CN100404215C - Single-degree-of-freedom linear translation space six-bar linkage mechanism - Google Patents

Abstract

Single-degree-of-freedom linear translation space six-bar linkage mechanism, the mechanism includes a fixed base, a front crank, a front rocker, an end member, a rear rocker and a rear crank; Composition, the two planes where the left and right ears are located are not coplanar and not parallel, the two planes where the left and right ears are located are parallel to the two intersection lines of the middle plate; the axes of the three rotating pairs A, B and C of the kinematic chain ABC are parallel to each other And they are all perpendicular to the plane determined by the connecting rods AB and BC; the axes of the three revolving pairs F, E and D of the kinematic chain FED are parallel to each other and are all perpendicular to the plane determined by the connecting rods FE and ED; the axes determined by the connecting rods AB and BC The included angle between the plane and the plane determined by the connecting rods FE and ED is equal to the included angle of the plane where the left and right lifting lugs are located. The invention can realize the single-degree-of-freedom linear translation motion of a certain component with only six connecting rods and six rotating pairs. It has the characteristics of simple structure and convenient engineering application. In the mechanical structure of the movement.

Description

Single-degree-of-freedom linear translation space six-bar linkage mechanism

technical field

The invention relates to a space multi-link mechanism, in particular to a space link mechanism with single-degree-of-freedom precise linear translation movement consisting of a link and a rotating pair, and belongs to the technical field of mechanical structure and mechanism.

Background technique

In the history of mechanism development, in order to achieve precise linear motion, people invented the crank-slider mechanism. However, such a mechanism requires a rigid guide rail. When the guide rail is straight, the slider will move in a straight line; when the guide rail is curved, the slider will move in a curved line. Thus, the linearity of the motion of the crank-slider mechanism is in a sense completely dependent on the guide rails.

In order to avoid the linearity of this type of mechanism movement being limited by the guide rail, mechanism scholars have conducted research on mechanisms that are completely connected by connecting rods and rotating pairs, trying to synthesize a linkage mechanism that can generate precise linear motion. Literature [US RL Norton (Robert L.Norton) translated by Chen Lizhou, Han Jianyou, Li Wei and Qiu Lifang, Mechanical Design—Synthesis and Analysis of Machines and Mechanisms, Mechanical Industry Press, March 2003 First Edition.] Point out in the 100th page that a If the mechanism wants to achieve precise linear motion, it needs at least six components and seven revolving pairs, that is, a Watt-type or Stephenson-type six-bar mechanism. The most representative one is the precise linear mechanism invented by Peaucellier as shown in Figure 1, which contains eight connecting rods and six rotating pairs. The components AB, BP, PD and DA form a rhombus of suitable size, and the components O ₂ B and O ₂ D can be of any length but must be equal. When O ₁ O ₂ is exactly equal to O ₁ A, point P will generate an infinite arc, that is, a precise straight line (as shown in Figure 1, the dotted line passing through point P and perpendicular to O ₁ O ₂ ). At the same time, it is also pointed out on page 100 of the book that if the rotating shaft _O1 is moved from the position shown in Figure 1 to the left or _right to change the length of the member _O1O2 , the trajectory of point P will no longer remain in a in a straight line. Therefore, the linearity of the motion of point P in the Peaucellier mechanism is relatively sensitive to the length perturbation of O ₁ O ₂ . Even if this point is discarded, Peaucellier's eight-bar mechanism only guarantees a point to do linear motion, which brings great difficulties to engineering applications. Document [Huang Zhen, Kong Lingfu, Fang Yuefa. Parallel Robot Mechanism Theory and Control. Beijing: Machinery Industry Press, December 1997] page 27 and document [Translated by Xu Wanchun, Higher Institution Design, Taiwan Beijing: Published by Xu Foundation : World Book Publishing Company, 1990.] The Sarrut mechanism shown in Figure 2 is provided in Figure 6.6 on page 653, and its moving platform has a degree of freedom of linear translational motion. However, the Sarrut mechanism shown in Figure 2 requires that its two kinematic chains A ₁ B ₁ C ₁ and A ₂ B ₂ C ₂ must connect the two adjacent sides of the rectangular moving platform C ₁ C ₂ and the static platform A ₁ A ₂ . The structural form of the rectangular moving platform C ₁ C ₂ and the static platform A ₁ A ₂ is not only difficult to guarantee but also occupies a large space, which is not conducive to engineering applications.

Later, although many mechanism scholars have also developed mechanisms that can make certain points on some connecting rods perform precise linear motion, most of them cannot do without sliders and guide rails. There are also many link mechanisms that approximate linear motion, so they will not be described here one by one.

Contents of the invention

The purpose of the present invention is to provide a single-degree-of-freedom linear translation space six-link mechanism suitable for engineering applications, which can be realized in a multi-link mechanism composed of only rotating pairs and connecting rods, so that one of the components can be strictly controlled. single-degree-of-freedom linear translational motion.

