(although most often the same).

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  The position of the center of mass (center of inertia) of a system of material points in classical mechanics is determined as follows:

  r → c = ∑ i m i r → i ∑ i m i , (\displaystyle (\vec (r))_(c)=(\frac (\sum \limits _(i)m_(i)(\vec (r))_ (i))(\sum \limits _(i)m_(i))),)

  where r → c (\displaystyle (\vec (r))_(c))- radius vector of the center of mass, r → i (\displaystyle (\vec(r))_(i))- radius vector i-th point of the system, m i (\displaystyle m_(i))- weight i-th point.

  For the case of continuous mass distribution:

  r → c = 1 M ∫ V ρ (r →) r → d V , (\displaystyle (\vec (r))_(c)=(1 \over M)\int \limits _(V)\rho ( (\vec (r)))(\vec (r))dV,) M = ∫ V ρ (r →) d V , (\displaystyle M=\int \limits _(V)\rho ((\vec (r)))dV,)

  where M (\displaystyle M) is the total mass of the system, V (\displaystyle V)- volume, ρ (\displaystyle \rho )- density. The center of mass thus characterizes the distribution of mass over a body or a system of particles.

  It can be shown that if the system does not consist of material points, but of extended bodies with masses M i (\displaystyle M_(i)), then the radius vector of the center of mass of such a system R c (\displaystyle R_(c)) associated with the radius vectors of the centers of mass of bodies R c i (\displaystyle R_(ci)) ratio:

  R → c = ∑ i M i R → c i ∑ i M i . (\displaystyle (\vec (R))_(c)=(\frac (\sum \limits _(i)M_(i)(\vec (R))_(ci))(\sum \limits _( i)M_(i))).)

  In other words, in the case of extended bodies, a formula is valid, which in its structure coincides with that used for material points.

  Centers of mass of flat homogeneous figures

  The coordinates of the center of mass of a homogeneous flat figure can be calculated by the formulas (a consequence of the Pappa–Guldin theorems):

  x s = V y 2 π S (\displaystyle x_(s)=(\frac (V_(y))(2\pi S))) and y s = V x 2 π S (\displaystyle y_(s)=(\frac (V_(x))(2\pi S))), where V x , V y (\displaystyle V_(x),V_(y))- the volume of the body obtained by rotating the figure around the corresponding axis, S (\displaystyle S) is the area of ​​the figure.

  Centers of Mass of Perimeters of Homogeneous Figures

  To avoid mistakes, it should be understood that in SRT the center of mass is characterized not by the distribution of mass, but by the distribution of energy. In the course on theoretical physics by Landau and Lifshitz, the term “center of inertia” is preferred. In Western literature on elementary particles, the term "center of mass" (English center-of-mass) is used: both terms are equivalent.

  The speed of the center of mass in relativistic mechanics can be found by the formula:

  v → c = c 2 ∑ i E i ⋅ ∑ i p → i . (\displaystyle (\vec (v))_(c)=(\frac (c^(2))(\sum \limits _(i)E_(i)))\cdot \sum \limits _(i) (\vec(p))_(i).) mass weight P = m g depends on the gravitational field parameter g), and, generally speaking, even located outside the rod.

  In a uniform gravitational field, the center of gravity always coincides with the center of mass. In non-cosmic problems, the gravitational field can usually be considered constant within the volume of the body, so in practice these two centers almost coincide.

  For the same reason, concepts center of gravity and center of gravity coincide when these terms are used in geometry, statics, and similar areas, where its application in comparison with physics can be called metaphorical and where the situation of their equivalence is implicitly assumed (since there is no real gravitational field, then taking into account its inhomogeneity does not make sense). In these uses, the two terms are traditionally synonymous, and often the second is preferred simply because it is older.

  center of gravity(or center of mass) of a certain body is called a point that has the property that if a body is suspended from this point, then it will retain its position.

  Below we consider 2D and 3D problems related to the search for various centers of mass, mainly from the point of view of computational geometry.

  In the solutions discussed below, there are two main fact. The first is that the center of mass of a system of material points is equal to the average of their coordinates, taken with coefficients proportional to their masses. The second fact is that if we know the centers of mass of two non-intersecting figures, then the center of mass of their union will lie on the segment connecting these two centers, and it will divide it in the same ratio as the mass of the second figure relates to the mass of the first.

