1. Structural analysis Mechanism
1.1 Determination of the degree of mechanism mobility
Where N.= 3 - the number of moving mechanism links
- the number of kinematic pairs of the fifth grade
- the number of fourth grade kinematic pairs
In the specified mechanism, four pairs of fifth grade
Rotational Couples
3.0 Progressive couples
Fourth Class pairs
1.2 Definition of the class of the mechanism
To do this, dismember the mechanism on the Assur group.
Determine the second-class assembly group formed by the links 2 and 3. The leading link forming the first class mechanism remains.
Mechanism I class I class II
Order 2.
Formula structure of the mechanism
I (0.1) II (2.3)
The class of the connecting group is the second, so the mechanism under consideration refers to the second class.
2 geometric mechanism synthesis
2.1 Draw a mechanism in extreme positions
2.2 Determine the linear dimensions of the crank and connecting rod
Rotation frequency of crank N1 \u003d 82 rpm
Stroke slider S \u003d 0.575 m
The ratio of the length of the crank to the length of the connecting rod
The attitude of the eccentricity to the length of the crank
2.3 During one turnover of crank with;
The slider will pass the distance S, with s \u003d 2A
Determine the length of the link;
Determine the length of the link;
Determine the position of the point M on the link Av from the relationship
; ATM.\u003d 0.18 × 1.15 \u003d 0.207 m;
3 Building a crystal-slider plan
To build a plan for a crank-slider mechanism, the circle with a radius of AB, then spend the horizontal of the AU. We divide the circumference of 12 parts (for 12 positions of the mechanism). Next, laying off the segments of the B0С0, B1C1 ... B11C11 on the horizontal of the AU. We connect the center of the circle and with points B0, B1 ... B11. At each of the 12 positions of the crank, we lay the segment of VMI (where I is the number of the crank position). Connecting the points M0, M1 ... M11 We get the trajectory of motion of M.
4 Determination of speeds of points O, A, B, M for four positions.
Position 1:
Determine the speed of the point in
Consider
Determine from the triangle ABC
Consider
Determine RS through
Through determining Ar.
Determine BP
Determine Ð J.
Determine MR.
Determine the speed of points A, C and M from the formula
Determine
Perform check:
Position 2:
Determine the speed of the point in
Consider
By the sinus theorem, we determine:
Determine the OAV triangle
By the sinus theorem, we define the speakers
Consider
Determine RS through
Through determining Ar.
Determine BP
Determine Ð J.
We define MR.
Determine Ð Y.
Perform check:
Position 3:
Since the speeds VV, VC and VM are parallel and the points B, C and M cannot lie on one perpendicular to the direction of these velocities, at the moment the instantaneous center of the connecting rod fleece lies in infinity, its angular velocity And he makes an instant progressive movement. Consequently, at the moment:
Position 4:
Determine the speed of the point in
Consider
By the sinus theorem, we determine:
Determine Ð B. From the triangle AVS.
By the sinus theorem, we define the speakers
Consider
Determine RS through
Through determining Ar.
Consider
Determine BP
Determine Ð J.
We define MR.
Determine the speed of points A, B and M of the formula
Determine Ð Y.
Perform check:
5. Construction of displacement charts, speeds and accelerations.
Let it be necessary to build a kinematic diagram of distances, speeds and accelerations of the slider from a crank-slider mechanism. Cracked Av long L \u003d 0,29m rotates with a constant angular velocity N1 \u003d 82Of.
Crank sleeping mechanism It serves to turn the rotational movement into the translational and vice versa. It consists of bearings 1, crank 2, rod 3 and a slider 4.
The crank makes the rotational movement, the rod-plane-plane, and the slider is reciprocating.
Two bodies connected to each other movably form a kinematic pair. The bodies constituting a pair are called links. Usually set the law of movement of the master (crank). The construction of kinematic diagrams is produced within one period (cycle), which has steadfit motion for several positions of the host.
We build on a scale of twelve positions corresponding to the consistent turns of the crank every 300.
Where S \u003d 2R is the actual magnitude of the slider, equal to the twice the magnitude of the crank.
