CN106126812A - A kind of Tape movement pair six-bar mechanism linear-elsatic buckling method - Google Patents

Abstract

The invention discloses a kind of linear-elsatic buckling method of Tape movement pair six-bar mechanism, the most respectively two loops relevant to input and output are analyzed, and then draw a kind of Tape movement pair six-bar mechanism linear-elsatic buckling method by its interaction.The present invention realizes identifying branch (loop) with mechanism's basic ring equation, identifies more accurately, intuitively, quickly；The present invention is easy to develop mathematics software and is simulated emulation, contributes to the research and analysis of parallel robot structure, has the highest use value；The present invention can be embedded in various Machine Design class business software, has good social value and economic worth.

Description

A kind of Tape movement pair six-bar mechanism linear-elsatic buckling method

Technical field

The invention belongs to Machine Design manufacturing technology field, relate to a kind of new Tape movement pair six-bar mechanism linear-elsatic buckling side Method, the method is applicable to all Tape movement pair plane six-bar linkages.

Background technology

For Tape movement pair six-bar mechanism, seek a kind of simple possible and accurately mobility method of discrimination be the heaviest Want.At present, both at home and abroad existing multiple scholar's linkage are studied, complete to quadric chain of Grashof criterion Rotatory has carried out judging accurately；Kwun-Lon Ting professor's N bar spin theory provides series of theories method to monocycle Plane mechanism and space mechanism are studied；Domestic, Guo Xiaoning professor and professor Chu Jinkui are by Stephenson six bar Mechanism regards a basic double leval jib and double-rod group mechanism as, is differentiated the mobility of Stephenson six-bar mechanism.This The most all research to research single-degree-of-freedom dicyclo linkage is had laid a good foundation, but, the object of study of said method The most single, the branch of different types of Tape movement pair six-bar mechanism effectively can not be identified.

Summary of the invention

In order to solve above-mentioned technical problem, the present invention proposes the branch of the Tape movement pair six-bar mechanism of a kind of simple possible Recognition methods, will combine with two rings of input, output, input, the dependent equation of output angle, thus to Tape movement pair The branch of six-bar mechanism effectively identifies.

The technical solution adopted in the present invention is: a kind of linear-elsatic buckling method of Tape movement pair six-bar mechanism, its feature exists In: being combined by two relevant to input and output for Tape movement pair six-bar mechanism monocycles, being interacted by it show that Tape movement is secondary The linear-elsatic buckling method of six-bar mechanism.

As preferably, implementing of the present invention comprises the following steps:

Step 1: Tape movement pair four-bar mechanism branch differentiates:

Any one a kind of Tape movement pair six-bar mechanism all comprises one or more Tape movement pair four-bar mechanism ABCDA, its Ring equation can be represented by Euler's formula, it may be assumed that

$a_2{e}^{{\mathrm{i\θ}}_2}+a_3{e}^{{\mathrm{i\θ}}_3}=a_1+s_1{e}^{{\mathrm{i\α}}_2}---\left(1\right)$

Wherein, a₁、a₂、a₃、s₁Referring to the length of DA, AB, BC, CD double leval jib in four-bar linkage ABCDA respectively, e is certainly So the end of logarithm, i is imaginary unit, θ₂、θ₃、α₂Refer to trunnion axis and rotate the angle arriving double leval jib AB, BC, CD respectively counterclockwise Degree.

According to its real part and imaginary part, equation (1) can be expressed as:

s₁=(a₂cosθ₂+a₃cosθ₃-a₁)/cosα₂ (2)

s₁=(a₂sinθ₂+a₃sinθ₃)/sinα₂ (3)

By eliminating s₁, equation (1) can be expressed as:

a₁sinα₂+a₂sin(θ₂-α₂)+a₃sin(θ₃-α₂)=0 (4)

Work as θ₂、θ₃When being input parameter, equation (4) can be expressed as Input output Relationship model.Utilize half-angle formulas:

x₃=tan (θ₃/2) (5)

Equation (4) can be expressed as:

$P_1{x}_3^2+Q_1{x}_3+R_1=0---\left(6\right)$

Wherein:

P₁=a₂sin(θ₂-α₂)+(a₁+a₃)sinα₂ (7.1)

Q₁=2a₃cosα₂ (7.2)

Q₁=a₂sin(θ₂-α₂)+(a₁-a₃)sinα₂ (7.3)

Exist to four-bar mechanism configuration, work as P₁It must is fulfilled for when ≠ 0

${\mathrm{\Δ}}_1=Q_1^2-4P_1{R}_1\≥0---\left(8\right)$

Abbreviation obtains:

△₁=4S₁S₂ (9)

Wherein:

S₁=a₃-a₂sin(θ₂-α₂)-a₁sinα₂ (10.1)

S₂=a₃+a₂sin(θ₂-α₂)+a₁sinα₂ (10.2)

Work as △₁When=0, represent that four-bar mechanism is in dead-centre position；

By equation (6), input angle θ can be passed through₂Obtain output angle θ₃:

θ₃Determined completely by equation (11.1) and (11.2), wherein arctan (x₃) change at (-pi/2, pi/2), the most one by one Corresponding θ₃Change at (-π, π)；

Utilize half-angle formulas y=tan (θ₂/ 2), equation (8) can be rewritten as:

F (y)=C₁y⁴+C₂y²+C₀ (12)

Wherein, the number of the root of y is determined by the configuration of four-bar mechanism；May there be 0,2,4 roots f (y)=0, each with Corresponding be the dead point of four-bar mechanism；

If θ₂Being input angle, branch and the sub-branch of its four-bar mechanism determine by the following method:

1. set up when f (y) > 0 is permanent, represent that f (y)=0 without real root, then, in this four-bar mechanism configuration, exists without dead point, root According to formula (11), θ can be inputted₂Corresponding two output θ₃, mechanism has Liang Ge branch.