  Two-dimensional case: polygons

  In fact, when speaking about the center of mass of a two-dimensional figure, one of the following three can be meant: tasks:

  • The center of mass of the system of points - i.e. the entire mass is concentrated only at the vertices of the polygon.
  • The center of mass of the frame - i.e. the mass of a polygon is concentrated on its perimeter.
  • The center of mass of a solid figure - i.e. the mass of the polygon is distributed over its entire area.

  Each of these problems has an independent solution, and will be considered below separately.

  Center of mass of point system

  This is the simplest of the three problems, and its solution is the well-known physical formula for the center of mass of a system of material points:

  where are the masses of the points, are their radius vectors (specifying their position relative to the origin), and is the desired radius vector of the center of mass.

  In particular, if all points have the same mass, then the coordinates of the center of mass are average point coordinates. For triangle this point is called centroid and coincides with the point of intersection of the medians:

  

  For proof of these formulas, it suffices to recall that equilibrium is reached at a point at which the sum of the moments of all forces is equal to zero. In this case, this turns into a condition for the sum of the radius vectors of all points relative to the point, multiplied by the masses of the corresponding points, to be equal to zero:

  

  and, expressing from here , we obtain the required formula.

  Frame center of gravity

  But then each side of the polygon can be replaced by one point - the middle of this segment (since the center of mass of a homogeneous segment is the middle of this segment), with a mass equal to the length of this segment.

  Now we have received the problem about the system of material points, and applying the solution from the previous paragraph to it, we find:

  

  where is the midpoint of the th side of the polygon, is the length of the th side, is the perimeter, i.e. the sum of the lengths of the sides.

  For triangle one can show the following statement: this point is bisector intersection point triangle formed by the midpoints of the sides of the original triangle. (to show this, we must use the above formula, and then notice that the bisectors divide the sides of the resulting triangle in the same ratio as the centers of mass of these sides).

  Center of mass of a solid figure

  We believe that the mass is uniformly distributed over the figure, i.e. the density at each point of the figure is equal to the same number.

  Triangle case

  It is argued that for a triangle the answer is still the same centroid, i.e. the point formed by the arithmetic mean of the coordinates of the vertices:

  

  Triangle Case: Proof

  We give here an elementary proof that does not use the theory of integrals.

  The first such, purely geometric, proof was given by Archimedes, but it was very complex, with a large number of geometric constructions. The proof given here is taken from the article by Apostol, Mnatsakanian "Finding Centroids the Easy Way".

  The proof boils down to showing that the center of mass of the triangle lies on one of the medians; repeating this process twice more, we thereby show that the center of mass lies at the point of intersection of the medians, which is the centroid.

  Let's divide this triangle into four, connecting the midpoints of the sides, as shown in the figure:

  

  The four resulting triangles are similar to a triangle with coefficient .

  Triangles No. 1 and No. 2 together form a parallelogram, the center of mass of which lies at the point of intersection of its diagonals (since this is a figure symmetrical with respect to both diagonals, which means that its center of mass must lie on each of the two diagonals). The point is in the middle of the common side of triangles No. 1 and No. 2, and also lies on the median of the triangle:

  

  Now let the vector be the vector drawn from the vertex to the center of mass of triangle No. 1, and let the vector be the vector drawn from to the point (which, recall, is the midpoint of the side on which it lies):

  

  Our goal is to show that the vectors and are collinear.

  Denote by and the points that are the centers of mass of triangles No. 3 and No. 4. Then, obviously, the center of mass of the aggregate of these two triangles will be the point , which is the midpoint of the segment . Moreover, the vector from point to point is the same as the vector .

  The desired center of mass of the triangle lies in the middle of the segment connecting the points and (since we have divided the triangle into two parts of equal areas: No. 1-No. 2 and No. 3-No. 4):

  

  Thus, the vector from the vertex to the centroid is . On the other hand, since triangle No. 1 is similar to a triangle with coefficient , then the same vector is equal to . From here we get the equation:

  

  from where we find:

  Thus, we have proved that the vectors and are collinear, which means that the desired centroid lies on the median emanating from the vertex .

  Moreover, along the way, we proved that the centroid divides each median with respect to , counting from the top.

  Polygon case

  Now let's move on to the general case - i.e. to the occasion polygon. For him, such reasoning is no longer applicable, so we reduce the problem to a triangular one: namely, we divide the polygon into triangles (i.e., triangulate it), find the center of mass of each triangle, and then find the center of mass of the resulting centers of mass of the triangles.

