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\title{About the Presence of Irregular Precession Motions in a Symmetric}
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             \author[1]{P.K.  Plotnikov}

             \affil[1]{  SSTU}

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\date{\small \em Received: 14 December 2019 Accepted: 4 January 2020 Published: 15 January 2020}

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\begin{abstract}
        


It is generally accepted that the only type of motion present in a symmetric Euler gyroscope (SEG) is regular precession. This paper proves that regular precession is not the only type of motion present, but corresponds only to the well-known initial coordinated Euler angles. At any other initial angles, motions that differ from regular precession occur. In the article, the problem is solved analytically in two stages: first, angular velocities of the gyroscope are determined using differential dynamic equations, at the second stage, as a result of integration of differential matrix kinematic and differential matrix Poisson equations (both with periodic coefficients), final relations about the SEG motion with arbitrary initial Euler angles are derived. Periodic coefficients are the SEG angular velocities that are found as a solution to the dynamic equations. From the obtained general formulas, special formulas of regular precession for particular coordinated initial Euler angles that coincide with the well-known ones are derived.

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\let\tabcellsep& 	 	 		 
\section[{P.K. Plotnikov}]{P.K. Plotnikov}\par
I. 
\section[{Annotation}]{Annotation}\par
t is generally accepted that the only type of motion present in a symmetric Euler gyroscope (SEG) is regular precession. This paper proves that regular precession is not the only type of motion present, but corresponds only to the well-known initial coordinated Euler angles. At any other initial angles, motions that differ from regular precession occur. In the article, the problem is solved analytically in two stages: first, angular velocities of the gyroscope are determined using differential dynamic equations, at the second stage, as a result of integration of differential matrix kinematic and differential matrix Poisson equations (both with periodic coefficients), final relations about the SEG motion with arbitrary initial Euler angles are derived. Periodic coefficients are the SEG angular velocities that are found as a solution to the dynamic equations. From the obtained general formulas, special formulas of regular precession for particular coordinated initial Euler angles that coincide with the well-known ones are derived. For other initial angles, formulas for irregular precession are obtained. In addition to the solutions for the Euler angles, solutions for the Euler-Krylov angles were found, which in some cases provide a more explicit geometric interpretation of motion. The analytical results are supported by mathematical modeling. In particular, certain conditions were found -the "strong impact" condition when irregular SEG precession for the Euler-Krylov angles occurs in the direction of the rotational pulse, and the sign of the angular velocity of the gyroscope proper rotation changes to the opposite. At the Euler angles, the motions of irregular precession during the "strong" and "weak" impact conditions are qualitatively identical. In relation to the case of regular precession under the "strong" impact conditions, the changes are significant: the angles of precession and nutation become oscillatory, and the angular velocity and the angle of proper rotation change their sign to the opposite. 
\section[{a) Relevance}]{a) Relevance}\par
Modern gyroscopic technology has achieved the highest accuracy in measuring angular motion parameters of moving objects (MO) in the field of classical symmetric Euler gyroscopes with electrostatic suspension. In the US Gravity Probe experiment, the four axially symmetric Euler gyroscopes with electrostatic cryogenic suspension mounted on the astronomical Earth satellite had values of drift angular velocities of less than 10 -11 angular deg/hr. This, together with the telescope readings, experimentally confirms the Einsteinian general theory of relativity (GTR) by detecting a gyro axis shift with the accuracy of 1\% equal to 6.6 angular seconds per year, which is effectively predicted by the GTR \hyperref[b0]{[1,}\hyperref[b1]{2]}. It is noted that classical symmetric Euler gyroscopes (SEG) with electrostatic suspensions have drift angular velocities values of 10 -5 angular deg/hr in terrestrial conditions, which is a better accuracy level than that of fiber optic (FOG) and laser (LG) gyroscopes, i.e. gyros based on new physical measurement principles in which drift angular velocities values are in the range of 10 -4 -10 -3 angular deg/hr, respectively \hyperref[b1]{[2]}. Considering the fact that rotary classical Euler gyroscopes with magnetic active and magnetic resonance suspensions are still being developed and manufactured, it can be stated that studies concerning angular motions of the rotor's axis of proper rotation, which characterize its errors, are relevant. In this aspect, for a symmetric Euler gyroscope designed for GTR validation \hyperref[b0]{[1,}\hyperref[b3]{4]}, the parameters of its regular precession are evaluated, i.e. its errors, including the Poinsot analysis. A fundamental presentation of the theory of symmetric Euler gyroscopes with the Poinsot and McCullagh analyses of motion is given in \hyperref[b4]{[5]}\hyperref[b5]{[6]}.\par
It should be recalled that elementary particleselectrons, protons, etc. are essentially Euler gyroscopes \hyperref[b2]{[3]} (one can say that the entire Universe consists of corpuscular Euler gyroscopes), which also emphasizes the relevance of this study. 