If 2. f (y)=0, and there are two different real roots f (y)=0, this four-bar mechanism only comprises a branch, and two are dead Point is divided into Liang Ge sub-branch it；

If 3. f (y)=0, and there are four different real roots f (y)=0, this four-bar mechanism comprises Liang Ge branch, four dead points Liang Ge branch is divided into four sub-branches；

Step 2: the linear-elsatic buckling of Tape movement pair five-rod；

Step 2.1: two moving sets six-bar mechanisms of a band comprise a Tape movement pair five-rod ABEFGA, its ring side Journey can be represented by Euler's formula, it may be assumed that

$a_2{e}^{{\mathrm{i\θ}}_2}+a_4{e}^{i({\mathrm{\θ}}_3+\mathrm{\β})}=a_1+a_7+a_5{e}^{{\mathrm{i\θ}}_5}+s_2{e}^{{\mathrm{i\α}}_1}---\left(13\right)$

Wherein, a₇、a₂、a₉、a₅、a₆Refer to the length of GA, AB, BF, EF, FG double leval jib in planar five-bar mechanism ABEFG respectively, E is the end of natural logrithm, and i is imaginary unit, θ₂、θ₃+β、θ₅、α₁Refer to trunnion axis rotate counterclockwise arrive respectively connecting rod AB, The angle of BF, EF, FG.

According to its real part and imaginary part, equation (13) can be expressed as:

s₂=[a₂cosθ₂+a₄cos(θ₃+β)-a₁-a₇-a₅cosθ₅]/cosα₁ (14)

s₂=[a₂sinθ₂+a₄sinθ₄-a₅sinθ₅]/sinα₁ (15)

Eliminate s₂, equation (13) is represented by:

a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁)-a₅sin(θ₅-α₁)+a₁+a₇=0 (16)

Utilize half-angle formulas:

x₅=tan (θ₅/2) (17)

Equation (16) can be expressed as:

$P_2{x}_5^2+Q_2{x}_5+R_2=0---\left(18\right)$

Wherein:

P₂=a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁)-a₅sinα₁+a₁+a₇ (19.1)

Q₂=-2a₅cosα₁ (19.2)

R₂=a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁)+a₅sinα₁+a₁+a₇ (19.3)

If five-rod exists, work as P₂≠ 0 must is fulfilled for:

${\mathrm{\Δ}}_2=Q_2^2-4P_2{R}_2\≥0---\left(20\right)$

Abbreviation obtains:

△₂=4S₁S₂≥0 (21)

Wherein:

S₁=a₅-a₁-a₇-a₂sin(θ₂-α₁)-a₄sin(θ₃+β-α₁) (22.1)

S₂=a₅+a₁+a₇+a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁) (22.2)

Equation (20), (21) produce θ₂With θ₃Between joint revolution space, wherein, S₁=0 or S₂=0 represents joint rotation Turn the border in space, when five-rod is at singular position or 3 non-output joint E, F, G conllinear, it may appear that above-mentioned situation；

By equation (15), try to achieve θ₅:

The corresponding a kind of mechanism configuration of each solution, wherein arctan (x₅) change at (-pi/2, pi/2), the most corresponding θ₅ (-π, π) change；

Step 2.2: three moving sets six-bar mechanisms of a band comprise one moving sets four-bar mechanism ABCDA of band and Two moving sets five-rod ABEFGA of one band, five rings is transformed into Fourth Ring by stretching rotation；Therefore, three moving sets six bars of band The linear-elsatic buckling of mechanism depends on Fourth Ring ABCDA and the interaction of five rings ABCEFGA, and its Fourth Ring ABCDA can be by above-mentioned right The linear-elsatic buckling of four-bar mechanism differentiates, and the ring equation at its five rings can represent by Euler's formula, it may be assumed that

$a_2{e}^{{\mathrm{i\θ}}_2}+a_4{e}^{i({\mathrm{\θ}}_3+\mathrm{\β})}=a_1+a_7+s_3{e}^{i({\mathrm{\α}}_1+\mathrm{\γ})}+s_2{e}^{{\mathrm{i\α}}_1}---\left(24\right)$

According to its real part and imaginary part, equation (24) is represented by:

s₂=[a₂cosθ₂+a₄cos(θ₃+β)-a₁-a₇-s₃cos(α₁+γ)]/cosα₁ (25.1)

s₂=[a₂sinθ₂+a₄sin(θ₃+β)-s₃sin(α₁+γ)]/sinα₁ (25.2)

Eliminate s₂, equation (24) is represented by:

s₃sinγ-a₂sin(θ₂-α₁)-a₇sinα₁-a₄sin(θ₃+β-α₁)=0 (26)

With s₃For unknown number, the discriminant of equation (26) is:

△₃=sin²γ (27)

In equation (27), sin²γ >=0 is permanent sets up.When sin γ ≠ 0, for every couple of θ₂And θ₃, s₃Have and uniquely determine Value, but as sin γ=0, s₃It is useless for having infinite having a bowel movement, mechanism in this case, and therefore, mechanism to be met exists, Must sin γ ≠ 0.

Step 3: Tape movement pair six-bar mechanism linear-elsatic buckling；

According to the presence or absence of branch point, Tape movement pair six-bar mechanism linear-elsatic buckling is divided into the following two kinds type:

Type I: Tape movement pair six-bar mechanism；Do not have △=0 in branch point, equation (4) and equation (20) it Between do not have the solution of public solution and equation (4) to meet the △ > 0 in equation (20)；The process that its branch differentiates is as follows:

Step A1: Fourth Ring branch differentiates；Use equation (12) that the branch of four-bar mechanism is differentiated；

Step A2: branch point；Exist without branch point；

Step A3: Tape movement pair six-bar mechanism branch differentiates；

For the branch of given four-bar mechanism, use equation (23) that the branch of Tape movement pair six-bar mechanism is sentenced Not, each solution of equation (23) represents a branch of Tape movement pair six-bar mechanism；If the configuration of given four-bar mechanism is full Foot equation (20), then the branching representation of this four-bar mechanism the Liang Ge branch of Tape movement pair six-bar mechanism；If be unsatisfactory for, then this The branch of four-bar mechanism is invalid, and this Tape movement pair six-bar mechanism cannot assemble；

Step A4: if four-bar mechanism has another one branch, uses above 3 steps to enter Tape movement pair six-bar mechanism Row differentiates；

Step A5: to Tape movement pair six-bar mechanism, with 3 arthrodial branch pattern always types I, dividing of Fourth Ring Prop up the branch that represent this mechanism；

Type II: Tape movement pair six-bar mechanism；Between △=0 in branch point, equation (4) and equation (20) There is public solution；Fourth Ring input-output curve is divided into some, and the part wherein meeting equation (20) is effective, each Effective part is a continuous print disaggregation and a kind of configuration representing Tape movement pair six-bar mechanism；The process that its branch differentiates is such as Under:

Step B1: Fourth Ring branch differentiates；Use equation (12) that Fourth Ring branch is differentiated；

Step B2: branch point；

Equation (4) and equation (20) is utilized to try to achieve branch point；The input-output curve part meeting equation (20) is effective Part, stop is branch point or dead point；

Step B3: Tape movement pair six-bar mechanism branch differentiates；

In the same sub-branch of four-bar mechanism, effective continuous part of two contiguous branch points represents secondary six bars of Tape movement One branch of mechanism；Different sub-branches in the branch of given four-bar mechanism form a continuous print according to public dead point to be had Effect part, therefore, as stop, branch point can differentiate that all live parts, the live part of each separation are branches；

Step B4: if four-bar mechanism has the branch containing other branch point, repeat above step, if it did not, reference Single-degree-of-freedom dicyclo mechanism is differentiated by type I；

Step B5: in the same sub-branch of four-bar mechanism, the live part of two contiguous branch points represents whole mechanism A branch；Live part in the different sub-branches of four-bar mechanism forms continuous print live part by public dead point, Thus define the branch of whole mechanism；

Step 4: the sub-branch of Tape movement pair six-bar mechanism identifies；

Work as θ₂During as input, dead-centre position can be passed through △=0 in equation (8) and try to achieve；If branch point exists, this is same Also it is Tape movement pair six-bar mechanism singular position；Therefore, two adjacent singular points, i.e. live part between dead point and branch point It it is the sub-branch of Tape movement pair six-bar mechanism structure；Its concrete following two situation that is divided into that identifies:

(1) there is no branch point；

For in the branch that a Tape movement pair six-bar mechanism is given, its sub-branch is identified by formula (11)；In equation Two solutions respectively correspond to a sub-branch of Tape movement pair six-bar mechanism；

(2) branch point exists；

In the input-output curve of four-bar mechanism, each point on all live parts represents the various configuration of mechanism, This can differentiate according to equation (23)；Therefore, for the branch of given Tape movement pair six-bar mechanism, the sub-branch of its live part It is identified, then the sub-branch of Tape movement pair six-bar mechanism also can be identified.

The beneficial effect of patent of the present invention:

1) present invention proposes and a kind of Tape movement pair six-bar mechanism branch is identified new method, the method profit automatically Realize branch (loop) is identified with mechanism's basic ring equation, identify more accurately, intuitively, efficiently；

2) method that the present invention provides is easy to develop mathematics software simulation emulation, contributes to Machine Design, There is the highest use value；

3) method that the present invention provides can be embedded in various Machine Design class business software, has good social value And economic worth.

Accompanying drawing explanation

Two moving sets six-bar mechanism schematic diagrams of the band of Fig. 1: the embodiment of the present invention；

Three moving sets six-bar mechanism schematic diagrams of the band of Fig. 2: the embodiment of the present invention；

The Liang Ge branch schematic diagram of the four-bar mechanism of Fig. 3: the embodiment of the present invention, when Tape movement pair four-bar mechanism is configured as a₁ =9, a₂=6, a₃=10, α₂When=15 °, f (y) > 0 is permanent to be set up, and this configuration exists without dead point；

The Liang Ge branch schematic diagram of the four-bar mechanism of Fig. 4: the embodiment of the present invention, when Tape movement pair four-bar mechanism is configured as a₁ =9, a₂=6, a₃=10, α₂When=45 °, there are two solutions f (y)=0, and this configuration is with the presence of 2 dead points；

The Liang Ge branch schematic diagram of the four-bar mechanism of Fig. 5: the embodiment of the present invention, when Tape movement pair four-bar mechanism is configured as a₁ =4, a₂=5, a₃=2.2, α₂When=45 °, there are four solutions f (y)=0, and this configuration is with the presence of 4 dead points；

The Tape movement pair six-bar mechanism linear-elsatic buckling flow chart of Fig. 6: the embodiment of the present invention；

Two, the band of Fig. 7: embodiment of the present invention moving sets six-bar mechanism branch schematic diagram；

Detailed description of the invention

Understand and implement the present invention for the ease of those of ordinary skill in the art, below in conjunction with the accompanying drawings and embodiment is to this Bright it is described in further detail, it will be appreciated that enforcement example described herein is merely to illustrate and explains the present invention, not For limiting the present invention.

Branch is by the FAQs that problem is that linkage analysis and synthesis runs into, and it includes the continuous of mechanism kinematic Property, (branch) assembling mode or the problem such as loop, sub-branch.Wherein branch refers to a kind of assembling form or the company of linkage The configuration space of linkage, i.e. its all possible position that can reach of moving continuously in the case of not taking mechanism apart.Point After identifying, motion continuity may just can be guaranteed and motion smoothing or sub-branch just can identify.In invention, Branch, branch point, sub-branch recognition methods general, a continuous print is contained and without unusual input domain in each sub-branch, i.e. Input and output one_to_one corresponding, first the branch of Tape movement pair monocycle is studied by the present invention, then research ring between mutual Relation, utilizes the method the branch of Tape movement pair six-bar mechanism quickly to be identified, design for mechanism provide new auxiliary Aid method.

The linear-elsatic buckling method of a kind of Tape movement pair six-bar mechanism that the present invention provides, by single-degree-of-freedom dicyclo linkage Two monocycles relevant to input and output combine, and being interacted by it draws the linear-elsatic buckling side of Tape movement pair six-bar mechanism Method.

As it is shown in figure 1, two moving sets six-bar mechanisms of band are by one moving sets four-bar mechanism ABCDA of band and a band One moving sets five-rod ABEFGA composition；As in figure 2 it is shown, three moving sets six-bar mechanisms of a band are by one shifting of a band Dynamic secondary four-bar mechanism ABCDA and one by one moving sets five-rod ABEFGA.And the linear-elsatic buckling of Tape movement pair six-bar mechanism Relevant to input and fixed-link, in view of this, the present invention is by two relevant to input and output for Tape movement pair six-bar mechanism lists Ring combines, and being interacted by it draws the linear-elsatic buckling of Tape movement pair six-bar mechanism.

Asking for an interview Fig. 6, implementing of the present invention comprises the following steps:

Step 1: the linear-elsatic buckling of Tape movement pair four-bar mechanism；

Linear-elsatic buckling problem is the FAQs that linkage analysis and synthesis runs into, and it includes the continuous of mechanism kinematic Property, (branch) assembling mode or the problem such as loop, sub-branch.Wherein branch refers to a kind of assembling form or the company of linkage The configuration space of linkage, i.e. its all possible position that can reach of moving continuously in the case of not taking mechanism apart.Point After identifying, motion continuity may just can be guaranteed and motion smoothing or sub-branch just can identify.