The single-degree-of-freedom linear translation space six-bar linkage mechanism provided by the present invention includes a rod and a rotating pair connecting the rod, and is characterized in that: the rod includes a fixed base, a front crank, a front rocker, and a terminal components, a rear rocker and a rear crank; the end component is composed of a middle plate and non-coplanar and non-parallel left and right lugs arranged at both ends of the middle plate, and the plane where the left and right lugs are located The two intersecting lines formed by the intersection with the middle plate are parallel; the left and right lifting lugs are respectively provided with through holes matching the rotating shaft of the revolving pair, and the left and right lifting lugs are connected to the fixed base through the kinematic chain ABC and FED respectively. Connected, these two kinematic chains are plane RRR kinematic chains; the axes of the three rotating pairs A, B and C of the kinematic chain ABC are parallel to each other and are all perpendicular to the plane determined by the connecting rods AB and BC; The axes of the three rotating pairs F, E and D of the kinematic chain FED are parallel to each other and are all perpendicular to the plane determined by the connecting rods FE and ED; the plane determined by the connecting rods AB and BC is the same as that determined by the connecting rods FE and ED The angle between the planes is equal to the angle between the planes of the left and right lugs on the end member.

The preferred technical solution of the present invention is: the planes where the non-coplanar left and right lifting lugs are arranged at both ends of the end member are perpendicular to each other.

Another preferred technical solution of the present invention is: the non-coplanar left and right lifting lugs arranged at both ends of the end member are located on the same side of the middle plate, and the angle between the plane where the left and right lifting lugs are located and the middle plate is equal. is 135°; the two ends of the fixed base are connected to the front and rear cranks, and the axes of the rotating shafts A and F are perpendicular to each other, and the intersection point o forms an isosceles right triangle with the rotating pair A and F.

The present invention not only realizes the single-degree-of-freedom linear translation motion of the mechanism through the connecting rod and the rotating pair, but also only needs six connecting rods and six rotating pairs to make a certain component in the mechanism realize the precise linear translation motion, and the structure is simple , to overcome the technical prejudice of "at least six components and seven revolving pairs are required to achieve precise linear motion" in the prior art, which is compared to the Peaucellier eight-bar mechanism that can only make a specific point do linear motion , the structure is simpler and more conducive to engineering applications; compared with the Sarrut mechanism, the single-degree-of-freedom linear translation space six-bar linkage mechanism provided by the present invention does not require two kinematic chains to connect the corresponding adjacent sides of the upper and lower rectangular platforms, Therefore, it has a wider range of angle adaptation, and the stroke of linear movement is larger, and the space occupied by the mechanism itself is relatively small. Therefore, it is also more conducive to engineering applications. The present invention can be used as a guide mechanism in any mechanical structure that requires linear translational movement, for example, it can be used as a guide device for automobile independent suspension and aircraft landing gear, a feeding mechanism for an automatic assembly machine, a stamping head for a riveting machine, Printers and stepping feeding mechanisms, etc.

Description of drawings

Figure 1 is a Peaucellier eight-bar linkage.

Fig. 2 is a schematic diagram of the spatial connection relationship of the Sarrut mechanism and its components.

Fig. 3 is a schematic diagram of the single-degree-of-freedom linear translation space six-bar linkage mechanism provided by the present invention.

Fig. 4 is a schematic diagram of an isosceles right triangle formed by the fixed base and the rotating shafts at both ends in the spatial six-bar linkage mechanism.

Fig. 5 is a structural schematic diagram of the front and rear cranks and the front and rear rockers of the spatial six-bar linkage mechanism.

Fig. 6 is a schematic diagram of the structural composition of the terminal member of the spatial six-bar linkage mechanism.

Fig. 7 is a cross-sectional view of an end member of a spatial six-bar linkage mechanism.

Fig. 8 is a top view of the assembly of the space six-bar linkage mechanism.

Fig. 9 is a schematic diagram when the end member of the spatial six-bar linkage mechanism moves to the lower limit position along the negative direction of the z-axis.

In the figure: 1-front crank; 2-fixed base; 3-front rocker; 4-end member; 5-rear rocker; 6-rear crank; 7-left lug; 8-left through hole; 9- Middle plate; 10-right through hole; 11-right lug.

Detailed ways

The structure, principle and specific implementation of the present invention will be further described below in conjunction with the accompanying drawings.

