Question 16.5: Determine the lengths of all the four links of a four-bar me...

x_{i}=1, x_{f}=11, n=3 .

Using Chebyshev’s precision points,

a=\frac{1}{2}\left(x_{i}+x_{f}\right)=\frac{1}{2}(1+11)=6 .

b=\frac{1}{2}\left(x_{f}-x_{i}\right)=\frac{1}{2}(11-1)=5 .

x_{m}=a+b \cos \left[\frac{(2 m-1) \pi}{2 n}\right], m=1,2,3 .

x_{1}=a+b \cos \left[\frac{(2 \times 1-1) \pi}{6}\right]=a+b \cos \frac{\pi}{6}=6+5 \cos \frac{\pi}{6}=10.33 .

x_{2}=a+b \cos \frac{\pi}{2}=6 .

x_{3}=a+b \cos \left(\frac{5 \pi}{6}\right)=6+5 \cos \left(\frac{5 \pi}{6}\right)=1.67 .

y_{1}=\log x_{1}=\log 10.33=1.014 .

y_{2}=\log x_{2}=\log 6=0.778 .

y_{3}=\log x_{3}=\log 1.67=0.223 .

y_{i}=\log x_{i}=\log 1=0 .

y_{f}=\log x_{f}=\log 11=1.0414 .

Scale factors are:

r_{x}=\frac{\theta_{f}-\theta_{i}}{x_{f}-x_{i}}=\frac{105-45}{11-1}=\frac{60}{10}=6 .

r_{y}=\frac{\phi_{f}-\phi_{i}}{y_{f}-y_{i}}=\frac{225-135}{1.0414-0}=\frac{90}{1.0414}=86.423 .

r_{x}=\frac{\theta-\theta_{i}}{x-x_{i}}, \frac{\theta_{1}-\theta_{i}}{x_{1}-x_{i}}=\frac{\theta_{1}-45^{\circ}}{10.33-1}=6, \theta_{1}=100.98^{\circ} .

\frac{\theta_{2}-45^{\circ}}{6-1}=6, \theta_{2}=75^{\circ}, \frac{\theta_{3}-45^{\circ}}{1.67-1}=6, \theta_{3}=49.02^{\circ} .

r_{y}=\frac{\phi-\phi_{i}}{y-y_{i}}, \frac{\phi_{1}-\phi_{i}}{y_{1}-y_{i}}=\frac{\phi_{1}-135^{\circ}}{1.014-0}=86.423, \phi_{1}=222.63^{\circ} .

\frac{\phi_{2}-135^{\circ}}{0.778-0}=86.423, \phi_{2}=202.24^{\circ} .

\frac{\phi_{3}-135^{\circ}}{0.223-0}=86.423, \phi_{3}=154.27^{\circ} .

d^{2}=(0.221+1.908+1+1.538)=b^{2} .

4.667 d^{2}=b^{2} .

\frac{d}{b}=0.463 .

∴    \frac{d}{a}=2.126, \frac{d}{b}=0.463, \frac{d}{c}=-0.724 .

∴    \frac{d}{a}>\frac{d}{c}>\frac{d}{b} \text { or } a \angle d \angle c \angle b .

∴      Link a is the smallest. Thus a=10 cm, d=21.26 cm.

b=\frac{21.26}{0.463}=45.92 cm , c=\frac{-21.26}{0.724}=-29.36 cm .

– ve sign indicates that the length C is to be drawn in the reverse direction.