Any one Tape movement pair six-bar mechanism all comprises one or more four-bar mechanism.As shown in Figure 1, 2, two shiftings of band Dynamic secondary six-bar mechanism and three moving sets six-bar mechanisms of band comprise four-bar mechanism ABCDA.Its ring equation can be according to Euler's formula table It is shown as:

$a_2{e}^{{\mathrm{i\θ}}_2}+a_3{e}^{{\mathrm{i\θ}}_3}=a_1+s_1{e}^{{\mathrm{i\α}}_2}---\left(1\right)$

According to its real part and imaginary part, equation (1) is represented by:

s₁=(a₂cosθ₂+a₃cosθ₃-a₁)/cosα₂ (2)

s₁=(a₂sinθ₂+a₃sinθ₃)/sinα₂ (3)

By eliminating s₁, equation (1) can be expressed as:

a₁sinα₂+a₂sin(θ₂-α₂)+a₃sin(θ₃-α₂)=0 (4)

Work as θ₂、θ₃When being input parameter, equation (4) is represented by Input output Relationship model, utilizes half-angle formulas:

x₃=tan (θ₃/2) (5)

Equation (4) is represented by:

$P_1{x}_3^2+Q_1{x}_3+R_1=0---\left(6\right)$

Wherein:

P₁=a₂sin(θ₂-α₂)+(a₁+a₃)sinα₂ (7.1)

Q₁=2a₃cosα₂ (7.2)

Q₁=a₂sin(θ₂-α₂)+(a₁-a₃)sinα₂ (7.3)

Exist to four-bar mechanism configuration, work as P₁≠ 0 must is fulfilled for:

${\mathrm{\Δ}}_1=Q_1^2-4P_1{R}_1\≥0---\left(8\right)$

Abbreviation obtains:

△₁=4S₁S₂ (9)

Wherein:

S₁=a₃-a₂sin(θ₂-α₂)-a₁sinα₂ (10.1)

S₂=a₃+a₂sin(θ₂-α₂)+a₁sinα₂ (10.2)

Work as △₁When=0, represent that four-bar mechanism is in dead-centre position；

By equation (6), input angle θ can be passed through₂Obtain output angle θ₃:

θ₃Determined completely by equation (11.1) and (11.2), wherein arctan (x₃) change at (-pi/2, pi/2), the most one by one Corresponding θ₃Change at (-π, π)；

Utilize half-angle formulas y=tan (θ₂/ 2), equation (8) can be rewritten as:

F (y)=C₁y⁴+C₂y²+C₀ (12)

Wherein, the number of the root of y is determined by the configuration of four-bar mechanism；May there be 0,2,4 roots f (y)=0, each with Corresponding be the dead point of four-bar mechanism；

If θ₂Being input angle, branch and the sub-branch of its four-bar mechanism determine by the following method:

1. set up when f (y) > 0 is permanent, represent that f (y)=0 without real root, then in this four-bar mechanism configuration, exists without dead point, side Journey (9) has two solutions, and corresponding is the Liang Ge branch of four-bar mechanism, and each branch only comprises only a sub-branch； Ask for an interview Fig. 3, when Tape movement pair four-bar mechanism is configured as a₁=9, a₂=6, a₃=10, α₂When=15 °, f (y) > 0 is permanent to be set up, this structure Type exists without dead point；

2. if f (y)=0, and there are two different real roots f (y)=0, then, in this four-bar mechanism configuration, have two dead points Existing, this four-bar mechanism only comprises a branch, and two dead points are divided into Liang Ge sub-branch it；Ask for an interview Fig. 4, when Tape movement pair four Linkage is configured as a₁=9, a₂=6, a₃=10, α₂When=45 °, there are two solutions f (y)=0, and this configuration is with the presence of 2 dead points；

3. if f (y)=0, and there are four different real roots f (y)=0, then, in this four-bar mechanism configuration, have four dead points Existing, and this four-bar mechanism comprises Liang Ge branch, four dead points are divided into four sub-branches Liang Ge branch；Ask for an interview Fig. 5, when band moves Dynamic secondary four-bar mechanism is configured as a₁=4, a₂=5, a₃=2.2, α₂When=45 °, there are four solutions f (y)=0, and this configuration has 4 extremely Point exists.

Step 2: the linear-elsatic buckling of Tape movement pair five-rod；

Two moving sets six-bar mechanisms of one band comprise a Tape movement pair five-rod ABEFGA, and its ring equation can basis Euler's formula is expressed as:

$a_2{e}^{{\mathrm{i\θ}}_2}+a_4{e}^{i({\mathrm{\θ}}_3+\mathrm{\β})}=a_1+a_7+a_5{e}^{{\mathrm{i\θ}}_5}+s_2{e}^{{\mathrm{i\α}}_1}---\left(13\right)$

Wherein, a₇、a₂、a₉、a₅、a₆Refer to the length of GA, AB, BF, EF, FG double leval jib in planar five-bar mechanism ABEFG respectively, E is the end of natural logrithm, and i is imaginary unit, θ₂、θ₃+β、θ₅、α₁Refer to trunnion axis rotate counterclockwise arrive respectively connecting rod AB, The angle of BF, EF, FG.

