t1 four-bar linkage analysis revised - mechanical design 101 · 2020. 6. 26. · linkage in the...

Four-bar Linkage Position Analysis

J. M. McCarthy and B. Roth

ME 322 Kinematic Synthesis of Mechanisms

Linkage in the World Frame


Four-bar LinkageA four-bar linkage consists of two interconnected levers. The

connecting link, or coupler, transfers the rotation of the input lever, or crank, to the output crank.

Locate the linkage in a reference position and let the coordinates of its fixed pivots be O, C and the coordinates of its moving pivots be A, B.

Introduce the vectors Z1=A-O, Z2=B-A, Z3=B-C and Z4=C-O, and ZO=O, ZC=C.

Closure ConstraintThe constant length of the coupler constrains the rotation

angles of the two cranks.

Rotation Operator


Rotation matrix

The planar rotation matrix rotates a vector Z=(x, y) by the angle 𝜃:

The matrix 𝜽 is an operator that rotates the vector Z from its initial position to its new position 𝜽Z.

To see this substitute (x, y) = R(cos𝛼, sin𝛼) and compute,

Linkage Closure Equation


The constant length of the coupler imposes a constraint on the input and output angles of the two cranks of the linkage.

The vector along the coupler link is given by

𝝋Z2 = (ZC + 𝜼Z3) - (ZO + 𝜸Z1),

or

𝝋Z2 = Z4 + 𝜼Z3 - 𝜸Z1.

This can also be written in the form,

𝜸Z1 + 𝝋Z2 = Z4 + 𝜼Z3.

known as the loop equation of the linkage.

The length of the coupler link is given by

h2 = (Z4 + 𝜼Z3 - 𝜸Z1).(Z4 + 𝜼Z3 - 𝜸Z1).

This is the linkage closure equation.

Expand the Linkage Closure Equation


Use Mathematica to expand the linkage closure equation.

Define the function Ei[𝜃_] to assemble the rotation matrix.

Then define the coordinates of the fixed and moving pivots in the reference position, Opt, Apt, But, and Cpt.

Use this data to define the link vectors, Z1, Z3, Z4, and compute Ei[𝛾]Z1, and Ei[𝜂]Z3.

Expand the Linkage Closure Equation


Use TrigExpand to compute the linkage closure equation.

Collect the coefficients of Cos[𝜂] to define Ac, the coefficients of Sin[𝜂] to define Bc, and all the remaining terms to define Cc.

The result is an equation of the form

A(𝛾) cos𝜂 + B(𝛾) sin𝜂 + C(𝛾) = 0.

For a given four-bar linkage we have the coordinates O, A, B, and C, so for a given value of the input angle 𝛾, this equation determines the output 𝜂.

Solving the Closure Equation


For a given value 𝛾, the linkage closure equation A(𝛾) cos𝜂 + B(𝛾) sin𝜂 + C(𝛾) = 0,

is solved to determine the angle 𝜂, as follows.

• Introduce the angle δ, defined such that cosδ = A/(A2+B2)1/2 and sinδ=B/(A2+B2)1/2, so

δ = arctan(B/A);• then the constraint equation becomes

cosδ cos𝜂 + sinδ sin𝜂 + C/(A2+B2)1/ 2= 0;• The trigonometric identity for the cosine of the difference of two angles yields

cos(𝜂 - δ) = -C/(A2+B2)1/ 2,• introduce 𝜀= arccos(-C/(A2+B2)1/ 2), then we have the two solutions,

𝜂- = δ - 𝜀 and 𝜂+ = δ + 𝜀

Notice, the arccos() can be evaluated only if -1≤ C/(A2+B2)1/ 2 ≤ 1, which is the condition on values for 𝛾 that allow the linkage to be assembled.

Compute the Coupler Angle


Once the input angle 𝛾 and output angle 𝜂 are known, we determine the coupler angle 𝜑 by the following formula,

cos 𝜑 = (B-A).(Z4 + 𝜼Z3 - 𝜸Z1) = xh, sin 𝜑 = (B-A)×(Z4 + 𝜼Z3 - 𝜸Z1) = yh.

The result is

𝜑 = arctan(xh, yh).

The cross product for two dimensional vectors is defined as

The Range of Crank Angles


• Limits to the input angle θ occur in the configurations shown in the figures below. Using the cosine law we obtain the two cases:

cosθmin = ((h-b)2-a2 - g2)/2ag, and cosθmax = ((h+b)2-a2 - g2)/2ag.

• It is possible that one, both or neither limit exists depending on the dimensions of the linkage. Also notice that arccos() has two values, therefore if θmin is a limit then -θmin is as well.

Thus, we have four cases: 1. Neither limit exists and the crank can make a full rotation; 2. only θmin exists and the crank oscillates through the range θmin≤θ≤π, and -π≤θ ≤-θmin;

3. only θmax exists and the crank oscillates though the range -θmax≤θ≤θmax; or

4. both limits exist and the crank oscillates in either range θmin≤θ≤θmax or -θmax≤θ≤θmin .