- The passage of the slider on the mechanism scheme.
From where the time scale
Segment 1 On the time axis, we divide the 12 equal parts of the corresponding ones in the selected scale to the rotation of the crank at the angles: 300, 600, 900, 1200, 1,500, 3000, and 3300, 3,600 (at points 1-12). We postpone from these points vertical segments: 1-1S \u003d B0V1, 2-2S \u003d B0V2, etc., until the extreme right position of the slider in the distance, these increase, and starting from the position in decreases. If the points 0s, 1s, 2s ... 12S connect the curve consistently, then the movement chart of the point V.
To construct charts of speeds and accelerations, use the method of graphic differentiation. Speed \u200b\u200bdiagram build as follows.
Under the movement diagram, we conclude the coordinates V and T and on the continuation of the axis V to the left arbitrarily lay the selected pole distance HV \u003d 20mm.
From the PV point we carry out direct, parallel with tangential crooked s, respectively, 0s, 1S, 2S ... 12S points. These straight cuts on the axis V segments: 0-0V, 0-1V, 0-2V ... proportional to the speeds at the appropriate points of the diagram. We demote points to the ordinates of the corresponding points. We connect a series of points obtained 0V, 1V, 2V ... a smooth curve, which is a speed chart. The time scale remains the same, speed scale:
The acceleration diagram is building similarly to the speed diagram. Scale of accelerations
Where HA \u003d 16mm is the selected pole distance for the acceleration chart.
Since the speed and acceleration are the 1st and 2nd derivative of movement by time, but relative to the upper chart of the lower the differential curve, and relative to the lower upper - the integral curve. So the speed diagram for the displacement diagrams is differential. When building kinematic charts to verify the properties of the derivative:
- the increasing graph of movements (speeds) corresponds to the positive values \u200b\u200bof the chart of the speed (equation), and the decreasing - negative;
- point of maximum and minimum, i.e. the extremal value of the movement schedule (speeds) corresponds to the zero values \u200b\u200bof the speed graph (acceleration);
- the intention of the movement graphics (speeds) corresponds to the extremal values \u200b\u200bof the speed graphics (acceleration);
- the intersection point on the movement diagram corresponds to the point where the acceleration is zero;
- The ordents of the beginning and end of the period of any kinematic diagram are equal, and the tangents carried out at these points parallel.
To build a slide movement schedule in the selected axis of the coordinate S, T. On the abscissa axis, we postpone the segment L \u003d 120mm, depicting the time T alone full turn Krivoship
They made a geometric calculation of the links of the crank-slider mechanism, identified the lengths of the crank and the slider, and also set their ratio. Calculated the crank-slider mechanism in four positions and determined the speed of points using the instantaneous speed center for four positions. Built displacement charts, speeds and accelerations. It was established that there is some error due to the construction and rounding during the calculations.
Dano (Fig.2.10): J 1, W 1 \u003d const, l. BD, l. DC, l. AB, l. BC, M. L [m / mm. ] .
Speed V B.\u003d W 1. l A B.points in aimed perpendicular to the link AB towards its rotation.
To determine the speed of the point with the composition of the vector equality:
C \u003d. B +. St.
The direction of the absolute speed of the point is known - parallel lines xh. The speed of the point is known, and the relative velocity V c is sent perpendicular to the link Sun.
Build the speed plan (Fig. 2.11) in accordance with the equation written above. With this m n \u003d V b / RV[m / s mm ].
Absolute acceleration point in equally normal acceleration and n v.(since w 1 \u003d const E 1 \u003d 0 and but T B \u003d 0) a b \u003d a n val \u003d w 2× l V.[m / s 2]
and directed on the link AV from point to to point A.
The large-scale coefficient of the acceleration plan M a \u003d and in /p. at[m / s mm], where p at - arbitrary length segment depicting acceleration on the plan a B..