  The final formula is as follows:

  

  where is the centroid of the -th triangle in the triangulation of the given polygon, is the area of ​​the -th triangle of the triangulation, is the area of ​​the entire polygon.

  Triangulation of a convex polygon is a trivial task: for this, for example, we can take triangles , where .

  Polygon case: alternative way

  On the other hand, the application of the above formula is not very convenient for non-convex polygons, since triangulating them is not an easy task in itself. But for such polygons, you can come up with a simpler approach. Namely, let's draw an analogy with how you can find the area of ​​an arbitrary polygon: an arbitrary point is selected, and then the sign areas of the triangles formed by this point and the points of the polygon are summed up: . A similar technique can be used to find the center of mass: only now we will sum the centers of mass of triangles taken with coefficients proportional to their areas, i.e. the final formula for the center of mass is:

  

  where is an arbitrary point, are the points of the polygon, is the centroid of the triangle , is the sign area of ​​this triangle, is the sign area of ​​the entire polygon (i.e. ).

  3D Case: Polyhedra

  Similarly to the two-dimensional case, in 3D we can talk about four possible problem statements at once:

  • The center of mass of the system of points - the vertices of the polyhedron.
  • The center of mass of the frame is the edges of the polyhedron.
  • Center of mass of the surface - i.e. the mass is distributed over the surface area of ​​the polyhedron.
  • The center of mass of a solid polyhedron - i.e. the mass is distributed over the entire polyhedron.

  Center of mass of point system

  As in the 2D case, we can apply the physical formula and get the same result:

  which, in the case of equal masses, turns into the arithmetic mean of the coordinates of all points.

  Center of mass of polyhedron frame

  Similarly to the two-dimensional case, we simply replace each edge of the polyhedron with a material point located in the middle of this edge, and with a mass equal to the length of this edge. Having received the problem of material points, we can easily find its solution as a weighted sum of the coordinates of these points.

  The center of mass of the surface of the polyhedron

  Each face of the surface of a polyhedron is a two-dimensional figure, the center of mass of which we can find. Finding these centers of mass and replacing each face with its center of mass, we get a problem with material points, which is already easy to solve.

  Center of mass of a solid polyhedron

  Tetrahedron case

  As in the two-dimensional case, we first solve the simplest problem - the problem for the tetrahedron.

  It is stated that the center of mass of a tetrahedron coincides with the point of intersection of its medians (the median of a tetrahedron is a segment drawn from its vertex to the center of mass of the opposite face; thus, the median of a tetrahedron passes through the vertex and through the point of intersection of the medians of a triangular face).

  Why is it so? Reasonings similar to the two-dimensional case are correct here: if we cut a tetrahedron into two tetrahedra using a plane passing through the vertex of the tetrahedron and some median of the opposite face, then both resulting tetrahedra will have the same volume (because the triangular face will be divided by the median into two triangle of equal area, and the height of the two tetrahedra does not change). Repeating this reasoning several times, we get that the center of mass lies at the intersection point of the medians of the tetrahedron.

  This point - the point of intersection of the medians of the tetrahedron - is called its centroid. It can be shown that it actually has coordinates equal to the arithmetic mean of the coordinates of the vertices of the tetrahedron:

  

  (this can be inferred from the fact that the centroid divides the medians with respect to )

  Thus, there is no fundamental difference between the cases of a tetrahedron and a triangle: a point equal to the arithmetic mean of the vertices is the center of mass in two formulations of the problem at once: both when the masses are only at the vertices, and when the masses are distributed over the entire area / volume. In fact, this result generalizes to an arbitrary dimension: the center of mass of an arbitrary simplex(simplex) is the arithmetic mean of the coordinates of its vertices.

  The case of an arbitrary polyhedron

  Let us now turn to the general case, the case of an arbitrary polyhedron.

  Again, as in the two-dimensional case, we reduce this problem to the already solved one: we divide the polyhedron into tetrahedra (i.e., we tetrahedronize it), find the center of mass of each of them, and obtain the final answer to the problem in the form of a weighted sum of the found centers wt.

  Definition

  When considering a system of particles, it is often convenient to find a point that characterizes the position and movement of the system under consideration as a whole. Such a point is center of gravity.

  If we have two particles of the same mass, then such a point is in the middle between them.

