VELOCITY AND ACCLERATION BY COMPLEX NUMBERS RAVEN’S APPROACH: The two dimensional kinematics problem can be easily solved by using complex algebra. The complex algebra formulation provides the advantages of increased accuracy and suitable for digital computer computation. The complex algebra approach to velocity and acceleration analysis leads to a set of linear equations, which can be solved in quite straight forward. The methods illustrated in the following examples were developed by Raven. In Raven’s method, the links of a mechanism are replaced by vectors, and the vector equations are obtained by equating the resultant of the vectors around every closed loop to zero.
Four-bar mechanism: To illustrate Raven’s approach, let us analyse a four-bar mechanism shown in fig. Let a, b, c, and d denote the length of links, and θ1,θ2, θ3 and θ4 denote the angular position of links 1,2,3 and 4 respectively. The angles are considered as positive when – – – – measured counterclockwise as shown. Let a, b, c and d be the position vectors of the links a, b, c and d respectively (refer fig. 3.1b). The loop-closure equation is – – – – b+c+d=a ……(1) – – Where a has constant magnitude and direction (fixed link). The vector b has constant magnitude and its direction θ2 varies, which is the input angle. If link 2 is the driving link, then θ2, ω2, and α2 are known quantities. – – The vector c and d has unknown magnitude and direction. Expressing the vectors in complex form, iθ2 iθ3 iθ4 iθ1 be + ce +de = ae = a (since θ1 = 0) …..(2) Differentiating the equation (2) with respect to time, we get iθ2 iθ3 iθ4 ib dθ2 e + ic dθ3 e + id dθ4 e =0 (since a is a fixed link) …. (3) dt dt dt
Let dθ2 = ω2, dt
dθ3 = ω3, and dt
dθ4 = ω4, dt
Equation (3) becomes iθ2 iθ3 iθ4 ib ω2 e + ic ω3 e + id ω4 e =0
….
(4)
iθ
Substitute e =cos θ + isin θ in equation (4), we get ib ω2 (cos θ2 + isin θ2) + ic ω3 ( cos θ3 + isin θ3 ) + id ω4 (cos θ4 + isin θ4) = 0
…. (5)
The real part of equation (5) is, Rv = -b ω2 sin θ2 – c ω3sin θ3 – dω4 sin θ4 = 0
…… (6)
The imaginary part of equation (5) is, Iv = b ω2 cos θ2 + c ω3 cos θ3 + d ω4 cos θ4 = 0
…… (7)
Solving the simultaneous equations (6) and (7), we get b sin (θ2 – θ4) Angular velocity of link 3, ω3 = × c sin (θ3 – θ4)
Angular velocity of link 4,
ω4 =
b sin (θ2 – θ3) d sin (θ3 – θ4)
ω2
…..(8)
× ω2
Differentiating the equation (4) with respect to time, we get iθ2 2 iθ3 2 iθ4 2 be i dω2 – ω2 + ce i dω3 – ω3 + de i dω4 – ω4 dt dt dt
……
= 0
(9)
….. (10)
Let dω2 = α2, dω3 = α 3, and dω4 = α4 dt dt dt Equation (10) becomes iθ2 2 iθ3 2 iθ4 2 be (i α2 - ω2 ) + ce (i α3 – ω3 ) + de (i α4 – ω4 ) = 0 ……. (11) iθ Substituting e =cos θ + isin θ in equation (11), we get 2 2 b(cos θ2 + isin θ2) (i α2 - ω2 ) + c(cos θ2 + isin θ2) (i α3 – ω3 ) + d(cos θ4 + isin θ4) 2 (i α4 – ω4 ) = 0 ….. (12) The real part of the equation (12) is 2 2 2 Ra = -b (ω2 cos θ2 + α2 sin θ2) –c (ω3 cos θ3 + α3 sin θ3) –d (ω4 cos θ4 + α4 sin θ4) = 0 … (13) The imaginary part of the equation (12) is 2 2 2 Ia = b (α2 cos θ2 - ω2 sin θ2) + c (α3 cos θ3 – ω3 sin θ3) + d (α4 cos θ4 – ω4 sin θ4) =0 …… (14) Solving the simultaneous equations (13) and (14), we get,
Angular acceleration of link 3 as, 2 2 2 ω3 b ω2 cos(θ2- θ4) + c ω3 cos(θ3- θ4) + d ω4 α3= α2 ω2 c sin (θ3 – θ4) Angular acceleration of link 4 as, 2 2 2 ω4 b ω2 cos(θ2- θ3) + c ω3 + d ω4 cos(θ3- θ4) α4= α2 + ω2 d sin (θ3 – θ4)
…… (15)
…. (16)