\section[{b) Formulation of the problem}]{b) Formulation of the problem}\par
The solution to the problem of inertial motion of a symmetric Euler gyroscope is well known and described in many works, in particular, in \hyperref[b0]{[1]}\hyperref[b1]{[2]}. This motion is regular precession, characterized by a constant angle of nutation between the kinetic moment axis, superimposed with the inertial basis axis, and the axis of SEG proper rotation. At the same time, the angular velocities of precession and nutation are constant.\par
The indicated properties have found application in \hyperref[b3]{[4]} in the process of preparation of an experiment to validate the general theory of relativity using a SEG and a telescope on an artificial Earth satellite when solving the problem of selection of relations between the primary moments of inertia that provide very low angular precession velocities. In the experiment \hyperref[b0]{[1]}, drift angular velocities values were less than 10 -11 angular deg/hr, which validated the Einsteinian general theory of relativity with an error of less than 1\%.\par
It should be noted that the solution to the problem of regular precession was possible with the following restrictions on the initial Euler angles [6, formulas (2.39), (2.41)]:\par
where G is the kinetic moment; r 0 is the SEG proper rotation angular velocity component; C is the primary moment of SEG inertia around the same axis.\par
This paper sets the task of finding the solution to the problem of SEG motion for arbitrary initial angles not only along the precession angle 0 ? , but also along the initial angles of nutation and proper rotation. The Poisson differential kinematic equations are used for this purpose. To clarify the problem formulation, let us cite a statement on this subject from the work  {\ref [6, p. 79}]. The first step in solving the problem is to determine the angular velocities of the body. This is solved analytically regardless of the Euler angles. The second step consists of determining the Euler angles by integrating the kinematic equations due to the angular velocities found in the first step. This long and arduous process is eased by applying the kinetic moment theorem and the method of selection of a coordinate system, one of the axes of which coincides with the kinetic moment vector \hyperref[b4]{[5]}\hyperref[b5]{[6]}, etc. For this article, we chose the way of integration of the matrix differential equations in quaternions, as well as of Poisson equations by means of solving the Cauchy problem with arbitrary initial angles, which is not related to the special selection of a coordinate system, one of the axes of which is directed along the kinetic moment vector of the SEG. 
\section[{II.}]{II.}\par
On the Influence of Initial Conditions for Kinematic Equations on the Nature of Motions in a Symmetric Euler Gyroscope\par
In this section, we set the task to clarify the range of values of the initial Euler angles for the kinematic equations of the symmetric Euler gyroscope, with which they are reduced to identities -after substituting their analytical solutions given in \hyperref[b6]{[7]}, as well as the solutions of dynamic equations given in \hyperref[b5]{[6]}. Since these solutions describe regular precession, we are talking about the initial conditions under which it is observed, and under which it is not.\par
Dynamic equations for a symmetric Euler gyroscope have the form  {\ref [7, p. 126}]: ? ? ? ? const r r dt dr dt dr C rp C A dt dq A qr A C dt dp A ? ? ? ? ? ? ? ? ? ? 0 ; 0 ; 0 0 0 (A.1)\par
The kinematic Euler equations  {\ref [7, p. 115}]: ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? cos sin cos sin cos sin sin r q p (A.2)\par
The solutions of these equations obtained in  {\ref [7, p. 37} ]: ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 0 0 1 0 1 0 0 0 cos ; ; cos ; ? ? ? ? ? ? ? ? n r n t n const G Cr G Cr A G n nt (A.3) G Cr const const 0 0 0 0 cos ; 0 ; ? ? ? ? ? ? ? , Where G r , 0 are the constants, ? ? 2 0 2 2 2 2 2 2 r C q A p A G ? ? ? . Equations (A.2), resolved in relation to ? ? ? ? ? ? , , [ [6, p. 46]: ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? C A G A n n H G C A G r r t q t p ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (A.5)\par