  Center of mass coordinates

  Let's assume that two material points having masses $m_1$ and $m_2$ are located on the x-axis and have coordinates $x_1$ and $x_2$. The distance ($\Delta x$) between these particles is:

  \[\Delta x=x_2-x_1\left(1\right).\]

  Definition

  Point C (Fig. 1), which divides the distance between these particles into segments inversely proportional to the masses of the particles, is called center of mass this system of particles.

  In accordance with the definition for Fig. 1, we have:

  \[\frac(l_1)(l_2)=\frac(m_2)(m_1)\left(2\right).\]

  where $x_c$ is the coordinate of the center of mass, then we get:

  From formula (4) we get:

  Expression (5) is easily generalized for a set of material points, which are located arbitrarily. In this case, the abscissa of the center of mass is equal to:

  Similarly, expressions for the ordinate ($y_c$) of the center of mass and its applicates ($z_c$) are obtained:

  \ \

  Formulas (6-8) coincide with the expressions that determine the center of gravity of the body. In the event that the dimensions of the body are small in comparison with the distance to the center of the Earth, the center of gravity is considered to coincide with the center of mass of the body. In most problems, the center of gravity coincides with the center of mass of the body.

  If the position of N material points of the system is given in vector form, then the radius - the vector that determines the position of the center of mass is found as:

  \[(\overline(r))_c=\frac(\sum\limits^N_(i=1)(m_i(\overline(r))_i))(\sum\limits^N_(i=1)( m_i))\left(9\right).\]

  Center of mass movement

  The expression for the center of mass velocity ($(\overline(v))_c=\frac(d(\overline(r))_c)(dt)$) is:

  \[(\overline(v))_c=\frac(m_1(\overline(v))_1+m_2(\overline(v))_2+\dots +m_n(\overline(v))_n)(m_1+m_2+ \dots +m_n)=\frac(\overline(P))(M)\left(10\right),\]

  where $\overline(P)$ is the total momentum of the system of particles; $M$ is the mass of the system. Expression (10) is valid for motions with velocities that are significantly less than the speed of light.

  If the system of particles is closed, then the sum of the momenta of its parts does not change. Therefore, the speed of the center of mass is a constant value. They say that the center of mass of a closed system moves by inertia, that is, in a straight line and uniformly, and this movement is independent of the movement of the constituent parts of the system. In a closed system, internal forces can act; as a result of their action, parts of the system can have accelerations. But this does not affect the movement of the center of mass. Under the action of internal forces, the speed of the center of mass does not change.

  Examples of problems with a solution

  Example 1

  Exercise. Write down the coordinates of the center of mass of the system of three balls located at the vertices and the center of an equilateral triangle, the side of which is equal to $b\ (m)$ (Fig. 2).

  

  Solution. To solve the problem, we use expressions that determine the coordinates of the center of mass:

  \ \

  From Fig. 2 we see that the abscissas of the points:

  \[\left\( \begin(array)(c) m_1=2m,\ \ x_1=0;;\ \ \\ (\rm \ )m_2=3m,\ \ \ \ x_2=\frac(b)( 2);; \\ m_3=m,\ \ x_3=\frac(b)(2);; \\ m_4=4m,\ \ x_4=b.\end(array) \right.\left(2.3\right ).\]

  Then the abscissa of the center mass is equal to:

  Let's find the ordinates of the points.

  \[ \begin(array)(c) m_1=2m,\ \ y_1=0;;\ \ \\ (\rm \ )m_2=3m,\ \ \ \ y_2=\frac(b\sqrt(3)) (2);; \\ m_3=m,\ \ y_3=\frac(b\sqrt(3))(6);; \\ m_4=4m,\ \ y_4=0. \end(array)\left(2.4\right).\]

  To find the ordinate $y_2$, let's calculate the height in an equilateral triangle:

  We find the ordinate $y_3$, remembering that the medians in an equilateral triangle are divided by the intersection point in a ratio of 2:1 from the top, we get:

  Calculate the ordinate of the center of mass:

  Answer.$x_c=0.6b\ (\rm \ )(\rm m)$; $y_c=\frac(b\sqrt(3)\ )(6)$ m

  Example 2

  Exercise. Write down the law of motion of the center of mass.