Velocity Loop Equations


The loop equations of a four-bar linkage are give byPosition: 𝜸Z1 + 𝝋Z2 = Z4 + 𝜼Z3.

The velocity loop equations are obtained by computing the time derivatives of this equation. Because vectors Z1, Z2, Z3 and Z4 are constant, we need only consider the derivative of a rotation operator 𝜽 is given by

where 𝜔𝜃 is the time derivative (angular velocity) of the angle 𝜃. The matrix J rotates a vector by 90 degrees and has the property that J2=-1, where 1 is the 2x2 identity matrix.

It is convenient to use vector notation, by introducing the unit vector k perpendicular to the x-y plane, so we have JZ3 = k×Z3 = Z3⊥, so the vector Z3⊥ is obtained from Z3 by a 90 degree rotation.

Therefore, we haveVelocity: 𝜔𝛾𝜸Z1⊥ + 𝜔𝜑 𝝋Z2⊥ = 𝜔𝜂𝜼Z3⊥,

where 𝜔𝛾, 𝜔𝜑, and 𝜔𝜂 are the angular velocities associated with the angles 𝛾, 𝜑, and 𝜂. Substitute Z1 = A-O, Z2 = B-A, Z3 = B-C, to obtain the equivalent version

𝜔𝛾𝜸(A-O)⊥ + 𝜔𝜑𝝋(B-A)⊥ = 𝜔𝜂𝜼(B-C)⊥.

Given the input angular velocity 𝜔𝛾 the velocity loop equations are solved to determine 𝜔𝜑, and 𝜔𝜂.

Velocity Pole (input side)


The movement of the coupler has the property that at every instant there is a point that has zero velocity, called the velocity pole.

Let the rotation of the input crank be parameterized by t, so we have 𝛾(t), then the coupler curve is given by

P(𝛾(t))= ZO + 𝜸(t)Z1 + 𝝋(t)Z5,or

P(t)= O + 𝜸(t)(A-O) + 𝝋(t)(P-A),

The velocity of the coupler point is given byVP = ω𝛾𝜸(t)(A-O)⊥ + ωφ 𝝋(t)(P-A)⊥.

Let A𝛾 = 𝜸(t)(A-O) + O and P𝜑 = 𝝋(t)(P-A) + A𝛾, so this equation becomesVP = ω𝛾(A𝛾-O)⊥ + ωφ (P𝜑-A)⊥.

There is a point I in the coupler that has zero velocityVI = ω𝛾(A𝛾-O)⊥ + ωφ(I-A𝛾)⊥ = 0,

that isI24 = A𝛾 - (ω𝛾/ωφ)(A𝛾-O).

This point, velocity pole, lies on the line through A𝛾 in the direction A𝛾-O.

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C

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O

Z3

Z4

ZO

ZC

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Z6

I24

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r24

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s24

Velocity Pole (output side)


The rotation of the output crank is also parameterized by t, so we have 𝜂(t), then the coupler curve is given by

P(𝜂(t))= ZC + 𝜼(t)Z3 + 𝝋(t)Z5,or

P(t)= C + 𝜼(t)(B-C) + 𝝋(t)(P-B),

The velocity of the coupler point is now given byVP = ω𝜂𝜼(t)(B-C)⊥ + ωφ𝝋(t)(P-B)⊥.

Introduce B𝜂 = 𝜼(t)(B-C) + C and P𝜑 = 𝝋(t)(P-B) + B𝜂, so this equation becomes

VP = ω𝜂(B𝜂-C)⊥ + ωφ(P𝜑-B𝜂)⊥

The coordinates of the velocity pole are now given byVI = ω𝜂(B𝜂-C)⊥ + ωφ (I24-B𝜂)⊥ = 0,

soI24 = B𝜂 + (ω𝜂/ωφ)(B𝜂-C).

The velocity pole I24 lies on the line through B𝜂 in the direction B𝜂-C.

The velocity pole lies on the intersection of the lines along the links OA𝛾 and CB𝜂.

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I24

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r24

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s24

Velocity of Points in the Coupler


The velocity of points in the coupler can calculated in terms of their position relative to the velocity pole.

The velocity of the coupler point P is given byVP = ω𝛾(A𝛾-O)⊥ + ωφ (Pφ-A𝛾)⊥,

andVP = ω𝜂(B𝜂-C)⊥ + ωφ(Pφ-B𝜂)⊥.

The definition of the velocity pole I24 is given byVI = ω𝛾(A𝛾-O)⊥ + ωφ (I-A𝛾)⊥ = 0,

andVI = ω𝜂(B𝜂-C)⊥ + ωφ (I24-B𝜂)⊥ = 0.

Compute VP - VI for both the input and output equations to obtain,

VP = ωφ (Pφ-I24)⊥.