Substitution of solutions into equations (A.4) Substituting (A.5) into (A.4), we obtain, in consideration of (A.3) for the third equation in (A.4):(A.6)\par
The equality (A.6) is reduced to an identity when0 0 ? ? ; ? ? ,... 3 , 2 , 1 2 ? ? m m ? . (A.7)\par
That is, the angle 0 ? should be zero. For the equation (A.2) of the system (A.4) we have:? ? ? ? ? ? 0 0 10 sin cos cos sin 0 ? ? ? ? ? ? ? ? ? ? ? ? ? t t t t ? ? ? ? ? ? ? ? 0 10 0 10 sin sin 0 ? ? ? ? ? ? ? ? ? ? ? t t ? ? (A.8)\par
The equality (A.8) is reduced to an identity at the angles 0? = 0; ? ? ,... 3 , 2 , 1 ? ? m m ? .\par
For the angle nt ? 0 ? from the first equation in (A.4) we have:? ? A G q p n ? ? ? ? ? ? sin cos sin (A.9)\par
In consideration of (A.3) and (A.5) we obtain: That is, we obtain the relations (A.7) once again. This means that the equation (A.9) is reduced to an identity with 0 ? ? ? ; t n 1 ? ? for any value of 0 ? . From (A.9) it also follows that when? ? 0 10 0 10 0 cos cos sin ? ? ? ? ? ? ? ? ? ? ? t t A G ? ? 0 10 10 0 10 10 cos cos ? ? ? ? ? ? ? ? ? ? G A A G © 2020 Global Journals? ? 0 ? ? of the initial value of the angle 0 ? is varied, i.e. for ? ? 0 0 ? ? ? ? ? ?\par
, the equality (A.9) is not reduced to an identity. From these calculations, we conclude that regular precession in a symmetric Euler gyroscope is possible only with the following values of the initial angles:G Cr const const 0 0 0 0 cos ; 0 ; ? ? ? ? ? ? ? (A.10)\par
With any other initial values of the Euler angles, the equations (A.4) are reduced to identities with other solutions that do not coincide with the functions (A.3).\par
The relevance of the article is further reinforced by publications \hyperref[b3]{[4]}\hyperref[b4]{[5]}\hyperref[b5]{[6]}\hyperref[b6]{[7]}.  
\section[{III.}]{III.} 
\section[{Problem Solution}]{Problem Solution}\par
In this article, we use the method of integration of quaternion and Poisson matrices that are non-degenerate for any angle value of the equation:\par
(2)\par
The choice of the two types of equations is related to their widespread use in science and technology, it also enables comparison of their solutions. The coefficients and variables included in the differential equations ( \hyperref[formula_31]{2}) are indicated below.\par
Following \hyperref[b5]{[6]}, we present the Euler rotation angles diagram depicting the inertialess frames of the cardan suspension according to Fig. \hyperref[fig_10]{1}. Let us associate the moving coordinate system Oxyz (corresponds to the coordinate system O1'2'3' in \hyperref[b5]{[6]}) with the gyroscope body, and also introduce inertial coordinate systems: the expanded O???, system, which coincides with the coordinate system Oxyz at the initial moment, and the original system ?? ? ? ? ? ? , relative to which the coordinate system ???? is rotated at the initial angles ? ? ,? ? ,? ? . Figure  {\ref 2} shows a similarly constructed diagram of the same gyroscope, but for the Euler -Krylov angles (?, ?, ?).? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? cos sin sin cos sin cos sin q p ctg r dt d q p dt d q p dt d ? ? ? ? . 0 ; 2 1 1 1 E N N t P dt dN ? ? ? ? ? ? ; ; 1 1 1 E t A A t P dt dA ? ? Fig.1\par
Fig.  {\ref 2} The following scheme \hyperref[b11]{[12]} corresponds to the rotation diagram of the introduced systems according to Fig. \hyperref[fig_10]{1}:xyz ? N ? N ? ? ? 1 1 1 1 1 0 0 0 0 0 , , ? ? ? ? ? ? ??? ? ? ? \textasciitilde xyz ? N ? ? ? ??? , ? ? ? , 0 1 N N N ? (3)\par
where ? ? , ? ? , ? ? , N ? =N(0) are the initial angles of SEGrotation and the corresponding quaternion matrix \hyperref[b9]{[10,}\hyperref[b10]{11]}; ? 1 , ? 1 , ? 1 , N 1 are the rotation angles corresponding to the matriciant N 1 , when N ? =? (? is the identity matrix); ?, ?, ?, N are the angles of the resulting rotation and the corresponding quaternion matrix of the resulting rotation.\par
