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Lesson 3: Velocity Analysis and Instantaneous Centers

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Position analysis found where every link sits for a given input angle. Velocity analysis answers the next question: how fast is each link moving at that instant? The answer comes almost for free, because velocity is the time derivative of position, and you already have the position equations. Differentiate the vector loop from the position analysis and you get a set of linear equations for the velocities. Velocity matters because it sets the kinetic energy of every part, the flow rate of a pump, and the bearing loads, and because a smooth-looking mechanism can still have a sharp velocity peak that drives vibration. In this lesson you differentiate the loop, locate instantaneous centers, and read the velocity ratio that becomes mechanical advantage. #VelocityAnalysis #InstantaneousCenters #MechanicalAdvantage

Learning Objectives

By the end of this lesson, you will be able to:

  1. Differentiate the vector loop to obtain the velocity equations of any planar linkage
  2. Solve for piston and angular velocities in closed form
  3. Locate instantaneous centers using Kennedy’s theorem
  4. Read the velocity ratio as mechanical advantage and verify every result in a simulator

Real-World System Problem: Piston Velocity in Engines and Compressors

The engine in a car, the compressor in an air conditioner, and the pump in a hydraulic system are all crank-sliders. Each converts steady rotation into a back-and-forth piston motion, and the velocity of that piston is never constant through the stroke. It starts at zero, rises to a peak somewhere past mid-stroke, and falls back to zero, and exactly where the peak lands decides the bearing loads, the lubrication demand, and the vibration the machine produces.

The Velocity Problem

Engineering Question: Given the input link’s angular velocity, how fast is every other point and link moving at this instant?

For the crank-slider the key output is the piston velocity. For the four-bar it is the angular velocities of the coupler and follower. For the scissor lift it is the platform velocity. All of them come from one operation: differentiating the position loop with respect to time.

Why Velocity Analysis Matters

Bearing and lubrication loads

Peak sliding speed sets the oil-film demand and the friction power lost at every joint.

Flow and delivery

In a pump or compressor the piston velocity is the volumetric flow rate, so its profile is the delivery curve.

Vibration

A sharp velocity peak means a large acceleration nearby (acceleration analysis), and that is what shakes the machine.

Mechanical advantage

The ratio of input speed to output speed is the velocity ratio, and it is the reciprocal of the force ratio (force analysis).

Fundamental Theory: Velocity by Differentiating the Loop

Differentiate the Position Loop

From Position to Velocity

The position loop is a statement that the link vectors close. Differentiating it with respect to time gives a statement that their velocities are consistent. For the four-bar position loop:

Position loop (x, y):

Differentiate (the ground is constant, ):

These are two linear equations in the unknown angular velocities and . The positions from the position analysis are the known coefficients.

This is the central idea of the lesson. Position analysis was nonlinear and needed the Freudenstein trick to solve. Velocity analysis is linear, because differentiation turns the trigonometric position terms into coefficients that multiply the unknown velocities. Once you have the positions, finding the velocities is just solving two linear equations, and the same step repeated in the acceleration analysis gives the accelerations.

Two Routes to the Same Answer

Differentiate the loop and solve the linear equations. This is exact, fast, and easy to program. It is the method used in our simulators and in Applications 1 to 4 below.

Kennedy’s Theorem

Kennedy's Theorem

For any three rigid bodies in plane motion, their three instantaneous centers lie on a single straight line.

A mechanism of links has instantaneous centers. The obvious ones are the fixed pivots and sliding contacts; Kennedy’s theorem locates the rest as intersections of the lines through known IC pairs. The most useful is often the IC between the input and output links, because it gives the velocity ratio directly.

Velocity Ratio and Mechanical Advantage

Velocity Ratio

The velocity ratio of a mechanism is the output speed divided by the input speed. For a four-bar it is .

By conservation of power (input power equals output power in an ideal mechanism), the force or torque ratio is the reciprocal of the velocity ratio:

This quantity is the mechanical advantage. It is defined here from velocities and used again in the force analysis from forces; the two views are the same number seen from opposite sides. When the output slows to a near stop (a limit position), the velocity ratio approaches zero and the mechanical advantage grows very large.

Application 1: Piston Velocity of the Slider-Crank

This is the central worked example. We differentiate the slider position from the position analysis and find where the piston velocity actually peaks, which is not where intuition first suggests.

Step 1: Draw the Space Diagram

Choose and mark a length scale before you start (for a page, 1 cm = 20 mm works well); the scale is yours to pick, but always state it on the drawing. Then construct the space diagram with a set square and compass at the instant you want, here the crank at .

