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Planar mechanisms: degrees of freedom and kinematics

A mechanism is a set of rigid parts (links) joined so that, when one moves, the others move in a definite way: a windshield wiper, the pedal of an exercise bike, a car engine. Analyzing it means answering how many inputs it needs, whether the crank can turn all the way around, and where each point is and how fast it moves. That is what the Planar mechanisms tool computes for any mechanism you draw.

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Links, pairs and degrees of freedom

Each free link in the plane has 3 degrees of freedom (two translations and one rotation). Each joint removes some: a pin (revolute) or a slider (prismatic) leaves only 1 relative motion and removes 2; a cam or gear-tooth contact leaves 2 (rolling and sliding) and removes 1. The Grübler–Kutzbach formula does the count:

M = 3 (n - 1) - 2 j_{1} - j_{2}

Lower pairs (

j_{1}

, they remove 2 DOF) and a higher pair (

j_{2}

, it removes 1).

$n$ counts every link, ground (the fixed frame) included. $M$ is the number of independent inputs (motors or actuators) the mechanism needs: with $M = 1$ one motor is enough, with $M = 0$ it is a rigid structure and with $M < 0$ it is statically indeterminate.

Mechanism	$n$	$j_{1}$	$M$
Four-bar	4	4	1
Slider-crank	4 (with the block)	4 (3 R + 1 P)	1
Five-bar	5	5	2
Watt six-bar	6	7	1
Triangle (3 bars)	3	3	0

Careful when counting pins: if

k

links meet at one pin, it counts as

k - 1

joints. The formula also fails for special geometries (parallelograms, redundant links): that is why the simulator compares the count with the rank of its equations.

Grashof condition

In a four-bar, can any link turn all the way around? With $s$ the shortest link, $l$ the longest and $p$ , $q$ the other two, the Grashof condition says that at least one link makes full turns if

s + l \leq p + q

If it is Grashof and the shortest is…	Type
the crank (next to ground)	crank-rocker: the crank turns, the rocker oscillates
ground	double crank (drag link): both turn
the coupler	double rocker: only the coupler turns
with $s + l = p + q$	change point: all links line up and it may switch branch
Non-Grashof ( $s + l > p + q$ )	triple rocker: none turns fully

Position: the vector loop

Each link is a vector with fixed length and variable angle. Going around the four-bar from $O_{2}$ to $O_{4}$ along both possible paths gives the loop equation, in complex form or in components:

a e^{i θ_{2}} + b e^{i θ_{3}} = d + c e^{i θ_{4}} ⟹ {a cos θ_{2} + b cos θ_{3} = d + c cos θ_{4} a sin θ_{2} + b sin θ_{3} = c sin θ_{4}

With $θ_{2}$ known (the input), two equations remain for $θ_{3}$ and $θ_{4}$ . Geometrically, joint D lies at the intersection of two circles: one of radius $b$ around C and another of radius $c$ around $O_{4}$ . There are two solutions, the open and the crossed configuration: the mechanism stays in the branch it was assembled in.

The four-bar of the examples at

θ_{2} = 90°

, with its angles and the transmission angle μ.

Velocity and acceleration

Differentiating the loop with respect to time brings in the angular velocities. It is the same as the relative-velocity relation of two points of a rigid body: the velocity of D is that of C plus that of the coupler turning about C,

v_{D} = v_{C} + ω_{3} \times r_{D / C} v_{D} = ω_{4} \times r_{D / O_{4}}

Equating both expressions gives two linear equations in $ω_{3}$ and $ω_{4}$ . The same happens with accelerations, but each term has two parts: the tangential one $α \times r$ and the normal (centripetal) one $- ω^{2} r$ , which points toward the center of rotation:

a_{D} = a_{C} + α_{3} \times r_{D / C} - ω_{3}^{2} r_{D / C} = α_{4} \times r_{D / O_{4}} - ω_{4}^{2} r_{D / O_{4}}

When a point slides along a rotating link (the slider of the quick-return mechanism), there is also the Coriolis acceleration,

2 ω \times v_{rel}

. The simulator includes it: it solves the constraint equations of the whole mechanism and differentiates them twice.

Transmission angle

The transmission angle $μ$ is the angle between the coupler and the rocker. It measures how well the coupler pushes: with $μ = 90°$ all of its force turns the rocker; near 0° or 180° most of the force just loads the pivot and the mechanism jams. Keeping it between about 40° and 140° is recommended.

cos μ = \frac{b ^{2} + c ^{2} - a ^{2} - d ^{2} + 2 a d cos θ _{2}}{2 b c}

The extremes occur with the crank lined up with ground ( $θ_{2} = 0°$ and $180°$ ).

