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

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

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CN106126812A
CN106126812A CN201610460879.7A CN201610460879A CN106126812A CN 106126812 A CN106126812 A CN 106126812A CN 201610460879 A CN201610460879 A CN 201610460879A CN 106126812 A CN106126812 A CN 106126812A
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bar mechanism
branch
equation
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branches
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王君
龚雅静
牛克佳
陈红杰
郑晓
梁斌
陈青欣
汪泉
任军
孙金风
魏琼
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Hubei University of Technology
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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

一种带移动副六杆机构分支识别方法A Branch Recognition Method of Six-bar Mechanism with Moving Auxiliary

技术领域technical field

本发明属于机械设计制造技术领域，涉及一种新的带移动副六杆机构分支识别方法，该方法适用于所有带移动副平面六杆机构。The invention belongs to the technical field of mechanical design and manufacture, and relates to a new method for identifying branches of a six-bar mechanism with a moving pair, which is applicable to all planar six-bar mechanisms with a moving pair.

背景技术Background technique

对于带移动副六杆机构来说，寻求一种简单可行且准确的可动性判别方法是很重要的。目前，国内外已有多个学者对连杆机构进行研究，Grashof准则对四连杆机构的完全旋转性进行了准确的判断；Kwun-Lon Ting教授N杆旋转理论提供了一系列理论方法对单环平面机构及空间机构进行了研究；国内，郭晓宁教授和褚金奎教授通过把Stephenson六杆机构看做一个基础四连杆和双杆组机构，对Stephenson六杆机构的可动性进行了判别。这些都对研究单自由度双环连杆机构的研究奠定了良好的基础，然而，上述方法的研究对象相对单一，并不能对不同类型的带移动副六杆机构的分支进行有效识别。For a six-bar mechanism with a moving pair, it is very important to find a simple, feasible and accurate method for judging the mobility. At present, many scholars at home and abroad have studied the linkage mechanism. The Grashof criterion has accurately judged the complete rotation of the four-bar linkage mechanism; Professor Kwun-Lon Ting's N-bar rotation theory has provided a series of theoretical methods for single The ring plane mechanism and the space mechanism have been studied; in China, Professor Guo Xiaoning and Professor Chu Jinkui judged the mobility of the Stephenson six-bar mechanism by considering the Stephenson six-bar mechanism as a basic four-bar linkage and double-bar group mechanism. These have laid a good foundation for the study of single-degree-of-freedom double-loop linkages. However, the research object of the above method is relatively single, and cannot effectively identify the branches of different types of six-bar linkages with moving pairs.

发明内容Contents of the invention

为了解决上述技术问题，本发明提出了一种简单可行的带移动副六杆机构的分支识别方法，将与输入、输出的两个环相结合，得到输入、输出角的相关方程，从而对带移动副六杆机构的分支进行有效识别。In order to solve the above-mentioned technical problems, the present invention proposes a simple and feasible branch recognition method with a moving auxiliary six-bar mechanism, which combines the two rings of input and output to obtain the correlation equation of the input and output angles, so as to Move the branches of the auxiliary six-bar mechanism for effective identification.

本发明所采用的技术方案是：一种带移动副六杆机构的分支识别方法，其特征在于：将带移动副六杆机构与输入输出相关的两个单环结合，通过其相互作用得出带移动副六杆机构的分支识别方法。The technical solution adopted in the present invention is: a branch recognition method with a six-bar mechanism with a moving pair, which is characterized in that: the six-bar mechanism with a moving pair is combined with two single loops related to input and output, and the result is obtained through their interaction. Branch recognition method with moving sub-six-bar mechanism.

作为优选，本发明的具体实现包括以下步骤：As preferably, the concrete realization of the present invention comprises the following steps:

步骤1：带移动副四杆机构分支判别：Step 1: Branch discrimination of four-bar mechanism with moving vice:

任何一个一种带移动副六杆机构都包含一个或多个带移动副四杆机构ABCDA，其环方程可由欧拉公式表示，即：Any six-bar mechanism with a moving pair includes one or more four-bar mechanisms ABCDA with a moving pair, and its loop equation can be expressed by Euler's formula, namely:

aa 22 ee iθiθ 22 ++ aa 33 ee iθiθ 33 == aa 11 ++ sthe s 11 ee iαiα 22 -- -- -- (( 11 ))

其中，a1、a2、a3、s1分别指平面四杆机构ABCDA中DA、AB、BC、CD四连杆的长度，e是自然对数的底，i是虚数单位，θ2、θ3、α2指的是水平轴逆时针旋转分别到四连杆AB、BC、CD的角度。Among them, a 1 , a 2 , a 3 , and s 1 refer to the lengths of the DA, AB, BC, and CD four-bar links in the planar four-bar linkage ABCDA respectively, e is the base of natural logarithm, i is the imaginary unit, θ 2 , θ 3 and α 2 refer to the angles when the horizontal axis rotates counterclockwise to the four links AB, BC, and CD respectively.

根据其实部和虚部，方程(1)能表示为：According to its real and imaginary parts, equation (1) can be expressed as:

s1＝(a2cosθ2+a3cosθ3-a1)/cosα2 (2)s 1 =(a 2 cosθ 2 +a 3 cosθ 3 -a 1 )/cosα 2 (2)

s1＝(a2sinθ2+a3sinθ3)/sinα2 (3)s 1 =(a 2 sinθ 2 +a 3 sinθ 3 )/sinα 2 (3)

通过消除s1，方程(1)能表示为：By eliminating s 1 , equation (1) can be expressed as:

a1sinα2+a2sin(θ2-α2)+a3sin(θ3-α2)＝0 (4)a 1 sinα 2 +a 2 sin(θ 2 -α 2 )+a 3 sin(θ 3 -α 2 )＝0 (4)

当θ2、θ3是输入参数时，方程(4)能表示为输入-输出关系模型。利用半角公式：When θ 2 and θ 3 are input parameters, Equation (4) can be expressed as an input-output relationship model. Use the half-width formula:

x3＝tan(θ3/2) (5)x 3 =tan(θ 3 /2) (5)

方程(4)能表示为：Equation (4) can be expressed as:

PP 11 xx 33 22 ++ QQ 11 xx 33 ++ RR 11 == 00 -- -- -- (( 66 ))

其中：in:

P1＝a2sin(θ2-α2)+(a1+a3)sinα2 (7.1)P 1 ＝a 2 sin(θ 2 -α 2 )+(a 1 +a 3 )sin α 2 (7.1)

Q1＝2a3cosα2 (7.2)Q 1 =2a 3 cos α 2 (7.2)

Q1＝a2sin(θ2-α2)+(a1-a3)sinα2 (7.3)Q 1 ＝a 2 sin(θ 2 -α 2 )+(a 1 -a 3 )sinα 2 (7.3)

若要四杆机构构型存在，当P1≠0时必须满足For the four-bar linkage configuration to exist, when P 1 ≠ 0 must satisfy

ΔΔ 11 == QQ 11 22 -- 44 PP 11 RR 11 ≥&Greater Equal; 00 -- -- -- (( 88 ))

化简得：Simplified:

△1＝4S1S2 (9)△ 1 ＝4S 1 S 2 (9)

其中：in:

S1＝a3-a2sin(θ2-α2)-a1sinα2 (10.1)S 1 ＝a 3 -a 2 sin(θ 2 -α 2 )-a 1 sin α 2 (10.1)

S2＝a3+a2sin(θ2-α2)+a1sinα2 (10.2)S 2 ＝a 3 +a 2 sin(θ 2 -α 2 )+a 1 sinα 2 (10.2)

当△1＝0时，表示四杆机构处在死点位置；When △ 1 = 0, it means that the four-bar mechanism is at the dead point;

由方程(6)，能通过输入角θ2得输出角θ3：From equation (6), the output angle θ 3 can be obtained through the input angle θ 2 :

θ3由方程(11.1)和(11.2)完全确定，其中arctan(x3)在(-π/2,π/2)变化，与之一一相对应的θ3在(-π,π)变化；θ 3 is completely determined by equations (11.1) and (11.2), where arctan(x 3 ) varies at (-π/2,π/2), and θ 3 corresponding to one of them varies at (-π,π) ;

利用半角公式y＝tan(θ2/2)，方程(8)能改写为：Using the half-angle formula y=tan(θ 2 /2), equation (8) can be rewritten as:

f(y)＝C1y4+C2y2+C0 (12)f(y)＝C 1 y 4 +C 2 y 2 +C 0 (12)

其中，y的根的个数由四杆机构的构型决定；f(y)＝0可能有0、2、4个根，每个根与之相对应的是四杆机构的死点；Wherein, the number of roots of y is determined by the configuration of the four-bar mechanism; f(y)=0 may have 0, 2, and 4 roots, and each root corresponds to the dead point of the four-bar mechanism;

如果θ2是输入角，其四杆机构的分支和子分支通过以下方法确定：If θ2 is the input angle, the branches and sub - branches of its four-bar mechanism are determined by:

1.当f(y)>0恒成立，表示f(y)＝0无实根，则在此四杆机构构型中，无死点存在，根据公式(11)，可得输入θ2对应两个输出θ3，机构具有两个分支。1. When f(y)>0 holds true, it means that f(y)=0 has no real root, then in this four-bar mechanism configuration, there is no dead point, according to formula (11), the input θ 2 can be obtained corresponding to Two outputs θ 3 , the mechanism has two branches.

2.若f(y)＝0，且f(y)＝0有两个不同的实根，此四杆机构只包含一个分支，两个死点把其分成两个子分支；2. If f(y)=0, and f(y)=0 has two different real roots, this four-bar mechanism only includes one branch, and two dead points divide it into two sub-branches;

3.若f(y)＝0，且f(y)＝0有四个不同的实根，此四杆机构包含两个分支，四个死点把两个分支分成四个子分支；3. If f(y)=0, and f(y)=0 has four different real roots, this four-bar mechanism comprises two branches, and four dead points divide the two branches into four sub-branches;

步骤2：带移动副五杆机构的分支识别；Step 2: Branch recognition with a mobile sub-five-bar mechanism;

步骤2.1：一个带两个移动副六杆机构包含一个带移动副五杆机构ABEFGA，其环方程可由欧拉公式表示，即：Step 2.1: A six-bar mechanism with two moving pairs includes a five-bar mechanism with moving pairs ABEFGA, and its ring equation can be expressed by Euler's formula, namely:

aa 22 ee iθiθ 22 ++ aa 44 ee ii (( θθ 33 ++ ββ )) == aa 11 ++ aa 77 ++ aa 55 ee iθiθ 55 ++ sthe s 22 ee iαiα 11 -- -- -- (( 1313 ))

其中，a7、a2、a9、a5、a6分别指平面五杆机构ABEFG中GA、AB、BF、EF、FG四连杆的长度，e是自然对数的底，i是虚数单位，θ2、θ3+β、θ5、α1指的是水平轴逆时针旋转分别到连杆AB、BF、EF、FG的角度。Among them, a 7 , a 2 , a 9 , a 5 , and a 6 respectively refer to the lengths of the GA, AB, BF, EF, and FG four-bar links in the planar five-bar mechanism ABEFG, e is the base of natural logarithm, and i is an imaginary number Units, θ 2 , θ 3 +β, θ 5 , and α 1 refer to the angles from the counterclockwise rotation of the horizontal axis to the connecting rods AB, BF, EF, and FG, respectively.