Following the technique described in \hyperref[b7]{[8,}\hyperref[b8]{9]} for matrices of directional cosines, we find the analytical solution for the quaternion matrix N 1 based on kinematic equations. Note that the quaternion matrices are related to the matrices of directional cosines of the angles by the relation A=M T N \hyperref[b9]{[10,}\hyperref[b10]{11]}. In the article \hyperref[b7]{[8]}, the formulas for the angular velocities p, q, r of the gyroscope are solutions of the dynamic equations of the SEG, which had the initial angular velocity p(0)=q(0)=0; r(0)=R, and which was affected by the impact to the axis of the gyroscope figure in the form of a rotational pulse ? ? around the axis Ox (hereinafter, ? ? =? ? is the kinetic moment from the impact). The dynamic Euler equations for a gyroscope with a dynamic axis of symmetry have the following form \hyperref[b7]{[8]}:? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? A A C r dt dr p dt dq t I dt d A M q dt dp ; 0 0 0 (4)\par
p, q, r are the components of the vector of angular velocity of rotation of the gyroscope in the axes associated with it; I (t) is the unit function. For initial conditions t=0; p(0)=0; q(0)=0; r(0)=R The solution to the system of differential equations (2) has the following form:. ; ; sin ; cos 0 A H A M a R r t a q t a p x ? ? ? ? ? ? ? (5)\par
The transformation of coordinate systems from theinertial O??? to the moving Oxyz, in consideration of the initial inertial coordinate system ?? ? ? ? ? ? , according to (3) is determined by the relations:? ? ? ?? ? ? ? T ? ? ? T ? ? ? T A A A xyz ? ? ? ? ? ? ? ? 0 1\par
or, equivalently, through quaternion matrices \hyperref[b9]{[10]}, \hyperref[b10]{[11]}:? ? ? ? ? ?? ? ; 0 0 1 1 ? ? ? ? T T T N M M N xyz ? ? ? ? (6) ? ? ? ? ? ? ? ?, 0 0 0 ; ; 0 ; 1 1 1 1 1 N M A N M A N N N N N N N T T ? ? ? ? ? ? ?\par
where N, A are the quaternion matrix and the matrix of directional cosines of the resulting rotation; N 1 , A 1 are the matriciants; N ? , N ? , N ? are the quaternion matrices of the corresponding simplest rotations. At the same time, M and N are the corresponding types of quaternion matrices \hyperref[b9]{[10,}\hyperref[b10]{11]}.\par
The matrix of directional cosines of the Euler angles for Fig. \hyperref[fig_10]{1}  1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? A (7)\par
The matrix of directional cosines of the Euler-Krylov angles (Fig.  {\ref 2}), which is equal to the matrix \hyperref[b6]{(7)} ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? k A (8)\par
The matrix N 1 corresponding to N 1 (0)=E, i.e. to theangles ?(0)=?(0)=?(0)=0 (that is, the matriciant), can be determined by integrating the quaternion matrix equation \hyperref[b9]{[10,}\hyperref[b10]{11]}:  \hyperref[b9]{(10)} Which means that it satisfies the condition P(t)=P(t+?); ?=2?/?. Therefore, the system (  {\ref 9}) is Lyapunov reducible \hyperref[b7]{[8]}. By means of substitution? ? ? ? . 0 ; 2 1 1 1 E N N t P dt dN ? ? (9) 12 Year 2020 Global Journal of Researches in Engineering ( ) Volume Xx X Issue I V ersion I D ? ? . ; 0 0 0 0 0 0 0 0 1 0 1 1 1 2 1 3 1 1 1 0 1 3 1 2 1 2 1 3 1 0 1 1 1 3 1 2 1 1 1 0 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? N p? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? t a t a R t a R t a t a R t a R t a t a t P1 N N N Z ? ? (11)\par
The system (  {\ref 9}), (  {\ref 10}) is reduced to an equivalent differential equation with constant coefficients .Z B Z N P dt dN ?\textbf{(12)}\par
. 0 00 0 0 0 0 0 ; 1 1 1 1 0 1 2 3 1 0 3 2 2 3 0 1 3 2 1 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? a R a R R a R a P N B ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (13) . 2 sin ; 0 ; 2 cos ; 3 2 1 0 1 t t A C R R ? ? ? ? ? ? ? ? ? ? ?\par
Given these formulas, we have:\par
.2 cos 0 0 2 sin 0 2 cos 2 sin 0 0 2 sin 2 cos 0 2 sin 0 0 2 cos ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? t t t t t t t t N (14)\par
The equivalence of equations (  {\ref 9}) and ( \hyperref[formula_24]{12}), (  {\ref 13}) is confirmed by the fulfillment of the identity? ? B P N PN N ? ? ? ? ? ? ? 1 1 ? (15)\par
The solution to the equation ( \hyperref[formula_24]{12}) with constant coefficients is the Cauchy formula:? ? ? ? ? ?, 0 0 1 Z Z N L t L N ? ? (16)\par
where L(t) is the fundamental matrix of solutions; N Z (0) is the matrix of initial values of the angles, equal, by condition, to the identity matrix: N Z (0)=E.\par