Click to reveal the space-diagram construction

  1. Centre-line and pivot. Draw the cylinder centre-line horizontally and mark the crank pivot on it. ✅

  2. Set the crank. From , draw at above the centre-line, length mm to scale. Point is the crank pin. ✅

  3. Swing the rod. With centre and radius mm, strike an arc cutting the centre-line at . The line is the connecting rod and is the piston. ✅

  4. Measure. The piston sits mm from , matching the position solution. Your drawing should match the figure below. ✅

Slider-crank space diagram: crank OA at 60 degrees, connecting rod AB to piston B on the centre-line

Step 2: Construct the Velocity Polygon

The velocity polygon solves graphically: draw the three velocity vectors head to tail to a chosen scale and read the answer off the figure.

Click to reveal the velocity-polygon construction

  1. Crank-pin velocity. Pick a pole and a velocity scale. Draw perpendicular to the crank , with length . Taking rad/s gives mm/s. ✅

  2. Direction of the piston velocity. The piston slides along the centre-line, so is horizontal. Draw a construction line through the pole parallel to the slide. ✅

  3. Direction of the relative velocity. is perpendicular to the connecting rod . From , draw a line perpendicular to . ✅

  4. Close the polygon. Where the two construction lines cross is point . Then is the piston velocity and is the relative velocity. ✅

  5. Measure. mm/s, so at . This matches the analytical value in Step 3 within drawing tolerance. ✅

Slider-crank velocity polygon: pole o, with v_A, v_B/A and v_B closing the triangle

Step 3: Confirm by Differentiation

Click to reveal the closed-form piston velocity

  1. Differentiate the slider position () with and :

    ✅

  2. Write it as harmonics (): the primary plus secondary terms

    The secondary (twice-per-revolution) term from the connecting rod is what engine balancing must handle. ✅

  3. Tabulate for :

    Crank
    0\degree0.000
    30\degree-0.646
    60\degree-1.017
    73\degree-1.055 (peak)
    90\degree-1.000
    120\degree-0.715
    180\degree0.000

    At the formula gives , confirming the mm/s measured from the polygon. ✅

  4. The peak is about near , not at : the secondary harmonic shifts it earlier in the stroke. The drawing gives one instant; the calculus gives the whole curve and the true maximum. ✅

Step 4: Cross-Check with the Instantaneous Center

Click to reveal the IC method and rod angular velocity

  1. Locate the connecting-rod IC. The crank pin moves perpendicular to the crank, so the rod’s instantaneous center lies on the crank line extended. The piston moves horizontally, so the IC also lies on the vertical line through the piston. Their intersection is the rod IC relative to ground, the same intersection you used to find in the polygon. ✅

  2. Rod angular velocity from differentiating :

    ✅

    At : (rod rotating fastest). At : (rod momentarily not rotating, only translating). ✅

Step 5: Verify in the Simulator

Click to reveal the simulator check

  1. Open the simulator (siwit.co/CSM), set , , , and run it. ✅

  2. Read the velocity chart. The reported maximum piston velocity divided by is about , with the peak before mid-stroke, matching the table. The mean of over a cycle is . ✅

  3. Add an offset. Set and the forward and return peaks become unequal (the quick-return of Experiment 2). Mobility stays one; only the velocity profile changes. ✅

Application 2: Angular Velocities of the Four-Bar

For the four-bar we draw the velocity polygon to read the coupler and follower angular velocities, then confirm with the velocity loop and read the velocity ratio that becomes mechanical advantage.

Step 1: Draw the Space Diagram

Construct the four-bar to scale at using the same arc-intersection method you used for position analysis.

Click to reveal the space-diagram construction

  1. Ground and crank. Draw the ground mm horizontally. From , draw the crank at , length mm. ✅

  2. Intersect the arcs. With centre and radius mm, and centre and radius mm, strike two arcs. Their intersection is the coupler-follower joint (the upper intersection is the open assembly). ✅

  3. Measure. The coupler sits at and the follower at , matching the position solution. ✅

Four-bar space diagram at crank angle 120 degrees, found by intersecting the coupler and follower arcs at B

Step 2: Construct the Velocity Polygon

Click to reveal the velocity-polygon construction

  1. Crank-pin velocity. Pick a pole and a velocity scale. Draw perpendicular to the crank , length . With rad/s, mm/s. ✅

  2. Direction of the follower velocity. Point moves perpendicular to the follower . Draw a construction line through the pole perpendicular to the follower. ✅

  3. Direction of the relative velocity. is perpendicular to the coupler . From , draw a line perpendicular to the coupler. ✅

  4. Close the polygon. The two construction lines cross at . Then is the velocity of on the follower and is the relative velocity. ✅

  5. Measure and convert. mm/s and mm/s, so

    ✅

Four-bar velocity polygon at 120 degrees: pole o with v_A, v_B/A and v_B closing the triangle

Step 3: Confirm by the Velocity Loop

Click to reveal the closed-form angular velocities

  1. Differentiating the position loop gives two linear equations whose Cramer’s-rule solution is:

    ✅

  2. At (, , ):

    ✅

    These confirm the and measured from the polygon.