Fig. 3 is a schematic structural diagram of the single-degree-of-freedom linear translation space six-bar linkage mechanism provided by the present invention. Wherein the structural form of the fixed base 2 and the angle between the axes of the rotating shafts A (x-axis) and F (y'-axis) at both ends are set as ψ, and the orthogonal relationship when ψ=90° is shown in Figure 4 . The space six-bar linkage mechanism includes a front crank 1, a fixed base 2, a front rocker 3, an end member 4, a rear rocker 5 and a rear crank 6 six rods and six rotation pairs A, B, C, F, E and D; the end member 4 is composed of the middle plate 9 and the left lug 7 and the right lug 11 arranged at both ends, and the two planes where the left and right lugs are located are non-coplanar and non-parallel, The two planes where the left and right lifting lugs are located are parallel to the two intersecting lines of the middle plate 9; the left and right lifting lugs are respectively provided with through holes matching the rotating shaft of the revolving pair, and the left and right lifting lugs pass through the kinematic chain respectively. ABC and FED are connected to the fixed base 2, and the two kinematic chains are planar RRR kinematic chains; the axes of the three revolving pairs A, B and C of the kinematic chain ABC are parallel to each other and perpendicular to the connecting rod AB and the plane determined by BC; the axes of the three rotating pairs F, E and D of the kinematic chain FED are parallel to each other and are all perpendicular to the plane determined by the connecting rods FE and ED; the plane determined by the connecting rods AB and BC The angle between the plane determined by the connecting rods FE and ED is equal to the angle between the planes of the left and right lugs on the end member 4 .

The structural diagrams of the front crank 1, the front rocker 3, the rear rocker 5, and the rear crank 6 are all shown in Figure 5, and their dimensions can be the same or different, and the structure of the end connecting rod 4 is shown in Figure 6, which consists of The middle plate 9 is composed of the left lug surface 7 and the right lug surface 11 protruding at both ends, and the planes where the left and right lugs are located are perpendicular to each other; the left lug surface 7 is provided with a left through hole for connecting the rotating pair 8. The right lug surface 11 is provided with a right through hole 10 for connecting the rotating pair, and the left lug surface 7 and the right lug surface 11 are located on the same side of the middle plate 9; as shown in Figure 7, when the left When the angle between the ear surface 7 and the middle plate 9 is 135° and the angle between the right lug surface 11 and the middle plate 9 is also 135°, the mechanism will be optimal in force and configuration stability. It will be analyzed later; the axes of the rotating shafts A and F connected between the two ends of the fixed base 2 and the front crank 1 and the rear crank 6 are perpendicular to each other, and the intersection point o forms an isosceles right triangle with the revolving pair A and F; each structure The specific size can be set according to engineering needs; Figure 8 shows the top view of the assembled single-degree-of-freedom linear translation space six-bar linkage mechanism provided by the present invention. According to Figure 8, it can be found that the middle plate 9 of the end member 4 is always connected with the fixed The sides of the base 2 remain parallel, ie in FIG. 8 CD//AF and the distance from CD to AF also remains constant throughout. This feature is determined by the structural characteristics of the single-degree-of-freedom linear translation space six-bar linkage mechanism provided by the present invention. Therefore, the next step is to analyze the kinematics principle of the mechanism to maintain this feature.

In order to facilitate the description of the kinematics principle of the single-degree-of-freedom linear translation space six-bar linkage mechanism provided by the present invention, here the plane where the two axes of the revolving pairs A and F are located is the xoy plane, and the intersection of these two axes is used as the coordinate At the origin, a right-handed coordinate system is established with the axis of rotating joint A as the x-axis. When the two axes of the rotating pair A and F are perpendicular to each other, as shown in Fig. 4 and Fig. 8, the y-axis coincides with the axis y' of the rotating pair F exactly. In this way, in the coordinate system shown in Figure 3, the plane where the front crank 1 and the front rocker 3 are located (the plane determined by the connecting rods FE and ED) is parallel to the xoz plane, and the plane where the rear crank 6 and the rear rocker 5 are located ( The plane determined by the links AB and BC) is parallel to the yoz plane. This corresponding relationship can be seen more clearly from the top view of the mechanism assembly shown in FIG. 8 .

Next, the free movement of the end member 4 will be analyzed. Without loss of generality, assume that the angle between the axis y′ of the rotating pair F and the x-axis is ψ (0°<ψ<180°), the distance between the rotating pair A and the coordinate origin is a, and the distance between the rotating pair F and the coordinate origin is is b, then the coordinates of swivel joint A can be obtained as (a 0 0), and the coordinates of swivel joint F are (bcosψ bsinψ 0). On this basis, the coordinates of rotating pairs B, C, D and E can be set as (a y _B z _B ), (a y _C z _C ), (x _D y _D z _D ) and (x _E y _E z _E ). The following analysis shows that when ψ=90° and a=b, ΔoAF forms an isosceles right triangle; at this time, the mechanism is optimal in force and configuration stability.

In the oxyz coordinate system, the direction vectors of axes of revolving joints A, B and C are all s ₁ =(1 0 0) ^T , and the direction vectors of axes of revolving joints F, E and D are all s ₂ =(cosψ sinψ 0) ^T. The top view of the mechanism assembly shown in Fig. 8 shows the situation when ψ = 90°.