After finding the fundamental matrix of solutions and a number of transformations, let us write down the expression (16) in the form:? ? ? ? ? ? ? ? . ; ; ; ; 0 2 sin 2 cos 2 / 1 2 1 2 0 1 3 0 1 nt n R a d R d a N D E N t t Z Z ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (17)\par
After transformations, the matriciant takes the form:? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 cos 2 sin 0 2 sin 2 sin 2 cos 2 sin 0 0 2 sin 2 cos 2 sin 2 sin 0 2 sin 2 cos 1 1 1 1 ? ? ? ? ? ? ? ? ? ? ? ? n a n R n a n R n R n a n R n a N Z (18)\par
From the expression \hyperref[b10]{(11)} we have:? ? ? ? 0 0 ; 1 1 N N N N N N N N N Z T Z T ? ? ? ? ? ; 1 N = ) ( 1 ?? N ? (k=0,1\textbf{,2,3). (19)}\par
In consideration of (  {\ref 13}), (  {\ref 14}) and (  {\ref 18}), the expanded expression for the quaternion matriciant N 1 is derived below.\par
Since? ? ? ? 0 0 1 N N N N N N Z T ? ? ?\par
, we have the following expression for the quaternion matrix of the resulting rotation N for nonzero initial conditions:\par
. 03 02 01 000 1 2 3 1 0 3 2 2 3 0 1 3 2 1 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? n n n n n n n n n n n n n n n n n n n n N a a a a a a a a a a a a a a a a\par
Formulas for the components of the quaternion matrix N: ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 03 0? ? 3 , 0 ? ? i n ai i a ? ,\textbf{(21)}\par
We have the explicit form of the formulas for the components of the quaternion matriciant N 1 : ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ,\textbf{(22)}\par
For regular precession, the angles of the initial orientation and the components of the initial quaternion are expressed by the formulas: ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? n n n n H H tg x ? ? ? ? (23)\par
In this regard, we have:  ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?0 1 0 0 2 0 1 0 0 1 0 0 1 0 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? nt t n a nt t n R nt t n nt t n R nt t nt t n a n nt t n R nt t nt t n a n nt t n a nt t n R nt t n (24)\par
After that, let us similarly determine the trigonometric functions for the Euler-Krylov angles ?, ?, on the basis of the matrix (8) and its quaternion counterpart \hyperref[b7]{[8,}\hyperref[b8]{9]}. We have: \par
These expressions coincide with formulas (18) \hyperref[b7]{[8]}, confirming the fidelity of the solutions to the problem for zero initial Euler-Krylov angles both in the quaternion form and in the form associated with the application of the Poisson differential kinematic equations.\par
For arbitrary initial Euler-Krylov angles, explicit solutions can be obtained from relations (24), (25) (in (25), the ? i must be replaced by values? ? 3 , 0 ? i n i ).\par
In turn, for the Euler angles we have the following solutions:(27)\par
In consideration of (22), we obtain the solutions in explicit form:= 2 2 ? 2 2 + 2 2 ? 2 2 2 2 + 2 2 2 2 2 2 + 2 2 + 2 2 ? 2 2 + 2 2 = 2 2 + 2 2 2 2 2 + 2 2 + 2 2 ? ?2 2 2 + 2 2 ? 2 2 ? 2 2 = (\textbf{28})\par
For regular precession in (24), (25), it is necessary to consider ? ? ?? 2 ?????? ????+ ?? 2 ?? 2 ?+ ?? 1 ?? ?????? ???? ?????? ???? ?????? ????? ?? 1 2 ?? 2 ?????? ????+ ?? 2 ?? 2 ?+ ?? 1 ?? ?????? ???? ?????? ???? ? = ? 0 + ????; tan? = ???? ?(????) 2 + ???? 1 2 = ?? ?? ? ; ??????? = ?? 32 ?? 31 = ?? 2 ?? 3 ? ?? 0 ?? 1 ?? 0 ?? 2 + ?? 1 ?? 3 ?????? ? = ?? 33 = ?? 0 2 + ?? 3 2 ? ?? 1 2 ? ?? 2 2 ??????? = ? ?? 23 ?? 13 = ?? 0 ?? 1 +?? 2 ?? 3 ?? 1 ?? 3 +?? 0 ?? 2 ???????? = ? ?? 21 ?? 11 = 2(?? 0 ?? 3 ??? 1 ?? 2 ) ?? 0 2 +?? 1 2 ??? 2 2 ??? 3\textbf{(29)}\par
The result coincided with the classical one, which is expressed by the formulas (A.3, A.5).\par
Let us now consider a variant of the solution to the problem for irregular precession. It corresponds to the initial angles = = 0; = that differ from the angles (23), which generate regular precession, only by the sign of the angle of nutation. After transformations, the formulas for determining the Euler angles for the SEG are:\par
(30)\par
The expressions (30) suggest that only the change of the sign of the initial angle of nutation -with the other two initial angles unchanged -caused the appearance of irregular precession motions in the Euler gyroscope. 
