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D given:d  12  12 find:a 12 a 23 a 34 input link output link Relative pole method. 1.

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Presentation on theme: "D given:d  12  12 find:a 12 a 23 a 34 input link output link Relative pole method. 1."— Presentation transcript:

1 d given:d  12  12 find:a 12 a 23 a 34 input link output link Relative pole method. 1

2 The pole is a center of rotation of the moving link relative to the fixed link of the mechanism. 2

3 position 1 position 2 3 pole –point about which link a 23 can rotate relative to link a 41 to move between the 2 positions

4 position 1 position 2 4

5 5 4 1 3 2 The fixed pivots (points 4 and 1) can lie anywhere on the perpendicular bisector lines that are shown.

6 6

7 7 Note that relative to the pole, the links a 41 and a 23 are ‘seen under equal angles’. 4 3 2 1

8 8 Note that relative to the pole, the links a 12 and a 34 are ‘seen under equal angles’. 4 3 2 1

9 If the motion of the link is considered relative to another moving link, the pole is known as a relative pole. The relative pole can be found by fixing the link of reference and observing the other link in the reverse direction. 9

10 position 1 position 2 Relative pole – link a 12 rotates about this point relative to link a 34 10  12  12

11 11 4 1 3 2 To move link a 12 from position 1 to position 2, point 3 at position 1 can be anywhere on this line. position 1 position 2

12 12 Now repeat the problem where you consider link a 12 to be fixed and consider how link a 34 moves relative to it. The relative pole will not change.  12  12 position 1 position 2

13 13

14 14 4 1 3 2 Point 2 (a “fixed pivot” from the vantage point of link a 12 ) at position 1 can lie anywhere on this line. position 1 position 2

15 15 So, in summary.  12  12 Can use these 2 lines to rapidly find the relative pole.

16 16 Point 3 at position 1 can lie anywhere on this line.  12  12 Point 2 at position 1 can lie anywhere on this line. 4 1 3 2 But where are these lines?

17 17 But where are these lines? Given d,  12, and  12, we can easily find the relative pole, but I don’t see the blue lines.  12 = 50  12 = 30  d = 10” 41

18 18 Remember, that relative to the pole, opposite links are ‘seen under equal angles’. 1.Pick an arbitrary line through the relative pole on which point 3 at position 1 can lie. We will call it line 3. 2.The line on which point 2 at position 1 can lie will be called line 2. The angle between line 1 and line 2 seen with respect to the relative pole must be the same as the angle between the lines through points 4 and 1.  12 = 50  12 = 30  d = 10” 41

19 19 Remember, that relative to the pole, opposite links are ‘seen under equal angles’. 1.Pick an arbitrary line through the relative pole on which point 3 at position 1 can lie. We will call it line 3. 2.The line on which point 2 at position 1 can lie will be called line 2. The angle between line 3 and line 2 as seen with respect to the relative pole must be the same as the angle between the lines through points 4 and 1.  12 = 50  12 = 30  d = 10” 41 line 3 line 2

20 20 Point 3 at position 1 can lie anywhere on line 3. Point 2 at position 1 can lie anywhere on line 2. There are  3 4-bar mechanisms that can perform the desired motion. free choice for angle of line 3 relative to a 41 free choice for point 3 at position 1 along line 3 free choice for point 2 at position 1 along line 2  12 = 50  12 = 30  d = 10” 41 line 3 line 2

21 21 So, given d,  12, and  12, find the relative pole between links a 34 and a 12.

22 22

23 23  12 =50 deg CW  12 = 30 deg CW d=10

24 24  12 =30 deg CW  12 = 50 deg CW d=10

25 25  12 =-50 deg CW  12 = 30 deg CW d=10


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