The Classical Transmission Angle Problem:

The classical transmission angle problem can be stated as follows:
“Determine the crank-rocker proportions of a four-bar mechanism with a given swing angle (ψ) and corresponding  crank rotation (ϕ), or time ratio, such that the maximum deviation of the transmission angle from 90° is minimum.”

The problem can be considered in two parts. The first part is the synthesis problem in which one must determine four-bar mechanisms with crank-rocker proportions that must have a given swing angle and a corresponding crank rotation. There is an infinite set of solutions for this part of the problem. The second part of the problem is concerned with the optimisation. Out of the infinite possible solutions obtained in the first part, one must determine a particular four-bar mechanism whose maximum transmission angle deviation from 90° is a minimum.

For the first part of the problem consider the two dead center positions of the crank-rocker mechanism (see figure above). We can write the loop closure equations for these two particular positions (note that the coupler and the crank are collinear at the dead-centers).

a2eiβ + a3eiβ = a1 + a4eiψ1	(7)
a2ei(β + ϕ) + a3ei(β + ϕ − π) = a1 + a4ei(ψ1 + ψ)	(8)

Rearranging:

(a2 + a3)eiβ − a4eiψ1 = a1	(9)
(a2 − a3)eiβeiϕ − a4eiψ1eiψ = a1	(10)

Let us define Z₁, Z₂ and λ as:

Z₁ = a₂e^iβ

Z₂ = a₄e^iψ₁

λ = a₃/a₂

Z₁ and Z₂  are complex numbers which represent the vectors A₀A_e and B₀B_e and λ is the ratio of the coupler link to the crank length. Without loss of generality we can let a₁ = 1 unit (this correspond to the scaling of the mechanism). Now the loop closure equations (9) and (10) for the dead-center positions can be written in “normalised form” as:

(1 + λ)Z1 − Z2 = 1	(11)
eiϕ(1 − λ)Z1 − Z2eiψ = 1	(12)

Unlike the use of the loop equations for kinematic analysis, now ϕ and ψ are the given swing angle and corresponding crank rotation for which we have to determine the mechanism proportions. The two complex equations are linear in terms of the unknowns Z₁ and Z₂ the solution of which is:

\displaystyle {{\text{Z}}_{1}}=\frac{{1-{\text{e}^{{\text{iψ}}}}}}{{{\text{e}^{{\text{iϕ}}}}-{\text{e}^{{\text{iψ}}}}-\text{λ}\left( {{\text{e}^{{\text{iϕ}}}}+{\text{e}^{{\text{iψ}}}}} \right)}}={{\text{a}}_{2}}{\text{e}^{{\text{iβ}}}}	(13)
\displaystyle {{\text{Z}}_{2}}=\frac{{1-{\text{e}^{{\text{iψ}}}}+\text{λ}\left( {1+{\text{e}^{{\text{iψ}}}}} \right)}}{{{\text{e}^{{\text{iϕ}}}}-{\text{e}^{{\text{iψ}}}}-\text{λ}\left( {{\text{e}^{{\text{iϕ}}}}+{\text{e}^{{\text{iψ}}}}} \right)}}={{\text{a}}_{4}}{\text{e}^{{{{\text{iψ}}_{1}}}}}	(14)

As λ changes from minus infinity to plus infinity, Z₁ and Z₂ describe a curve which is the loci of the tip of the vectors A₀A_e and B₀B_e. According to equations (13) and (14), these loci are two circles for any given value of ϕ and ψ. For ϕ = 160° and ψ = 80° these two circles are as shown below (in order to draw the two circles with the same reference frame, A₀ as the origin and the real axis is along A₀B₀; instead of Z₂, the vector 1 + Z₂ is drawn). There is an infinite set of solutions and these can be obtained by drawing a line from A₀ making  any angle β with respect to A₀B₀. The intersection of this line and the circles are A_e and B_e respectively. Geometrically, the link lengths can be found from the figure as |A₀A_e| = a₂, |A_eB_e| = a₃, |B₀B_e| = a₄, |A₀B₀| = a₁ = 1.

