Graphical Linkage Synthesis Lecture 5
Time ratio The timing of the crank-rocker can be assessed using the two stationary positions. Time ratio = OR Time ratio = Time it takes to complete forward motion/ Time it takes to complete return motion Crank angle to complete forward motion/ Crank angle to complete return motion
Time ratio and return mechanisms Even return crank-rocker For a given constant input speed, It takes the same time to complete the forward and the return motion, therefore: Time ratio =1 The two stationary positions of the crank line up with each other, and the angles swept between these stationary positions are equal.
Time ratio and the return mechanisms Quick-return crank-rocker For a given constant input speed, the Time it takes to complete the forward motion is more than the time it takes to complete the return motion, therefore: Time ratio >1 The two stationary positions of the crank do not line up with each other, and the angles swept between the stationary positions are not equal.
Time ratio and geometry of QRM Quick-return crank-rocker α + β =360 Time ratio: T R = α/β δ = 180 - α = β - 180 Or the time ratio is: T R = (180 + δ)/(180 - δ) These angular values and relationships will be used to synthesize the linkage.
Transmission Angle Transmission angle, μ: Angle between the coupler link and the output link. μ =θ 4 θ 3 As the input link 2 rotates, all of the angles change. Therefore you can plot μ as a function of input angle θ 2.
Min. and Max.Transmission Angle Max. transmission angle µ Min. transmission angle Minimum and maximum transmission angle occur when link 2 (crank) becomes collinear with link 1 (ground link)
Transmission Angle and Efficiency Transmission angle of a crank-rocker gives an indication of the efficiency of power transmission from the input to the output link. At μ = 90 o all of the force from the coupler link is being transmitted to the output link to produce the maximum output torque. μ = 90 o gives maximum efficiency (torque transmission). In practice, μ = [40,140 ] is preferred.
Design of a CR Quick-return Mechanism Design a Grashof four-bar, crank-rocker, quickreturn mechanism with the following requirements: The mechanism will have: 1) a time ratio of 1.35. 2) an output rocker angle of 50. while maintaining a transmission angle between 40 and 140, and, 3) must fit and operate within a design space of 8 x 11 inches.
Design Solution Step 1: Draw the output link in both extreme positions (pivot O4, end positions B1 and B2). Step 2: Draw a construction line through B1 at any convenient angle. Note: The orientation of the line drawn in Step 2 is arbitrary and an iterative trial and error process will have to be employed for making design corrections, if necessary.
Design Solution Step 3: Calculate δ using the given time ratio equation. Step 4: Draw a construction line through B 2 at an angle of δ. Step 5: Label the intersection point O 2.
Design Solution Step 6: The line O 2 O 4 now defines the ground link. Step 7: Locate point C on the extension of O 2 B 2 so that O 2 C = O 2 B 1
Design Solution Step 8: Determine the length of the crank by Measuring length B 2 C, and solving for Link 2 = r 2 = (B 2 C) /2 Step 9: Calculate the lengths of the coupler: length of link 3, r 3 = O 2 B 1 r 2
Design Solution Step 10: Check if Grashof s condition is satisfied. If non-grashof, repeat steps 3 to 8 with O 2 further from O 4. Step 11: Check the transmission angles. Note: to promote smooth running and good force transmission, μ = [40,140] degrees
Design Conclusion and Specifications Step 12: Specify the mechanism: - List the link lengths. Label the links. - Draw the mechanism in at least one position. Ground link, R1 = Input link, R2 = Coupler link, R3 = Output link, R4 =