2.1 General case

Assume a fixed joint point $A_{\mathrm{0}}=(A_{\mathrm{0}x},A_{\mathrm{0}y})$ and a point $P_{\mathrm{1}}=(P_{\mathrm{1}x},P_{\mathrm{1}y})$ on a straight line, where the direction of the straight line is β₁ and the displacement is T. Vector T (T≠0) points from the Ball point P₁ to the polar point T. We note that the parameter λ at point P between the t axis and x axis is known during the synthesis of the five-point contact mechanism that approximates a straight line, as compared with the case of four-point contact (Yin et al., 2012). Parameters α_a0,α_b0, α_b0, α_c0, and α₁₀ are defined as the angles from the positive direction of the T axis to the vectors B, B₀P, C₀P, and PP₁, respectively. Vectors α_a0,α_b0, α_b, α_c, and α₁₀ are defined as ${\mathit{\alpha }}_a={\mathit{\alpha }}_{a\mathrm{0}}-\mathit{\lambda }$, ${\mathit{\alpha }}_a={\mathit{\alpha }}_{a\mathrm{0}}-\mathit{\lambda }$, ${\mathit{\alpha }}_a={\mathit{\alpha }}_{a\mathrm{0}}-\mathit{\lambda }$, and ${\mathit{\alpha }}_a={\mathit{\alpha }}_{a\mathrm{0}}-\mathit{\lambda }$, respectively. For simplicity of calculation, let diameter D of the inflection point circle be 1. The expected mechanism can be obtained by scaling the synthesized mechanism by the diameter of the inflection point circle. Furthermore, the coordinates of another fixed joint B₀ and two motion joints A and B must be determined to synthesize four-bar mechanisms that approximate a straight line. Details regarding the definition of parameters are taken from Figs. 1 and 2 in reference of Yin et al. (2012), as shown in Fig. 1.

According to the kinematic geometric theory of infinitely close positions, several points can be selected from a motion plane at any instantaneous position, whose trajectory has fourth-order contact with its curvature circle. This means that these points are the circle points of five infinitely close positions, namely Burmester points, which are higher-order stationary points of the trajectory curvature. Using the Euler-Savary equation (Yin et al., 2012) to solve the two-order derivative equation $\frac{d^{\mathrm{2}}{\mathit{\rho }}_m}{\mathrm{d}{\mathit{\sigma }}^{\mathrm{2}}}=\mathrm{0}$, or using the curvature stagnation point curve equation to obtain the first derivative equation $\frac{d{\mathit{\rho }}_m}{\mathrm{d}\mathit{\sigma }}=\mathrm{0}$, we obtain:

where N and M are two auxiliary variables defined as follows:

Equation (1) is a quartic equation that has four real roots at most. These four roots have the value α_i of the polar coordinates of the Burmester point. Motion joints A and B correspond to roots α_a and α_b, respectively, and the connecting rod point P₁ corresponds to root α₁. We assume the other motion joint is C, which corresponds to root α_c.

After we determine the instantaneous center P and parameter λ, by definition, we can obtain angles α₁ and α_a directly. α_b and α_c are solved as follows.

Figure 2$M_{B_{\mathrm{0}}BC{C}_{\mathrm{0}}}$solution region diagram of mechanism type.

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According the relation between the fourth-order equation root and coefficient of Eq. (1), we obtain:

or

Assuming tanα₁ in Eq. (4) and tanα_c in Eq. (5) are the two roots of a quadratic equation, and according to the Vieta theorem, we have the following:

Similarly, Eq. (6) can be regarded as the quadratic equation of tanα, and according to the Vieta theorem, the roots of Eq. (6) are imaginary when:

Therefore, the other two Burmester points are imaginary. Combining Eq. (6) with the cubics of stationary curvature equations, we have the following:

By substituting PP₁=Dsinα₁ into Eq. (10), and assuming D equals 1, we obtain:

By substituting Eq. (11) into Eq. (6) and doing some rearranging, we have the following:

where U=tanα_atanα_b and $V=\tan {\mathit{\alpha }}_a+\tan {\mathit{\alpha }}_b$. By substituting Eq. (12) into Eq. (11) and doing some rearranging, we obtain:

Furthermore, by substituting Eqs. (11) and (13) into the Euler–Savary equation: $\left\{\begin{array}{l}PA=\frac{P{A}_{\mathrm{0}}\cdot D\sin {\mathit{\alpha }}_a}{P{A}_{\mathrm{0}}+D\sin {\mathit{\alpha }}_a}\\ PB=\frac{P{B}_{\mathrm{0}}\cdot D\sin {\mathit{\alpha }}_b}{P{B}_{\mathrm{0}}+D\sin {\mathit{\alpha }}_b}\end{array}\right.$,

we obtain the following:

Since the diameter D of the inflection-point circle equals 1, by combining it with Eq. (14), we obtain:

By substituting Eq. (16) into Eq. (4) and doing some rearranging, we have:

Considering that motion joints C and B are Burmester points, we replace joint B in Eq. (15) with joint C and substitute Eq. (17) into Eq. (15). Then, PC can be obtained similarly, as follows:

According to the Euler-Savary equation:

By substituting Eq. (21) into Eq. (14), we obtain:

Equation (22) can be regarded as an equation for the unknown α_b. After rearranging, we have:

where: $E=\tan {\mathit{\alpha }}_a\left(PA-\sin {\mathit{\alpha }}_a\right)$; $F=\mathrm{2}PA\tan {\mathit{\alpha }}_a\tan {\mathit{\alpha }}_{\mathrm{1}}+\mathrm{2}PA{\tan }^{\mathrm{2}}{\mathit{\alpha }}_a-{\tan }^{\mathrm{2}}{\mathit{\alpha }}_a\sin {\mathit{\alpha }}_a-\mathrm{3}\tan {\mathit{\alpha }}_a\tan {\mathit{\alpha }}_{\mathrm{1}}\sin {\mathit{\alpha }}_a$; G=PAtanα_atanα₁sinα_a.