  Solution. The law of change in the momentum of a system of particles is the law of motion of the center of mass. From the formula:

  \[(\overline(v))_c=\frac(\overline(P))(M)\to \overline(P)=M(\overline(v))_c\left(2.1\right)\]

  for a constant mass $M$, differentiating both parts of expression (2.1), we obtain:

  \[\frac(d\overline(P))(dt)=M\frac(d(\overline(v))_c)(dt)\left(2.2\right).\]

  Expression (2.2) means that the rate of change of the momentum of the system is equal to the product of the mass of the system and the acceleration of its center of mass. Because

  \[\frac(d\overline(P))(dt)=\sum\limits^N_(i=1)((\overline(F))_i\left(2.3\right),)\]

  In accordance with expression (2.4), we find that the center of mass of the system moves in the same way as one material point of mass M would move if it is acted upon by a force equal to the sum of all external forces acting on the particles that are included in the system under consideration. If $\sum\limits^N_(i=1)((\overline(F))_i=0,)$ then the center of mass moves uniformly and rectilinearly.

  The concept of an integral is widely applicable in life. Integrals are used in various fields of science and technology. The main tasks calculated using integrals are tasks for:

  1. Finding the volume of the body

  2. Finding the center of mass of the body.

  Let's consider each of them in more detail. Here and below, to denote a definite integral of some function f(x), with integration limits from a to b, we will use the following notation ∫ a b f(x).

  Finding the volume of a body

  Consider the following figure. Suppose there is some body whose volume is equal to V. There is also a straight line such that if we take a certain plane perpendicular to this straight line, the cross-sectional area S of this body by this plane will be known.

  Each such plane will be perpendicular to the x-axis, and therefore will intersect it at some point x. That is, each point x from the segment will be assigned the number S (x) - the cross-sectional area of ​​\u200b\u200bthe body, the plane passing through this point.

  It turns out that some function S(x) will be given on the segment. If this function is continuous on this segment, then the following formula will be valid:

  V = ∫ a b S(x)dx.

  The proof of this statement is beyond the scope of the school curriculum.

  Calculating the center of mass of a body

  The center of mass is most often used in physics. For example, there is some body which moves with any speed. But it is inconvenient to consider a large body, and therefore in physics this body is considered as the movement of a point, on the assumption that this point has the same mass as the whole body.

  And the task of calculating the center of mass of the body is the main one in this matter. Because the body is large, and which point should be taken as the center of mass? Maybe the one in the middle of the body? Or maybe the closest point to the leading edge? This is where integration comes in.

  The following two rules are used to find the center of mass:

  1. Coordinate x’ of the center of mass of some system of material points A1, A2,A3, … An with masses m1, m2, m3, … mn, respectively, located on a straight line at points with coordinates x1, x2, x3, … xn is found by the following formula:

  x’ = (m1*x1 + ma*x2 + … + mn*xn)/(m1 + m2 + m3 +… + mn)

  2. When calculating the coordinates of the center of mass, any part of the figure under consideration can be replaced by a material point, while placing it in the center of mass of this separate part of the figure, and the mass can be taken equal to the mass of this part of the figure.

  For example, if a mass of density p(x) is distributed along the rod - a segment of the Ox axis, where p(x) is a continuous function, then the coordinate of the center of mass x' will be equal to.

  Any body can be considered as a set of material points, which, for example, can be taken as molecules. Let the body consist of n material points with masses m1, m2, ...mn.

  center of mass of the body, consisting of n material points, is called a point (in the geometric sense), the radius vector of which is determined by the formula:

  Here R1 is the radius vector of the point with number i (i = 1, 2, ... n).

  This definition looks unusual, but in fact it gives the position of the very center of mass, about which we have an intuitive idea. For example, the center of mass of the rod will be in its middle. The sum of the masses of all points included in the denominator of the above formula is called the mass of the body. body weight called the sum of the masses of all its points: m = m1 + m2 + ... + mn .

  

  In symmetrical homogeneous bodies, the CM is always located at the center of symmetry or lies on the axis of symmetry if the figure does not have a center of symmetry. The center of mass can be located both inside the body (disk, square, triangle) and outside it (ring, frame, square).

  For a person, the position of the CM depends on the adopted posture. In many sports, an important component of success is the ability to maintain balance. So, in gymnastics, acrobatics

  a large number of elements will include different types of balance. The ability to maintain balance is important in figure skating, in skating, where the support has a very small area.

  The equilibrium conditions for a body at rest are the simultaneous equality to zero of the sum of forces and the sum of moments of forces acting on the body.

  Let us find out what position the axis of rotation should occupy in order for the body fixed on it to remain in equilibrium under the action of gravity. To do this, we will break the body into many small pieces and draw the forces of gravity acting on them.

