• Velocity is ‘how fast something is moving’.
• One form of velocity is translation. In proper terms it is the first derivative of position,
• A second form of velocity is rotational. This is in terms of orientation angle,
• When we deal with 3D rotation and velocity we must consider the effects of the entire body. If we consider two points (P and Q) on a single rigid body, then we can write the two following relationships,
• Consider that the apparent position equation from before can be extended to apparent velocity,
Problem 23.1 (Shigley and Uicker, 1995) If the crank EA has been rotated 30° counterclockwise about E and has a rotational velocity of 36 rad/sec, what is the angular velocity of the angled rod BD?
• While we have considered apparent velocity in translation before we may also consider apparent angular velocity.
• Consider that in most cases seen so far the joints are of the pin type, and therefore are fixed points on each member.
• If we have a rolling contact we will have the following conditions, (Note: even though the rollers are round below, they could be other shapes)
• We can expand the tools for analysis of position to find velocities.
• The most important factor to remember is that velocity is the first derivative of position.
• Recall the position vector in complex polar notation. And, after finding the first derivative we get the simple form,
Problem 23.2 The crank is 20cm long, and the driver is 80cm long. If the crank is rotating at 60rpm, what are the velocities of each of the links? Use a complex approach.
Problem 23.3 Next, consider a four bar linkage, in the double rocker configuration. Find the velocity of the connector, if the left-hand rocker is the input. Assume that the input position and velocity is known. Use a vector approach.
• Basically use techniques to develop equations for position, and then differentiate
Problem 23.4 Consider the mechanism pictured below. Relate the position of the yoke to the leftmost crank.
• If we look at any body in a mechanism it will be rotating and translating.
• And, if we pick the right point to watch, it will appear to be rotating, we will call this the center of rotation.
• And, because the translation and rotation often changes as the mechanism moves through different positions we say that it is instantaneous.
• For a binary link the instant center is defined by the velocities at the joints.
• The easiest way to find an instantaneous center is to find the intersection (P) of vectors perpendicular to the velocity vectors at the joints (A and B).
Problem 23.5 Try the example below.
• We can also find an instantaneous center for a chain of links.
• If we consider an entire mechanism with ‘n’ links, it will generally have ‘N’ instantaneous centers.
• We can help keep track of which centers are to be found using points mapped out on a circle. In this technique we lay out a circle, and then put on a point for each of the links (including the ground). Draw on lines to connect each point where links exist. Also draw on lines for the Kennedy theorem.
Problem 23.6 Try the example below.
• The instantaneous center has no velocity (VP=0), but there is also a rotational velocity about this point. This can be found using the relationship,
• The theorem states that for three rigid links in a mechanisms, the instantaneous centers for each of the three mechanisms lie on the same straight line.
• We can consider a linkage with three members as pictured below,
• Consider the following mechanism: use the Aronhold-Kennedy theorem,
Problem 23.7 Consider the following example. Find all of the instant centers.
Problem 23.8 Find the instant centers for the mechanism below.
• “The angular velocity ratio of any two bodies in planar motion relative to a third body is inversely proportional to the segments into which the common instant center cuts the line of centers”
• In general, after finding the instant centers, we can find rations between angular velocities using,
• Keeping in mind that energy is conserved we can develop equations for mechanical advantage.
• These two relationships can then be used with the results of earlier calculations to find the mechanical advantage.
Problem 23.9 Find the velocity of the connector, if the left-hand rocker is the input. Assume that the input position and velocity is known. Use a vector approach.
• If in a 4 bar linkage we find instant centers for adjacent links (i.e. 1 & 3 and 2 & 4). These form a collinear axis if there is a perpendicular to the connecting rod then that position has the minimum mechanical advantage.
• Instant Centers are valid for only a single point in time. If we watch the locations of these centers evolve, we will see them move through space. Below the centrode for the four bar linkage shows how P13 sweeps through space as the mechanism moves.
• In the example above the centrode is called ‘fixed’. If we had used the kinematic inversion where 3 was the frame, then the instant center would trace out another called a ‘moving’ centrode.
Problem 23.10 Develop position and velocity equations for the following mechanisms in complex polar notation. If other calculations are required, write these out also. Show the equations needed to calculate any unknown angles. Do not attempt to solve any equations, but indicate which variables are unknown by underlining them in the final equation.
a) Find all of the positions and velocities.
b) In addition to the basic equations, also write the equations for the position and velocity of point G?
c) Find all of the positions and velocities. Assume points A, C and F lie on a horizontal line.
Problem 23.11 Find the orientations and velocities of the joints and point G in the mechanism using trigonometry, or any other non complex method. Assume that link 2 is being driven at 2 rad/sec
23.1 Erdman, A.G. and Sandor, G.N., Mechanism Design Analysis and Synthesis, Vol. 1, 3rd Edition, Prentice Hall, 1997.
23.2 Shigley, J.E., Uicker, J.J., “Theory of Machines and Mechanisms, Second Edition, McGraw-Hill, 1995.