根据其实部和虚部，方程(13)能表示为：According to the real and imaginary parts, equation (13) can be expressed as:

s2＝[a2cosθ2+a4cos(θ3+β)-a1-a7-a5cosθ5]/cosα1 (14)s 2 =[a 2 cosθ 2 +a 4 cos(θ 3 +β)-a 1 -a 7 -a 5 cosθ 5 ]/cosα 1 (14)

s2＝[a2sinθ2+a4sinθ4-a5sinθ5]/sinα1 (15)s 2 =[a 2 sinθ 2 +a 4 sinθ 4 -a 5 sinθ 5 ]/sinα 1 (15)

消除s2，方程(13)可表示为：Eliminating s 2 , equation (13) can be expressed as:

a2sin(θ2-α1)+a4sin(θ3+β-α1)-a5sin(θ5-α1)+a1+a7＝0 (16)a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 )-a 5 sin(θ 5 -α 1 )+a 1 +a 7 ＝0 (16)

利用半角公式：Use the half-width formula:

x5＝tan(θ5/2) (17)x 5 =tan(θ 5 /2) (17)

方程(16)能表示为：Equation (16) can be expressed as:

PP 22 xx 55 22 ++ QQ 22 xx 55 ++ RR 22 == 00 -- -- -- (( 1818 ))

其中：in:

P2＝a2sin(θ2-α1)+a4sin(θ3+β-α1)-a5sinα1+a1+a7 (19.1)P 2 ＝a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 )-a 5 sinα 1 +a 1 +a 7 (19.1)

Q2＝-2a5cosα1 (19.2)Q 2 =-2a 5 cos α 1 (19.2)

R2＝a2sin(θ2-α1)+a4sin(θ3+β-α1)+a5sinα1+a1+a7 (19.3)R 2 ＝a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 )+a 5 sinα 1 +a 1 +a 7 (19.3)

若五杆机构存在，当P2≠0必须满足：If the five-bar mechanism exists, when P 2 ≠0 must satisfy:

ΔΔ 22 == QQ 22 22 -- 44 PP 22 RR 22 ≥&Greater Equal; 00 -- -- -- (( 2020 ))

化简得：Simplified:

△2＝4S1S2≥0 (21)△ 2 = 4S 1 S 2 ≥ 0 (21)

其中：in:

S1＝a5-a1-a7-a2sin(θ2-α1)-a4sin(θ3+β-α1) (22.1)S 1 ＝a 5 -a 1 -a 7 -a 2 sin(θ 2 -α 1 )-a 4 sin(θ 3 +β-α 1 ) (22.1)

S2＝a5+a1+a7+a2sin(θ2-α1)+a4sin(θ3+β-α1) (22.2)S 2 ＝a 5 +a 1 +a 7 +a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 ) (22.2)

方程(20)、(21)产生θ2与θ3之间的关节旋转空间，其中，S1＝0或S2＝0表示关节旋转空间的边界，当五杆机构在奇异位置时或者3个非输出关节E、F、G共线，会出现上述情况；Equations (20), (21) generate the joint rotation space between θ 2 and θ 3 , where S 1 = 0 or S 2 = 0 represents the boundary of the joint rotation space, when the five-bar mechanism is in a singular position or 3 The above situation will occur when the non-output joints E, F, and G are collinear;

通过方程(15)，求得θ5：Through equation (15), get θ 5 :

每个解对应一种机构构型，其中arctan(x5)在(-π/2,π/2)变化，与之一一相对应θ5的(-π,π)在变化；Each solution corresponds to a mechanism configuration, where arctan(x 5 ) changes in (-π/2, π/2), corresponding to one-to-one (-π, π) of θ 5 is changing;

步骤2.2：一个带三个移动副六杆机构包含一个带一个移动副四杆机构ABCDA以及一个带两个移动副五杆机构ABEFGA，五环通过伸旋转转变成四环；因此，带三个移动副六杆机构的分支识别取决于四环ABCDA和五环ABCEFGA的相互作用，其四环ABCDA可通过上述对四杆机构的分支识别进行判别，而其五环的环方程可用欧拉公式表示，即：Step 2.2: A six-bar mechanism with three moving pairs includes a four-bar mechanism ABCDA with one moving pair and a five-bar mechanism ABEFGA with two moving pairs. The branch recognition of the auxiliary six-bar mechanism depends on the interaction between the four-ring ABCDA and the five-ring ABCEFGA. The four-ring ABCDA can be discriminated through the above-mentioned branch recognition of the four-bar mechanism, and the ring equation of the fifth ring can be expressed by Euler's formula. which is:

aa 22 ee iθiθ 22 ++ aa 44 ee ii (( θθ 33 ++ ββ )) == aa 11 ++ aa 77 ++ sthe s 33 ee ii (( αα 11 ++ γγ )) ++ sthe s 22 ee iαiα 11 -- -- -- (( 24twenty four ))

根据其实部和虚部，方程(24)可表示为：According to its real and imaginary parts, equation (24) can be expressed as:

s2＝[a2cosθ2+a4cos(θ3+β)-a1-a7-s3cos(α1+γ)]/cosα1 (25.1)s 2 =[a 2 cosθ 2 +a 4 cos(θ 3 +β)-a 1 -a 7 -s 3 cos(α 1 +γ)]/cosα 1 (25.1)

s2＝[a2sinθ2+a4sin(θ3+β)-s3sin(α1+γ)]/sinα1 (25.2)s 2 =[a 2 sinθ 2 +a 4 sin(θ 3 +β)-s 3 sin(α 1 +γ)]/sinα 1 (25.2)

消除s2，方程(24)可表示为：Eliminating s 2 , equation (24) can be expressed as:

s3sinγ-a2sin(θ2-α1)-a7sinα1-a4sin(θ3+β-α1)＝0 (26)s 3 sinγ-a 2 sin(θ 2 -α 1 )-a 7 sinα 1 -a 4 sin(θ 3 +β-α 1 )＝0 (26)

以s3为未知数，方程(26)的判别式为：Taking s 3 as the unknown, the discriminant of equation (26) is:

△3＝sin2γ (27)△ 3 = sin 2 γ (27)

在方程(27)中，sin2γ≥0恒成立。当sinγ≠0时，对于每对θ2和θ3，s3有唯一确定值，但当sinγ＝0，s3有无穷大解，在这种情况下的机构是无用的，因此，要满足机构存在，必须sinγ≠0。In equation (27), sin 2 γ≥0 is always established. When sinγ≠0, for each pair of θ 2 and θ 3 , s 3 has a unique definite value, but when sin γ = 0, s 3 has an infinite solution, the mechanism in this case is useless, therefore, to satisfy the mechanism Exist, must sinγ≠0.

步骤3：带移动副六杆机构分支识别；Step 3: Branch identification of the six-bar mechanism with moving vice;

根据分支点的存在与否，带移动副六杆机构分支识别分成如下两种类型：According to the existence of the branch point, the branch recognition of the six-bar mechanism with moving vice is divided into the following two types:

类型I：带移动副六杆机构；没有分支点存在，方程(4)以及方程(20)中的△＝0之间没有公共解并且方程(4)的解满足方程(20)中的△>0；其分支判别的过程如下：Type I: Six-bar mechanism with moving pair; there is no branch point, there is no common solution between equation (4) and △=0 in equation (20), and the solution of equation (4) satisfies △> in equation (20) 0; the process of branch discrimination is as follows:

步骤A1：四环分支判别；采用方程(12)对四杆机构的分支进行判别；Step A1: four-ring branch discrimination; use equation (12) to discriminate the branches of the four-bar mechanism;

步骤A2：分支点；无分支点存在；Step A2: branch point; no branch point exists;

步骤A3：带移动副六杆机构分支判别；Step A3: Judging the branch of the six-bar mechanism with a moving pair;

对于给定的四杆机构的分支，采用方程(23)对带移动副六杆机构的分支进行判别，方程(23)的每个解代表带移动副六杆机构的一个分支；如果给定的四杆机构的构型满足方程(20)，则此四杆机构的分支表示着带移动副六杆机构的两个分支；如果不满足，则此四杆机构的分支是无效的，此带移动副六杆机构无法装配；For a given branch of a four-bar mechanism, use equation (23) to discriminate the branch of a six-bar mechanism with a moving pair, and each solution of equation (23) represents a branch of a six-bar mechanism with a moving pair; if given The configuration of the four-bar mechanism satisfies equation (20), then the branch of the four-bar mechanism represents the two branches of the six-bar mechanism with a moving pair; if not satisfied, the branch of the four-bar mechanism is invalid, and the belt moves The auxiliary six-bar mechanism cannot be assembled;

步骤A4：如果四杆机构有另外一个分支，采用以上3个步骤对带移动副六杆机构进行判别；Step A4: If the four-bar mechanism has another branch, use the above three steps to discriminate the six-bar mechanism with a moving pair;

步骤A5：对带移动副六杆机构，带有3个滑动关节的分支类型总是类型I，四环的分支代表着此机构的分支；Step A5: For a six-bar mechanism with a moving pair, the branch type with 3 sliding joints is always type I, and the branch of the fourth ring represents the branch of this mechanism;

类型II：带移动副六杆机构；有分支点存在，方程(4)以及方程(20)中的△＝0之间存在公共解；四环输入输出曲线被分成多个部分，其中满足方程(20)的部分是有效的，每个有效的部分是个连续的解集并且代表带移动副六杆机构的一种构型；其分支判别的过程如下：Type II: six-bar mechanism with moving pair; there are branch points, and there is a common solution between equation (4) and △=0 in equation (20); the four-ring input and output curve is divided into multiple parts, which satisfy the equation ( 20) is effective, and each effective part is a continuous solution set and represents a configuration of a six-bar mechanism with a moving pair; the process of branch discrimination is as follows:

步骤B1：四环分支判别；采用方程(12)对四环分支进行判别；Step B1: Discrimination of the four-ring branch; use equation (12) to distinguish the four-ring branch;

步骤B2：分支点；Step B2: branch point;

利用方程(4)和方程(20)求得分支点；满足方程(20)的输入输出曲线部分是有效部分，止点是分支点或死点；Utilize equation (4) and equation (20) to obtain branch point; Satisfy the input-output curve part of equation (20) is effective part, and stop point is branch point or dead point;