  3. Full profile across the crank rotation:

    Crank
    30\degree-0.262+0.122
    60\degree-0.040+0.457
    90\degree+0.064+0.539
    120\degree+0.139+0.514

    At (the instant used for position analysis) the coupler is almost in pure translation, , so its velocity triangle is very thin. That is why we drew the polygon at instead. ✅

Step 4: Velocity Ratio and Mechanical Advantage

Click to reveal the mechanical advantage

  1. Velocity ratio at : the follower turns at of the crank. ✅

  2. Mechanical advantage is the reciprocal:

    ✅

    An ideal crank torque appears amplified about twice at the follower here. The value changes through the cycle and grows large near the limit positions, which the force analysis uses.

Step 5: Cross-Check with Instantaneous Centers

Click to reveal Kennedy’s construction

  1. Six instantaneous centers. The fixed pivots give (crank-ground) at and (follower-ground) at . Kennedy’s theorem places , the input-output IC, where the coupler line extended meets the ground line. ✅

  2. Velocity ratio from the IC. The ratio of the distances from to and to gives directly, reproducing with no equations. The polygon, the loop, and the IC all agree because they describe the same motion. ✅

Application 3: Platform Velocity of the Scissor Lift

The scissor-lift height was a one-line expression in the position analysis, so its velocity is one differentiation away.

Step 1: Draw the Space Diagram

Draw the scissor to scale at to fix the geometry before finding the velocity.

Click to reveal the space-diagram construction

  1. Base and arms. Draw the base horizontally. From the fixed bottom pin draw one arm of length mm at ; draw the second arm from the sliding bottom pin so the two cross at their midpoints. ✅

  2. Platform. Join the two upper arm ends with the platform line, which stays parallel to the base. ✅

  3. Measure. The platform height is mm, matching the position solution. ✅

Single-stage scissor lift drawn at 30 degrees: base, crossed arms of length L, and platform at height h = L sin theta

Step 2: Differentiate the Height

Click to reveal the platform velocity

  1. From the platform height , differentiate with respect to time:

    ✅

  2. Read the behaviour. Near the flat position (), , so the platform rises quickly for a given . Near the top (), , so the platform velocity falls to zero even while the arms keep closing. The lift slows as it reaches full height. ✅

  3. The actuator side. A constant actuator speed does not give a constant platform speed, because the geometry between actuator length and angle is itself nonlinear. The simulator’s velocity chart shows the actual platform-velocity curve for the chosen actuator type. ✅

Step 3: Verify in the Simulator

Click to reveal the simulator check

  1. Open the simulator (siwit.co/SLM), set , one stage, and run it at a fixed actuator speed. ✅

  2. Confirm that the platform velocity is largest at low angle and tapers toward zero near full height, matching . ✅

Application 4: Velocity Ratio of the Toggle Clamp

The toggle clamp shows velocity analysis at its most dramatic: at the toggle position the output velocity ratio collapses to zero, which is the exact mechanism behind self-locking.

Step 1: Draw the Space Diagram at the Toggle Position

Sketch the four-bar skeleton at top-dead-centre, where the geometry behind self-locking becomes visible.

Click to reveal the toggle-position construction

  1. Ground and handle. Draw the base line . From draw the handle to joint . ✅

  2. Collinear main link. Draw the main link from to in line with the handle, so , , and lie on one straight line. This collinear configuration is top-dead-centre. ✅

  3. Clamp arm. Join to . Near this position moves almost perpendicular to the clamp arm, so the pad velocity per unit handle velocity drops toward zero. ✅

Toggle clamp four-bar skeleton at top-dead-centre, with the handle and main link collinear

Step 2: Velocity Ratio Near the Toggle

Click to reveal the vanishing velocity ratio

  1. Apply the four-bar velocity solution. As the handle approaches top-dead-center, the handle link and main link become collinear. In the velocity-ratio expression, the term in the denominator does not vanish, but the geometry drives the output pad velocity per unit handle velocity toward zero: the pad momentarily stops while the handle still moves. ✅

  2. The consequence. A vanishing velocity ratio means, by the power balance of the theory section, that the mechanical advantage grows very large:

    ✅

    A modest handle force produces a very large clamping force. This is the quantitative form of the self-locking seen in the mobility analysis.