Acceleration point C:
(1 way),
where and p sv \u003d v 2 sv / l[m / s 2]
Cut depicting this acceleration on the acceleration plan:
p sv \u003d a p sv /m. but[mm]
We choose a pole P of the acceleration plan. From the pole carrying a line along which acceleration is directed a B. (// AV) and postpone the selected segment P atdepicting this acceleration on the plan (Fig. 2.12). From the end of the resulting vector, we carry out the line of direction of the normal component a nparallel to the link of St. and lay off the segment pdepicting M butthis is normal acceleration. From the end of the normal acceleration vector, we carry out the direction of the direction of the tangential component a t of St., and from the pole p - Direction of absolute acceleration point C ( ïï xx). In the intersection of these two directions we get a point C; In this case, the vector PC depicts the desired acceleration.
The module of this acceleration is:
and C \u003d (p. with)m. but[m / s 2]
Corner acceleration E 2 will be determined as:
e. 2 = a T CV / L SV= (t CB)m. A / L sv[1 / C 2]
E. 2 showing the mechanism scheme.
To find the speed point D you need to use similarity on the likenesswhich is used to determine the speeds and accelerations of the points of one link when speed (acceleration) of the two other points of this link are known: relative speeds (acceleration) of the points of one link form on the rates of speeds (accelerations) of the shapes similar to the diagram of the same name in the mechanism scheme. These figures are similarly located, i.e. When reading alphabetic designations in one direction on the mechanism scheme, the letters on the speed plan (accelerations) follow in the same direction.
To find the speed point D, it is necessary to build a triangle similar to a triangle on the mechanism scheme.
Triangles D. cVD (on the speed plan) and DSD (on the plan of the mechanism) are triangles with mutually perpendicular parties. Therefore, to build a triangle D cVD We will carry out perpendicular to the CD and to the VD from the points C and at respectively. In their intersection, we obtain a point D, which we connect with the pole.
The acceleration of the point D is also determined by the similarity theorem, since the acceleration of other two points 2 is known, namely but In and but C. Required in terms of acceleration triangle D atcd, similar to a DBCD triangle on a mechanism scheme.
To do this, we construct it first on the mechanism scheme, and then we transfer to the acceleration plan.
Section " sun."The acceleration plan is transferred to the segment of the SV on the mechanism diagram, laying it on the link of SV from any point (C or B) (Fig. 2.10). Then by cut " sun.»The mechanism is built triangle D atdS, similar to the triangle DBDS, for which from the point "C" is carried out direct "DC", parallel to the direct DC, up to the intersection with a straight CD. We get D. atdC ~ DBDC.
The resulting sides of the triangle R 1 and R 2 are equal in size to the parties of the desired
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the triangle on the acceleration plan, which can be constructed using sneakers (Fig.2.12). Next, it is necessary to check the similarity of the location of the figures. So, when reading the alphabetic designations of the vertices of the triangle DBDs on the mechanism scheme clockwise, we get order letters in d-s; in terms of accelerations in the same direction, i.e. clockwise, we must get the same order of letters at-d-s. Consequently, the solution satisfies the left point of intersection of the circles R 1 and R 2.
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All the themes of this section:
Graphic method of kinematic research
2.1.1 Basic equations for determining speeds and accelerations .................................................................................................25 2.1.2 Kinematics of quarrel mechanisms ..............................
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1. Structural analysis mechanism
The crispically-slider mechanism is presented.
The number of degrees of the studied mechanism Determine by the Chebyshev formula:
(1)
where n - The number of moving links in the composition of the investigated kinematic chain; p 4. and p 5. - respectively, the number of pairs of the fourth and fifth grade.
To determine the magnitude of the coefficient n. We analyze the structural scheme of the mechanism (Figure 1):
Picture 1 - Structural scheme Mechanism
The structural scheme of the mechanism consists of four units:
1 - crank,
2 - Shatun Av,
3 - slider in,
0 - rack,
at the same time, the links 1 - 3 are moving links, and a rack 0 is a fixed link. It is represented as part of the structural circuit with two hinge-fixed supports and the guide slider 3.
Hence, n \u003d 3.
To determine the values \u200b\u200bof the coefficients P 4. and p 5. We will find all the kinematic pairs that are part of the kinematic chain under consideration. Research results We enter in Table 1.
