Question
kinematic design of machinery

Problem #4. Considering the drag link quick-return mechanism shown below, show the linkage in its limiting positions (corresp
0 0
Add a comment Improve this question Transcribed image text
Answer #1

B 12.50 A 04 02 118.25 C D A C 04 02 R 24.50 8.80이\theta_1 = 118.25

U=8.8

left-right 01

left-right 118.25 8.8 127.05

\theta_{right-left} = 360^0-(\theta_1+\theta_2)

\theta_{right-left} = 360 - 127.05^0 = 232.95^0

Time\;Ratio:TR=\frac{232.95}{127.05}

TR=1.8335

Stroke:

At right most position O_4D = 24.5''

At left most position O_4D = 12.5''

Stroke = 24.5-12.5 = 12''

Average Piston velocity:

w = 2*\pi*100/60 = 10.47rad/s

Vavg,le ft-right Stroke

v_{avg,left-right} = 12*\frac{10.47}{127.05*(\pi/180)}

v_{avg,left-right} = 56.66in/s

v_{avg,right-left} = Stroke*\frac{w}{\theta}

v_{avg,right-left} = 12*\frac{10.47}{232.95*(\pi/180)}

Vavg.right-left 30.902in/s​​​​​​​

Add a comment
Know the answer?
Add Answer to:
kinematic design of machinery Problem #4. Considering the drag link quick-return mechanism shown below, show the li...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT