Chapter 3 Velocity nalysis The position of point with respect to 0, Fig 1 may be defined mathematically in either polar or Cartesian form. Two scalar quantities, the length R and the angle θ with respect to the x axis, define the vector locating point. Polar form: R = R e iθ (1) Using Euler s equation, we obtain the Cartesian form: R = R (cos θ + i sin θ ) (2 ) The absolute linear velocity of a point is the time rate of change of position vector of that point with respect to the ground reference frame or polar form or Cartesian form V ir dθ dt e V iθ dr dt R ω 2 i e iθ i R ω 2 The polar form of V is very informative. The scalar value of the velocity is the radius R times the angular velocity ω 2, while the direction is 90 from the unit position vector e iθ in the same sense as ω 2. The notation followed when using complex numbers is that counterclockwise rotations are positive; thus ω 2 is positive here and the absolute velocity V is to the left in Fig. 3.1. Notice that the magnitude of the linear velocity and angular velocity are related: Figure 3.1 bsolute velocities V of point on rotating link 2. Figure 3.2 bsolute velocities V and VB of points and B on rotating link 2. Dr. Mostafa S. Habib MEG 373 Velocity 1
Figure 3.2 bsolute velocities V and VB of points and B on rotating link 2. lso, the direction of the velocity vector is always perpendicular to the position vector originating at the point of reference since, if a component of the velocity were to be along the position vector, the link would deform, which contradicts the rigid link assumption. Suppose that link 2 contained another point of interest, point B (see Fig. 3.34). The absolute velocity of point B would be Polar form: V B =i R B ω 2 e iθb = R B iω 2 (3.15) Cartesian form: Thus far, we have discussed only absolute velocities of a point on a link pinned to ground. What is the difference between the absolute velocity at point B and the absolute velocity at point (still case 2 motion); that is, if you were to sit on link 2 at point, keep looking in the fixed x direction and out of the corner of your eye watched point B, what would be the apparent velocity of point B with respect to you at? You would observe the velocity difference, which is the difference of two absolute velocities of two points on the same link. The velocity difference is in this case V B where the second subscript is the point of reference and the first subscript is the point of interest. It follows that from which vector solution to this equation appears in Fig. 3.3. Notice that the absolute velocities have only one subscript. The absent second subscript is understood to be ground. In Fig. 3.3, point O V is an arbitrary origin for drawing the velocity diagram. ll absolute velocities are laid out starting at this origin so that V B [from the left side of Eq. (3.17)] is drawn to a convenient scale, parallel to V B in Fig. 3.3. Then the right side of the equation is drawn in by starting at O V and drawing V to the same scale. The vector difference between V B and V is V B closing the vector polygon. Note that V B is perpendicular to R B (as it should be since link 2 is rigid) and that and Dr. Mostafa S. Habib MEG 373 Velocity 2
Solving Eq. (3.19) for ω 2 we obtain B Figure 3.3 Finding the velocity difference V B of point B with respect to point by means of the vector triangle. Both and B are points of rotating link 2 from Fig. 3.2. Example 3.2 Link J in Fig. 3.4 is moving with respect to ground. Points P and Q of link J are locations of known absolute velocities V P and V Q. Find the angular velocity ω 2, of this link with respect to ground, or, which is the same thing, with respect to the fixedly oriented xpiy system, attached to link J at P and moving along with it while remaining parallel with the fixed x 0 Oiy 0 system. Solution Using Eq. (3.17) yields Figure 3.5 shows the graphical solution of this equation. Finally, from Eq. (3.20), Figure 3.4 Motion of link Jknown only by way of known velocity vectors V and VQ of its points P and Q. Dr. Mostafa S. Habib MEG 373 Velocity 3
Figure 3.5 Graphical construction to find velocity difference of point Q with respect to point Pof link J of Fig. 3.4 in preparation for calculating the unknown angular velocity w. Thus the preceding example demonstrates that the complex-number notation will determine the angular velocity with the correct algebraic sign to indicate whether it is clockwise (cw) or counterclockwise (ccw), without rules of thumb or visual inspection of the geometry involved. This approach is therefore well adapted for automatic digital computation, where there is no opportunity for visual verification. It is now possible to tackle the velocity analysis of a linkage made up of several links. Graphical solution of complex-number equations (such as Figs. 3.35 and 3.37) will be emphasized initially due to the inherent visual feedback offered. nalytical solutions for the same examples are included later. The slider-crank linkage in Fig. 3.38 is a good mechanism to begin with. The objective is to determine the velocity of point B on the slider (link 4), given the input angular velocity (ω2). Since this mechanism is being analyzed for velocity considerations, the displacement information should already be known (i.e., the positions of point, point B, θ2, and θ3 are given). Dr. Mostafa S. Habib MEG 373 Velocity 4
Step 1. Find the absolute velocity of point on link 2 (case 2 analysis): B Figure 3.38 The motion of this slider- crank mechanism is known by way of given input angular velocity (02. The velocity of slider 4 is to be found. Step 2. Find the absolute velocity of point of link 3 (case 3 analysis). This is a trivial step since a pin joint connects link 2 and link 3 at, and V (3) = V (2). Step 3. Find the velocity of point B 3 by using point 3 (both points on link 3) and Eq. (3.17) (case 2 analysis): V B =V +V B (3.22) Recall that the vector equation (3.22) is equivalent to two independent scalar equations: the summation of the x components and the summation of the y components. lso (using polar notation this time), each velocity vector has two scalar unknowns: its magnitude and its direction. useful accounting scheme that keeps track of knowns and unknowns in a vector equation is to place a D under the vector if the direction is known (accompanied with an arrow showing the approximate direction) and an M if the magnitude is known. fter this is done for each vector in the equation, at most two scalars can be left as unknowns, and the unknowns can be found either graphically or analytically. In this case, Eq. (3.22) becomes Both components of V are known (the magnitude is V = R ω 2 ). The direction of V B is vertical since the slider is constrained to move in the vertical slot. lso, the direction of V B is known to be perpendicular to link B. With just two unknowns remaining, Eq. (3.23) can be solved graphically, as in Fig. 3.39, by choosing an appropriate scale for V. Step 4. Find the velocity of point B on link 4 (case 3 analysis). gain, this is a trivial step. Thus, the velocity of the slider is found by simply measuring the length of V B in Fig. 3.39. Example 3.3 The four-bar linkage shown in Fig. 3.40 is driven by a motor connected to link 2 at 600 rpm clockwise. Determine the linear velocities of points and B and the angular velocities of links 3 and 4 in the position shown in the figure. Dr. Mostafa S. Habib MEG 373 Velocity 5
Solution Step 1. Calculate V, as part of link 2. To obtain ω 2 in radians per second, we use the relationship and give it the algebraic sign + for ccw and - for cw rotation. Thus Step 2. Find V B. Using Eq. (3.17), Measuring from the velocity diagram (Fig. 3.41), 1. Measure the lengths of the linear velocities according to the velocity scale; 2. Divide these by their respective radius vectors. Dr. Mostafa S. Habib MEG 373 Velocity 6
Example 3.4 Figure 3.42 shows the same four-bar mechanisms as that of Example 3.3 (Fig. 3.40) with the addition of point P on the coupler link. Input velocity is the same as in Example 3.3. Calculate V P. Solution: Using the velocity-difference equation between P and, Dr. Mostafa S. Habib MEG 373 Velocity 7
There are not enough knowns to solve this equation, but we have not made use of all pertinent information. The velocity-difference equation between P and B can be expressed as This equation also contains three unknowns and cannot be solved by itself, but the two equations can be solved simultaneously: There are now two scalar equations and two scalar unknowns (the magnitudes of V PB and V P ). Figure 3.43 shows that the intersection of the directions of the velocity-difference vectors V PB and V P yields point P, and thus: It should be noted that in the vector diagram of Fig. 3.43, triangle abp is similar to coupler triangle BP (Fig. 3.42), because ab B, bp BP, and ap P. Triangle abp is called the velocity image of link BP. lso, the velocity image (triangle in this case) is rotated 90 deg from the original link in the direction of the angular velocity of that link. The reason is that all velocity-difference components are related to the original link vectors by V = iωr. The relationship between a rigid link with three or more points of interest and the corresponding velocity diagram yields a very useful shortcut analysis procedure. Once velocities of two points on a link are calculated, the velocity difference of any other point can be obtained by similar triangles. For example, the absolute velocity of point E may be directly measured (Fig. 3.43). V E = 144 cm/sec. Example 3.5 Figure 3.44 shows a six-bar linkage which is actually a four-bar connected to an inverted slider-crank mechanism (see Fig. 3.10). With ω 2 = -186 rpm cw, find V D, V (F5) (velocity of F as a point of link 5), and ω 5. Solution: Figure 3.45 shows the graphical solution of this problem based on the successive solutions of the following equations: Dr. Mostafa S. Habib MEG 373 Velocity 8
From the velocity polygon, Example 3.6 Use complex-number arithmetic on the same problem as Example 3.3. Solution: where ω 2 is known, and ω 3 and ω 4 are unknown. Separating real and imaginary parts, we have Dr. Mostafa S. Habib MEG 373 Velocity 9
from which The slight differences in numerical values between the results of the two approaches are due to the inaccuracies in graphical construction in Fig. 3.41 and or rounding-off errors in computation. Dr. Mostafa S. Habib MEG 373 Velocity 10