According to its real part and imaginary part, equation (13) can be expressed as:

s₂=[a₂cosθ₂+a₄cos(θ₃+β)-a₁-a₇-a₅cosθ₅]/cosα₁ (14)

s₂=[a₂sinθ₂+a₄sin(θ₄+β)-a₅sinθ₅]/sinα₁ (15)

Eliminate s₂, equation (13) is represented by:

a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁)-a₅sin(θ₅-α₁)+(a₁+a₇)sinα₁=0 (16)

Utilize half-angle formulas:

x₅=tan (θ₅/2) (17)

Equation (16) can be expressed as:

$P_2{x}_5^2+Q_2{x}_5+R_2=0---\left(18\right)$

Wherein:

P₂=a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁)+a₅sinα₁+(a₁+a₇)sinα₁ (19.1)

Q₂=2a₅cosα₁ (19.2)

R₂=a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁)-a₅sinα₁+(a₁+a₇)sinα₁ (19.3)

If five-rod exists, work as P₂≠ 0 must is fulfilled for:

${\mathrm{\Δ}}_2=Q_2^2-4P_2{R}_2\≥0---\left(20\right)$

Abbreviation obtains:

△₂=4S₁S₂≥0 (21)

Wherein:

S₁=a₅-(a₁+a₇)sinα₁-a₂sin(θ₂-α₁)-a₄sin(θ₃+β-α₁) (22.1)

S₂=a₅+(a₁+a₇)sinα₁+a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁) (22.2)

Equation (20), (21) produce θ₂With θ₃Between joint revolution space, wherein, S₁=0 or S₂=0 represents joint rotation Turn the border in space, when five-rod is at singular position or 3 non-output joint E, F, G conllinear, it may appear that above-mentioned situation；

By equation (15), try to achieve θ₅:

The corresponding a kind of mechanism configuration of each solution, wherein arctan (x₅) change at (-pi/2, pi/2), the most corresponding θ₅ (-π, π) change；

As shown in Figure 2 three moving sets six-bar mechanisms of band, by one moving sets four-bar mechanism ABCDA of band and Two moving sets five-rod ABEFGA compositions of one band；Therefore, the linear-elsatic buckling with three moving sets six-bar mechanisms depends on Fourth Ring ABCDA and the interaction of five rings ABCEFGA, its Fourth Ring ABCDA can be entered by the above-mentioned linear-elsatic buckling to four-bar mechanism Row differentiates, and the ring equation at its five rings can represent by Euler's formula, it may be assumed that

$a_2{e}^{{\mathrm{i\θ}}_2}+a_4{e}^{i({\mathrm{\θ}}_3+\mathrm{\β})}=a_1+a_7+s_3{e}^{i({\mathrm{\α}}_1+\mathrm{\γ})}+s_2{e}^{{\mathrm{i\α}}_1}---\left(24\right)$

According to its real part and imaginary part, equation (24) is represented by:

s₂=[a₂cosθ₂+a₄cos(θ₃+β)-a₁-a₇-s₃cos(α₁+γ)]/cosα₁ (25.1)

s₂=[a₂sinθ₂+a₄sin(θ₃+β)-s₃sin(α₁+γ)]/sinα₁ (25.2)

Eliminate s₂, equation (24) is represented by:

s₃sinγ-a₂sin(θ₂-α₁)-a₇sinα₁-a₄sin(θ₃+β-α₁)=0 (26)

With s₃For unknown number, the discriminant of equation (26) is:

△₃=sin²γ (27)

In equation (27), sin²γ >=0 is permanent sets up.When sin γ ≠ 0, for every couple of θ₂And θ₃, s₃Have and uniquely determine Value, but as sin γ=0, s₃It is useless for having infinite having a bowel movement, mechanism in this case, and therefore, mechanism to be met exists, Must sin γ ≠ 0.

Step 3: Tape movement pair six-bar mechanism linear-elsatic buckling；

According to the presence or absence of branch point, Tape movement pair six-bar mechanism linear-elsatic buckling is divided into the following two kinds type:

Type I: Tape movement pair six-bar mechanism；Do not have △=0 in branch point, equation (4) and equation (20) it Between do not have the solution of public solution and equation (4) to meet the △ > 0 in equation (20)；The process that its branch differentiates is as follows:

Step A1: Fourth Ring branch differentiates；Use equation (12) that the branch of four-bar mechanism is differentiated；

Step A2: branch point；Exist without branch point；

Step A3: Tape movement pair six-bar mechanism branch differentiates；

For the branch of given four-bar mechanism, use equation (23) that the branch of Tape movement pair six-bar mechanism is sentenced Not, each solution of equation (23) represents a branch of Tape movement pair six-bar mechanism；If the configuration of given four-bar mechanism is full Foot equation (20), then the branching representation of this four-bar mechanism the Liang Ge branch of Tape movement pair six-bar mechanism；If be unsatisfactory for, then this The branch of four-bar mechanism is invalid, and this Tape movement pair six-bar mechanism cannot assemble；

Step A4: if four-bar mechanism has another one branch, uses above 3 steps to enter Tape movement pair six-bar mechanism Row differentiates；

Step A5: to Tape movement pair six-bar mechanism, with 3 arthrodial branch pattern always types I, dividing of Fourth Ring Prop up the branch that represent this mechanism；

Type II: Tape movement pair six-bar mechanism；Between △=0 in branch point, equation (4) and equation (20) There is public solution；Fourth Ring input-output curve is divided into some, and the part wherein meeting equation (20) is effective, each Effective part is a continuous print disaggregation and a kind of configuration representing Tape movement pair six-bar mechanism；The process that its branch differentiates is such as Under:

Step B1: Fourth Ring branch differentiates；Use equation (12) that Fourth Ring branch is differentiated；

Step B2: branch point；

Equation (4) and equation (20) is utilized to try to achieve branch point；The input-output curve part meeting equation (20) is effective Part, stop is branch point or dead point；

Step B3: Tape movement pair six-bar mechanism branch differentiates；

In the same sub-branch of four-bar mechanism, effective continuous part of two contiguous branch points represents secondary six bars of Tape movement One branch of mechanism；Different sub-branches in the branch of given four-bar mechanism form a continuous print according to public dead point to be had Effect part, therefore, as stop, branch point can differentiate that all live parts, the live part of each separation are branches；

Step B4: if four-bar mechanism has the branch containing other branch point, repeat above step, if it did not, reference Single-degree-of-freedom dicyclo mechanism is differentiated by type I；

Step B5: in the same sub-branch of four-bar mechanism, the live part of two contiguous branch points represents whole mechanism A branch；Live part in the different sub-branches of four-bar mechanism forms continuous print live part by public dead point, Thus define the branch of whole mechanism；

Step 4: the sub-branch of Tape movement pair six-bar mechanism identifies；

Work as θ₂During as input, dead-centre position can be passed through △=0 in equation (8) and try to achieve；If branch point exists, this is same Also it is Tape movement pair six-bar mechanism singular position；Therefore, two adjacent singular points, i.e. live part between dead point and branch point It it is the sub-branch of Tape movement pair six-bar mechanism structure；Its concrete following two situation that is divided into that identifies:

(3) there is no branch point；

For in the branch that a Tape movement pair six-bar mechanism is given, its sub-branch is identified by formula (11)；In equation Two solutions respectively correspond to a sub-branch of Tape movement pair six-bar mechanism；

(4) branch point exists；

In the input-output curve of four-bar mechanism, each point on all live parts represents the various configuration of mechanism, This can differentiate according to equation (23)；Therefore, for the branch of given Tape movement pair six-bar mechanism, the sub-branch of its live part It is identified, then the sub-branch of Tape movement pair six-bar mechanism also can be identified.

