introduction

The topic of this experiment is velocity analysis of planar mechanisms Through velocity analysis we obtain the velocity characteristics of all the links of a mechanism when the input velocity is specified Two points must be mentioned at this stage one to carry out velocity analysis must be completed first Second is that velocity analysis is carried out for a particular configuration of the mechanism Just like in displacement analysis both graphical and analytical methods can be used to carry out the velocity analysis

To study the motion of a simple amp compound mechanism knowledge of velocity analysis is required The analysis of velocity depend upon the graphical as well as analytical methods The graphical approach is suitable for finding out the velocity of the links of a mechanism in one or two positions of the crank However if it is required to find these values at various configurations of the mechanism or to find the maximum values of maximum velocity it is not convenient to draw velocity diagrams again and again In that case analytical expressions for displacement velocity in terms of general parameters are derived In this paper a Analytical method using simple trigonometry is used for getting values of linear velocities amp Angular velocities of a simple or compound mechanisms at any position of the crank Graphical method is very monotonous as compare to Analytical method because it is not feasible to draw at one rotation of crank angle

4bar-linkage analysis

The vector loop equation is written as

r1 ei θ1 + r2 ei θ2 + r3 ei θ3 + r4 ei θ4 = 0

r1(cos θ1 + isin θ1) + r2(cos θ2 + isin θ2) + r3(cos θ3 + isin θ3) + r4(cos θ4 + isin θ4) = 0

Collecting real and imaginary parts yields

Real r1cos θ1 + r2cos θ2 + r3cos θ3 + r4cos θ4 = 0 Imag r1sin θ1 + r2sin θ2 + r3sin θ3 + r4sin θ4 = 0

Velocity Analysis Taking time derivative of the position vector r = reiθ in general yieldsd td r

= r = (r + irω) e i θ

where ω is known as the angular velocity of the link Hence taking time derivative of thevector loop equation results inr1048581 + ( 2 r + ir 2 ω2) 2 θ i e + ( 3r + ir 3 ω3) 3 θ i e + ( 4 r + ir 4 ω4) e iθ 4

= 0

where r 1 = 0 and also r 2 = r 3 = r 4 = 0 since r1 is frame and the lengths of links 2 3and 4 are fixed Collecting real and imaginary parts then yields

real minusr2 ω2sin θ2 minus r3 ω3sin θ3 minus r4 ω4sin θ4 = 0Imag r2 ω2cos θ2 + r3 ω3cos θ3 + r4 ω4cos θ4 = 0

Slider-Crank analysis Position Analysis The vector loop equation is written as

r2 eiθ2 + r3 ei θ3 minus r4 eiθ4 = 0

Since θ4 = 0 then

r2 eiθ2 + r3 ei θ3 minus r4 = 0 orr2(cos θ2 + isin θ2) + r3(cos θ3 + isin θ3) minus r4 = 0

Collecting real and imaginary parts yields

Real r2cos θ2 + r3cos θ3 minus r4 = 0 Imag r2sin θ2 + r3sin θ3 = 0

Velocity Analysis Taking time derivative of the vector loop equation results in( r2

+ ir 2 ω2) e iθ2 + (r3+ ir 3 ω3) e iθ3

minus r 4= 0

where r2 = r3

= 0 since the lengths of links 2 and 3 are fixed Collecting real and

imaginary parts then yields

Real minusr2 ω2sin θ2 minus r3 ω3sin θ3 ndash rrsquo4 = 0Imag r2 ω2cos θ2 + r3 ω3cos θ3 = 0

4bar-linkage analysis

The vector loop equation is written as

r1 ei θ1 + r2 ei θ2 + r3 ei θ3 + r4 ei θ4 = 0

r1(cos θ1 + isin θ1) + r2(cos θ2 + isin θ2) + r3(cos θ3 + isin θ3) + r4(cos θ4 + isin θ4) = 0

Collecting real and imaginary parts yields

Real r1cos θ1 + r2cos θ2 + r3cos θ3 + r4cos θ4 = 0 Imag r1sin θ1 + r2sin θ2 + r3sin θ3 + r4sin θ4 = 0

Velocity Analysis Taking time derivative of the position vector r = reiθ in general yieldsd td r

= r = (r + irω) e i θ

where ω is known as the angular velocity of the link Hence taking time derivative of thevector loop equation results inr1048581 + ( 2 r + ir 2 ω2) 2 θ i e + ( 3r + ir 3 ω3) 3 θ i e + ( 4 r + ir 4 ω4) e iθ 4

= 0

where r 1 = 0 and also r 2 = r 3 = r 4 = 0 since r1 is frame and the lengths of links 2 3and 4 are fixed Collecting real and imaginary parts then yields

real minusr2 ω2sin θ2 minus r3 ω3sin θ3 minus r4 ω4sin θ4 = 0Imag r2 ω2cos θ2 + r3 ω3cos θ3 + r4 ω4cos θ4 = 0

Slider-Crank analysis Position Analysis The vector loop equation is written as

r2 eiθ2 + r3 ei θ3 minus r4 eiθ4 = 0

Since θ4 = 0 then

r2 eiθ2 + r3 ei θ3 minus r4 = 0 orr2(cos θ2 + isin θ2) + r3(cos θ3 + isin θ3) minus r4 = 0

Collecting real and imaginary parts yields

Real r2cos θ2 + r3cos θ3 minus r4 = 0 Imag r2sin θ2 + r3sin θ3 = 0

Velocity Analysis Taking time derivative of the vector loop equation results in( r2

+ ir 2 ω2) e iθ2 + (r3+ ir 3 ω3) e iθ3

minus r 4= 0

where r2 = r3

= 0 since the lengths of links 2 and 3 are fixed Collecting real and

imaginary parts then yields

Real minusr2 ω2sin θ2 minus r3 ω3sin θ3 ndash rrsquo4 = 0Imag r2 ω2cos θ2 + r3 ω3cos θ3 = 0

Slider-Crank analysis Position Analysis The vector loop equation is written as

r2 eiθ2 + r3 ei θ3 minus r4 eiθ4 = 0

Since θ4 = 0 then

r2 eiθ2 + r3 ei θ3 minus r4 = 0 orr2(cos θ2 + isin θ2) + r3(cos θ3 + isin θ3) minus r4 = 0

Collecting real and imaginary parts yields

Real r2cos θ2 + r3cos θ3 minus r4 = 0 Imag r2sin θ2 + r3sin θ3 = 0

Velocity Analysis Taking time derivative of the vector loop equation results in( r2

+ ir 2 ω2) e iθ2 + (r3+ ir 3 ω3) e iθ3

minus r 4= 0

where r2 = r3

= 0 since the lengths of links 2 and 3 are fixed Collecting real and

imaginary parts then yields

Real minusr2 ω2sin θ2 minus r3 ω3sin θ3 ndash rrsquo4 = 0Imag r2 ω2cos θ2 + r3 ω3cos θ3 = 0