Analytically, we can determine a₂, a₄, and a₃ (= λa₂) for any value of λ, ψ and ϕ using equations (13) and (14) as:

\displaystyle {{\text{a}}_{2}}^{2}={{\text{Z}}_{1}}\overline{{{{\text{Z}}_{1}}}}

\displaystyle {{\text{a}}_{4}}^{2}={{\text{Z}}_{2}}\overline{{{{\text{Z}}_{2}}}}

a₃ = λa₂

a₁ = 1

The link lengths are functions of the variable parameter λ, the given swing angle and the corresponding crank rotation (ϕ and ψ). For crank rocker proportions a necessary (but not sufficient) condition is λ > 1 (why??). One can use the initial crank angle b as the free parameter rather than λ. In such a case the link lengths are given by the following equations:

\displaystyle {{\text{a}}_{2}=-\sin\left( \frac{\text{ψ}}{2}\right)\frac{{\cos\left( {\frac{\text{ϕ}}{2}+\text{β}} \right)}}{{\sin\left( {\frac{{\text{ϕ}-\text{ψ}}}{2}} \right)}}}	(15)
\displaystyle {{\text{a}}_{3}}=+\sin\left( \frac{\text{ψ}}{2}\right)\frac{{\sin\left( {\frac{\text{ϕ}}{2}+\text{β}} \right)}}{{\cos\left( {\frac{{\text{ϕ}-\text{ψ}}}{2}} \right)}}	(16)
a42 = (a2 + a3)2 + 1 − 2(a2 + a3)cosβ	(17)
a1 = 1

One can either use λ or β as a free parameter and obtain different four-bar mechanism proportions. The mechanism proportions that are of crank rocker type will all have the given swing angle and corresponding crank rotation. In order to obtain crank rocker proportions we have limits on swing angle and corresponding crank rotation as:

0° < ψ < 180°

90° + ψ/2 < ϕ < 270° + ψ/2

If we let:

t = tan(ϕ/2)  ,  u = tan[(ϕ − ψ)/2]  ,  v = tan(ψ/2)

The link lengths can be expressed as:

\displaystyle {{\text{a}}_{1}}^{2}=\frac{{{{\text{u}}^{2}}+{{\text{λ}}^{2}}}}{{1+{{\text{u}}^{2}}}}	(18)
\displaystyle {{\text{a}}_{2}}^{2}=\frac{{{{\text{v}}^{2}}}}{{1+{{\text{v}}^{2}}}}	(19)
\displaystyle {{\text{a}}_{3}}^{2}={\text{λ}}^{2}{{\text{a}}_{2}}^{2}=\frac{{\text{λ}}^{2}{\text{v}}^{2}}{{1+{{\text{v}}^{2}}}}	(20)
\displaystyle {{\text{a}}_{4}}^{2}=\frac{{{{\text{t}}^{2}}+{{\text{λ}}^{2}}}}{{1+{{\text{t}}^{2}}}}	(21)

Example:

Determine the proportions of a four-bar mechanism of 120 mm fixed link length and  with ψ = 40° and ϕ = 160°.

• Let us select the initial crank angle β = 60°. Using equations (15)-(17):

\displaystyle {{\text{a}}_{2}}=-\sin \left( {20{}^\circ } \right)\frac{{\cos \left( {80{}^\circ +60{}^\circ } \right)}}{{\sin \left( {80{}^\circ -20{}^\circ } \right)}}=0.30254

\displaystyle {{\text{a}}_{3}}=\sin \left( {20{}^\circ } \right)\frac{{\sin \left( {80{}^\circ +60{}^\circ } \right)}}{{\cos \left( {80{}^\circ -20{}^\circ } \right)}}=0.43969

and

a₄² = (0.30254 + 0.43969)² + 1 − 2(0.30254 + 0.43969)cos60° = 0.80867    ⇒     a₄ = 0.89926

For a₁ = 120 mm, the link lengths are a₂ = 36.3 mm, a₃ = 52.76 and a₄ = 107.91 mm.