There are two roots for α_b in Eq. (23), which are assumed to be α_b1 and α_b2. So, we have:

PB can be obtained by substituting α_b into Eq. (15), and furthermore, PB₀ can be calculated using Eq. (20). We note that there is no real root for α_b when $\mathrm{\Delta }=F^{\mathrm{2}}-\mathrm{4}EG<\mathrm{0}$.

The values of PC and PC₀ can be determined similarly, and the diameter of the actual inflexion circle can be calculated using the equation PP₁=Dsinα₁. After calculating PA, PA₀, PB, PB₀, PC, and PC₀, we can obtain the actual size of the mechanism by multiplying these values by the actual diameter of the inflection point circle.

The coordinates of motion joints A, B, and C and fixed joints C₀ and B₀ can be calculated as follows:

For a given set of conditions including a fixed joint A₀, a point P₁ on a straight line, the direction of the straight line β₁, displacement T, and parameters λ, we can calculate three sets of connecting rods AA₀, BB₀, and CC₀ using the proposed method. By arbitrarily selecting two connecting rods from AA₀, BB₀, and CC₀, we obtain a four-bar linkage that approximates a straight line. Since α_b has two roots α_b1 and α_b2, a set of given conditions corresponds to six five-point-contact four-bar linkages that approximate a straight line. However, the fact that connecting rod BB₀ corresponds to α_b1 is the same as connecting rod CC₀ corresponding to α_b2. Also, the fact that connecting rod CC₀ corresponds to α_b1 is the same as connecting rod BB₀ corresponding to α_b2. Therefore, there are only three four-bar linkages.

Table 1Mechanism parameters.

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Figure 3Example diagrams of mechanisms.

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2.2 Special condition: Burmester point located at the pole of the inflection circle

When the Burmester point is located at the pole of the inflection circle, parameter λ cannot be determined arbitrarily. Below, we calculate the coordinates of another fixed joint B₀ and two motion joints A and B.

Figure 4Mechanism solution region diagrams for all mechanism solutions.

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When the Burmester point coincides with the Ball point at the pole of the inflection circle, α₁ equals 90^∘. The coordinates of the instantaneous center P can be determined by its definition. According to the definition of parameter λ, we have:

By substituting ${\mathit{\alpha }}_{\mathrm{1}}=\mathrm{90}{}^{\circ }$ into Eq. (14), and assuming that tanα₁ approaches positive infinity, we have:

where U=tanα_atanα_b and $V=\tan {\mathit{\alpha }}_a+\tan {\mathit{\alpha }}_b$. Another Burmester point can be obtained using Eqs. (17) and (18). Using Eq. (4), we have:

By substituting Eq. (34) into the equation for the curvature of a point on the curve, we obtain:

We can then derive PA, PA₀, and PB₀ from the formula in reference (Yin et al., 2011). The coordinates of joints A, B, and B₀ can be calculated using Eqs. (26), (27), and (28), respectively, and the coordinates of joints C and C₀ are as follows:

Unlike the general condition, when the Burmester point coincides with the Ball point at the pole of the inflection circle, α_b has only one root. Therefore only three five-point-contact four-bar linkages that approximate a straight line can be obtained with a given set of conditions.

Figure 5Synthesized mechanisms.

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3.1 General case

3.1.2 Example

The task is to design four-bar linkages that approximate a straight line with the following conditions: fixed joint $A_{\mathrm{0}}=(\mathrm{0},\mathrm{0})$, a point $P_{\mathrm{1}}=(\mathrm{10},\mathrm{10})$ on the straight line, and the displacement T=5. Also, the synthesized four-bar linkages should be five-point-contact linkages that approximate a straight line.

There are three solution-region diagrams. In the mechanism-type solution region diagram of mechanism $M_{B_{\mathrm{0}}BC{C}_{\mathrm{0}}}$ shown in Fig. 2, we assume that direction β₁ equals 120^∘ and angle λ equals −85^∘, and then we have three synthesized linkages that approximate a straight line, as shown in Fig. 3d, the parameters of which are shown in Table 1. The other two synthesized mechanisms $M_{A_{\mathrm{0}}AB{B}_{\mathrm{0}}}$ and $M_{A_{\mathrm{0}}AC{C}_{\mathrm{0}}}$ are shown in Fig. 3b and c. The classification of planar four-bar linkages is performed with reference to the method developed by Barker (1985):

• 0.

  No solution,

• 1.

  Crank rocker,

• 2.

  Double rocker,

• 3.

  Rocker crank,

• 4.

  Double crank,

• 5.

  Triple rocker (in–out),

• 6.

  Triple rocker (out–out),

• 7.

  Triple rocker (out-in), and

• 8.

  Triple rocker (in-in).

Figure 6Mechanism solutions with different displacements T.

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Figure 7Optimal mechanism and two cognate mechanisms.

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Table 2Mechanism parameters.

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3.2 Special condition: Burmester point located at the pole of the inflection circle