Table 1 - Cinematic Couples
| № | Cinematic pair (KP) | Cinema scheme pary |
Kinema class pary |
Degree of mobile |
||||||
| 1 | 0 – 1 | rotational |
||||||||
| 2 | 1 – 2 |  |
rotational |
1 | ||||||
| 3 | 2 – 3 |  |
rotational |
1 | ||||||
| 4 | 3 – 0 |  |
rotational |
1 | ||||||
From the data analysis of Table 1 it follows that the study mechanism of DVS With an increased stroke of the piston consists of seven pairs of the fifth grade and forms a closed kinematic chain. Hence, p 5 \u003d 4, but p 4 \u003d 0.
Substituting the found values \u200b\u200bof the coefficients n, P 5 and p 4. In the expression (1), we get:
To identify the structural composition of the mechanism, we divide the scheme to the structural groups of the Assur.
The first group of links 0-3-2 (Figure 2).
Figure 2 - Structural Assessura Group
This group consists of two moving links:
rod 2 and slider 3;
two leashes:
and three kinematic pairs:
1-2 - rotational pair of fifth grade;
2-3 - rotational steam of the fifth grade;
3-0 - progressive pair of fifth grade;
then n \u003d 2; P 5 \u003d 3, a p 4 \u003d 0.
Substituting the identified values \u200b\u200bof the coefficients in the expression (1),
Consequently, the group of links 4-5 is a structural group of Assur 2 class 2 of about 2 types.
The second group of units 0-1 (Figure 3).
Figure 3 - Primary mechanism
This group of units consists of a rolling link - crank 1, racks 0 and one kinematic pair:
0 - 1 - rotational pair of fifth grade;
then n \u003d 1; p 5 \u003d 1, a p 4 \u003d 0.
Substitting the found values \u200b\u200bin the expression (1), we get:
Consequently, the group of links 1 - 2 is indeed the primary mechanism with mobility 1.
Structural formula of the mechanism
Mechanism \u003d PM (w \u003d 1) + SGA (2 class, 2 Order, 2 species)
2. Synthesis kinematic scheme
For the synthesis of the kinematic scheme, it is first necessary to establish a large-scale length coefficient μ. To find μ ℓ, it is necessary to take the natural size of the crystal size of the option and divide it on the size of the arbitrary length of │ ° C│:
After that, with the help of a large-scale length coefficient, we translate all the natural dimensions of the links into segments, with which we will build a kinematic scheme:
After calculating the sizes, we proceed to the construction of one position of the mechanism (Figure 4) using the resort method.
To do this, first draw the rack 0 on which the crank is fixed. Then spend through the center of the circle, which was drawn to build a rack, horizontal direct XX. It is necessary for the next location of the center of the crawl 3. Further, we carry out two other radius from the center of the same circle.
and. Then, from there, we build diameters with a length at an angle to the horizontal direct XX. The intersection points of this segment with constructed circles will be points A and C, respectively. Then from the point and we build a circle with a radius.The intersection point of this circle with a straight XX will be point V. Draw a guide for a slider, which will coincide with direct XX. We build a crawler and all other are necessary Drawing details. We denote all points. The synthesis of the kinematic scheme is completed.
3. Kinematic analysis Flat mechanism
We proceed to the construction of the speed plan for the position of the mechanism. To simplify the calculations, you should calculate speeds and directions for all points of the position of the mechanism, and then build speed plan.
Figure 4 - one of the positions of the mechanism
We analyze the scheme of the crank-slider mechanism: the point O and O 1 are immobile dots, therefore, the velocity modules of these points are zero (
).The velocity vector of the point A is the geometric sum of the velocity vector of the point O and the rate of the relative rotational motion of the point and around the point O:
. (2)
Speed \u200b\u200bvector action line
It is a perpendicular to the axis of crank 1, and the direction of the action of this vector coincides with the direction of rotation of the crank.Module Speed \u200b\u200bPoint A:
, (3)
- the angular velocity of the OA; - Length is. Angular speed
Perm State Technical University
Department of "Mechanics of Composite Materials and Designs".