In this way, according to the literature [Jing-Shan Zhao, Kai Zhou and et al, A New Method to Study the Degree of Freedom of Spatial Parallel Mechanisms, The International Journal of Advanced Manufacturing Technology, Vol.23, No.3-4, February 2004, pp .288-294.] [Jing-Shan Zhao, Zhi-Jing Feng, KaiZhou and De-Wen Jin, Re-analysis on the Degree-of-Freedom Configuration of the Platforms in Spatial Parallel Mechanisms with Constraints Spaces, The International Journal of Advanced Manufacturing Technology, Vol.28, No.1-2, February 2006, pp.190-196.] [Jing-Shan Zhao, KaiZhou and Zhi-Jing Feng, A Theory of Degrees of Freedom For Mechanisms, Mechanism and Machine Theory, Vol. 39, No.6, June 2004, pp.621-643.] [Jing-Shan Zhao, Zhi-Jing Feng and Jing-Xin Dong, Computation of the Configuration Degree of Freedom of a Spatial Parallel Mechanism by Using Reciprocal Screw Theory, Mechanism and Machine Theory, DOI: 10.1016/j.mechmachtheory.2006.01.006.] proposed the analysis theory of the degree of freedom of the mechanism, and the kinematic helical system of the kinematic chain ABC of the end member 4 can be written as:

$ _ABC ＝[$ _A $ _B $ _C ] (1)

where $ _A = (1 0 0 0 0 0) ^T

$ _B ＝(1 0 0 0 z _B -y _B ) ^T

$ _C ＝(1 0 0 0 z _C -y _C ) ^T

Obviously, the conditions for $ _ABC to rank down are:

Link BC is collinear with link AB (C ₁ )

Further analysis shows that the condition (C ₁ ) can be expressed as:

$\left|\begin{array}{cc}{\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; B</span>}}& <span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; the\; y</span>}}_{\mathrm{<span\; class="notranslate">\; B</span>}}\\ {\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}}& <span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; the\; y</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}}\end{array}\right|<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 2</span>}<span\; class="notranslate">\; )</span>$

When the condition (C ₁ ) does not hold, the terminal constraint $ _ABC ^r of the kinematic chain ABC can be calculated by the reciprocal spiral theory, namely

$E$ ^r = 0 (3)

Where $ is the motion spiral system, $\mathrm{E.}=\left[\begin{array}{cc}0& I_3\\ I_3& 0\end{array}\right],$ $I_3=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],$ $ ^r is the anti-helical system of $.

From formula (3), $ _ABC ^r can be calculated as:

${<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; ABC</span>}}^{\mathrm{<span\; class="notranslate">\; r</span>}}<span\; class="notranslate">\; =</span>{\left[\begin{array}{cccccc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right]}^{\mathrm{<span\; class="notranslate">\; T</span>}}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 4</span>}<span\; class="notranslate">\; )</span>$

When the condition (C ₁ ) is established, that is, when the formula (2) is established, the terminal constraint of the kinematic chain ABC is:

${<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; <figure-callout\; id="1"\; label="ABC\; r\; C"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">ABC</figure-callout></span><figure-callout\; id="1"\; label="ABC\; r\; C"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">\; </figure-callout>}}^{{\mathrm{<span\; class="notranslate">\; r</span>}}_{{}_{}}}<span\; class="notranslate">\; =</span>{\left[\begin{array}{cccccc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& {\mathrm{<span\; class="notranslate">\; the\; y</span>}}_{\mathrm{<span\; class="notranslate">\; B</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; the\; y</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}}& {\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; B</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\end{array}\right]}^{\mathrm{<span\; class="notranslate">\; T</span>}}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 5</span>}<span\; class="notranslate">\; )</span>$

Similarly, the kinematic helical system of another kinematic chain FED of the end member (4) can be written as:

$ _FED = [$ _F $ _E $ _D ] (6)

where $ _F = (cosψ sinψ 0 0 0 0) ^T

$ _E ＝(cosψ sinψ 0 -z _E sinψ z _E cosψ x _E sinψ-y _E cosψ) ^T

$ _D ＝(cosψ sinψ 0 -z _D sinψ z _D cosψ x _D sinψ-y _D cosψ) ^T

And the conditions for $ _FED to rank down are:

ED is collinear with FE (C ₂ )

Condition (C ₂ ) can be expressed as:

$\left\{\begin{array}{c}\left|\begin{array}{cc}{\mathrm{<span\; class="notranslate">\; x</span>}}_{\mathrm{<span\; class="notranslate">\; E.</span>}}& {\mathrm{<span\; class="notranslate">\; the\; y</span>}}_{\mathrm{<span\; class="notranslate">\; E.</span>}}\\ {\mathrm{<span\; class="notranslate">\; x</span>}}_{\mathrm{<span\; class="notranslate">\; D.</span>}}& {\mathrm{<span\; class="notranslate">\; the\; y</span>}}_{\mathrm{<span\; class="notranslate">\; D.</span>}}\end{array}\right|<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \left|\begin{array}{cc}{\mathrm{<span\; class="notranslate">\; x</span>}}_{\mathrm{<span\; class="notranslate">\; E.</span>}}& {\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; E.</span>}}\\ {\mathrm{<span\; class="notranslate">\; x</span>}}_{\mathrm{<span\; class="notranslate">\; D.</span>}}& {\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; D.</span>}}\end{array}\right|<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\end{array}\right.<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 7</span>}<span\; class="notranslate">\; )</span>$

