Synthesis of mechanismsFreudenstein’s equation:This equation is used to compute the length of links of a four bar mechanism.The equation can be written as

where a,b,c,d are lengths of links of the four bar mechanism with θ2 and θ4 as the input and output link angles.

The equation (i) can be written as for three pairs of values (θ21, θ41), (θ22, θ42 ) and (θ23, θ43) as:

Subtracting equation (iii) and (iv) from (ii), we get

Solving these two equations for K1 and K2, we get

Where Substituting the values of K1 and K2 in equation (ii), we get the value of K3.

Example: A four bar mechanism is required such that the input and out angles are coordinated as given in the table.Synthesize the 4-bar mechanism:Input crank angle (θ2) 300 500 800

Output follower angle (θ4) 00 300 600

Solution:The displacement equation given by Freudenstein is written as:

Substituting the values of input (θ2) and (θ4) in above equation, we get

Solving equation (i), (ii) and (iii), K1=1.8321, K2=1.4294, K3=0.2718Assuming the length of one link as unity i.e. a = 1We know that

b= 2.3039The length of various links are: a = 1.0b= 2.3039c= 1.2817d = 1.8321

Function generation:In function generation, the motion of input link is correlated to motion of output link.

Figure shows a four bar mechanism The input link AB rotates through an angle θ2 and represents independent variable x. The output link CD rotates through angle θ4 and indicates independent variable y.

Here, Y is dependent on x.

Let us assume X1 and X2 are initial and final positions of independent variable X. Y1 and Y2 are initial and final positions of dependent variable Y. There is linear relationship between θ2 and X There is linear relationship between θ4 and Y

Mathematically,

and

where θ2f and θ2i = final and initial angular positions of input link θ4f and θ4i = final and initial angular positions of output link The constants r1 and r2 are called scale factors. If link ratios are known i.e. a/b, a/c, a/d. , the mechanism can be designed.

Precision points and structural errorLet us draw a graph between input variables and output variables as shown in figure:

Figure 4.11

The motion of input is represented by X0, X0, X0 ……..Xn, Xn+1

The motion of input is represented by Y0, Y0, Y0 ……..Yn, Yn+1

Desired function and generated function meet at certain points. These are called precision points.

The difference between generated function and desire function is called structural error.

Chebyshev spacing of precision points Chebyshev spacing of precision points is useful to minimize the structural error. For n points in range , the precision points xj according to chebyshev

spacing are given as: where

Example: Synthesize a four bar linkage using Freudenstein’s equation to generate the function Y = X1.5 for the interval . The input crank is to start form θ2=300 and is to have range of 900. Take three accuracy points. Take output crank angle from 0 to 900.

Solution:

Step1: Find out three accuracy points(n=3):

where Here X0=1 and Xn+1 = 4 (Given)

Put the value of a, b, θj in above equation

XS =1 YS =1 θ2s = 300 θ4s = 00

X1 =1.20 Y1 =1.314 θ21= 360 θ41= 4.030

X2 = 2.5 Y2 = 3.953 θ22 =? θ42 =?

X3 =3.799 Y3 =7.404 θ23 =? θ43 =?

Xf = 4 Yf = 8 θ2f =1200(Given) θ4f =900 (Given)

ΔX=4-1=3 ----- ΔY=8-1=7 Δθ2 =900 (Given) Δθ4= 900 (Given)

Let the input crank angles corresponding to three accuracy points X1, X2 and X3 be θ21, θ22, θ23 respectively. Let corresponding output crank angles be θ41, θ42, and θ43.

We know that

and

From equation (i) and (ii)

θ2s= 300 θ4s= 00

θ21= 360 θ41= 4.030

θ22 =750 θ42 =37.960

θ23 =113.970 θ43 =82.330

θ2f= 1200 θ4f= 900

The displacement equation given by Freudenstein is written as:

When θ2=360 and θ4 = 4.030

Substituting the values of input (θ2) and (θ4) in above equation, we get

When θ2=750 and θ4 = 37.960

When θ2=113.970 and θ4 = 82.30

Solving the above equationsK1 = 0.0896K2= 0.1279K3= 0.7265Let us assume d = 1.0

Thus,a= 11.16, b= 7.76, c= 7.82, d=1.0

Example 2: A four bar function generator is used to generate the function Y = (1/X) for between the input angle of a crank and angle of follower makes with the frame.

Find the three precision points from Chebyshev’s spacing if the initial value of input i.e. crank angle and output i.e. the follower angle are 300 and 2000 respectively. The difference between the final and initial values of the crank and follower angles are each equal to 900. Solution: Given: X0 = 1, Xn+1=3, n=3(Three precision points)Step1: Find out three accuracy points:

where Here X0=1 and Xn+1 = 3 (Given)

Put the value of a, b, θj in above equation

XS =1 YS =1 θ2s = 300 θ4s =2000

X1 =1.134 =1/1.134 Y1 =0.0881 θ21= ? θ41= ?

X2 = 2.0 =1/2.0 Y2 = 0.5 θ22 =? θ42 =?

X3 =2.866 =1/2.866 Y3 =0.349 θ23 =? θ43 =?

Xf = 3 =1/3 Yf = 0.333 θ2f =1200(Given) θ4f =1100 (Given)

ΔX=3-1=2.0

---- ΔY=0.333-1= - 0.666

Δθ2 =900 (Given) Δθ4= 900 (Given)

Ys is reducing so angle θ4 will also reduce

Let the input crank angles corresponding to three accuracy points X1, X2 and X3 be θ21, θ22, θ23 respectively. Let corresponding output crank angles be θ41, θ42, and θ43. We know that

and

From equation (i) and (ii)

θ2s= 300 θ4s= 2000

θ21= 36.030 θ41= 183.920

θ22 =750 θ42 =132.380

θ23 =113.970 θ43 =112.010

θ2f =1200 θ43 =1100

The displacement equation given by Freudenstein is written as:

When θ2=36.030 and θ4 = 183.920

Substituting the values of input (θ2) and (θ4) in above equation, we get

When θ2=750 and θ4 = 132.380

When θ2=113.970 and θ4 = 112.010

Solving the above equationsK1 = 5.371K2= -13.5K3= 8.272Let us assume d = 1.0

(Negative sign indicates that the direction of C has reversed itself.)

Thus,a= 0.186, b= 1.033, c= -0.074, d=1.0