The maximum deviation of the transmission angle from 90° will be (Eq. (4)):

\displaystyle {\cos \text{μ}_{\begin{smallmatrix} \text{min} \\ \text{max} \end{smallmatrix}}=\frac{{107.91}^{2}+{52.76}^{2}-{120}^{2}-{36.30}^{2}}{2 \cdot 107.91 \cdot 52.76} \pm \frac{120 \cdot 36.30}{107.91 \cdot 52.76}=-0.113247 \pm 0.765106}

from which we obtain μ_max = 151.44° (D₁ = 61.44°) and μ_min = 49.32° (D₂ = 40.68°).  Since μ_max  deviates by 61.44° from right angle, μ_max is the critical transmission angle.

b) If we select λ = 1.4, using equations (18)-(21):

t = tan(80°) = 5.671282  ,  u = tan(60°) = 1.732051  ,  v = tan(20°) = 0.36397

\displaystyle {{\text{a}}_{1}}^{2}=\frac{{{{{1.732051}}^{2}}+{{{1.4}}^{2}}}}{{1+{{{1.732051}}^{2}}}}=1.24\text{ }\Rightarrow \text{ }{{\text{a}}_{1}}=1.113553

\displaystyle {{\text{a}}_{2}}^{2}=\frac{{{{{0.363970}}^{2}}}}{{1+{{{0.363970}}^{2}}}}=0.116978\text{ }\Rightarrow \text{ }{{\text{a}}_{2}}=0.342020

\displaystyle {{\text{a}}_{3}}=1.4\cdot 0.342020=0.478828

\displaystyle {{\text{a}}_{4}}^{2}=\frac{{{{{5.671282}}^{2}}+{{{1.4}}^{2}}}}{{1+{{{5.671282}}^{2}}}}=1.028948\text{ }\Rightarrow \text{ }{{\text{a}}_{4}}=1.0114371

For a₁ = 120 mm; a₂ = (0.342020/1.113553)×120 = 36.86 mm, a₃ = 51.60 mm, a₄ = 109.31 mm. The critical transmission angles will be:

\displaystyle {\cos \text{μ}_{\begin{smallmatrix} \text{min} \\ \text{max} \end{smallmatrix}}=\frac{{{{{109.31}}^{2}}+{{{51.60}}^{2}}-{{{120}}^{2}}-{{{36.86}}^{2}}}}{{2\cdot 109.31\cdot 51.60}}\pm \frac{{120\cdot 36.86}}{{109.31\cdot 51.60}}=-0.101715 \pm 0.784200}

from which we obtain μ_max = 152.36° (D₁ = 62.36°) and μ_min = 49.96° (D₂ = 43.04°). Since μ_max deviates by 62.36° from right angle, μ_max is the critical transmission angle.

In the second part of the problem, amongst the infinite set of solutions for a given swing angle and crank rotation, one must select the one for which the maximum deviation of the transmission angle from 90° is a minimum. Substituting the link lengths expressed in terms of a single parameter (λ), into the equation for minimum (or maximum) transmission angle (equation (4)):

\displaystyle {\cos \text{μ}_{\begin{smallmatrix} \text{min} \\ \text{max} \end{smallmatrix}}=\frac{{{{\text{a}}_{3}}^{2}+{{\text{a}}_{4}}^{2}-{{\text{a}}_{1}}^{2}-{{\text{a}}_{2}}^{2}}}{{2{{\text{a}}_{3}}{{\text{a}}_{4}}}} \pm \frac{{{{\text{a}}_{1}}{{\text{a}}_{2}}}}{{{{\text{a}}_{3}}{{\text{a}}_{4}}}}}

(5)

which will be a function of λ only. One can determine the value of λ (λ_opt) which minimises the maximum deviation of the transmission angle by setting the derivative of the transmission angle μ with respect to λ in equation (4) to zero. The value of λ_opt is then obtained from the solution of the cubic equation:

Q3 + 2Q2 − t2Q − t2(1 + t2)/u2 = 0

(22)

where Q = t²/λ_opt². The positive real root between 1/u² < Q < t² yields the four-bar mechanism with crank rocker proportions for which the maximum deviation of the transmission angle is a minimum and which satisfies the given swing angle and corresponding crank rotation.