\section[{b) Solution for the Poisson matrix differential equation}]{b) Solution for the Poisson matrix differential equation}\par
The transformation of the coordinate system Oxyz from the initial position O??? is characterized by the formulas:? ? ? ? , ; 1 1 ? ? ? ? ? A A A A A xyz T ??? (31)\par
Where ? ? , ? ? , ? ? are the transformation matrices of the coordinates of the simplest rotations. On the other hand, this matrix can be determined by integrating the Poisson matrix kinematic equation: ? ? ? ? ; ; 1 1 1 E t A A t P dt dA ? ? (32) 
\section[{? ? ?}]{? ? ?}\par
-form (32), and the matrix of directional cosines of the Euler-Krylov angles (Fig.  {\ref 2}) -form (33). The angular velocity tensor for gyroscopes with a dynamic axis of symmetry has the form:? ? 0 cos sin cos 0 sin 0 t a t a t a R t a R t P ? ? ? ? ? ? ? ? , (\textbf{34}) That is, it satisfies the condition ? ? ? ? ? ? ? ? ? ? ? 2 ; t P t P .\par
As a result of this condition, the system (32) -(  {\ref 33}) is Lyapunov reducible \hyperref[b12]{[13]}. Indeed, by substitution ??????? * = ? ?????? ?? ?? 2 ?????? 2 ?? 0 ? ?????? 2 ?? 0 ?????? ?? ?? ?????? ? * = ?????? ?? 0 ?????? 2 ?? 0 ?????? 2 ?? 0 + 2 ?????? 2 ?? 0 ?????? ?? 0 ?????? ?? ?? ??????? * = ?????? ?? 0 ?????? 2?? 0 ?????? ????+?????? 2?? 0 ?????? ???? ?????? ???? ?????? ?? 0 ?????? 2?? 0 ?????? ?????2 ?????? 2?? 0 ?????? ???? ?????? ???? ; 1 0 0 0 cos sin 0 sin cos ) ( ? ? ? ? ? ? t t t t t ?\par
The equivalence of the equations (32) and (  {\ref 36}) is confirmed by the validity of the identity? ? ? t ? t P? t ? ? ? ? ? ? ) ( ) ( ) ( 1 1\par
. The differential linear homogeneous equation (  {\ref 36}) is solved by the Cauchy formula Where Q(t) is the fundamental matrix of solutio provided by the condition, Z(0)=E. After finding the fundamental matrix and performing a number of transformations, the solution (38) takes the form:   ?1 cos 2 2 2 1 1 2 2 1 R a n n aR n R n R Z ? ? ? ? ? From (37) it follows that ? ? ? ?nt n aR t nt n R nt n R t t nt n R nt n R t A ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?; ; 0 0 0 0 0 ; 1 1 1 ? ? ? ? ? ? ? R R Z Z a a R R B ij\par
The equivalence of the equations (32) and (  {\ref 36}) is confirmed by the validity of the identity . The differential linear homogeneous equation (  {\ref 36}) is solved by the Cauchy formula Q(t) is the fundamental matrix of solutions; Z(0) is the matrix of initial values of directional cosines, and as provided by the condition, Z(0)=E. After finding the fundamental matrix and performing a number of transformations,? ? ? . ; ; cos sin cos sin cos sin cos 1 sin cos 1 2 1 2 2 1 2 2 1 2 1 1 2 2 A C R R R n R nt n a nt n a nt nt n a nt nt nt n aR nt n R n a nt ? ? ? ? ? ? ? Z t ? ? ) ( 1\par
, as a result, the solution to the equation (32) for a gyroscope with a dynamic axis of symmetry is the matrix (matriciant):  ? ? ? ? ? 2 1 2 2 2 1 1 2 2 2? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?0 0 0 0 0 0 0 0 0 0 0 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?\par
Formulas for determining the Euler angles:? ? 3 ; 2 ; 1 , ? j i . (37)\par
The equivalence of the equations (32) and (  {\ref 36}) is confirmed by the validity of the identity . The differential linear homogeneous equation (  {\ref 36}) is solved by the Cauchy formula (38) ns; Z(0) is the matrix of initial values of directional cosines, and as provided by the condition, Z(0)=E. After finding the fundamental matrix and performing a number of transformations, (39) , as a result, the solution to the equation (32) for a gyroscope with ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? A (44)\par
In consideration of this we obtain: After calculations we have:? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?\par
(52)\par
The obtained formulas coincide with the formulas of the classical solution, but with zero initial angles of precession and proper rotation. Let us now consider a variant of the solution to the problem for irregular precession.\par
For the initial angles = = 0; = that differ from the angles (45), which generate regular precession, only by the sign of the angle of nutation. After transformations, the formulas for determining the Euler angles for the SEG are:\par
(53)\par
The expressions (53) suggest that only the change of the sign of the initial angle of nutation -with the other two initial angles unchanged -caused the appearance of irregular precession motions in the Euler gyroscope.\par
IV.   