Course project
By theoryMechanisms and machines
Topic:
The task:
Option:
Performed:group student
Checked:professor
Piazzhava E.V.
Perm 2005
Structural analysis of the mechanism ...................................................... 3
Kinematic analysis of the mechanism ................................................ ..4
Kinetostatic analysis of the mechanism ........................................... ... 9
Calculation of the flywheel ............................................................... ............ 12
Profiling cam ............................................................ 17
Designing of the toothed gear ............................................. ... 20
Guidelines for the execution of calculations for the course project on TMM ...... .23
References .................................................................. ... 24
Structural Anali3 crank-slider
1. Show the structural scheme of the mechanism
OA - Krivoship - performs a rotational movement;
Av - Shatun - Makes a plane-parallel movement;
B - the slider - communal movement.
2. We will find the degree of mobility of the Chebyshev formula:
3. Spread on the structural groups of the Assur
4. Write structural formula Mechanism i \u003d\u003e II 2 2
5. We define the class, the order of the entire mechanism.
The studied mechanism consists of a first-class mechanism and the second-class second-class structural group (connecting rod and slider), therefore, the OAV hydraulic pump - mechanism second class second order.
Kinematic analysis of the mechanism
Initial data: 
Oa \u003d m, 
AB \u003d MM.
With kinematic analysis, three tasks are solved:
the problem of positions;
speed task;
the task of accelerations.
The task of positions
Designing a crank-slider mechanism, we will find the extreme position of the mechanism: the beginning and end of the working stroke. The beginning of the workstop will find by the formula:
l-crank OA
g - Shatun Length Av
The end of the workstop will find by the formula:
Working
S \u003d s "- s" \u003d 2r [m];
Build a mechanism on scale
1
=
AB / OA \u003d [m / mm]
Find the length of AV:
AB \u003d. 
AB / 1 \u003d [mm]
Let us show the movement of points at the twelve positions of the mechanism. To do this, we divide the circle at 12 equal parts (using the seed method).
We construct a connecting rod curve. To do this, find the center of gravity of each link and connect the smooth line.
Mechanism provisions are used to determine the speeds and accelerations in the specified provisions.
The task of speeds
Kinematic analysis is performed by a graph-analytical method, which reflects the visibility of changes in speeds and provides sufficient accuracy. Lead speed:
[MS -1]
We write vector equations:
V B \u003d V A + V AB; V B \u003d V x + V BX
where V x \u003d 0; V a oa; V AB AB; V BX BX
Vectors of V BA, V B, V S 2 We define the construction. Choose speed plan scale
[MS -1 / mm].
GE PA is a segment characterizing the speed in the drawing \u003d mm. From an arbitrary point P - the pole of the speed plan, we postpone the vector of RA,
perpendicular oa. After t. And we carry out perpendicular to AV direct. The intersection point of the x axis (selected in directions T. c) with this direct will give. B, connecting t. In the pole, we obtain the vector of speed t. We define the speed of T. in:
[MS -1]
Position t. On the speed plan, we define from the proportion:
Connecting t. S 2 with a pole p, we obtain the value and direction of the speed T. S2:
[MS -1]
[MS -1]
We define:
[MS -1]
[MS -1]
[MS -1]
We define:
[s -1]
The direction 2 is determined by the transfer of the VBA vector in T.V.
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Parameter |
Position of the mechanism |
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ipno-slider mechanism
2.1. Structural scheme of the mechanism
Figure 2.1 Structural scheme of the crystal-slider mechanism
2.2. Detection of complex and separated kinematic couples
In the crank-and-slide mechanism of separated kinematic pairs there is no. Couple ATcomplex, so we will consider it for two kinematic pairs.
2.3. Classification of kinematic pairs of mechanism
Table 2.1
|
No. p / p |
Rooms of links forming a couple |
Symbol |
Name |
Mobility |
Top / Lost |
Circuit (Geometric / Strength) |
Open / Closed |
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Rotational |
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Rotational |
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Rotational |
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Rotational |
|||||||
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Rotational |
|||||||
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Rotational |
||||||
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Progress |
The mechanism under study consists only of single-duty kinematic pairs ( r 1 = 7, r\u003d 7) where r 1 - the number of single-duty kinematic pairs in the mechanism, r- The total number of kinematic couples in the mechanism.