Analytical solution of cubic equations is possible. Alternatively, one can use the root finding routines available in MathCAD or MATLAB or use Goal Seek and Solver tools in Excel to determine the correct root of this cubic. A simple numerical solution for the root of the cubic equation can be obtained by applying Newton-Raphson method. Making an initial guess for the root at the mid point of the interval  1/u² < Q < t²  (i.e. Q₀ = (1/u² + t²)/2) and refining the estimate on the root Q by the recursion formula:

\displaystyle {{Q}_{{i+1}}}=\frac{{2{{Q}_{i}}\left( {{{Q}_{i}}+1} \right)+{{\text{t}}^{2}}\left( {1+{{\text{t}}^{2}}} \right)/{{\text{u}}^{2}}}}{{{{Q}_{i}}\left( {3{{Q}_{i}}+4} \right)-{{\text{t}}^{2}}}}

(23)

The iteration can be stopped when (Q_i+1 – Q_i)/Q_i   is a sufficiently small number (say 10^-6).

The solution for optimum transmission angle for given swing angle corresponding crank rotation is shown in a chart form in Chart 1 (this chart is known as Alt chart and is given in VDI 2130 standard). From these charts the value of β and max(μ_min) can be obtained and the link lengths can also be found using the equations which express the link lengths in terms of β.

You can download the Excel file for the design of crank rocker mechanisms when the swing angle and the corresponding crank rotations are given.

Example:

Determine the proportions of a four-bar mechanism of 120 mm fixed link length and with ψ = 40° and ϕ = 160° and for which the maximum deviation of the transmission angle from 90° is a minimum.

Noting that t = 5.671282 and u = 1.732051, 0.333333 < Q < 32.163437. Let us assume Q₀ = 16.248. The values for seven iterations are:

i	0	1	2	3	4	5	6	7
Q	16.248	11.47207	8.095663	7.982059	7.857866	7.855707	7.855706	7.855706

Therefore Q = 7.855706 is correct to at least six significant digits. Using λ_opt = 2.023432, the link lengths are:

\displaystyle {{\text{a}}_{1}}^{2}=\frac{{{{{1.732051}}^{2}}+{{{2.023432}}^{2}}}}{{1+{{{1.732051}}^{2}}}}=1.773569\text{ }\Rightarrow \text{ }{{\text{a}}_{1}}=1.331754

\displaystyle {{\text{a}}_{2}}^{2}=\frac{{{{{0.363970}}^{2}}}}{{1+{{{0.363970}}^{2}}}}=0.116978\text{ }\Rightarrow \text{ }{{\text{a}}_{2}}=0.342020

\displaystyle {{\text{a}}_{3}}=2.023432\cdot 0.342020=0.692054

\displaystyle {{\text{a}}_{4}}^{2}=\frac{{{{{5.671282}}^{2}}+{{{2.023432}}^{2}}}}{{1+{{{5.671282}}^{2}}}}=1.093304\text{ }\Rightarrow \text{ }{{\text{a}}_{4}}=1.045612

For a₁ = 120 mm:  a₂ = (0.342020/1.331754)×120 = 30.82 mm a₃ = 62.36 mm, a₄ = 94.22 mm. The maximum deviation of the transmission angle from 90° will be:

\displaystyle {\cos \text{μ}_{\begin{smallmatrix} \text{min} \\ \text{max} \end{smallmatrix}}=\frac{{{{{94.22}}^{2}}+{{{62.36}}^{2}}-{{{120}}^{2}}-{{{30.82}}^{2}}}}{{2\cdot 94.22\cdot 62.36}}\pm \frac{{120\cdot 30.82}}{{94.22\cdot 62.36}}=0.219938\pm 0.629455}

from which we obtain μ_max = 114.17° (D₁ = 24.17°) and μ_min = 31.85° (D₂ = 58.15°). Since μ_min deviates most, μ_min is the critical transmission angle. If we refer to Chart 1, we see that for ψ = 40° and ϕ = 160°, max(μ_min) = 32° and β = 50.5°. Using the formula for the link lengths: a₁ = 1, a₂ = 0.2565, a₃ = 0.5201 and a₄ = 0.784. For a₁ = 120 mm; a₂ = 30.78 mm, a₃ = 62.41 mm and a₄ = 94.08 mm. The final mechanism is shown below.