Having found the angular velocities of links 3 and 4, we can compute the linear velocity at any point on the links as follows. Let C be point whose velocity is to be found (refer fig. 3.1c).Join AC. The position vector g is then g = b +c = be
iθ2
+ ce
iθ3
…… (17)
Differentiating equation (17) with respect to time, iθ2
+ ic ω3 e
iθ3
Velocity
Vc = = ib ω2 e
Substitute
e =cos θ + isin θ in equation 18, we get
……. (18)
iθ
Vc = ib ω2 (cos θ2 + isin θ2) + ib ω3 (cos θ3 + isin θ3) The real and imaginary components are; Rv = -b ω2 sin θ2 - c ω3 sin θ3 Iv = b ω2 cos θ2 + c ω3 cos θ3 The magnitude of velocity V c = √ Rv + Iv 2
and direction is,
θc= tan
-1
2
Iv Rv
The linear acceleration of the point C on the link 3 is obtained by differentiating the equation (18) with respect to time. Therefore, acceleration Ac, =
ĝ = be iθ2(i α2 - ω22) + ce iθ3(i α3 – ω32)
iθ
Substitute e =cos θ + isin θ in above equation, 2 2 Ac = b (cos θ2 + isin θ2) (i α2 - ω2 ) + c (cos θ3 + isin θ3) (i α3 – ω3 ) The real and the imaginary components are: Ra = -b (ω2 cos θ2 + α2 sin θ2) - c (ω3 cos θ3 + α3 sin θ3) 2
2
….. (19)
Ia = b (α2 cos θ2 - ω2 sin θ2) + c (α3 cos θ3 – ω3 sin θ3) 2
Acceleration
2
Ac = √Ra + Ia 2
2
and
θc = tan
-1
Ia Ra
Example 1: A four bar mechanism ABCD is made up of four links, pin jointed at the ends. AD is a fixed link which is 120 mm long. The links AB, BC, and CD are 60 mm, 80 mm, and 80 mm long respectively. At certain instant, the link AB makes an angle of 60º with the link AD. If the link AB rotates at uniform speed of 10 rpm clockwise direction, determine analytically (i) Angular velocity of the link BC and CD (ii) Angular acceleration of the link BC and CD
Data: AD = a = 120 mm, AB = b = 60 mm, θ2 = 60º, n2 = -10 rpm (clockwise)
BC = c = 80 mm, CD = d = 80 mm,
Solution: Draw the configuration diagram as shown in the fig below. We have to determine first the angles θ3 and θ4 made by the links 3 and 4.
Consider the triangle ABD, BD = s = √ a +b – 2ab cos θ2 2
2
= √120 + 60 – 2 × 120 × 60 cos 60 = 103.923 mm 2
2
sin β
sin θ2
Also,
= b
s
sin β
sin 60
i.e.,
= 60
103.923
β = 30º
Therefore, angle
From the triangle BCD, CD = d = √ c + s – 2cs cos ψ 2
2
i.e., 80 = √ 80 + 103.923 - 2 × 80 × 103.923 cos ψ 2
2
Squaring both sides, we get 80 = 80 + 103.923 – 2 × 80 × 103.923 cos ψ 2
2
2
Therefore angle
ψ = 49.49º
The angle made by the link BC, sin λ Similarly,
θ3 = ψ – β = 49.49 – 30 = 19. 49º
sin ψ =
c
d
sin λ i.e.,
sin 49.49 =
80 Therefore, angle
80
γ = 49.49º
The angle made by the link CD,
θ4 = 360 – (β + λ ) = 360 – (30 + 49.49) = 280.50º
Angular velocity of link AB,
ω2 = 2πn2 60 = 2π × (-10) = - 1.047 rad / sec 60
Angular velocity of link BC,
ω3 = -
b sin (θ2 – θ4) c sin (θ3 – θ4)
×
ω2
60 sin (60 – 280.5) =-
× (-1.047) = .5163 rad /sec 80 sin (19.49 – 280.5)
Angular velocity of link CD,
ω4 =
b sin (θ2 – θ3) d sin (θ3 – θ4)
× ω2
60 sin (60 – 19.49) =
× (-1.047) = - .5164 rad /sec 80 sin (19.49 – 280.5)
Angular acceleration of the link BC,
α3 =
ω3 ω2
b ω2 cos(θ2- θ4) + c ω3 cos(θ3- θ4) + d ω4 2
α2 -
2
2
c sin (θ3 – θ4)
Since there is no acceleration for link AB,
α2 = 0
α3 = -60 × (-1.047)2 cos (60 – 280.5) + 80 × .5163 2 cos (19.49 – 280.5) + 80 × (-.5164) 2 = .8608 rad / sec
80 × sin (19.49 -280.5)
2
Angular acceleration of the link CD,
α4=
ω4 ω2
b ω2 cos(θ2- θ3) + c ω3 + d ω4 cos(θ3- θ4) 2
α2 +
2
2
d sin (θ3 – θ4) 2
2
2
= 60 × (-1.047) cos (60 – 19.49) + 80 × .5163 + 80 × (-.5164) cos (19.49 – 280.5) 80 × sin (19.49 -280.5) 2 =.8608 rad /sec