\section[{Mathematical Modeling}]{Mathematical Modeling}? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (M.1)\par
That is, corresponding to the conditions (23) of regular precession in the Euler angles. The relationship between the Euler and the Euler-Krylov angles is established due to the equality of the respective elements of the matrices \hyperref[b6]{(7)} and \hyperref[b7]{(8)}. The graphs in Fig. \hyperref[fig_16]{3} depict the change of the Euler angles for regular precession. The same cannot be said about the graphs in Fig. \hyperref[fig_17]{4} for the Euler-Krylov angles -where one can see harmonic oscillations for the angles ? and ? with a frequency slightly higher than 500 Hz, and for the angle ? , its increscent property is evident.  When applying a stronger rotational pulse around the axis Ox for which = 4000, unchanged other conditions for Fig. \hyperref[fig_20]{3 and 4}, the nature Fig. \hyperref[fig_17]{4} When applying a stronger rotational pulse / > , with unchanged other conditions for Fig. \hyperref[fig_20]{3 and 4}, the nature of the motion does not change (therefore, the graphs are not shown), however, for the Euler angles we Additionally, with unchanged parameters of modeling of SEG motions according to (M.1), (M.2) (figures 3 and 4), but with the sign of the initial angles of nutation reversed and equal to = motion patterns shown in figures 5 and 6 were obtained.   In Fig. \hyperref[fig_24]{5}, for the Euler angles, the motion has acquired the character of irregular precession, namely, along ? a vibrational pattern with frequencies slightly above 500 Hz of different amplitudes with oscillation centers shifted by about 0.3 rad. For the angle ? , the velocity sign in Fig. \hyperref[fig_16]{3} has changed to the opposite, and the angle become increscent. The graphs confirm the derived formulas (30).\par
For the Euler -Krylov angles, the motion is of a qualitatively similar character.     At the same time, the motion for the Euler Krylov angles has changed dramatically (Fig. \hyperref[fig_26]{8}).\par
angle ? began to increase monotonically in the up to 0.45 rad, and the oscillation increased 820 Hz. The angle ? remains to be in crescent with superimposed oscillations.\par
At the same time, the motion for the Euler-Krylov angles has changed dramatically (Fig. \hyperref[fig_26]{8}). The began to increase monotonically in the direction of the rotational pulse action, which is novel.\par
The angle ? is still oscillatory in nature with a frequency of 820 Hz around the shifted center of oscillations, and the angle ? has changed the sign to the opposite in relation to Fig.  {\ref 6}.\par
V. 
\section[{Conclusion}]{Conclusion}\par
According to the results of mathematical modeling, it is shown that the motions that correspond to regular precession in the Euler angles are independent of the magnitude of the angular velocity a , which is caused by the action of the rotational pulse. However, a change of the sign of the initial angle of nutation leads to a sharp change in the nature of motion -it becomes irregular, which is reflected in the explanation for Fig. \hyperref[fig_24]{5}. The motion along the Euler-Krylov angles radically depends on a : with R a ? , the angle ? becomes monotonically increscent in the direction of the pulse action, and the angle of proper rotation changes the sign of its monotonic rotation to the opposite. Additionally, in the article: As for corpuscular gyroscopes, based on this study, it can be assumed that depending on the application of an external magnetic field over time, not only Larmor precession \hyperref[b13]{[14]}, but also "pseudo-Larmor" precession is possible in them.\begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-2.png}
\caption{\label{fig_0}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-3.png}
\caption{\label{fig_1}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-4.png}
\caption{\label{fig_2}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-5.png}
\caption{\label{fig_7}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-6.png}
\caption{\label{fig_8}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-7.png}
\caption{\label{fig_9}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{1}\includegraphics[]{image-8.png}
\caption{\label{fig_10}1 a}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-9.png}
\caption{\label{fig_11}For}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-10.png}
\caption{\label{fig_12};}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{12122}\includegraphics[]{image-11.png}
