes
	is box full?	stop pushed?
	yes
	turn off conveyor
	turn on light

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.

	( 0, 500, 0)
	A	y
	photo	x
	sensor
D	127/2mm
( –500, 0, 0)
B	C
	A = (0, 500-101.6/2, 0) = (0, 449.2, 0)
	B = (-350, -1.5*127, 0) = (-350, -190.5, 0)
	C = (-350 + 101.6, -1.5*127, 0) = (-248.4, -190.5, 0)
	D = (-350, 1.5*127, 0) = (-350, 190.5, 0)

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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

1lb3		lb
	= 3------Amin
		in2
Amin	= 1in2
		dmin	2
		---------
Amin ≤ π		2	2
2		dmin
1in		---------
	≤ π	2

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)

	Start
	reset pallet values
no	start
	button pushed?
	yes
	pick up small box
	index pallet
	move above box
		no
	no	yes
	is box full?	stop pushed?
	yes
	turn off conveyor
	turn on light

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

13. Given the scenario below, find the minimum angular resolution of the rotating sensor.

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-the robot has +/- 0.5” accuracy

-the pallet can slide +/- 0.1” on the belt

4.8”	belt travels

5”

-the driving motor is continuous, and can be run to any angle

-the rotating sensor is an incremental encoder, every rotation of some small angle it issues a pulse. But, because of the construction of the device, it has a minimum resolution for angular measurements

-the robot must be able to touch the part to pick it up

-the tool on the end of the robot is a 1” magnet, and it must be able to touch the part completely to pick it up.

-pulley size is 10” dia.

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 mass of 1.5 kg with its center of gravity at 3cm from the TCP and transfers a mass of 4kg with its C.G. at 5cm from the TCP.

a) determine the inertia torque about the theta1 axis showing all correct units

b) compare the value in a) with a maximum inertia torque estimated from decelerating a 6kg mass from 1100mm/s to zero in 0.5 sec.

c) Estimate the combined error at the CG of the load due to theta1 and theta 2 resolution

15.Consider a double jointed manipulator as shown below. It is subjected to a loading at the tip of 8 lbs, and works in a heated environment (i.e. T0(room temp.) = 60°F and T1 (working temp.)

=80°F.

a)Determine the elongation of the manipulator.

b)Determine the total linear deflection of the manipulator.

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c) Determine the total final accuracy of the manipulator of the tip of the manipulator.

50”

10”

cross section is 1” wide by 2” high solid square aluminum stock

16. For the robot pictured below, assume the that a maximum payload of 10kg is specified. The joints are controlled by stepper motors with 200 steps per revolution. Each of the joints slides, and the gearing is such that 1 revolution of the stepper motor will result in 1” of travel. What is the accuracy of the robot?

Assume the joints are solid, and

maximum 10”

to robot links are made from 1”

solid aluminum stock.

maximum 15”

17. Consider a double jointed manipulator as shown below. It is subjected to a loading at the tip of 8 lbs, and works in a heated environment (i.e. T0(room temp.) = 60°F and T1 (working temp.)

=80°F.

a)Determine the elongation of the manipulator.

b)Determine the total linear deflection of the manipulator.

c)Determine the total final accuracy of the manipulator of the tip of the manipulator.

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50”

10”

cross section is 1” wide by 2” high solid square aluminum stock