步骤B3：带移动副六杆机构分支判别；Step B3: Judging the branches of the six-bar mechanism with a moving pair;

在四杆机构同一个子分支中，两个相邻分支点的有效连续部分表示带移动副六杆机构的一个分支；在给定的四杆机构的分支的不同子分支根据公共死点形成一个连续的有效部分，因此，分支点作为止点可判别所有有效部分，每个分离的有效部分是一个分支；In the same sub-branch of the four-bar mechanism, the effective continuous part of two adjacent branch points represents a branch of the six-bar mechanism with a moving sub-branch; different sub-branches of the branch of a given four-bar mechanism form a continuous Therefore, the branch point can be used as the stop point to distinguish all valid parts, and each separated valid part is a branch;

步骤B4：如果四杆机构有含有其它分支点的分支，重复以上步骤，如果没有，参照类型I对单自由度双环机构进行判别；Step B4: If the four-bar mechanism has branches containing other branch points, repeat the above steps, if not, refer to type I to distinguish the single-degree-of-freedom double-loop mechanism;

步骤B5：在四杆机构的同一子分支中，两个相邻分支点的有效部分代表整个机构的一个分支；四杆机构的不同子分支中的有效部分通过公共的死点形成连续的有效部分，从而形成了整个机构的分支；Step B5: In the same sub-branch of the four-bar mechanism, the effective parts of two adjacent branch points represent a branch of the whole mechanism; the effective parts in different sub-branches of the four-bar mechanism form a continuous effective part through the common dead point , thus forming a branch of the whole institution;

步骤4：带移动副六杆机构的子分支识别；Step 4: Identification of sub-branches with moving auxiliary six-bar mechanism;

当θ2作为输入时，死点位置可通过方程(8)中△＝0求得；如果分支点存在，这同样也是带移动副六杆机构奇异位置；因此，两个相邻奇异点，即死点和分支点之间的有效部分是带移动副六杆机构构的子分支；其具体识别分为以下两种情况：When θ 2 is used as input, the position of the dead point can be obtained by △=0 in equation (8); if the branch point exists, this is also the singular position of the six-bar mechanism with moving vice; therefore, two adjacent singular points, that is, the dead point The effective part between the point and the branch point is a sub-branch with a moving auxiliary six-bar mechanism; its specific identification is divided into the following two situations:

(1)没有分支点存在；(1) No branch point exists;

对于一个带移动副六杆机构给定的分支中，其子分支通过公式(11)识别；方程中的两个解各对应着带移动副六杆机构的一个子分支；For a given branch of a six-bar mechanism with a moving pair, its sub-branches are identified by formula (11); the two solutions in the equation correspond to a sub-branch of the six-bar mechanism with a moving pair;

(2)分支点存在；(2) The branch point exists;

四杆机构的输入输出曲线中所有有效部分上的每个点都代表着机构的不同构型，这能根据方程(23)判别；因此，对于给定的带移动副六杆机构的分支，其有效部分的子分支被识别了，那么带移动副六杆机构的子分支也可被识别。Each point on all effective parts of the input-output curve of the four-bar mechanism represents a different configuration of the mechanism, which can be judged according to equation (23); therefore, for a given branch of the six-bar mechanism with a moving pair, its The sub-branches of the effective part are identified, so the sub-branches with the moving auxiliary six-bar mechanism can also be identified.

本发明专利的有益效果：The beneficial effects of the patent of the present invention:

1)本发明提出了一种对带移动副六杆机构分支进行自动识别新的方法，该方法利用机构基本环方程实现对分支(回路)识别，识别更精确、直观、高效；1) The present invention proposes a new method for automatically identifying branches of a six-bar mechanism with a moving pair. The method utilizes the basic ring equation of the mechanism to identify branches (loops), and the identification is more accurate, intuitive and efficient;

2)本发明提供的方法便于开发计算机数学软件进行模拟仿真，有助于机械设计，具有很高的使用价值；2) The method provided by the invention is convenient to develop computer mathematics software to carry out simulation simulation, contributes to mechanical design, and has very high use value;

3)本发明提供的方法能嵌入到各种机械设计类商业软件中，具有良好的社会价值和经济价值。3) The method provided by the present invention can be embedded in various mechanical design commercial software, and has good social value and economic value.

附图说明Description of drawings

图1：本发明实施例的带两个移动副六杆机构示意图；Fig. 1: a schematic diagram of a six-bar mechanism with two moving pairs according to an embodiment of the present invention;

图2：本发明实施例的带三个移动副六杆机构示意图；Figure 2: a schematic diagram of a six-bar mechanism with three moving pairs in an embodiment of the present invention;

图3：本发明实施例的四杆机构的两个分支示意图，当带移动副四杆机构构型为a1＝9，a2＝6，a3＝10，α2＝15°时，f(y)>0恒成立，此构型无死点存在；Fig. 3: Schematic diagram of two branches of the four-bar mechanism of the embodiment of the present invention, when the configuration of the four-bar mechanism with moving pair is a 1 =9, a 2 =6, a 3 =10, α 2 =15°, f (y)>0 is always established, and there is no dead point in this configuration;

图4：本发明实施例的四杆机构的两个分支示意图，当带移动副四杆机构构型为a1＝9，a2＝6，a3＝10，α2＝45°时，f(y)＝0有两个解，此构型有2个死点存在；Figure 4: Schematic diagram of the two branches of the four-bar mechanism of the embodiment of the present invention, when the configuration of the four-bar mechanism with a moving pair is a 1 =9, a 2 =6, a 3 =10, α 2 =45°, f (y)=0 has two solutions, and this configuration has 2 dead points to exist;

图5：本发明实施例的四杆机构的两个分支示意图，当带移动副四杆机构构型为a1＝4，a2＝5，a3＝2.2，α2＝45°时，f(y)＝0有四个解，此构型有4个死点存在；Figure 5: Schematic diagram of the two branches of the four-bar mechanism of the embodiment of the present invention, when the configuration of the four-bar mechanism with a moving pair is a 1 =4, a 2 =5, a 3 =2.2, α 2 =45°, f (y)=0 has four solutions, and this configuration has 4 dead points to exist;

图6：本发明实施例的带移动副六杆机构分支识别流程图；Fig. 6: Flow chart of branch recognition of six-bar mechanism with moving pair according to the embodiment of the present invention;

图7：本发明实施例的带两个移动副六杆机构分支示意图；Figure 7: Schematic diagram of the branches of the six-bar mechanism with two moving pairs in the embodiment of the present invention;

具体实施方式detailed description

为了便于本领域普通技术人员理解和实施本发明，下面结合附图及实施例对本发明作进一步的详细描述，应当理解，此处所描述的实施示例仅用于说明和解释本发明，并不用于限定本发明。In order to facilitate those of ordinary skill in the art to understand and implement the present invention, the present invention will be described in further detail below in conjunction with the accompanying drawings and embodiments. It should be understood that the implementation examples described here are only used to illustrate and explain the present invention, and are not intended to limit this invention.

分支是被问题是连杆机构分析和综合遇到的常见问题，它包括机构运动的连续性、(分支)装配方式或回路、子分支等问题。其中分支指的是连杆机构的一种装配形式或连杆机构的构型空间，即在不拆开机构的情况下其连续运动所能达到的所有可能的位置。分支识别后，运动连续性的可能就能得到保证并且运动平滑性或子分支就能识别。在发明中，分支、分支点、子分支的识别方法通用，每个子分支含有一个连续的且无奇异的输入域，即输入输出一一对应，本发明首先对带移动副单环的分支进行研究，继而研究环之间的相互关系，利用此方法可将带移动副六杆机构的分支很快识别出来，为机构设计提供了新的辅助方法。Branching is a common problem encountered in the analysis and synthesis of linkage mechanisms, which includes the continuity of mechanism motion, (branch) assembly methods or circuits, sub-branches and other issues. The branch refers to an assembly form of the linkage mechanism or the configuration space of the linkage mechanism, that is, all possible positions that can be reached by its continuous movement without disassembling the mechanism. After branch identification, the possibility of motion continuity can be guaranteed and motion smoothness or sub-branches can be identified. In the invention, the identification methods of branches, branch points, and sub-branches are common, and each sub-branch contains a continuous and non-singular input field, that is, one-to-one correspondence between input and output. The present invention first studies the branch with a moving sub-single ring , and then study the interrelationships between the rings, using this method can quickly identify the branches of the six-bar mechanism with moving sub, and provide a new auxiliary method for mechanism design.

本发明提供的一种带移动副六杆机构的分支识别方法，将单自由度双环连杆机构与输入输出相关的两个单环结合，通过其相互作用得出带移动副六杆机构的分支识别方法。The invention provides a branch recognition method of a six-bar mechanism with a moving pair, which combines a single-degree-of-freedom double-loop linkage mechanism with two single-rings related to input and output, and obtains a branch of a six-bar mechanism with a moving pair through their interaction recognition methods.

如图1所示，带两个移动副六杆机构由一个带一个移动副四杆机构ABCDA和一个带一个移动副五杆机构ABEFGA组成；如图2所示，一个带三个移动副六杆机构由一个带一个移动副四杆机构ABCDA和一个一个移动副五杆机构ABEFGA。而带移动副六杆机构的分支识别与输入和固定链接相关，有鉴于此，本发明将带移动副六杆机构与输入输出相关的两个单环结合，通过其相互作用得出带移动副六杆机构的分支识别。As shown in Figure 1, a six-bar mechanism with two moving pairs consists of a four-bar mechanism with one moving pair ABCDA and a five-bar mechanism with one moving pair ABEFGA; as shown in Figure 2, a six-bar mechanism with three moving pairs The mechanism consists of a four-bar mechanism ABCDA with a movable sub-bar and an ABEFGA with a movable sub-five-bar mechanism. And the branch recognition of the six-bar mechanism with moving pair is related to the input and the fixed link. In view of this, the present invention combines the six-bar mechanism with moving pair with the two single loops related to input and output, and obtains the link with moving pair through their interaction. Branch identification of a six-bar mechanism.

请见图6，本发明的具体实现包括以下步骤：See also Fig. 6, the concrete realization of the present invention comprises the following steps:

步骤1：带移动副四杆机构的分支识别；Step 1: Branch recognition with a mobile sub-four-bar mechanism;

分支识别问题是连杆机构分析和综合遇到的常见问题，它包括机构运动的连续性、(分支)装配方式或回路、子分支等问题。其中分支指的是连杆机构的一种装配形式或连杆机构的构型空间，即在不拆开机构的情况下其连续运动所能达到的所有可能的位置。分支识别后，运动连续性的可能就能得到保证并且运动平滑性或子分支就能识别。The problem of branch recognition is a common problem encountered in the analysis and synthesis of linkage mechanisms, which includes the continuity of mechanism motion, (branch) assembly method or circuit, sub-branch and so on. The branch refers to an assembly form of the linkage mechanism or the configuration space of the linkage mechanism, that is, all possible positions that can be reached by its continuous movement without disassembling the mechanism. After branch identification, the possibility of motion continuity can be guaranteed and motion smoothness or sub-branches can be identified.