The slider-crank

It is the mechanism of combustion engines and piston compressors. With crank $r$ , connecting rod $l$ and no offset, the position of the block and its velocity are

x = r cos θ + l^{2} - r^{2} sin^{2} θ \overset{x}{˙} = - r ω sin θ (1 + \frac{r cos θ}{l ^{2} - r ^{2} sin ^{2} θ})

The stroke is $2 r$ and, with no offset, the forward and return strokes take the same time (time ratio 1). With the slider axis offset, one stroke is slower than the other: that is the principle of the quick return of shapers.

Slider-crank of example 3 and the block velocity over one turn: it is not a pure sine, because of the connecting-rod term.

Common mistakes

Counting ground twice, or not at all, in $n$ .
Counting a pin with three links as a single joint (it is two).
Taking the other branch of the position solution (crossed configuration) without noticing.
Forgetting the normal acceleration $ω^{2} r$ even when the crank turns at constant speed ( $α_{2} = 0$ does not mean zero acceleration).
Using rpm instead of rad/s: $ω = 2 π n /60$ .

Worked examples

Example 1Degrees of freedom, Grashof and transmission angle

A four-bar has ground

d = 100

mm, crank

a = 40

mm, coupler

b = 120

mm and rocker

c = 80

mm. Find its degrees of freedom, its Grashof type and the range of the transmission angle.

Solution

$n = 4$ (ground + 3), $j_{1} = 4$ pins: $M = 3 \cdot 3 - 2 \cdot 4 = 1$ . One motor is enough.
$s = 40$ , $l = 120$ , $p + q = 100 + 80 = 180$ . Since $160 \leq 180$ it is Grashof, and the shortest is the crank: crank-rocker.
With $θ_{2} = 0°$ : $cos μ = \frac{12 0 ^{2} + 8 0 ^{2} - ( 100 - 40 ) ^{2}}{2 \cdot 120 \cdot 80} = 0.896$ , $μ = 26.4°$ .
With $θ_{2} = 180°$ : $cos μ = \frac{12 0 ^{2} + 8 0 ^{2} - 14 0 ^{2}}{19 200} = 0.0625$ , $μ = 86.4°$ . The minimum, 26.4°, is below the recommended 40°: near $θ_{2} = 0°$ the mechanism transmits poorly.

M = 1

, crank-rocker (Grashof: 160 ≤ 180) and

μ

between 26.4° and 86.4°.

Solve it in Simuveliq

In Engineering → Planar mechanisms, in Examples, «Load mechanism» → Four-bar (crank-rocker).
The cards give Degrees of freedom M = 1, Type (Grashof) «160 ≤ 180 · Crank-rocker» and Transmission angle 26.4° – 86.4°, «outside the recommended range». Below, the Grübler–Kutzbach block shows the count of links and pairs.

Open in the simulator →

Example 2Position, velocity and acceleration at θ₂ = 90°

The same four-bar turns at 60 rpm counter-clockwise at constant speed. For

θ_{2} = 90°

find

θ_{3}

θ_{4}

, the angular velocities

ω_{3}

ω_{4}

and the angular accelerations

α_{3}

α_{4}

Solution

Position: C = (0, 40). D satisfies $x^{2} + (y - 40)^{2} = 12 0^{2}$ and $(x - 100)^{2} + y^{2} = 8 0^{2}$ ; subtracting, $y = 2.5 x - 205$ , and the open branch gives D = (113.54, 78.85) mm. Then $θ_{3} = arctan \frac{38.85}{113.54} = 18.89°$ and $θ_{4} = arctan \frac{78.85}{13.54} = 80.26°$ .
Velocity: $ω_{2} = 2 π \cdot 60/60 = 6.283$ rad/s and $v_{C} = ω_{2} k \times (0, 40) = (- 251.3, 0)$ mm/s. With $k \times (x, y) = (- y, x)$ , the equality $v_{C} + ω_{3} k \times (113.54, 38.85) = ω_{4} k \times (13.54, 78.85)$ gives: in $y$ , $113.54 ω_{3} = 13.54 ω_{4}$ ; in $x$ , $- 251.3 - 38.85 ω_{3} = - 78.85 ω_{4}$ .
Solving: $ω_{4} = 3.387$ rad/s and $ω_{3} = 0.4038$ rad/s, both counter-clockwise. $∣ v_{D} ∣ = 80 ω_{4} = 270.9$ mm/s.
Acceleration ( $α_{2} = 0$ ): $a_{C} = - ω_{2}^{2} (0, 40) = (0, - 1579)$ mm/s². The acceleration equation, with the normal terms $ω_{3}^{2} = 0.163$ and $ω_{4}^{2} = 11.47$ , gives two linear equations: $38.85 α_{3} = 78.85 α_{4} + 136.7$ and $113.54 α_{3} = 13.54 α_{4} + 681.4$ .
Solving: $α_{4} = 1.298$ rad/s² and $α_{3} = 6.155$ rad/s². Although the crank turns at constant speed, the coupler and the rocker accelerate.