Below by way of being embodied as the present invention is further elaborated；

Embodiment 1: provide two moving sets six-bar mechanism length of the band shown in Fig. 1 and related angle angle, as it is shown in figure 1, a₁=2.4, a₂=5.5, a₃=4.8, a₄=4.2, a₅=2.5, a₇=2.4, β=45.0 °, α₁=30.0 °, α₂=50.0 °.

The input-output curve of Tape movement pair four-bar mechanism ABCD in Fig. 1 can be obtained, according to equation according to equation (4) (20) the joint revolution space of Tape movement pair five-rod, dash area the most as shown in the figure can be obtained.

(1) Tape movement pair four-bar mechanism branch differentiates: work as θ₂For input angle, θ₃During for output angle, according to equation (8), can obtain Four-bar mechanism dead-centre position u (-162.6 ° ,-39.9 °), v (82.6 ° ,-39.9 °).According to equation (11.1) and equation (11.2), You Liangge sub-branch of its branch.As shown in Figure 7:

Sub-branch u-2-3-4-v: θ₂∈[-162.6°,82.6°]；

Sub-branch u-1-6-5-v: θ₂∈[-162.6°,82.6°]；

(2) branch point: work as θ₂For input angle, θ₃During for output angle, four bar machines can be obtained according to equation (4) and equation (20) Structure input-output curve and the intersecting point coordinate on revolution space border, joint, 1 (-141.6 ° ,-92.2 °), 2 (-161.4 ° ,- 28.6 °), 3 (-105.2 °, 55.3 °), 4 (-15.6 °, 91.3 °), 5 (72.6 ° ,-74.6 °), 6 (7.9 ° ,-152.7 °)

(3) live part of four-bar mechanism input-output curve: four-bar mechanism input/output relation must simultaneously meet five The joint revolution space of linkage, such six-bar mechanism just has mobility.As it is shown in fig. 7, in sub-branch u-2-3-4-v, Live part is: u-2 part, wherein θ₂[-161.4 ° ,-28.6 °] and 3-4 part, wherein θ₂[-105.2°,-15.6°]；? In sub-branch u-1-6-5-v, live part is: u-1 part, wherein θ₂[-162.6 ° ,-141.6 °] and 6-5 part, wherein θ₂ [7.9°,-141.6°]。

(4) with two moving sets six bar branches: the full curve meeting dash area is the branch of six-bar mechanism, band moves The dead point of dynamic secondary four-bar mechanism is that branch is divided into Liang Ge sub-branch, and therefore u-1 and u-2 forms a continuous effective point Prop up 1-u-2, it addition, also have other 2 live parts separated, i.e. 3-4,6-5.

(5) sub-branch differentiates: according to equation (23.1) and equation (23.2), it is known that the effective portion on input-output curve Each point on Fen represents two kinds of various configurations of mechanism, and each live part represents two moving sets six-bar mechanisms of band Liang Ge sub-branch in branch.

(6) as follows, list the branch with two moving sets six-bar mechanisms and sub-branch thereof:

Branch 1-u-2: comprise live part 1-u and u-2.

Sub-branch 1-u: wherein θ₂[-162.6 ° ,-141.6 °], meet equation (11.1), and θ₅Can be according to equation (23.1) Obtain；

Sub-branch u-1: wherein θ₂[-162.6 ° ,-141.6 °], meet equation (11.1), and θ₅Can be according to equation (23.2) Obtain；

Sub-branch u-2: wherein θ₂[-105.2 ° ,-15.6 °], meet equation (11.1), and θ₅Can obtain according to equation (23.1) Arrive；

Sub-branch 2-u: wherein θ₂[-105.2 ° ,-15.6 °], meet equation (11.1), and θ₅Can obtain according to equation (23.2) Arrive；

Branch 3-4: comprise live part 3-4.

Sub-branch 3-4: wherein θ₂[-105.2 ° ,-15.6 °], meet equation (11.1), and θ₅Can obtain according to equation (23.1) Arrive；

Sub-branch 4-3: wherein θ₂[-105.2 ° ,-15.6 °], meet equation (11.1), and θ₅Can obtain according to equation (23.2) Arrive；

Branch 6-5: comprise live part 6-5.

Sub-branch 6-5: wherein θ₂[7.9 °, 72.6 °], meet equation (11.1), and θ₅Can obtain according to equation (23.1)；

Sub-branch 5-6: wherein θ₂[7.9 °, 72.6 °], meet equation (11.1), and θ₅Can obtain according to equation (23.2).

Embodiment 2: to three moving sets six-bar mechanism length of the band shown in Fig. 2 and related angle angle, be given in embodiment 1 Identical size, and make sin γ ≠ 0.

According to equation (24), it is known that when sin γ ≠ 0, △₂> 0 permanent establishment, therefore, with three moving sets six-bar mechanisms θ₂-θ₃Input-output curve necessarily meets the joint revolution space (dash area) of Tape movement pair five-rod, and at one a pair The θ answered₂、θ₃And under the conditions of sin γ ≠ 0, s₃Only only one of which solution.Generally speaking, under the conditions of sin γ ≠ 0, Fig. 2 mechanism Only only one of which branch, and fully rotating.

It should be appreciated that the part that this specification does not elaborates belongs to prior art.

It should be appreciated that the above-mentioned description for preferred embodiment is more detailed, can not therefore be considered this The restriction of invention patent protection scope, those of ordinary skill in the art, under the enlightenment of the present invention, is weighing without departing from the present invention Profit requires under the ambit protected, it is also possible to make replacement or deformation, within each falling within protection scope of the present invention, this The bright scope that is claimed should be as the criterion with claims.

Claims (2)

1. a Tape movement pair six-bar mechanism linear-elsatic buckling method, it is characterised in that: respectively to single-degree-of-freedom six bar mechanism Two loops that input and output are relevant are analyzed, and then are interacted by it and draw a kind of Tape movement pair six-bar mechanism branch Recognition methods.