When the condition (C ₂ ) is not established, the terminal constraint $ _EED ^r of the kinematic chain FED can also be obtained according to formula (3):

${<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; FED</span>}}^{\mathrm{<span\; class="notranslate">\; r</span>}}<span\; class="notranslate">\; =</span>{\left[\begin{array}{cccccc}\mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& <span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right]}^{\mathrm{<span\; class="notranslate">\; T</span>}}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 8</span>}<span\; class="notranslate">\; )</span>$

When the condition (C ₂ ) is established, that is, the formula (7) is established, the terminal constraint of the kinematic chain FED is:

${<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; <figure-callout\; id="2"\; label="FED\; r\; C"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">FED</figure-callout></span><figure-callout\; id="2"\; label="FED\; r\; C"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">\; </figure-callout>}}^{{\mathrm{<span\; class="notranslate">\; r</span>}}_{{}_{}}}<span\; class="notranslate">\; =</span>{\left[\begin{array}{cccccc}\mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& <span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}\\ {\mathrm{<span\; class="notranslate">\; x</span>}}_{\mathrm{<span\; class="notranslate">\; D.</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; x</span>}}_{\mathrm{<span\; class="notranslate">\; E.</span>}}& \mathrm{<span\; class="notranslate">\; 0</span>}& {\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; D.</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; E.</span>}}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\end{array}\right]}^{\mathrm{<span\; class="notranslate">\; T</span>}}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 9</span>}<span\; class="notranslate">\; )</span>$

Therefore, when neither the condition (C ₁ ) nor the condition (C ₂ ) holds true, the constraints on the end member 4 are:

${<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; 6</span>}}^{\mathrm{<span\; class="notranslate">\; r</span>}}<span\; class="notranslate">\; =</span>\left[\begin{array}{cc}{<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; ABC</span>}}^{\mathrm{<span\; class="notranslate">\; r</span>}}& {<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; FED</span>}}^{\mathrm{<span\; class="notranslate">\; r</span>}}\end{array}\right]<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 10</span>}<span\; class="notranslate">\; )</span>$

Substituting Equation (10) into Equation (3), the free motion of the end member 4 can be calculated as:

${<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; 4</span>}}^{\mathrm{<span\; class="notranslate">\; f</span>}}<span\; class="notranslate">\; =</span>{\left(\begin{array}{cccccc}\mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right)}^{\mathrm{<span\; class="notranslate">\; T</span>}}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 11</span>}<span\; class="notranslate">\; )</span>$

Therefore, equation (11) indicates that the end member 4 has a free movement in translation along the z-axis direction. Obviously, when ψ=90°, that is, when y' and y coincide, (11) is also established. Therefore, the end member 4 of the spatial six-bar linkage mechanism can perform translational motion in a single-degree-of-freedom fixed line at any ψ value of ψ∈(0°, 180°); while the Sarrut mechanism is only one when ψ=90° a special form.

Next, discuss the situation when the conditions (C ₁ ) and (C ₂ ) are established.

1. When the condition (C ₁ ) is true but the condition (C ₂ ) is not true, the constraints on the end member 4 are:

${<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; 4</span>}}^{\mathrm{<span\; class="notranslate">\; r</span>}}<span\; class="notranslate">\; =</span>\left[\begin{array}{cc}{<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; <figure-callout\; id="1"\; label="ABC\; r\; C"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">ABC</figure-callout></span><figure-callout\; id="1"\; label="ABC\; r\; C"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">\; </figure-callout>}}^{{\mathrm{<span\; class="notranslate">\; r</span>}}_{{}_{}}}& {<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; FED</span>}}^{\mathrm{<span\; class="notranslate">\; r</span>}}\end{array}\right]<span\; class="notranslate">\; =</span>\left[\begin{array}{cccccc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& {\mathrm{<span\; class="notranslate">\; the\; y</span>}}_{\mathrm{<span\; class="notranslate">\; B</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; the\; y</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& {\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; B</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; C</span>}}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& <span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\end{array}\right]<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 12</span>}<span\; class="notranslate">\; )</span>$

Since 0°<ψ<180°, according to formula (12), as long as z _B -z _C ≠ 0, the rank of $ ₄ ^r is equal to 6, namely $R\left({\$}_4^r\right)=6.$ Therefore, the free motion of the end member 4 can be calculated to be zero from equation (3).