任何一个带移动副六杆机构都包含一个或多个四杆机构。如图1、2所示，带两个移动副六杆机构和带三个移动副六杆机构包含四杆机构ABCDA。其环方程可根据欧拉公式表示为：Any six-bar mechanism with movable sub-joint all comprises one or more four-bar mechanisms. As shown in Figures 1 and 2, the six-bar mechanism with two moving pairs and the six-bar mechanism with three moving pairs include the four-bar mechanism ABCDA. Its ring equation can be expressed according to Euler's formula as:

aa 22 ee iθiθ 22 ++ aa 33 ee iθiθ 33 == aa 11 ++ sthe s 11 ee iαiα 22 -- -- -- (( 11 ))

根据其实部和虚部，方程(1)可表示为：According to its real and imaginary parts, equation (1) can be expressed as:

s1＝(a2cosθ2+a3cosθ3-a1)/cosα2 (2)s 1 =(a 2 cosθ 2 +a 3 cosθ 3 -a 1 )/cosα 2 (2)

s1＝(a2sinθ2+a3sinθ3)/sinα2 (3)s 1 =(a 2 sinθ 2 +a 3 sinθ 3 )/sinα 2 (3)

通过消除s1，方程(1)能表示为：By eliminating s 1 , equation (1) can be expressed as:

a1sinα2+a2sin(θ2-α2)+a3sin(θ3-α2)＝0 (4)a 1 sinα 2 +a 2 sin(θ 2 -α 2 )+a 3 sin(θ 3 -α 2 )＝0 (4)

当θ2、θ3是输入参数时，方程(4)可表示为输入-输出关系模型，利用半角公式：When θ 2 and θ 3 are input parameters, Equation (4) can be expressed as an input-output relationship model, using the half-angle formula:

x3＝tan(θ3/2) (5)x 3 =tan(θ 3 /2) (5)

方程(4)可表示为：Equation (4) can be expressed as:

PP 11 xx 33 22 ++ QQ 11 xx 33 ++ RR 11 == 00 -- -- -- (( 66 ))

其中：in:

P1＝a2sin(θ2-α2)+(a1+a3)sinα2 (7.1)P 1 ＝a 2 sin(θ 2 -α 2 )+(a 1 +a 3 )sin α 2 (7.1)

Q1＝2a3cosα2 (7.2)Q 1 =2a 3 cos α 2 (7.2)

Q1＝a2sin(θ2-α2)+(a1-a3)sinα2 (7.3)Q 1 ＝a 2 sin(θ 2 -α 2 )+(a 1 -a 3 )sinα 2 (7.3)

若要四杆机构构型存在，当P1≠0必须满足：For the four-bar linkage configuration to exist, when P 1 ≠ 0 must satisfy:

ΔΔ 11 == QQ 11 22 -- 44 PP 11 RR 11 ≥&Greater Equal; 00 -- -- -- (( 88 ))

化简得：Simplified:

△1＝4S1S2 (9)△ 1 ＝4S 1 S 2 (9)

其中：in:

S1＝a3-a2sin(θ2-α2)-a1sinα2 (10.1)S 1 ＝a 3 -a 2 sin(θ 2 -α 2 )-a 1 sin α 2 (10.1)

S2＝a3+a2sin(θ2-α2)+a1sinα2 (10.2)S 2 ＝a 3 +a 2 sin(θ 2 -α 2 )+a 1 sinα 2 (10.2)

当△1＝0时，表示四杆机构处在死点位置；When △ 1 = 0, it means that the four-bar mechanism is at the dead point;

由方程(6)，能通过输入角θ2得输出角θ3：From equation (6), the output angle θ 3 can be obtained through the input angle θ 2 :

θ3由方程(11.1)和(11.2)完全确定，其中arctan(x3)在(-π/2,π/2)变化，与之一一相对应的θ3在(-π,π)变化；θ 3 is completely determined by equations (11.1) and (11.2), where arctan(x 3 ) varies at (-π/2,π/2), and θ 3 corresponding to one of them varies at (-π,π) ;

利用半角公式y＝tan(θ2/2)，方程(8)能改写为：Using the half-angle formula y=tan(θ 2 /2), equation (8) can be rewritten as:

f(y)＝C1y4+C2y2+C0 (12)f(y)＝C 1 y 4 +C 2 y 2 +C 0 (12)

其中，y的根的个数由四杆机构的构型决定；f(y)＝0可能有0、2、4个根，每个根与之相对应的是四杆机构的死点；Wherein, the number of roots of y is determined by the configuration of the four-bar mechanism; f(y)=0 may have 0, 2, and 4 roots, and each root corresponds to the dead point of the four-bar mechanism;

如果θ2是输入角，其四杆机构的分支和子分支通过以下方法确定：If θ2 is the input angle, the branches and sub - branches of its four-bar mechanism are determined by:

1.当f(y)>0恒成立，表示f(y)＝0无实根，则在此四杆机构构型中，无死点存在，方程(9)有两个解，与之相对应的是四杆机构的两个分支，每个分支仅仅只含有一个子分支；请见图3，当带移动副四杆机构构型为a1＝9，a2＝6，a3＝10，α2＝15°时，f(y)>0恒成立，此构型无死点存在；1. When f(y)>0 holds true, it means that f(y)=0 has no real roots, then in this four-bar mechanism configuration, there is no dead point, and equation (9) has two solutions, corresponding to Corresponding to the two branches of the four-bar mechanism, each branch contains only one sub-branch; please see Figure 3, when the configuration of the four-bar mechanism with a moving pair is a 1 =9, a 2 =6, a 3 =10 , when α 2 =15°, f(y)>0 is always established, and there is no dead point in this configuration;

2.若f(y)＝0，且f(y)＝0有两个不同的实根，则在此四杆机构构型中，有两个死点存在，此四杆机构只包含一个分支，两个死点把其分成两个子分支；请见图4，当带移动副四杆机构构型为a1＝9，a2＝6，a3＝10，α2＝45°时，f(y)＝0有两个解，此构型有2个死点存在；2. If f(y)=0, and f(y)=0 has two different real roots, then in this four-bar mechanism configuration, there are two dead points, and this four-bar mechanism contains only one branch , two dead points divide it into two sub-branches; see Figure 4, when the configuration of the four-bar mechanism with moving pair is a 1 =9, a 2 =6, a 3 =10, α 2 =45°, f (y)=0 has two solutions, and this configuration has 2 dead points to exist;

3.若f(y)＝0，且f(y)＝0有四个不同的实根，则在此四杆机构构型中，有四个死点存在，且此四杆机构包含两个分支，四个死点把两个分支分成四个子分支；请见图5，当带移动副四杆机构构型为a1＝4，a2＝5，a3＝2.2，α2＝45°时，f(y)＝0有四个解，此构型有4个死点存在。3. If f(y)=0, and f(y)=0 has four different real roots, then in this four-bar mechanism configuration, there are four dead points, and this four-bar mechanism contains two Branches, two branches are divided into four sub-branches by four dead points; please see Fig. 5, when the configuration of the four-bar mechanism with a moving pair is a 1 =4, a 2 =5, a 3 =2.2, α 2 =45° When f(y)=0, there are four solutions, and there are four dead points in this configuration.

步骤2：带移动副五杆机构的分支识别；Step 2: Branch recognition with a mobile sub-five-bar mechanism;

一个带两个移动副六杆机构包含一个带移动副五杆机构ABEFGA，其环方程可根据欧拉公式表示为:A six-bar mechanism with two moving pairs includes a five-bar mechanism with moving pairs ABEFGA, and its loop equation can be expressed according to Euler's formula:

aa 22 ee iθiθ 22 ++ aa 44 ee ii (( θθ 33 ++ ββ )) == aa 11 ++ aa 77 ++ aa 55 ee iθiθ 55 ++ sthe s 22 ee iαiα 11 -- -- -- (( 1313 ))

其中，a7、a2、a9、a5、a6分别指平面五杆机构ABEFG中GA、AB、BF、EF、FG四连杆的长度，e是自然对数的底，i是虚数单位，θ2、θ3+β、θ5、α1指的是水平轴逆时针旋转分别到连杆AB、BF、EF、FG的角度。Among them, a 7 , a 2 , a 9 , a 5 , and a 6 respectively refer to the lengths of the GA, AB, BF, EF, and FG four-bar links in the planar five-bar mechanism ABEFG, e is the base of natural logarithm, and i is an imaginary number Units, θ 2 , θ 3 +β, θ 5 , and α 1 refer to the angles from the counterclockwise rotation of the horizontal axis to the connecting rods AB, BF, EF, and FG, respectively.

根据其实部和虚部，方程(13)能表示为：According to the real and imaginary parts, equation (13) can be expressed as:

s2＝[a2cosθ2+a4cos(θ3+β)-a1-a7-a5cosθ5]/cosα1 (14)s 2 =[a 2 cosθ 2 +a 4 cos(θ 3 +β)-a 1 -a 7 -a 5 cosθ 5 ]/cosα 1 (14)

s2＝[a2sinθ2+a4sin(θ4+β)-a5sinθ5]/sinα1 (15)s 2 =[a 2 sinθ 2 +a 4 sin(θ 4 +β)-a 5 sinθ 5 ]/sinα 1 (15)

消除s2，方程(13)可表示为：Eliminating s 2 , equation (13) can be expressed as:

a2sin(θ2-α1)+a4sin(θ3+β-α1)-a5sin(θ5-α1)+(a1+a7)sinα1＝0 (16)a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 )-a 5 sin(θ 5 -α 1 )+(a 1 +a 7 )sinα 1 ＝0 (16)

利用半角公式：Use the half-width formula:

x5＝tan(θ5/2) (17)x 5 =tan(θ 5 /2) (17)

方程(16)能表示为：Equation (16) can be expressed as:

PP 22 xx 55 22 ++ QQ 22 xx 55 ++ RR 22 == 00 -- -- -- (( 1818 ))

其中：in:

P2＝a2sin(θ2-α1)+a4sin(θ3+β-α1)+a5sinα1+(a1+a7)sinα1 (19.1)P 2 ＝a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 )+a 5 sinα 1 +(a 1 +a 7 )sinα 1 (19.1)

Q2＝2a5cosα1 (19.2)Q 2 =2a 5 cos α 1 (19.2)

R2＝a2sin(θ2-α1)+a4sin(θ3+β-α1)-a5sinα1+(a1+a7)sinα1 (19.3)R 2 ＝a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 )-a 5 sinα 1 +(a 1 +a 7 )sinα 1 (19.3)

若五杆机构存在，当P2≠0必须满足：If the five-bar mechanism exists, when P 2 ≠0 must satisfy:

ΔΔ 22 == QQ 22 22 -- 44 PP 22 RR 22 ≥&Greater Equal; 00 -- -- -- (( 2020 ))

化简得：Simplified:

△2＝4S1S2≥0 (21)△ 2 = 4S 1 S 2 ≥ 0 (21)

其中：in:

S1＝a5-(a1+a7)sinα1-a2sin(θ2-α1)-a4sin(θ3+β-α1) (22.1)S 1 ＝a 5 -(a 1 +a 7 ) sinα 1 -a 2 sin(θ 2 -α 1 )-a 4 sin(θ 3 +β-α 1 ) (22.1)

S2＝a5+(a1+a7)sinα1+a2sin(θ2-α1)+a4sin(θ3+β-α1) (22.2)S 2 ＝a 5 +(a 1 +a 7 ) sinα 1 +a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 ) (22.2)

方程(20)、(21)产生θ2与θ3之间的关节旋转空间，其中，S1＝0或S2＝0表示关节旋转空间的边界，当五杆机构在奇异位置时或者3个非输出关节E、F、G共线，会出现上述情况；Equations (20), (21) generate the joint rotation space between θ 2 and θ 3 , where S 1 = 0 or S 2 = 0 represents the boundary of the joint rotation space, when the five-bar mechanism is in a singular position or 3 The above situation will occur when the non-output joints E, F, and G are collinear;

通过方程(15)，求得θ5：Through equation (15), get θ 5 :

每个解对应一种机构构型，其中arctan(x5)在(-π/2,π/2)变化，与之一一相对应θ5的(-π,π)在变化；Each solution corresponds to a mechanism configuration, where arctan(x 5 ) changes in (-π/2, π/2), corresponding to one-to-one (-π, π) of θ 5 is changing;

如图2中所示的带三个移动副六杆机构，由一个带一个移动副四杆机构ABCDA以及一个带两个移动副五杆机构ABEFGA组成；因此，带三个移动副六杆机构的分支识别取决于四环ABCDA和五环ABCEFGA的相互作用，其四环ABCDA可通过上述对四杆机构的分支识别进行判别，而其五环的环方程可用欧拉公式表示，即：As shown in Figure 2, the six-bar mechanism with three moving pairs is composed of a four-bar mechanism with one moving pair ABCDA and a five-bar mechanism with two moving pairs ABEFGA; therefore, the six-bar mechanism with three moving pairs The branch recognition depends on the interaction between the four-ring ABCDA and the five-ring ABCEFGA. The four-ring ABCDA can be discriminated through the above-mentioned branch recognition of the four-bar mechanism, and the ring equation of the five-ring can be expressed by Euler's formula, namely:

aa 22 ee iθiθ 22 ++ aa 44 ee ii (( θθ 33 ++ ββ )) == aa 11 ++ aa 77 ++ sthe s 33 ee ii (( αα 11 ++ γγ )) ++ sthe s 22 ee iαiα 11 -- -- -- (( 24twenty four ))

根据其实部和虚部，方程(24)可表示为：According to its real and imaginary parts, equation (24) can be expressed as:

s2＝[a2cosθ2+a4cos(θ3+β)-a1-a7-s3cos(α1+γ)]/cosα1 (25.1)s 2 =[a 2 cosθ 2 +a 4 cos(θ 3 +β)-a 1 -a 7 -s 3 cos(α 1 +γ)]/cosα 1 (25.1)

s2＝[a2sinθ2+a4sin(θ3+β)-s3sin(α1+γ)]/sinα1 (25.2)s 2 =[a 2 sinθ 2 +a 4 sin(θ 3 +β)-s 3 sin(α 1 +γ)]/sinα 1 (25.2)

消除s2，方程(24)可表示为：Eliminating s 2 , equation (24) can be expressed as:

s3sinγ-a2sin(θ2-α1)-a7sinα1-a4sin(θ3+β-α1)＝0 (26)s 3 sinγ-a 2 sin(θ 2 -α 1 )-a 7 sinα 1 -a 4 sin(θ 3 +β-α 1 )＝0 (26)

以s3为未知数，方程(26)的判别式为：Taking s 3 as the unknown, the discriminant of equation (26) is:

△3＝sin2γ (27)△ 3 = sin 2 γ (27)

在方程(27)中，sin2γ≥0恒成立。当sinγ≠0时，对于每对θ2和θ3，s3有唯一确定值，但当sinγ＝0，s3有无穷大解，在这种情况下的机构是无用的，因此，要满足机构存在，必须sinγ≠0。In equation (27), sin 2 γ≥0 is always established. When sinγ≠0, for each pair of θ 2 and θ 3 , s 3 has a unique definite value, but when sin γ = 0, s 3 has an infinite solution, the mechanism in this case is useless, therefore, to satisfy the mechanism Exist, must sinγ≠0.

步骤3：带移动副六杆机构分支识别；Step 3: Branch identification of the six-bar mechanism with moving vice;

根据分支点的存在与否，带移动副六杆机构分支识别分成如下两种类型：According to the existence of the branch point, the branch recognition of the six-bar mechanism with moving vice is divided into the following two types:

类型I：带移动副六杆机构；没有分支点存在，方程(4)以及方程(20)中的△＝0之间没有公共解并且方程(4)的解满足方程(20)中的△>0；其分支判别的过程如下：Type I: Six-bar mechanism with moving pair; there is no branch point, there is no common solution between equation (4) and △=0 in equation (20), and the solution of equation (4) satisfies △> in equation (20) 0; the process of branch discrimination is as follows:

步骤A1：四环分支判别；采用方程(12)对四杆机构的分支进行判别；Step A1: four-ring branch discrimination; use equation (12) to discriminate the branches of the four-bar mechanism;

步骤A2：分支点；无分支点存在；Step A2: branch point; no branch point exists;

步骤A3：带移动副六杆机构分支判别；Step A3: Judging the branch of the six-bar mechanism with a moving pair;

对于给定的四杆机构的分支，采用方程(23)对带移动副六杆机构的分支进行判别，方程(23)的每个解代表带移动副六杆机构的一个分支；如果给定的四杆机构的构型满足方程(20)，则此四杆机构的分支表示着带移动副六杆机构的两个分支；如果不满足，则此四杆机构的分支是无效的，此带移动副六杆机构无法装配；For a given branch of a four-bar mechanism, use equation (23) to discriminate the branch of a six-bar mechanism with a moving pair, and each solution of equation (23) represents a branch of a six-bar mechanism with a moving pair; if given The configuration of the four-bar mechanism satisfies equation (20), then the branch of the four-bar mechanism represents the two branches of the six-bar mechanism with a moving pair; if not satisfied, the branch of the four-bar mechanism is invalid, and the belt moves The auxiliary six-bar mechanism cannot be assembled;

步骤A4：如果四杆机构有另外一个分支，采用以上3个步骤对带移动副六杆机构进行判别；Step A4: If the four-bar mechanism has another branch, use the above three steps to discriminate the six-bar mechanism with a moving pair;

步骤A5：对带移动副六杆机构，带有3个滑动关节的分支类型总是类型I，四环的分支代表着此机构的分支；Step A5: For a six-bar mechanism with a moving pair, the branch type with 3 sliding joints is always type I, and the branch of the fourth ring represents the branch of this mechanism;

类型II：带移动副六杆机构；有分支点存在，方程(4)以及方程(20)中的△＝0之间存在公共解；四环输入输出曲线被分成多个部分，其中满足方程(20)的部分是有效的，每个有效的部分是个连续的解集并且代表带移动副六杆机构的一种构型；其分支判别的过程如下：Type II: six-bar mechanism with moving pair; there are branch points, and there is a common solution between equation (4) and △=0 in equation (20); the four-ring input and output curve is divided into multiple parts, which satisfy the equation ( 20) is effective, and each effective part is a continuous solution set and represents a configuration of a six-bar mechanism with a moving pair; the process of branch discrimination is as follows:

步骤B1：四环分支判别；采用方程(12)对四环分支进行判别；Step B1: Discrimination of the four-ring branch; use equation (12) to distinguish the four-ring branch;

步骤B2：分支点；Step B2: branch point;

利用方程(4)和方程(20)求得分支点；满足方程(20)的输入输出曲线部分是有效部分，止点是分支点或死点；Utilize equation (4) and equation (20) to obtain branch point; Satisfy the input-output curve part of equation (20) is effective part, and stop point is branch point or dead point;

步骤B3：带移动副六杆机构分支判别；Step B3: branch discrimination of the six-bar mechanism with moving pair;

在四杆机构同一个子分支中，两个相邻分支点的有效连续部分表示带移动副六杆机构的一个分支；在给定的四杆机构的分支的不同子分支根据公共死点形成一个连续的有效部分，因此，分支点作为止点可判别所有有效部分，每个分离的有效部分是一个分支；In the same sub-branch of the four-bar mechanism, the effective continuous part of two adjacent branch points represents a branch of the six-bar mechanism with a moving sub-branch; different sub-branches of the branch of a given four-bar mechanism form a continuous Therefore, the branch point can be used as the stop point to distinguish all valid parts, and each separated valid part is a branch;

步骤B4：如果四杆机构有含有其它分支点的分支，重复以上步骤，如果没有，参照类型I对单自由度双环机构进行判别；Step B4: If the four-bar mechanism has branches containing other branch points, repeat the above steps, if not, refer to type I to distinguish the single-degree-of-freedom double-loop mechanism;

步骤B5：在四杆机构的同一子分支中，两个相邻分支点的有效部分代表整个机构的一个分支；四杆机构的不同子分支中的有效部分通过公共的死点形成连续的有效部分，从而形成了整个机构的分支；Step B5: In the same sub-branch of the four-bar mechanism, the effective parts of two adjacent branch points represent a branch of the whole mechanism; the effective parts in different sub-branches of the four-bar mechanism form a continuous effective part through the common dead point , thus forming a branch of the whole institution;

步骤4：带移动副六杆机构的子分支识别；Step 4: Identification of sub-branches with moving auxiliary six-bar mechanism;

当θ2作为输入时，死点位置可通过方程(8)中△＝0求得；如果分支点存在，这同样也是带移动副六杆机构奇异位置；因此，两个相邻奇异点，即死点和分支点之间的有效部分是带移动副六杆机构构的子分支；其具体识别分为以下两种情况：When θ 2 is used as input, the position of the dead point can be obtained by △=0 in equation (8); if the branch point exists, this is also the singular position of the six-bar mechanism with moving vice; therefore, two adjacent singular points, that is, the dead point The effective part between the point and the branch point is a sub-branch with a moving auxiliary six-bar mechanism; its specific identification is divided into the following two situations:

(3)没有分支点存在；(3) No branch point exists;

对于一个带移动副六杆机构给定的分支中，其子分支通过公式(11)识别；方程中的两个解各对应着带移动副六杆机构的一个子分支；For a given branch of a six-bar mechanism with a moving pair, its sub-branches are identified by formula (11); the two solutions in the equation correspond to a sub-branch of the six-bar mechanism with a moving pair;

(4)分支点存在；(4) The branch point exists;

四杆机构的输入输出曲线中所有有效部分上的每个点都代表着机构的不同构型，这能根据方程(23)判别；因此，对于给定的带移动副六杆机构的分支，其有效部分的子分支被识别了，那么带移动副六杆机构的子分支也可被识别。Each point on all effective parts of the input-output curve of the four-bar mechanism represents a different configuration of the mechanism, which can be judged according to equation (23); therefore, for a given branch of the six-bar mechanism with a moving pair, its The sub-branches of the effective part are identified, so the sub-branches with the moving auxiliary six-bar mechanism can also be identified.