θ_{3} = 18.89°

θ_{4} = 80.26°

;

ω_{3} = 0.4038

ω_{4} = 3.387

rad/s;

α_{3} = 6.155

α_{4} = 1.298

rad/s².

Solve it in Simuveliq

The button opens this four-bar stopped at $θ_{2} = 90°$ . The one from example 1 works too: pause the animation (❚❚) and move the slider until it reads θ₂ = 90°.
In Values at the current position, the links table gives θ, ω and α of each one (18.89°, 0.4038, 6.155 for link 3; 80.26°, 3.387, 1.298 for link 4) and the points table, D = (113.5, 78.85) with |v| = 270.9 mm/s.
Turn on the v and a arrows on the drawing and the ω and α charts of «Kinematics and dynamics over one cycle» to see how they change over the turn.

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Example 3Slider-crank at 60°

An in-line slider-crank has a 40 mm crank and a 140 mm connecting rod, and turns at 120 rpm. For

θ = 60°

find the position, velocity and acceleration of the block.

Solution

$r sin θ = 34.64$ , $r cos θ = 20$ and $14 0^{2} - 34.6 4^{2} = 135.65$ mm: $x = 20 + 135.65 = 155.65$ mm.
$ω = 2 π \cdot 120/60 = 12.566$ rad/s: $\overset{x}{˙} = - 40 \cdot 12.566 \cdot 0.866 (1 + \frac{20}{135.65}) = - 499.5$ mm/s (the block moves back toward the crank).
Differentiating again with constant $ω$ : $\overset{x}{¨} = ω^{2} [- r cos θ - r ^{2} cos 2 θ - \frac{r ^{4} sin ^{2} θ cos ^{2} θ}{( ) ^{3}}] = 157.9 (- 20 + 5.898 - 0.192) = - 2257$ mm/s².
The rod turns with $ω_{3} = - \frac{r ω cos θ}{l ^{2} - r ^{2} sin ^{2} θ} = - \frac{251.3}{135.65} = - 1.853$ rad/s (clockwise).

x = 155.6

mm,

\overset{x}{˙} = - 499.5

mm/s,

\overset{x}{¨} = - 2257

mm/s² and

ω_{3} = - 1.853

rad/s; 80 mm stroke.

Solve it in Simuveliq

The button opens the Slider-crank from the examples menu stopped at θ₂ = 60°.
The sliders table gives ṡ = −499.5 mm/s and s̈ = −2257 mm/s², the points table C = (155.6, 0), and the links table ω = −1.853 rad/s for the rod. The Time ratio card says 1 with stroke = 80 mm.

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Example 4Five-bar: two inputs

A five-bar mechanism has two cranks pinned to ground and two links joined to each other at B. How many inputs does it need for its motion to be determined?

Solution

$n = 5$ (ground + 4 links) and $j_{1} = 5$ pins (the two pivots, the two crank–link joints and B).
$M = 3 \cdot 4 - 2 \cdot 5 = 2$ : with a single motor, point B is not determined (the mechanism «collapses»).
With two motors, one on each crank, B traces a curve that depends on both speeds: that is how many planar robots and parallel plotters work.

M = 2

: two inputs are needed.

Solve it in Simuveliq

In Examples, «Load mechanism» → Five-bar (2 inputs).
The Degrees of freedom card gives M = 2 and the Inputs (motors and actuators) panel has two motors. Delete one: the simulator warns that the mechanism needs 2 inputs and stops moving it.

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Planar mechanisms: degrees of freedom and kinematics

Links, pairs and degrees of freedom

Grashof condition

Position: the vector loop

Velocity and acceleration

Transmission angle

The slider-crank

Common mistakes

Worked examples

Example 1Degrees of freedom, Grashof and transmission angle

Solution

Solve it in Simuveliq

Example 2Position, velocity and acceleration at θ₂ = 90°

Solution

Solve it in Simuveliq

Example 3Slider-crank at 60°

Solution

Solve it in Simuveliq

Example 4Five-bar: two inputs

Solution

Solve it in Simuveliq

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