  3. Mobility is unchanged. The clamp still has one degree of freedom throughout. What changes at the toggle is the instantaneous velocity ratio, not the number of inputs. This is the difference between a singular configuration and a change in mobility. ✅

Step 3: Verify in the Simulator

Click to reveal the simulator check

  1. Open the simulator (siwit.co/TCM) and drive the handle toward top-dead-center. ✅

  2. Watch the mechanical-advantage chart rise sharply as the links approach collinear, while the pad velocity per handle increment falls toward zero. Past the toggle by the lock margin, the clamp holds itself closed. ✅

Programming Velocity Analysis

The whole lesson reduces to differentiating the loop and solving a linear system, which is a few lines of Python.

import numpy as np


def crank_slider_velocity(theta, r, l, omega):
    """Piston velocity for an in-line slider-crank (e = 0)."""
    phi = np.arcsin((r/l)*np.sin(theta))
    return -r*omega*(np.sin(theta) + (r*np.sin(theta)*np.cos(theta))/(l*np.cos(phi)))


def four_bar_omega(a, b, c, theta2, theta3, theta4, omega2):
    """Coupler and follower angular velocities from the velocity loop."""
    w3 = a*omega2*np.sin(theta4 - theta2) / (b*np.sin(theta3 - theta4))
    w4 = a*omega2*np.sin(theta2 - theta3) / (c*np.sin(theta4 - theta3))
    return w3, w4


# Slider-crank: locate the peak (l/r = 3)
r, l, omega = 0.050, 0.150, 1.0
th = np.linspace(0, 2*np.pi, 100000)
v = crank_slider_velocity(th, r, l, omega)
i = np.argmax(np.abs(v))
print(f"peak |Vp|/(r*omega) = {abs(v[i])/(r*omega):.3f} at {np.degrees(th[i]):.1f} deg")
# peak |Vp|/(r*omega) = 1.055 at 73.2 deg


# Four-bar at theta2 = 60 deg (positions from the position analysis)
w3, w4 = four_bar_omega(40, 120, 80,
                        np.radians(60), np.radians(18.4), np.radians(64.9), 1.0)
print(f"w3/w2 = {w3:.3f}, w4/w2 = {w4:.3f}")   # w3/w2 = -0.040, w4/w2 = 0.457

Design Guidelines for Velocity Analysis

Differentiate, don't restart

Velocity equations are the time derivative of the position loop. Reuse the positions from the position analysis rather than setting up a new problem.

Find the real peak

Do not assume the maximum speed is at mid-stroke. The secondary harmonic shifts the peak, and components must be sized for the true maximum.

Use ICs for a quick check

Instantaneous centers and velocity polygons give the velocity ratio geometrically, a fast sanity check on the analytical result.

Watch the limit positions

Where the velocity ratio approaches zero, mechanical advantage grows large. Place these positions deliberately, as in a toggle clamp.

Summary and Next Steps

Key Concepts Mastered

  1. Velocity loop: differentiating the position loop gives linear equations for the unknown velocities.
  2. Piston velocity: the slider-crank peak is about near for , ahead of mid-stroke, because of the secondary harmonic.
  3. Four-bar angular velocities: solved in closed form from the velocity loop, with instantaneous centers and velocity polygons as geometric cross-checks.
  4. Kennedy’s theorem: locates the input-output instantaneous center, which gives the velocity ratio directly.
  5. Mechanical advantage: the reciprocal of the velocity ratio, large near limit positions, the basis of self-locking.

Velocity Results at a Glance

MechanismWhat you solve forKey relationSimulator
Slider-crankpiston velocity siwit.co/CSM
Four-bar, velocity loop (linear)siwit.co/FBL
Scissor liftplatform velocitysiwit.co/SLM
Toggle clampvelocity ratio at the togglesiwit.co/TCM

A Note on Tools

Every velocity here was found by hand and reproduced with a few lines of Python (NumPy). The simulators confirm the same profiles interactively. No specialised motion software is involved; differentiating the loop is the whole method.

Next, Acceleration Analysis and Dynamic Forces differentiates once more. The velocity equations become acceleration equations, the piston acceleration reveals the primary and secondary inertia forces that shake an engine, and Newton’s second law turns those accelerations into the dynamic loads on bearings and links.

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