2. 4. Classification of mechanism units
Table 2.2.
|
No. p / p |
Rooms Rooms |
Symbol |
Name |
Traffic |
Number of vertices |
|
Is absent |
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|
Crank |
Rotational |
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Rotational |
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Progress |
The mechanism has: four () bullshish () linear lines 1,2,4,5; One (N 3 \u003d 1) three vertex link, which is the base link; Five () moving links.
We find the number of connections to the rack. The conveyor mechanism has three () rack connections.
In the examined mechanism, one elementary mechanism can be distinguished
Fig. 2.4 Cracked-slider mechanism.
There are no mechanisms with crushed kinematic chains in the exploded crank-slider mechanism.
The mechanism has only ordinary stationary mechanisms.
In the study mechanism, there is no consolidation links. The link 3 simultaneously enters two simple mechanisms - a hinge clear-visitor and a crank-slider. So for this link
We classify the mechanism. The studied mechanism has a permanent structure, is complex and the same type. It consists of one elementary mechanism and two stationary simple, which in their composition only closed kinematic chains.
The mechanism exists in three-trackspace.
Formulas for determining the mobility of these mechanisms will take a view accordingly:
We define the mobility of the hinge quadruple. This mechanism has: three () moving links 1,2,3; Four () single-diving kinematic pairs O, A, B, C.
Find the mobility of the crank-slider mechanism. It has: () moving links 3,4,5 and four () kinematic pairs C, B, D, K. It is determined similar to:
We determine the mobility of the complex mechanism by the formula:
We carry out the analysis of the structural model of the machine mechanism. We check whether the mechanism under study corresponds to the structure of the mathematical model. The mechanism has: seven () single-living kinematic pairs; five () movable bullshish () links, basic is; Three attachments to the rack () and no fastening links ().
Mathematical model:
;
;
Since the equations of the model have become identity, the device under study has the correct structure and is a mechanism.
We highlight and carry out the classification of structural groups. The elementary mechanism is conditionally attributed to the class I mechanism.
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The class of the structural group is determined by the number of kinematic couples included in the closed circuit formed by internal kinematic pairs. The order of the group is determined by the number of external kinematic pairs. The type of group is determined depending on the placement of rotational and translational kinematic pairs on it.
2-order
It can be seen that the allocated structural groups are completely similar to the species and quantitative composition of links and kinematic pairs. Each of the structural groups has: two moving links (), and the links of bullshine () and, it means, the baseline also has two vertices (); Three () single-diving kinematic pairs, of which two external ().
We check whether the allocated structural groups correspond to mathematical models. Since the groups are similar, then check only on one group, for example, OAB. Mathematical models of structural groups have the form:
The crank-slider mechanism refers to the second class.
3. Kinematic analysis of the mechanism
The kinematic analysis of any mechanism is to define: extreme (dead) machine positions, including the definition of trajectories of individual points; Speed \u200b\u200band accelerations of characteristic points of links according to the well-known law of the initial traffic (generalized coordinate).
3.1 Defining the extreme (dead) provisions of the mechanism
The extreme (dead) position of the mechanism can be determined analytically or graphically. Since the analytics gives higher accuracy, then when determining the extreme positions, it is preferred.
For a crank-slider and hinge crystrum-fysive four-fiscal crank, there will be such provisions when the crank and the connecting rod are pulled out (), then consur () in one line.
Fig. 3.1 Determination of the extreme positions of the mechanism.
3.2 Determination of the positions of the mechanism of the mechanism graphically.
Fig. 3.3 Construction of closed vector contours.
The structural scheme of the mechanism is located in a rectangular coordinate system, the beginning of which is placed in the point O. with the links of the mechanism vectors we associate so that their sequence is two closed circuits: OABCO and CBDC.
For OABCO contour: 
(3.1)
Imagine an equation in projections on the axis of coordinates.