A kind of Tape movement pair six-bar mechanism linear-elsatic buckling method the most according to claim 1, it is characterised in that include following Step:

Step 1: the linear-elsatic buckling of Tape movement pair four-bar mechanism；

Any one a kind of Tape movement pair six-bar mechanism all comprises one or more Tape movement pair four-bar mechanism ABCDA, its ring side Journey is represented by Euler's formula, it may be assumed that

$a_2{e}^{{\mathrm{i\θ}}_2}+a_3{e}^{{\mathrm{i\θ}}_3}=a_1+s_1{e}^{{\mathrm{i\α}}_2}---\left(1\right)$

Wherein, a₁、a₂、a₃、s₁Referring to the length of DA, AB, BC, CD double leval jib in four-bar linkage ABCDA respectively, e is nature pair The end of number, i is imaginary unit, θ₂、θ₃、α₂Refer to trunnion axis and rotate the angle arriving double leval jib AB, BC, CD respectively counterclockwise；

According to its real part and imaginary part, equation (1) is expressed as:

s₁=(a₂cosθ₂+a₃cosθ₃-a₁)/cosα₂ (2)

s₁=(a₂sinθ₂+a₃sinθ₃)/sinα₂ (3)

By eliminating s₁, equation (1) is expressed as:

a₁sinα₂+a₂sin(θ₂-α₂)+a₃sin(θ₃-α₂)=0 (4)

Work as θ₂、θ₃When being input parameter, equation (4) is expressed as Input output Relationship model；Utilize half-angle formulas:

x₃=tan (θ₃/2) (5)

Equation (4) is expressed as:

$P_1{x}_3^2+Q_1{x}_3+R_1=0---\left(6\right)$

Wherein:

P₁=a₂sin(θ₂-α₂)+(a₁+a₃)sinα₂ (7.1)

Q₁=2a₃cosα₂ (7.2)

R₁=a₂sin(θ₂-α₂)+(a₁-a₃)sinα₂ (7.3)

Exist to four-bar mechanism configuration, work as P₁It must is fulfilled for when ≠ 0

${\mathrm{\Δ}}_1=Q_1^2-4P_1{R}_1\≥0---\left(8\right)$

Abbreviation obtains:

Δ₁=4S₁S₂ (9)

Wherein:

S₁=a₃-a₂sin(θ₂-α₂)-a₁sinα₂ (10.1)

S₂=a₃+a₂sin(θ₂-α₂)+a₁sinα₂ (10.2)

Work as Δ₁When=0, represent that four-bar mechanism is in dead-centre position；

By equation (6), by input angle θ₂Obtain output angle θ₃:

Wherein θ₃=2arctan (x₃) (11.1)

Wherein θ₃=2arctan (x₃) (11.2)

θ₃Determined completely by equation (11.1) and (11.2), wherein arctan (x₃) change at (-pi/2, pi/2), the most relative The θ answered₃Change at (-π, π)；

Utilize half-angle formulas y=tan (θ₂/ 2), equation (8) is rewritten as:

F (y)=C₁y⁴+C₂y²+C₀ (12)

Wherein, the number of the root of y is determined by the configuration of four-bar mechanism；May there be 0,2,4 roots f (y)=0, each phase therewith Corresponding is the dead point of four-bar mechanism；

If θ₂Being input angle, branch and the sub-branch of its four-bar mechanism determine by the following method:

1, as f (y) > 0 permanent establishment, represent that f (y)=0 without real root, then, in this four-bar mechanism configuration, exists without dead point, according to public affairs Formula (11), can input θ₂Corresponding two output θ₃, mechanism has Liang Ge branch；

If 2 f (y)=0, and there are two different real roots f (y)=0, this four-bar mechanism only comprises a branch, two dead point handles It is divided into Liang Ge sub-branch；

If 3 f (y)=0, and have the different real root in f (y)=0 four, this four-bar mechanism comprises Liang Ge branch, and four dead points are two Individual branch is divided into four sub-branches；

Step 2: the linear-elsatic buckling of Tape movement pair five-rod；

Step 2.1: one band two moving sets six-bar mechanisms comprise a Tape movement pair five-rod ABEFGA, its ring equation by Euler's formula represents, it may be assumed that

$a_2{e}^{{\mathrm{i\θ}}_2}+a_4{e}^{i({\mathrm{\θ}}_3+\mathrm{\β})}=a_1+a_7+a_5{e}^{{\mathrm{i\θ}}_5}+s_2{e}^{{\mathrm{i\α}}_1}---\left(13\right)$

Wherein, a₇、a₂、a₉、a₅、a₆Referring to the length of GA, AB, BF, EF, FG double leval jib in planar five-bar mechanism ABEFG respectively, e is The end of natural logrithm, i is imaginary unit, θ₂、θ₃+β、θ₅、α₁Refer to trunnion axis rotate counterclockwise arrive respectively connecting rod AB, BF, The angle of EF, FG；

According to its real part and imaginary part, equation (13) is expressed as:

s₂=[a₂cosθ₂+a₄cos(θ₃+β)-a₁-a₇-a₅cosθ₅]/cosα₁ (14)

s₂=[a₂sinθ₂+a₄sin(θ₄+β)-a₅sinθ₅]/sinα₁ (15)

Eliminate s₂, equation (13) is expressed as:

a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁)-a₅sin(θ₅-α₁)+(a₁+a₇)sinα₁=0 (16)

Utilize half-angle formulas:

x₅=tan (θ₅/2) (17)

Equation (16) is expressed as:

$P_2{x}_5^2+Q_2{x}_5+R_2=0---\left(18\right)$

Wherein:

P₂=a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁)+a₅sinα₁+(a₁+a₇)sinα₁ (19.1)

Q₂=2a₅cosα₁ (19.2)

R₂=a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁)-a₅sinα₁+(a₁+a₇)sinα₁ (19.3)

If five-rod exists, work as P₂≠ 0 must is fulfilled for:

${\mathrm{\Δ}}_2=Q_2^2-4P_2{R}_2\≥0---\left(20\right)$

Abbreviation obtains:

Δ₂=4S₁S₂≥0 (21)

Wherein:

S₁=a₅-(a₁+a₇)sinα₁-a₂sin(θ₂-α₁)-a₄sin(θ₃+β-α₁) (22.1)

S₂=a₅+(a₁+a₇)sinα₁+a₂sin(θ₂-α₁)+a₄sin(θ₃+β-α₁) (22.2)