2. When the conditions (C ₂ ) are all true but the condition (C ₁ ) is not true, the constraints on the end member 4 are:

${<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; 4</span>}}^{\mathrm{<span\; class="notranslate">\; r</span>}}<span\; class="notranslate">\; =</span>\left[\begin{array}{cc}{<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; ABC</span>}}^{\mathrm{<span\; class="notranslate">\; r</span>}}& {<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; <figure-callout\; id="2"\; label="FED\; r\; C"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">FED</figure-callout></span><figure-callout\; id="2"\; label="FED\; r\; C"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">\; </figure-callout>}}^{{\mathrm{<span\; class="notranslate">\; r</span>}}_{{}_{}}}\end{array}\right]<span\; class="notranslate">\; =</span>\left[\begin{array}{cccccc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& {\mathrm{<span\; class="notranslate">\; x</span>}}_{\mathrm{<span\; class="notranslate">\; D.</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; x</span>}}_{\mathrm{<span\; class="notranslate">\; E.</span>}}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& {\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; D.</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; z</span>}}_{\mathrm{<span\; class="notranslate">\; E.</span>}}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& <span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\end{array}\right]<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 13</span>}<span\; class="notranslate">\; )</span>$

Since 0°<ψ<180°, according to formula (13), as long as z _D -z _E ≠ 0, the rank of $ ₄ ^r is equal to 6, namely $R\left({\$}_4^r\right)=6.$ Therefore, the free motion of the end member 4 can be calculated to be zero from equation (3).

3. Obviously, when the above two conditions 1 and 2 are both true, the free movement of the end member 4 must be zero.

To sum up, as long as one of the above three conditions is established, the free movement of the end member 4 must be zero. Therefore, in order to make the end member 4 have the free movement described in formula (11), it is necessary to ensure that the space six-bar linkage mechanism avoids the joints where the revolving pairs A, B and C are collinear or the revolving pairs F, E and D are collinear positional movement. This actually defines the range or stroke of the single-degree-of-freedom fixed-line motion of the end member 4 , and the following analyzes the range of the single-degree-of-freedom fixed-line motion of the terminal member 4 according to this constraint.

In the mechanism shown in Figure 3, a symmetrical structural arrangement is adopted, that is, the rod lengths of the front crank 1 and the rear crank 6 are equal and set to l ₁ , and the rod lengths of the front rocker 3 and the rear rocker 5 are also equal and set to is l ₂ , the angle between the projection line ABC and the base AF and the angle between the projection line FED and the base AF are both 45°. At the same time, it is assumed that the center distance CD between the left and right through holes of the end member 4 is l ₃ , and the length of the side AF of the fixed base 2 is l ₄ .

The relationship between the above structural parameters and the linear travel that the end member 4 can do will be analyzed below. Assuming that in the top view 8, the distance from the end member 4 to the fixed base 2 is h. It is not difficult to find that when the revolving pairs A, B, and C move downward to the limit position where the three points are collinear (as shown in Figure 9), due to the symmetry of the structure, the revolving pairs F, E, and D also just move downward to the extreme position where the three points are collinear. At this time, the distance S that the end member 4 translates downward from the horizontal position coplanar with the base can be expressed according to the Pythagorean theorem of right triangles as:

$\mathrm{<span\; class="notranslate">\; S</span>}<span\; class="notranslate">\; =</span>\sqrt{{<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; <figure-callout\; id="1"\; label="l"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">l</figure-callout></span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; l</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>{<span\; class="notranslate">\; (</span>\frac{{\mathrm{<span\; class="notranslate">\; l</span>}}_{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; <figure-callout\; id="3"\; label="l"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">l</figure-callout></span>}}_{\mathrm{<span\; class="notranslate">\; 3</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; h</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 14</span>}<span\; class="notranslate">\; )</span>$

Similarly, the end member (4) can also be moved upwards by a distance S from the horizontal position to reach the upper limit position. Therefore, the movement stroke z of the end member 4 of the mechanism should satisfy in the coordinate system shown in Fig. 3:

$<span\; class="notranslate">\; -</span>\sqrt{{<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; <figure-callout\; id="1"\; label="l"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">l</figure-callout></span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; l</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>{<span\; class="notranslate">\; (</span>\frac{{\mathrm{<span\; class="notranslate">\; l</span>}}_{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; <figure-callout\; id="3"\; label="l"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">l</figure-callout></span>}}_{\mathrm{<span\; class="notranslate">\; 3</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; <figure-callout\; id="2"\; label="h"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">h</figure-callout></span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}}<span\; class="notranslate">\; <</span>\mathrm{<span\; class="notranslate">\; z</span>}<span\; class="notranslate">\; <</span>\sqrt{{<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; <figure-callout\; id="1"\; label="l"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">l</figure-callout></span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; l</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>{<span\; class="notranslate">\; (</span>\frac{{\mathrm{<span\; class="notranslate">\; l</span>}}_{\mathrm{<span\; class="notranslate">\; 4</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; <figure-callout\; id="3"\; label="l"\; filenames="C20061011311200131.png,C20061011311200162.png"\; state="\{\{state\}\}">l</figure-callout></span>}}_{\mathrm{<span\; class="notranslate">\; 3</span>}}}{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; h</span>}}^{\mathrm{<span\; class="notranslate">\; 2</span>}}}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 15</span>}<span\; class="notranslate">\; )</span>$

According to the above analysis, it can be found that the end member 4 has a strictly single-degree-of-freedom straight-line free movement within the range defined by formula (15). Since the moving stroke of the Sarrut mechanism is limited to the distance from the static platform A ₁ A ₂ to the moving platform, therefore, relative to the Sarrut mechanism, the moving range z of the end member 4 of the single-degree-of-freedom linear translation type space six-bar linkage mechanism of the present invention If it is larger, its moving range z also depends entirely on the structural parameters l ₁ , l ₂ , l ₃ , l ₄ and h of the connecting rod. Therefore, according to the range z required by the project, the optimum values of the above five main parameters can be designed and optimized by using formula (15).