以下通过具体实施对本发明做进一步阐述；The present invention will be further elaborated below by specific implementation;

实施例1：给出图1所示的带两个移动副六杆机构长度及相关角角度，如图1所示，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°。Embodiment 1: Given the length and relative angles of the six-bar mechanism with two moving pairs shown in Figure 1, as shown in Figure 1, a 1 =2.4, a 2 =5.5, a 3 =4.8, a 4 =4.2 , a 5 =2.5, a 7 =2.4, β=45.0°, α 1 =30.0°, α 2 =50.0°.

根据方程(4)可以得到图1中带移动副四杆机构ABCD的输入输出曲线，根据方程(20)可得带移动副五杆机构的关节旋转空间，即如图所示阴影部分。According to equation (4), the input and output curves of the four-bar mechanism with moving pair ABCD in Figure 1 can be obtained. According to equation (20), the joint rotation space of the five-bar mechanism with moving pair can be obtained, which is the shaded part as shown in the figure.

(1)带移动副四杆机构分支判别：当θ2为输入角，θ3为输出角时，根据方程(8)，可得四杆机构死点位置u(-162.6°,-39.9°)，v(82.6°,-39.9°)。根据方程(11.1)及方程(11.2)，其分支有两个子分支。如图7所示：(1) Branch discrimination of the four-bar mechanism with moving pair: when θ 2 is the input angle and θ 3 is the output angle, according to equation (8), the dead point position u of the four-bar mechanism can be obtained (-162.6°, -39.9°) , v(82.6°,-39.9°). According to equation (11.1) and equation (11.2), its branch has two sub-branches. As shown in Figure 7:

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

子分支u-1-6-5-v：θ2∈[-162.6°,82.6°]；Sub-branch u-1-6-5-v: θ 2 ∈ [-162.6°,82.6°];

(2)分支点：当θ2为输入角，θ3为输出角时，根据方程(4)以及方程(20)可得四杆机构输入输出曲线以及关节旋转空间边界的交点坐标，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°)(2) Branch point: when θ 2 is the input angle and θ 3 is the output angle, according to Equation (4) and Equation (20), the coordinates of the intersection of the input-output curve of the four-bar mechanism and the boundary of the joint rotation space can be obtained, 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)四杆机构输入输出曲线的有效部分：四杆机构输入输出关系必须同时满足五杆机构的关节旋转空间，这样六杆机构才具有可动性。如图7所示，在子分支u-2-3-4-v中，有效部分为：u-2部分，其中θ2[-161.4°,-28.6°]以及3-4部分，其中θ2[-105.2°,-15.6°]；在子分支u-1-6-5-v中，有效部分为：u-1部分，其中θ2[-162.6°,-141.6°]以及6-5部分，其中θ2[7.9°,-141.6°]。(3) The effective part of the input-output curve of the four-bar mechanism: the input-output relationship of the four-bar mechanism must satisfy the joint rotation space of the five-bar mechanism at the same time, so that the six-bar mechanism has mobility. As shown in Figure 7, in the sub-branch u-2-3-4-v, the effective parts are: u-2 part, where θ 2 [-161.4°,-28.6°] and 3-4 part, where θ 2 [-105.2°,-15.6°]; in the sub-branch u-1-6-5-v, the effective part is: u-1 part, where θ 2 [-162.6°,-141.6°] and 6-5 part , where θ 2 [7.9°,-141.6°].

(4)带两个移动副六杆分支：满足阴影部分的连续曲线即为六杆机构的分支，带移动副四杆机构的死点只是将分支分成两个子分支，因此u-1以及u-2形成一个连续有效分分支1-u-2，另外，还有其它2个分离的有效部分，即3-4，6-5。(4) Six-bar branch with two moving pairs: the continuous curve that satisfies the shaded part is the branch of the six-bar mechanism, and the dead point of the four-bar mechanism with moving pair just divides the branch into two sub-branches, so u-1 and u- 2 forms a continuous valid sub-branch 1-u-2, in addition, there are 2 other separate valid parts, namely 3-4, 6-5.

(5)子分支判别：根据方程(23.1)以及方程(23.2)，可知输入输出曲线上的有效部分上的每个点都代表着机构的两种不同构型，每个有效部分都表示带两个移动副六杆机构分支上的两个子分支。(5) Discrimination of sub-branches: According to Equation (23.1) and Equation (23.2), it can be known that each point on the effective part of the input-output curve represents two different configurations of the mechanism, and each effective part represents Two sub-branches on the branch of a mobile six-bar mechanism.

(6)如下所示，列出了带两个移动副六杆机构的分支及其子分支：(6) As shown below, the branch with two moving auxiliary six-bar mechanisms and its sub-branches are listed:

分支1-u-2：包含有效部分1-u及u-2。Branch 1-u-2: contains valid parts 1-u and u-2.

子分支1-u：其中θ2[-162.6°,-141.6°]，满足方程(11.1)，而θ5可根据方程(23.1)得到；Sub-branch 1-u: where θ 2 [-162.6°,-141.6°] satisfies equation (11.1), and θ 5 can be obtained according to equation (23.1);

子分支u-1：其中θ2[-162.6°,-141.6°]，满足方程(11.1)，而θ5可根据方程(23.2)得到；Sub-branch u-1: where θ 2 [-162.6°, -141.6°] satisfies equation (11.1), and θ 5 can be obtained according to equation (23.2);

子分支u-2：其中θ2[-105.2°,-15.6°]，满足方程(11.1)，而θ5可根据方程(23.1)得到；Sub-branch u-2: where θ 2 [-105.2°, -15.6°] satisfies equation (11.1), and θ 5 can be obtained according to equation (23.1);

子分支2-u：其中θ2[-105.2°,-15.6°]，满足方程(11.1)，而θ5可根据方程(23.2)得到；Sub-branch 2-u: where θ 2 [-105.2°, -15.6°] satisfies equation (11.1), and θ 5 can be obtained according to equation (23.2);

分支3-4：包含有效部分3-4。Branch 3-4: Contains active part 3-4.

子分支3-4：其中θ2[-105.2°,-15.6°]，满足方程(11.1)，而θ5可根据方程(23.1)得到；Sub-branch 3-4: where θ 2 [-105.2°, -15.6°] satisfies equation (11.1), and θ 5 can be obtained according to equation (23.1);

子分支4-3：其中θ2[-105.2°,-15.6°]，满足方程(11.1)，而θ5可根据方程(23.2)得到；Sub-branch 4-3: where θ 2 [-105.2°, -15.6°] satisfies equation (11.1), and θ 5 can be obtained according to equation (23.2);

分支6-5：包含有效部分6-5。Branch 6-5: Contains valid part 6-5.

子分支6-5：其中θ2[7.9°,72.6°]，满足方程(11.1)，而θ5可根据方程(23.1)得到；Sub-branch 6-5: where θ 2 [7.9°,72.6°] satisfies equation (11.1), and θ 5 can be obtained according to equation (23.1);

子分支5-6：其中θ2[7.9°,72.6°]，满足方程(11.1)，而θ5可根据方程(23.2)得到。Sub-branch 5-6: where θ 2 [7.9°,72.6°] satisfies Equation (11.1), and θ 5 can be obtained according to Equation (23.2).

实施例2：对图2所示的带三个移动副六杆机构长度及相关角角度，给出实施例1中相同的尺寸，并且令sinγ≠0。Embodiment 2: For the length and relative angle of the six-bar mechanism with three moving pairs shown in Figure 2, the same dimensions as in Embodiment 1 are given, and sinγ≠0.

根据方程(24)，可知当sinγ≠0时，△2>0恒成立，因此，带三个移动副六杆机构的θ2-θ3输入输出曲线一定满足带移动副五杆机构的关节旋转空间(阴影部分)，并且在一一对应的θ2、θ3以及sinγ≠0条件下，s3仅仅只有一个解。总的来说，在sinγ≠0条件下，图2机构仅仅只有一个分支，并且完全旋转。According to equation (24), it can be seen that when sinγ≠0, △ 2 >0 is always established. Therefore, the θ 2 -θ 3 input and output curves of the six-bar mechanism with three moving pairs must satisfy the joint rotation of the five-bar mechanism with moving pairs space (shaded part), and under the condition of one-to-one correspondence of θ 2 , θ 3 and sinγ≠0, s 3 has only one solution. In general, under the condition of sinγ≠0, the mechanism in Figure 2 has only one branch, and it rotates completely.

应当理解的是，本说明书未详细阐述的部分均属于现有技术。It should be understood that the parts not described in detail in this specification belong to the prior art.

应当理解的是，上述针对较佳实施例的描述较为详细，并不能因此而认为是对本发明专利保护范围的限制，本领域的普通技术人员在本发明的启示下，在不脱离本发明权利要求所保护的范围情况下，还可以做出替换或变形，均落入本发明的保护范围之内，本发明的请求保护范围应以所附权利要求为准。It should be understood that the above-mentioned descriptions for the preferred embodiments are relatively detailed, and should not therefore be considered as limiting the scope of the patent protection of the present invention. Within the scope of protection, replacements or modifications can also be made, all of which fall within the protection scope of the present invention, and the scope of protection of the present invention should be based on the appended claims.