Thus, the velocity of points in the coupler can be computed directly from their position relative to the velocity pole Pφ-I24

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Z4

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r24

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Speed Ratio (reference position)


The closure equation of the linkage can be used to determine the input-output speed ratio, R= 𝜔𝛾/𝜔𝜂, by computing its derivative in the reference position,

h2 = (Z4 + Z3 - Z1).(Z4 + Z3 - Z1), that is

0 =(𝜔𝜂Z3⊥ - 𝜔𝛾Z1⊥).Z2.

For convenience introduce the notation Z3⊥ = k×Z3, then we have0 = (𝜔𝜂 k×(B-C) - 𝜔𝛾 k×(A-O)). (B-A) .

Factor out k×, then interchange the dot and cross products to obtaink.(𝜔𝜂(B-C) - 𝜔𝛾(A-O)) × (B-A)= 0 .

Notice that B × (B-A) = A× (B-A), which means A and B can be replaced by I13 = A + t(B-A), where I13 lies on the the line through OC. Thus, we have

k.(𝜔𝜂(I13-C) - 𝜔𝛾(I13-O)) × (B-A)=0.

Therefore, 𝜔𝛾/𝜔𝜂 = (r13+g)/r13, where r=|I13-O| and g=|I13-C|.

W

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Z4

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I24

I13

s13

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r13

Speed Ratio (any position)


The closure equation in any configuration can be used to determine the associated input-output speed ratio, R= 𝜔𝛾/𝜔𝜂, by computing the derivative of,

h2 = (Z4 + 𝜼Z3 - 𝜸Z1).(Z4 + 𝜼Z3 - 𝜸Z1), that is

(𝜔𝜂k×𝜼(B - C) - 𝜔𝛾k×𝜸(A - O)).𝝋(B - A) = 0,where for convenience we have [J]Z3 = k×Z3.

Introduce the notation A𝛾 = 𝜸(A-O)) + O, and B𝜂 = 𝜼(B-C) + C, then 𝝋(B-A) = B𝜂 - A𝛾, so we have

k×(𝜔𝜂(B𝜂 - C) - 𝜔𝛾(A𝛾 - O)).(B𝜂 - A𝛾) = 0.or

k.(𝜔𝜂(B𝜂 -C) - 𝜔𝛾(A𝛾-O)) × (B𝜂 - A𝛾)= 0 .As we have seen before the points A𝛾 and B𝜂 can be replaced by

I13 = A𝛾 + s (B𝜂 - A𝛾) = r13 eOC + O = (r13 + g) eOC + C. to define,

k.(𝜔𝜂(I13 - C) - 𝜔𝛾(I13- O))×(B𝜂 - A𝛾) = 0

Therefore, 𝜔𝛾/𝜔𝜂 = (r13+g)/r13.

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ZO

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I13

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r13

Coordinates of the Instant Center


The coordinates of the instant center I13 can be computed directly from the condition that it is the intersection of lines along the links OC and AB.

Introduce the unit vectors eOC= (cos𝜃0, sin𝜃0) and eAB= (cos𝜑0, sin𝜑0) obtained by normalizing the vectors Z4=C-A and Z2=B-A.

Then we solve for the distances s and t defined by the equationI13 = O + r13 eOC = A + s13 eAB.

Introduce A-O = a eOA = a (cos𝛾0, sin𝛾0) and rearrange to obtain,r13 eOC - s13 eAB = a eOA .

Solve these equations to obtain,r13 = a sin(𝛾0 - 𝜑0) / sin(𝜃0 - 𝜑0)

and r13 = a sin(𝛾0 - 𝜃0) / sin(𝜃0 - 𝜑0).

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Mechanical Advantage


The principle of virtual work states that if there are no energy losses in a machine, known as an ideal machine, then the “power in equals power out.” For a four-bar linkage the power in and power out are given by Pin=Tin𝜔𝛾 and Pout=Tout𝜔𝜂.

The mechanical advantage of the linkage is the ratio of output torque to input torque, so we have

MA = Tout/Tin = 𝜔𝛾/𝜔𝜂.

Thus, for an ideal system the mechanical advantage is given by the speed ratio, R=𝜔𝛾/𝜔𝜂, that is

MA = Tout/Tin = 𝜔𝛾/𝜔𝜂 = (r13+g)/r13. W

C

B

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O

Z3

Z4

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ZC

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I13

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r13

Summary


• An ideal linkage is a collection of rigid links connected by ideal joints. These assumptions allow the use of geometry to analyze the movement of the linkage and evaluate its mechanical advantage. Real linkage systems flex and lose energy through friction and wear.

• The closure equation of a linkage is defined by the requirement that input and output links must move in a way that maintains a constant length for the coupler link. Differentiation of this constraint yields the input-output speed ratio and mechanical advantage.

• The position loop equations are used to compute the coupler angle.

• The first derivative of the loop equations define the velocity loop equations, which are used to compute the angular velocities of the output crank and coupler link.

• The ratio of the input to output angular velocities of the linkage define its mechanical advantage.

t1 four-bar linkage analysis revised - mechanical design 101 · 2020. 6. 26. · linkage in the...

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