When the lengths of the front crank 1 and the rear crank 6 and the front rocker 3 and the rear rocker 5 are all different, the movement stroke of the end member 4 can also be similarly released.

Therefore, the present invention not only realizes the single-degree-of-freedom linear translation movement of the mechanism through the connecting rod and the rotating pair, but also only needs six connecting rods and six rotating pairs to realize the precise linear translation movement of the end member 4 in the mechanism, Its linear movement stroke can be expressed by (15) formula. This overcomes the technical prejudice in the prior art that "at least six components and seven revolving pairs are required to achieve precise linear motion". The reason for this bias is the use of the traditional flawed mechanism degree of freedom calculation formula in the mechanism synthesis. For the analysis of this aspect, please refer to the literature [Jing-Shan Zhao, Kai Zhou and etal, A New Method to Study the Degree of Freedom of Spatial Parallel Mechani sms, The International Journal of Advanced Manufacturing Technology, Vol.23, No.3-4, February 2004, pp.288-294.] [Jing-Shan Zhao, Zhi-Jing Feng, Kai Zhou and De- Wen Jin, Re-analysis on the Degree-of-Freedom Configuration of the Platforms in Spatial Parallel Mechanisms with ConstraintsSpaces, The International Journal of Advanced Manufacturing Technology, Vol.28, No.1-2, February 2006, pp.190-196. 】【Jing-Shan Zhao, Kai Zhou and Zhi-Jing Feng, A Theory of Degrees of Freedom For Mechanisms, Mechanism and Machine Theory, Vol.39, No.6, June 2004, pp.621-643.】【Jing-Shan Zhao, Zhi-Jing Feng and Jing-Xin Dong, Computation of the Configuration Degree of Freedom of a Spatial Parallel Mechanism by Using Reciprocal Screw Theory, Mechanism and Machine Theory, DOI: 10.1016/j.mechmachtheory.2006.01.0].

At the end of this part, let us analyze the free motion of the end member 4 described by formula (11). According to formula (11), it can be known that the end member 4 has a translation degree of freedom along the z-axis direction (vertical direction). Therefore, in order to ensure the establishment of this conclusion, the rank of the coefficient matrix $ ₆ ^r described in (10) must be 5. According to formula (10), we have:

${<span\; class="notranslate">\; \$</span>}_{\mathrm{<span\; class="notranslate">\; 6</span>}}^{\mathrm{<span\; class="notranslate">\; r</span>}}<span\; class="notranslate">\; =</span>\left[\begin{array}{ccccc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& <span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right]<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 16</span>}<span\; class="notranslate">\; )</span>$

A necessary and sufficient condition for the rank of matrix $ _6r to be 5 is ^that the determinant of its submatrix A is not zero, where

$\mathrm{<span\; class="notranslate">\; A</span>}<span\; class="notranslate">\; =</span>\left[\begin{array}{ccccc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; sin</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& <span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; cos</span>}\mathrm{<span\; class="notranslate">\; \&psi;</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right]<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 17</span>}<span\; class="notranslate">\; )</span>$

Set |A|=0 to get

ψ＝0° or ψ＝180° (18)

Equation (18) shows the reason why we limit 0°<ψ<180°. Further, since cond(A) ₂ ≥1, if let

$\mathrm{<span\; class="notranslate">\; cond</span>}{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; A</span>}<span\; class="notranslate">\; )</span>}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; =</span>{<span\; class="notranslate">\; |</span><span\; class="notranslate">\; |</span>\mathrm{<span\; class="notranslate">\; A</span>}<span\; class="notranslate">\; |</span><span\; class="notranslate">\; |</span>}_{\mathrm{<span\; class="notranslate">\; 2</span>}}{<span\; class="notranslate">\; |</span><span\; class="notranslate">\; |</span>{\mathrm{<span\; class="notranslate">\; A</span>}}^{<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; |</span><span\; class="notranslate">\; |</span>}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; =</span>\sqrt{\frac{{\mathrm{<span\; class="notranslate">\; \&lambda;</span>}}_{\mathrm{<span\; class="notranslate">\; max</span>}}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; A</span>}}^{\mathrm{<span\; class="notranslate">\; T</span>}}\mathrm{<span\; class="notranslate">\; A</span>}<span\; class="notranslate">\; )</span>}{{\mathrm{<span\; class="notranslate">\; \&lambda;</span>}}_{\mathrm{<span\; class="notranslate">\; min</span>}}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; A</span>}}^{\mathrm{<span\; class="notranslate">\; T</span>}}\mathrm{<span\; class="notranslate">\; A</span>}<span\; class="notranslate">\; )</span>}}<span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 19</span>}<span\; class="notranslate">\; )</span>$