Equation (20), (21) produce θ₂With θ₃Between joint revolution space, wherein, S₁=0 or S₂=0 represents joint revolution space Border, when five-rod is at singular position or 3 non-output joint E, F, G conllinear, it may appear that above-mentioned situation；

By equation (15), try to achieve θ₅:

Wherein θ₅=2arctan (x₅) (23.1)

Wherein θ₅=2arctan (x₅) (23.2)

The corresponding a kind of mechanism configuration of each solution, wherein arctan (x₅) change at (-pi/2, pi/2), the most corresponding θ₅(- π, π) in change；

Step 2.2: three moving sets six-bar mechanisms of a band comprise one moving sets four-bar mechanism ABCDA of band and one Carrying two moving sets five-rod ABEFGA, five rings is transformed into Fourth Ring by stretching rotation；Therefore, three moving sets six-bar mechanisms of band Linear-elsatic buckling depend on Fourth Ring ABCDA and the interaction of five rings ABCEFGA, its Fourth Ring ABCDA is by above-mentioned to four bar machines The linear-elsatic buckling of structure differentiates, and the ring equation Euler's formula at its five rings represents, it may be assumed that

$a_2{e}^{{\mathrm{i\θ}}_2}+a_4{e}^{i({\mathrm{\θ}}_3+\mathrm{\β})}=a_1+a_7+s_3{e}^{i({\mathrm{\α}}_1+\mathrm{\γ})}+s_2{e}^{{\mathrm{i\α}}_1}---\left(24\right)$

According to its real part and imaginary part, equation (24) is expressed as:

s₂=[a₂cosθ₂+a₄cos(θ₃+β)-a₁-a₇-s₃cos(α₁+γ)]/cosα₁ (25.1)

s₂=[a₂sinθ₂+a₄sin(θ₃+β)-s₃sin(α₁+γ)]/sinα₁ (25.2)

Eliminate s₂, equation (24) is expressed as:

s₃sinγ-a₂sin(θ₂-α₁)-a₇sinα₁-a₄sin(θ₃+β-α₁)=0 (26)

With s₃For unknown number, the discriminant of equation (26) is:

Δ₃=sin²γ (27)

In equation (27), sin²γ >=0 is permanent sets up；When sin γ ≠ 0, for every couple of θ₂And θ₃, s₃Have and uniquely determine value, but As sin γ=0, s₃It is useless for having infinite having a bowel movement, mechanism in this case, and therefore, mechanism to be met exists, it is necessary to sinγ≠0；

Step 3: Tape movement pair six-bar mechanism linear-elsatic buckling；

According to the presence or absence of branch point, Tape movement pair six-bar mechanism linear-elsatic buckling is divided into the following two kinds type:

Type I: Tape movement pair six-bar mechanism；Do not have between Δ=0 in branch point, equation (4) and equation (20) not have The solution having public solution and equation (4) meets the Δ in equation (20) > 0；The process that its branch differentiates is as follows:

Step A1: Fourth Ring branch differentiates；Use equation (12) that the branch of four-bar mechanism is differentiated；

Step A2: branch point；Exist without branch point；

Step A3: Tape movement pair six-bar mechanism branch differentiates；

For the branch of given four-bar mechanism, use equation (23) that the branch of Tape movement pair six-bar mechanism is differentiated, side Each solution of journey (23) represents a branch of Tape movement pair six-bar mechanism；If the configuration of given four-bar mechanism meets equation (20), then the branching representation of this four-bar mechanism the Liang Ge branch of Tape movement pair six-bar mechanism；If be unsatisfactory for, then this four bars machine The branch of structure is invalid, and this Tape movement pair six-bar mechanism cannot assemble；

Step A4: if four-bar mechanism has another one branch, uses above 3 steps to sentence Tape movement pair six-bar mechanism Not；

Step A5: to Tape movement pair six-bar mechanism, with 3 arthrodial branch pattern always types I, branch's generation at Fourth Ring The branch of this mechanism by table；

Type II: Tape movement pair six-bar mechanism；Exist with the presence of between Δ=0 in branch point, equation (4) and equation (20) Public solution；Fourth Ring input-output curve is divided into some, and the part wherein meeting equation (20) is effective, each effectively Part be a continuous print disaggregation and a kind of configuration representing Tape movement pair six-bar mechanism；The process that its branch differentiates is as follows:

Step B1: Fourth Ring branch differentiates；Use equation (12) that Fourth Ring branch is differentiated；

Step B2: branch point；

Equation (4) and equation (20) is utilized to try to achieve branch point；The input-output curve part meeting equation (20) is live part, Stop is branch point or dead point；

Step B3: Tape movement pair six-bar mechanism branch differentiates；

In the same sub-branch of four-bar mechanism, effective continuous part of two contiguous branch points represents Tape movement pair six-bar mechanism A branch；Different sub-branches in the branch of given four-bar mechanism form an effective portion of continuous print according to public dead point Point, therefore, as stop, branch point can differentiate that all live parts, the live part of each separation are branches；

Step B4: if four-bar mechanism has the branch containing other branch point, repeat above step, if it did not, with reference to type I Single-degree-of-freedom dicyclo mechanism is differentiated；

Step B5: in the same sub-branch of four-bar mechanism, the live part of two contiguous branch points represents the one of whole mechanism Individual branch；Live part in the different sub-branches of four-bar mechanism forms continuous print live part by public dead point, thus Define the branch of whole mechanism；

Step 4: the sub-branch of Tape movement pair six-bar mechanism identifies；

Work as θ₂During as input, dead-centre position can be passed through Δ=0 in equation (8) and try to achieve；If branch point exists, this is also Tape movement pair six-bar mechanism singular position；Therefore, two adjacent singular points, i.e. live part between dead point and branch point are bands The sub-branch of moving sets six-bar mechanism structure；Its concrete following two situation that is divided into that identifies:

(1) there is no branch point；

For in the branch that a Tape movement pair six-bar mechanism is given, its sub-branch is identified by formula (11)；In equation two Individual solution respectively correspond to a sub-branch of Tape movement pair six-bar mechanism；

(2) branch point exists；

In the input-output curve of four-bar mechanism, each point on all live parts represents the various configuration of mechanism, this energy Differentiate according to equation (23)；Therefore, for the branch of given Tape movement pair six-bar mechanism, the sub-branch of its live part is known Not, then the sub-branch of Tape movement pair six-bar mechanism also can be identified.