  

  In accordance with the rule of moments, for equilibrium it is necessary that the sum of the moments of all these forces about the axis be equal to zero.

  It can be shown that for each body there is a unique point where the sum of the moments of gravity about any axis passing through this point is equal to zero. This point is called the center of gravity (usually coincides with the center of mass).

  Center of gravity of the body (CG) called the point about which the sum of the moments of gravity acting on all particles of the body is equal to zero.

  Thus, the forces of gravity do not cause the body to rotate around the center of gravity. Therefore, all the forces of gravity could be replaced by a single force that is applied to this point and is equal to the force of gravity.

  To study the movements of an athlete's body, the term common center of gravity (CGG) is often introduced. Main properties of the center of gravity:

  If the body is fixed on an axis passing through the center of gravity, then gravity will not cause it to rotate;

  The center of gravity is the point of application of gravity;

  In a uniform field, the center of gravity coincides with the center of mass.

  Equilibrium is the position of the body in which it can remain at rest for an arbitrarily long time. When the body deviates from the equilibrium position, the forces acting on it change, and the balance of forces is disturbed.

  There are various types of equilibrium (Fig. 9). It is customary to distinguish three types of equilibrium: stable, unstable and indifferent.

  Stable equilibrium (Fig. 9, a) is characterized by the fact that the body returns to its original position when it is deflected. In this case, forces arise, or moments of forces, tending to return the body to its original position. An example is the position of the body with an upper support (for example, hanging on the crossbar), when, with any deviations, the body tends to return to its original position.

  Indifferent equilibrium (Fig. 9, b) is characterized by the fact that when the position of the body changes, there are no forces or moments of forces that tend to return the body to its original position or further remove the body from it. This is a rare occurrence in humans. An example is the state of weightlessness on a spaceship.

  Unstable equilibrium (Fig. 9, c) is observed when, with small deviations of the body, forces or moments of forces arise that tend to deviate the body even more from its initial position. Such a case can be observed when a person, standing on a support of a very small area (much smaller than the area of ​​his two legs or even one leg), deviates to the side.

  Figure 9 Body balance: stable (a), indifferent (b), unstable (c)

  

  Along with the listed types of equilibrium of bodies in biomechanics, one more type of equilibrium is considered - limited-stable. This type of equilibrium is distinguished by the fact that the body can return to its initial position if it deviates from it up to a certain limit, for example, determined by the boundary of the support area. If the deviation exceeds this limit, the equilibrium becomes unstable.

  The main task in ensuring the balance of the human body is to ensure that the projection of the GCM of the body is within the area of ​​support. Depending on the type of activity (maintaining a static position, walking, running, etc.) and the requirements for stability, the frequency and speed of corrective actions change, but the processes of maintaining balance are the same.

  The distribution of mass in the human body

  The mass of the body and the masses of individual segments are very important for various aspects of biomechanics. In many sports, it is necessary to know the distribution of mass in order to develop the correct technique for performing exercises. To analyze the movements of the human body, the segmentation method is used: it is conventionally divided into certain segments. For each segment, its mass and the position of the center of mass are determined. In table. 1 defines the masses of body parts in relative units.

  Table 1. Masses of body parts in relative units

  Often, instead of the concept of the center of mass, another concept is used - the center of gravity. In a uniform field of gravity, the center of gravity always coincides with the center of mass. The position of the center of gravity of the link is indicated as its distance from the axis of the proximal joint and is expressed relative to the length of the link taken as a unit.

  In table. 2 shows the anatomical position of the centers of gravity of various parts of the body.

  Table 2. Centers of gravity of body parts

  Part of the body	Center of gravity position
  Hip	0.44 link length
  Shin	0.42 link length
  Shoulder	0.47 link length
  Forearm	0.42 link length
  torso
  Head
  Brush
  Foot
  Shoulder	0.47 link length
  Forearm	0.42 link length
  torso	0.44 distance from the transverse axis of the shoulder joints to the axis of the hip
  Head	Located in the region of the Turkish saddle of the sphenoid bone (projection from the front between the eyebrows, from the side - 3.0 - 3.5 above the external auditory canal)
  Brush	In the region of the head of the third metacarpal bone
  Foot	On a straight line connecting the calcaneal tubercle of the calcaneus with the end of the second finger at a distance of 0.44 from the first point
  The general center of mass of gravity in the vertical position of the body	Located at the main stance in the pelvic area, in front of the sacrum