Where cond(A) ₂ represents the condition number of matrix A, ‖A‖ ₂ represents the 2-norm of matrix A, and λ(A ^T A) represents the eigenvalue of A ^T A.

then you can find

ψ＝90° (20)

Formula (20) shows that when ψ=90°, the condition number of matrix A just takes the minimum value of 1. This shows that when ψ=90°, the spatial six-link has the best configuration stability. This is why the axes of the swivel joint A and the swivel joint B are required to be orthogonal in the dependent claims. In the dependency, we also require: the non-coplanar and non-parallel left lug 7 and right lug 11 arranged at both ends of the end member 4 are located on the same side of the middle plate 9, and the left and right The angle between the plane where the lifting lug is located and the middle plate is 135°; the axes of the rotating shafts A and F connecting the two ends of the fixed base 2 with the front crank 1 and the rear crank 6 are perpendicular to each other, and the intersection o and the rotating pair A and F form an isosceles right triangle. This is a further requirement derived from the perspective of optimal component stress. The specific derivation process is omitted here.

Compared with the Peaucellier eight-bar mechanism that can only make a specific point do linear motion, the end member 4 of the single-degree-of-freedom linear translation space six-bar linkage mechanism provided by the present invention can perform an overall fixed linear translation, and the structure is more For simplicity; compared with the Sarrut mechanism, the single-degree-of-freedom linear translation space six-bar linkage mechanism provided by the present invention does not need to require two kinematic chains to connect the corresponding adjacent sides of the upper and lower rectangular platforms, so it has a wider range of ψ angle adaptation, Moreover, the stroke of linear movement is larger, and the space occupied by the mechanism itself is relatively small. Therefore, it is also more conducive to engineering applications. In addition, it is also pointed out in the subordination of the present invention that when the angle between the plane determined by the connecting rods AB and BC and the plane determined by the connecting rods FE and ED and the plane where the left and right lifting lugs are located are all equal to a right angle, and The angle between the plane where the left and right lifting lugs are located and the middle plate is 135°; the axes of the rotating shafts A and F connecting the two ends of the fixed base 2 with the front crank 1 and the rear crank 6 are perpendicular to each other, When the intersection point o forms an isosceles right triangle with the rotary joints A and F, the mechanism has the best configuration stability and force bearing. The present invention can be used as a guide mechanism in any mechanical structure that requires linear translational movement.

Claims (3)

1. a single-degree-of-freedom straight line translation spacing connecting rod mechanism comprises rod member and the revolute pair that connects rod member, and it is characterized in that: described six bar mechanism comprises preceding crank (1), fixed pedestal (2), preceding rocking bar (3), terminal link (4), back rocking bar (5) and rear crank (6); Described terminal link (4) is made up of middle dull and stereotyped (9) and the left and right hanger (7,11) that is arranged on two ends, the non-coplane in two planes at described left and right two hanger places and not parallel, two planes at left and right hanger place are parallel with two intersections of middle dull and stereotyped (9); Be respectively equipped with the through hole that the rotating shaft with revolute pair is complementary on left and right hanger, left and right hanger is connected with fixed pedestal (2) with FED by kinematic chain ABC respectively, and these two kinematic chains are plane RRR kinematic chain; Three revolute pair A, the B of described kinematic chain ABC and the axis of C is parallel to each other and all perpendicular to the plane of connecting rod AB and BC decision; Three revolute pair F, the E of described kinematic chain FED and the axis of D is parallel to each other and all perpendicular to the plane of connecting rod FE and ED decision; The angle on left and right hanger (7, the 11) plane on the angle between the plane of described connecting rod AB and BC decision and the plane that connecting rod FE and ED determine and the described terminal link (4) equates.

2. according to the described single-degree-of-freedom straight line translation spacing connecting rod mechanism of claim 1, it is characterized in that: the plane that is arranged on the non-coplane at described terminal link (4) two ends and uneven left and right hanger (7,11) place is vertical mutually.

3. according to the described single-degree-of-freedom straight line translation spacing connecting rod mechanism of claim 2, it is characterized in that: be arranged on the non-coplane at described terminal link (4) two ends and the homonymy that uneven left and right hanger (7,11) is positioned at described middle dull and stereotyped (9), the plane at described left and right hanger place is 135 ° with middle dull and stereotyped angle; Described fixed pedestal (2) two ends are orthogonal with forward and backward crank (1,6) revolute pair A that is connected and the axis of F, and its intersection point o and revolute pair A and F constitute an isosceles right angled triangle.

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