Claims (2)

1.一种带移动副六杆机构分支识别方法，其特征在于：分别对与单自由度六连杆机构输入输出相关的两个环路进行分析，进而通过其相互作用得出一种带移动副六杆机构分支识别方法。1. A method for identifying branches of a six-bar linkage with a moving pair, characterized in that: respectively analyzing two loops related to the input and output of a single-degree-of-freedom six-bar linkage, and then obtaining a belt movement through their interaction Branch identification method of auxiliary six-bar mechanism. 2.根据权利要求1所述的一种带移动副六杆机构分支识别方法，其特征在于，包括以下步骤：2. A method for identifying branches of a six-bar mechanism with a moving pair according to claim 1, comprising the following steps: 步骤1：带移动副四杆机构的分支识别；Step 1: Branch recognition with a mobile sub-four-bar mechanism; 任何一个一种带移动副六杆机构都包含一个或多个带移动副四杆机构ABCDA，其环方程由欧拉公式表示，即：Any six-bar mechanism with moving pair contains one or more four-bar mechanism ABCDA with moving pair, and its ring equation is expressed by Euler's formula, namely: aa 22 ee iθiθ 22 ++ aa 33 ee iθiθ 33 == aa 11 ++ sthe s 11 ee iαiα 22 -- -- -- (( 11 )) 其中，a1、a2、a3、s1分别指平面四杆机构ABCDA中DA、AB、BC、CD四连杆的长度，e是自然对数的底，i是虚数单位，θ2、θ3、α2指的是水平轴逆时针旋转分别到四连杆AB、BC、CD的角度；Among them, a 1 , a 2 , a 3 , and s 1 refer to the lengths of the DA, AB, BC, and CD four-bar links in the planar four-bar linkage ABCDA respectively, e is the base of natural logarithm, i is the imaginary unit, θ 2 , θ 3 and α 2 refer to the angles from the counterclockwise rotation of the horizontal axis to the four linkages AB, BC, and CD respectively; 根据其实部和虚部，方程(1)表示为：According to its real and imaginary parts, Equation (1) is expressed as: s1＝(a2cosθ2+a3cosθ3-a1)/cosα2 (2)s 1 =(a 2 cosθ 2 +a 3 cosθ 3 -a 1 )/cosα 2 (2) s1＝(a2sinθ2+a3sinθ3)/sinα2 (3)s 1 =(a 2 sinθ 2 +a 3 sinθ 3 )/sinα 2 (3) 通过消除s1，方程(1)表示为：By eliminating s 1 , equation (1) is expressed as: a1sinα2+a2sin(θ2-α2)+a3sin(θ3-α2)＝0 (4)a 1 sinα 2 +a 2 sin(θ 2 -α 2 )+a 3 sin(θ 3 -α 2 )＝0 (4) 当θ2、θ3是输入参数时，方程(4)表示为输入-输出关系模型；利用半角公式：When θ 2 and θ 3 are input parameters, equation (4) is expressed as an input-output relationship model; using the half-angle formula: x3＝tan(θ3/2) (5)x 3 =tan(θ 3 /2) (5) 方程(4)表示为：Equation (4) is expressed as: PP 11 xx 33 22 ++ QQ 11 xx 33 ++ RR 11 == 00 -- -- -- (( 66 )) 其中：in: P1＝a2sin(θ2-α2)+(a1+a3)sinα2 (7.1)P 1 ＝a 2 sin(θ 2 -α 2 )+(a 1 +a 3 )sin α 2 (7.1) Q1＝2a3cosα2 (7.2)Q 1 =2a 3 cos α 2 (7.2) R1＝a2sin(θ2-α2)+(a1-a3)sinα2 (7.3)R 1 ＝a 2 sin(θ 2 -α 2 )+(a 1 -a 3 )sinα 2 (7.3) 若要四杆机构构型存在，当P1≠0时必须满足For the four-bar linkage configuration to exist, when P 1 ≠ 0 must satisfy ΔΔ 11 == QQ 11 22 -- 44 PP 11 RR 11 ≥&Greater Equal; 00 -- -- -- (( 88 )) 化简得：Simplified: Δ1＝4S1S2 (9)Δ 1 =4S 1 S 2 (9) 其中：in: S1＝a3-a2sin(θ2-α2)-a1sinα2 (10.1)S 1 ＝a 3 -a 2 sin(θ 2 -α 2 )-a 1 sin α 2 (10.1) S2＝a3+a2sin(θ2-α2)+a1sinα2 (10.2)S 2 ＝a 3 +a 2 sin(θ 2 -α 2 )+a 1 sinα 2 (10.2) 当Δ1＝0时，表示四杆机构处在死点位置；When Δ 1 =0, it means that the four-bar mechanism is at the dead point; 由方程(6)，通过输入角θ2得输出角θ3：From equation (6), the output angle θ 3 is obtained through the input angle θ 2 : 其中θ3＝2arctan(x3) (11.1) where θ 3 =2arctan(x 3 ) (11.1) 其中θ3＝2arctan(x3) (11.2) where θ 3 =2arctan(x 3 ) (11.2) θ3由方程(11.1)和(11.2)完全确定，其中arctan(x3)在(-π/2,π/2)变化，与之一一相对应的θ3在(-π,π)变化；θ 3 is completely determined by equations (11.1) and (11.2), where arctan(x 3 ) varies at (-π/2,π/2), and θ 3 corresponding to one of them varies at (-π,π) ; 利用半角公式y＝tan(θ2/2)，方程(8)改写为：Using the half-angle formula y=tan(θ 2 /2), equation (8) is rewritten as: f(y)＝C1y4+C2y2+C0 (12)f(y)＝C 1 y 4 +C 2 y 2 +C 0 (12) 其中，y的根的个数由四杆机构的构型决定；f(y)＝0可能有0、2、4个根，每个根与之相对应的是四杆机构的死点；Wherein, the number of roots of y is determined by the configuration of the four-bar mechanism; f(y)=0 may have 0, 2, and 4 roots, and each root corresponds to the dead point of the four-bar mechanism; 如果θ2是输入角，其四杆机构的分支和子分支通过以下方法确定：If θ2 is the input angle, the branches and sub - branches of its four-bar mechanism are determined by: 1、当f(y)>0恒成立，表示f(y)＝0无实根，则在此四杆机构构型中，无死点存在，根据公式(11)，可得输入θ2对应两个输出θ3，机构具有两个分支；1. When f(y)>0 holds true, it means that f(y)=0 has no real root, then in this four-bar mechanism configuration, there is no dead point, according to the formula (11), the input θ 2 can be obtained corresponding to Two outputs θ 3 , the mechanism has two branches; 2、若f(y)＝0，且f(y)＝0有两个不同的实根，此四杆机构只包含一个分支，两个死点把其分成两个子分支；2. If f(y)=0, and f(y)=0 has two different real roots, this four-bar mechanism only includes one branch, and two dead points divide it into two sub-branches; 3、若f(y)＝0，且有f(y)＝0四个不同的实根，此四杆机构包含两个分支，四个死点把两个分支分成四个子分支；3. If f(y)=0, and there are four different real roots of f(y)=0, this four-bar mechanism comprises two branches, and four dead points divide the two branches into four sub-branches; 步骤2：带移动副五杆机构的分支识别；Step 2: Branch recognition with a mobile sub-five-bar mechanism; 步骤2.1：一个带两个移动副六杆机构包含一个带移动副五杆机构ABEFGA，其环方程由欧拉公式表示，即：Step 2.1: A six-bar mechanism with two moving pairs contains a five-bar mechanism with moving pairs ABEFGA, and its ring equation is expressed by Euler's formula, namely: aa 22 ee iθiθ 22 ++ aa 44 ee ii (( θθ 33 ++ ββ )) == aa 11 ++ aa 77 ++ aa 55 ee iθiθ 55 ++ sthe s 22 ee iαiα 11 -- -- -- (( 1313 )) 其中，a7、a2、a9、a5、a6分别指平面五杆机构ABEFG中GA、AB、BF、EF、FG四连杆的长度，e是自然对数的底，i是虚数单位，θ2、θ3+β、θ5、α1指的是水平轴逆时针旋转分别到连杆AB、BF、EF、FG的角度；Among them, a 7 , a 2 , a 9 , a 5 , and a 6 respectively refer to the lengths of the four-bar links GA, AB, BF, EF, and FG in the planar five-bar mechanism ABEFG, e is the base of natural logarithm, and i is an imaginary number Units, θ 2 , θ 3 +β, θ 5 , α 1 refer to the angles from the counterclockwise rotation of the horizontal axis to the connecting rods AB, BF, EF, and FG; 根据其实部和虚部，方程(13)表示为：According to its real and imaginary parts, Equation (13) is expressed as: s2＝[a2cosθ2+a4cos(θ3+β)-a1-a7-a5cosθ5]/cosα1 (14)s 2 =[a 2 cosθ 2 +a 4 cos(θ 3 +β)-a 1 -a 7 -a 5 cosθ 5 ]/cosα 1 (14) s2＝[a2sinθ2+a4sin(θ4+β)-a5sinθ5]/sinα1 (15)s 2 =[a 2 sinθ 2 +a 4 sin(θ 4 +β)-a 5 sinθ 5 ]/sinα 1 (15) 消除s2，方程(13)表示为：Eliminating s 2 , equation (13) is expressed as: a2sin(θ2-α1)+a4sin(θ3+β-α1)-a5sin(θ5-α1)+(a1+a7)sinα1＝0 (16)a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 )-a 5 sin(θ 5 -α 1 )+(a 1 +a 7 )sinα 1 ＝0 (16) 利用半角公式：Use the half-width formula: x5＝tan(θ5/2) (17)x 5 =tan(θ 5 /2) (17) 方程(16)表示为：Equation (16) is expressed as: PP 22 xx 55 22 ++ QQ 22 xx 55 ++ RR 22 == 00 -- -- -- (( 1818 )) 其中：in: P2＝a2sin(θ2-α1)+a4sin(θ3+β-α1)+a5sinα1+(a1+a7)sinα1 (19.1)P 2 ＝a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 )+a 5 sinα 1 +(a 1 +a 7 )sinα 1 (19.1) Q2＝2a5cosα1 (19.2)Q 2 =2a 5 cos α 1 (19.2) R2＝a2sin(θ2-α1)+a4sin(θ3+β-α1)-a5sinα1+(a1+a7)sinα1 (19.3)R 2 ＝a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 )-a 5 sinα 1 +(a 1 +a 7 )sinα 1 (19.3) 若五杆机构存在，当P2≠0必须满足：If the five-bar mechanism exists, when P 2 ≠0 must satisfy: ΔΔ 22 == QQ 22 22 -- 44 PP 22 RR 22 ≥&Greater Equal; 00 -- -- -- (( 2020 )) 化简得：Simplified: Δ2＝4S1S2≥0 (21)Δ 2 =4S 1 S 2 ≥ 0 (21) 其中：in: S1＝a5-(a1+a7)sinα1-a2sin(θ2-α1)-a4sin(θ3+β-α1) (22.1)S 1 ＝a 5 -(a 1 +a 7 ) sinα 1 -a 2 sin(θ 2 -α 1 )-a 4 sin(θ 3 +β-α 1 ) (22.1) S2＝a5+(a1+a7)sinα1+a2sin(θ2-α1)+a4sin(θ3+β-α1) (22.2)S 2 ＝a 5 +(a 1 +a 7 ) sinα 1 +a 2 sin(θ 2 -α 1 )+a 4 sin(θ 3 +β-α 1 ) (22.2) 方程(20)、(21)产生θ2与θ3之间的关节旋转空间，其中，S1＝0或S2＝0表示关节旋转空间的边界，当五杆机构在奇异位置时或者3个非输出关节E、F、G共线，会出现上述情况；Equations (20), (21) generate the joint rotation space between θ 2 and θ 3 , where S 1 =0 or S 2 =0 represents the boundary of the joint rotation space, when the five-bar mechanism is in a singular position or three The above situation will occur when the non-output joints E, F, and G are collinear; 通过方程(15)，求得θ5：Through equation (15), get θ 5 : 其中θ5＝2arctan(x5) (23.1) where θ 5 =2arctan(x 5 ) (23.1) 其中θ5＝2arctan(x5) (23.2) where θ 5 =2arctan(x 5 ) (23.2) 每个解对应一种机构构型，其中arctan(x5)在(-π/2,π/2)变化，与之一一相对应θ5的(-π,π)在变化；Each solution corresponds to a mechanism configuration, where arctan(x 5 ) changes in (-π/2, π/2), corresponding to one-to-one (-π, π) of θ 5 is changing; 步骤2.2：一个带三个移动副六杆机构包含一个带一个移动副四杆机构ABCDA以及一个带两个移动副五杆机构ABEFGA，五环通过伸旋转转变成四环；因此，带三个移动副六杆机构的分支识别取决于四环ABCDA和五环ABCEFGA的相互作用，其四环ABCDA通过上述对四杆机构的分支识别进行判别，而其五环的环方程用欧拉公式表示，即：Step 2.2: A six-bar mechanism with three moving pairs includes a four-bar mechanism ABCDA with one moving pair and a five-bar mechanism ABEFGA with two moving pairs. The branch recognition of the auxiliary six-bar mechanism depends on the interaction between the four-ring ABCDA and the fifth-ring ABCEFGA. The four-ring ABCDA is discriminated through the above-mentioned branch recognition of the four-bar mechanism, and the ring equation of the fifth ring is expressed by Euler's formula, namely : aa 22 ee iθiθ 22 ++ aa 44 ee ii (( θθ 33 ++ ββ )) == aa 11 ++ aa 77 ++ sthe s 33 ee ii (( αα 11 ++ γγ )) ++ sthe s 22 ee iαiα 11 -- -- -- (( 24twenty four )) 根据其实部和虚部，方程(24)表示为：According to its real and imaginary parts, Equation (24) is expressed as: s2＝[a2cosθ2+a4cos(θ3+β)-a1-a7-s3cos(α1+γ)]/cosα1 (25.1)s 2 =[a 2 cosθ 2 +a 4 cos(θ 3 +β)-a 1 -a 7 -s 3 cos(α 1 +γ)]/cosα 1 (25.1) s2＝[a2sinθ2+a4sin(θ3+β)-s3sin(α1+γ)]/sinα1 (25.2)s 2 =[a 2 sinθ 2 +a 4 sin(θ 3 +β)-s 3 sin(α 1 +γ)]/sinα 1 (25.2) 消除s2，方程(24)表示为：Eliminating s 2 , equation (24) is expressed as: s3sinγ-a2sin(θ2-α1)-a7sinα1-a4sin(θ3+β-α1)＝0 (26)s 3 sinγ-a 2 sin(θ 2 -α 1 )-a 7 sinα 1 -a 4 sin(θ 3 +β-α 1 )＝0 (26) 以s3为未知数，方程(26)的判别式为：Taking s 3 as the unknown, the discriminant of equation (26) is: Δ3＝sin2γ (27)Δ 3 = sin 2 γ (27) 在方程(27)中，sin2γ≥0恒成立；当sinγ≠0时，对于每对θ2和θ3，s3有唯一确定值，但当sinγ＝0，s3有无穷大解，在这种情况下的机构是无用的，因此，要满足机构存在，必须sinγ≠0；In equation (27), sin 2 γ≥0 holds true; when sinγ≠0, s 3 has a unique definite value for each pair of θ 2 and θ 3 , but when sin γ=0, s 3 has an infinite solution, in The organization in this case is useless, therefore, to satisfy the existence of the organization, sinγ≠0 must be satisfied; 步骤3：带移动副六杆机构分支识别；Step 3: Branch identification of the six-bar mechanism with moving vice; 根据分支点的存在与否，带移动副六杆机构分支识别分成如下两种类型：According to the existence of the branch point, the branch recognition of the six-bar mechanism with moving vice is divided into the following two types: 类型I：带移动副六杆机构；没有分支点存在，方程(4)以及方程(20)中的Δ＝0之间没有公共解并且方程(4)的解满足方程(20)中的Δ>0；其分支判别的过程如下：Type I: six-bar mechanism with moving pair; no branch point exists, there is no common solution between equation (4) and Δ=0 in equation (20), and the solution of equation (4) satisfies Δ> in equation (20) 0; the process of branch discrimination is as follows: 步骤A1：四环分支判别；采用方程(12)对四杆机构的分支进行判别；Step A1: four-ring branch discrimination; use equation (12) to discriminate the branches of the four-bar mechanism; 步骤A2：分支点；无分支点存在；Step A2: branch point; no branch point exists; 步骤A3：带移动副六杆机构分支判别；Step A3: Judging the branch of the six-bar mechanism with a moving pair; 对于给定的四杆机构的分支，采用方程(23)对带移动副六杆机构的分支进行判别，方程(23)的每个解代表带移动副六杆机构的一个分支；如果给定的四杆机构的构型满足方程(20)，则此四杆机构的分支表示着带移动副六杆机构的两个分支；如果不满足，则此四杆机构的分支是无效的，此带移动副六杆机构无法装配；For a given branch of a four-bar mechanism, use equation (23) to discriminate the branch of a six-bar mechanism with a moving pair, and each solution of equation (23) represents a branch of a six-bar mechanism with a moving pair; if given The configuration of the four-bar mechanism satisfies equation (20), then the branch of the four-bar mechanism represents the two branches of the six-bar mechanism with a moving pair; if not satisfied, the branch of the four-bar mechanism is invalid, and the belt moves The auxiliary six-bar mechanism cannot be assembled; 步骤A4：如果四杆机构有另外一个分支，采用以上3个步骤对带移动副六杆机构进行判别；Step A4: If the four-bar mechanism has another branch, use the above three steps to discriminate the six-bar mechanism with a moving pair; 步骤A5：对带移动副六杆机构，带有3个滑动关节的分支类型总是类型I，四环的分支代表着此机构的分支；Step A5: For a six-bar mechanism with a moving pair, the branch type with 3 sliding joints is always type I, and the branch of the fourth ring represents the branch of this mechanism; 类型II：带移动副六杆机构；有分支点存在，方程(4)以及方程(20)中的Δ＝0之间存在公共解；四环输入输出曲线被分成多个部分，其中满足方程(20)的部分是有效的，每个有效的部分是个连续的解集并且代表带移动副六杆机构的一种构型；其分支判别的过程如下：Type II: a six-bar mechanism with a moving pair; there are branch points, and there is a common solution between equation (4) and Δ=0 in equation (20); the four-ring input-output curve is divided into multiple parts, which satisfy the equation ( 20) is effective, and each effective part is a continuous solution set and represents a configuration of a six-bar mechanism with a moving pair; the process of branch discrimination is as follows: 步骤B1：四环分支判别；采用方程(12)对四环分支进行判别；Step B1: Discrimination of the four-ring branch; use equation (12) to distinguish the four-ring branch; 步骤B2：分支点；Step B2: branch point; 利用方程(4)和方程(20)求得分支点；满足方程(20)的输入输出曲线部分是有效部分，止点是分支点或死点；Utilize equation (4) and equation (20) to obtain branch point; Satisfy the input-output curve part of equation (20) is effective part, and stop point is branch point or dead point; 步骤B3：带移动副六杆机构分支判别；Step B3: Judging the branches of the six-bar mechanism with a moving pair; 在四杆机构同一个子分支中，两个相邻分支点的有效连续部分表示带移动副六杆机构的一个分支；在给定的四杆机构的分支的不同子分支根据公共死点形成一个连续的有效部分，因此，分支点作为止点可判别所有有效部分，每个分离的有效部分是一个分支；In the same sub-branch of the four-bar mechanism, the effective continuous part of two adjacent branch points represents a branch of the six-bar mechanism with a moving sub-branch; different sub-branches of the branch of a given four-bar mechanism form a continuous Therefore, the branch point can be used as the stop point to distinguish all valid parts, and each separated valid part is a branch; 步骤B4：如果四杆机构有含有其它分支点的分支，重复以上步骤，如果没有，参照类型I对单自由度双环机构进行判别；Step B4: If the four-bar mechanism has branches containing other branch points, repeat the above steps, if not, refer to type I to distinguish the single-degree-of-freedom double-loop mechanism; 步骤B5：在四杆机构的同一子分支中，两个相邻分支点的有效部分代表整个机构的一个分支；四杆机构的不同子分支中的有效部分通过公共的死点形成连续的有效部分，从而形成了整个机构的分支；Step B5: In the same sub-branch of the four-bar mechanism, the effective parts of two adjacent branch points represent a branch of the whole mechanism; the effective parts in different sub-branches of the four-bar mechanism form a continuous effective part through the common dead point , thus forming a branch of the whole institution; 步骤4：带移动副六杆机构的子分支识别；Step 4: Identification of sub-branches with moving auxiliary six-bar mechanism; 当θ2作为输入时，死点位置可通过方程(8)中Δ＝0求得；如果分支点存在，这同样也是带移动副六杆机构奇异位置；因此，两个相邻奇异点，即死点和分支点之间的有效部分是带移动副六杆机构构的子分支；其具体识别分为以下两种情况：When θ 2 is used as input, the position of the dead point can be obtained by Δ=0 in equation (8); if the branch point exists, this is also the singular position of the six-bar mechanism with moving vice; therefore, two adjacent singular points, that is, the dead point The effective part between the point and the branch point is a sub-branch with a moving auxiliary six-bar mechanism; its specific identification is divided into the following two situations: (1)没有分支点存在；(1) No branch point exists; 对于一个带移动副六杆机构给定的分支中，其子分支通过公式(11)识别；方程中的两个解各对应着带移动副六杆机构的一个子分支；For a given branch of a six-bar mechanism with a moving pair, its sub-branches are identified by formula (11); the two solutions in the equation correspond to a sub-branch of the six-bar mechanism with a moving pair; (2)分支点存在；(2) The branch point exists; 四杆机构的输入输出曲线中所有有效部分上的每个点都代表着机构的不同构型，这能根据方程(23)判别；因此，对于给定的带移动副六杆机构的分支，其有效部分的子分支被识别了，那么带移动副六杆机构的子分支也可被识别。Each point on all effective parts of the input-output curve of the four-bar mechanism represents a different configuration of the mechanism, which can be judged according to equation (23); therefore, for a given branch of the six-bar mechanism with a moving pair, its The sub-branches of the effective part are identified, so the sub-branches with the moving auxiliary six-bar mechanism can also be identified.
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Application publication date: 20161116