In the solution of the transmission angle problem, there are two special cases. One special case occurs when ϕ = 180° (centric four-bar), for which the maximum and minimum values of the transmission angle is:

\displaystyle {\cos \text{μ}_{\begin{smallmatrix} \text{min} \\ \text{max} \end{smallmatrix}}= \pm \frac{{{{\text{a}}_{1}}{{\text{a}}_{2}}}}{{{{\text{a}}_{3}}{{\text{a}}_{4}}}}}

If the maximum deviation of the transmission angle is to be minimised, then a₂ = 0 (since a₁ = 1), a₄ = 0 and a₃ = 1, which is physically impossible four-bar mechanism proportion. In this case, we search for a crank-rocker mechanism for which the transmission angle deviation is of reasonable magnitude while the ratio of the link lengths are of acceptable proportions (the transmission angle will improve as λ or β increases). Using equations (15)-(17) we obtain the link lengths in terms of initial crank angle and the swing angle (ϕ = 180°) as:

\displaystyle \frac{\text{a}_2}{\text{a}_1}=\text{tan}(\text{ψ}/2)\text{sinβ}

\displaystyle \frac{\text{a}_3}{\text{a}_1}=\text{cosβ}

\displaystyle \frac{\text{a}_4}{\text{a}_1}=\frac{\text{sinβ}}{\text{cos}(\text{ψ}/2)}

Substituting the link ratios into into the expression for cosμ_min:

\displaystyle \cos \text{μ}_\text{min}=\frac{\text{tan}(\text{ψ}/2)\text{sinβ}}{\text{cosβ}\frac{\text{sinβ}}{\text{cos}(\text{ψ}/2)}}=\frac{\text{sin}(\text{ψ}/2)}{\text{cosβ}}

Now the link lengths ratios can be expressed in terms of minimum transmission angle μ_min as:

\displaystyle \text{cosβ}=\frac{\text{sin}(\text{ψ}/2)}{\cos \text{μ}_\text{min}}

In the chart below, the change of the link length ratios a₄/a₁ and a₂/a₁ as a function of the swing angle is shown for different minimum transmission angle values. For example, for ψ = 90° the best minimum transmission angle (with a₄/a₁ = 0 and a₂/a₁ = 0) is 45°. By reducing the transmission angle to 40° from the chart a₄/a₁ ≈ 0.54 and a₂/a₄ ≈ 0.38. Using the equations for μ_min = 40°, ψ = 90° yield (to five digit accuracy): a₄/a₁ = 0.54398, a₃/a₁ = 0.92306 and a₂/a₁ = 0.38463. If the transmission angle is 43° then a₄/a₁ = 0.25970, a₃/a₁ = 0.98299 and a₂/a₁ = 0.18363. A higher transmission angle will result with a smaller crank and rocker and the coupler and the fixed link ratio close to unity. Note that for any crank-rocker mechanism μ_min < π/2 – ψ/2 i.e. for 90° swing angle the transmission angle cannot be greater than 45° (this statement is true for any corresponding crank angle rotation, ϕ. For a given swing angle centric four-bar will have a better force transmission characteristics).

Another special case is when ϕ – ψ = 180°. The transmission angle deviation is optimum when \displaystyle {{\text{λ}}_{{\text{opt}}}}=\sqrt {1+\frac{1}{{\sin \left( {\text{ψ}/2} \right)}}} . The link lengths are:

a₁ = 1

\displaystyle {\text{a}_{2}}=\sin \left( {\text{ψ}/2} \right)

\displaystyle {\text{a}_{3}}=\sqrt{{{\text{a}_{2}}\left( {1+{\text{a}_{2}}} \right)}}

\displaystyle {\text{a}_{4}}=\sqrt{{1+{\text{a}_{2}}}}