\caption{\label{fig_15}1 ?? 2 ( 1 ? 2 ?? 2}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{3}\includegraphics[]{image-12.png}
\caption{\label{fig_16}Figures 3 -}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{4}\includegraphics[]{image-13.png}
\caption{\label{fig_17}Figures 3 and 4}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-14.png}
\caption{\label{fig_19}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{3}\includegraphics[]{image-15.png}
\caption{\label{fig_20}Fig. 3 ©}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-16.png}
\caption{\label{fig_21}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-17.png}
\caption{\label{fig_22}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-18.png}
\caption{\label{fig_23}?}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{5}\includegraphics[]{image-19.png}
\caption{\label{fig_24}Fig. 5 velocity}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{7}\includegraphics[]{image-20.png}
\caption{\label{fig_25}Fig. 7}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{8}\includegraphics[]{image-21.png}
\caption{\label{fig_26}Figures 7 and 8}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{8}\includegraphics[]{image-22.png}
\caption{\label{fig_27}Fig. 8}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{7}\includegraphics[]{image-23.png}
\caption{\label{fig_28}Figures 7}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-24.png}
\caption{\label{fig_29}?}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-25.png}
\caption{\label{figure25}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-26.png}
\caption{\label{figure26}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{} \par 
\begin{longtable}{P{0.058333333333333334\textwidth}P{0.008333333333333333\textwidth}P{0.03333333333333333\textwidth}P{0.03333333333333333\textwidth}P{0.008333333333333333\textwidth}P{0.008333333333333333\textwidth}P{0.03333333333333333\textwidth}P{0.03333333333333333\textwidth}P{0.075\textwidth}P{0.075\textwidth}P{0.049999999999999996\textwidth}P{0.049999999999999996\textwidth}P{0.049999999999999996\textwidth}P{0.075\textwidth}P{0.075\textwidth}P{0.075\textwidth}P{0.049999999999999996\textwidth}P{0.049999999999999996\textwidth}P{0.008333333333333333\textwidth}}
??? and\tabcellsep ?\tabcellsep ?\tabcellsep ?\tabcellsep ?\tabcellsep ?\tabcellsep ?\tabcellsep :\tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \\
\tabcellsep \tabcellsep \multicolumn{3}{l}{cos}\tabcellsep \tabcellsep \multicolumn{2}{l}{cos}\tabcellsep cos\tabcellsep sin\tabcellsep sin\tabcellsep sin\tabcellsep cos\tabcellsep cos\tabcellsep cos\tabcellsep sin\tabcellsep sin\tabcellsep cos\\
\tabcellsep \tabcellsep \tabcellsep \multicolumn{3}{l}{cos}\tabcellsep \tabcellsep cos\tabcellsep sin\tabcellsep sin\tabcellsep cos\tabcellsep sin\tabcellsep cos\tabcellsep sin\tabcellsep cos\tabcellsep cos\tabcellsep sin\tabcellsep sin\tabcellsep .\\
\tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep \tabcellsep cos\tabcellsep sin\tabcellsep \tabcellsep \tabcellsep \tabcellsep sin\tabcellsep sin\tabcellsep cos\end{longtable} \par
 
\caption{\label{tab_1}}\end{figure}
 \begin{figure}[htbp]
\noindent\textbf{} \par 
\begin{longtable}{P{0.08693181818181818\textwidth}P{0.057954545454545446\textwidth}P{0.057954545454545446\textwidth}P{0.057954545454545446\textwidth}P{0.08693181818181818\textwidth}P{0.08693181818181818\textwidth}P{0.057954545454545446\textwidth}P{0.057954545454545446\textwidth}P{0.08693181818181818\textwidth}P{0.057954545454545446\textwidth}P{0.028977272727272723\textwidth}P{0.057954545454545446\textwidth}P{0.057954545454545446\textwidth}P{0.009659090909090909\textwidth}}
cos\tabcellsep cos\tabcellsep sin\tabcellsep sin\tabcellsep cos\tabcellsep cos\tabcellsep sin\tabcellsep cos\tabcellsep sin\tabcellsep \multicolumn{2}{l}{cos}\tabcellsep sin\tabcellsep sin\\
sin\tabcellsep cos\tabcellsep sin\tabcellsep sin\tabcellsep sin\tabcellsep cos\tabcellsep cos\tabcellsep cos\tabcellsep sin\tabcellsep sin\tabcellsep \tabcellsep sin\tabcellsep cos\tabcellsep .\\
\multicolumn{2}{l}{sin}\tabcellsep \tabcellsep \tabcellsep sin\tabcellsep cos\tabcellsep \tabcellsep \tabcellsep \multicolumn{2}{l}{cos}\tabcellsep cos\tabcellsep \tabcellsep \end{longtable} \par
 
\caption{\label{tab_2}}\end{figure}
 			\footnote{© 2020 Global Journals} 			\footnote{© 2020 Global Journals About the Presence of Irregular Precession Motions in a Symmetric Euler Gyroscope} 		 		\backmatter  			 \par
Let us now apply the obtained formulas to the case of regular precession. We use the initial values = = 0; = ? in the matrix ? ? associated with this type of precession			 			  				\begin{bibitemlist}{1}
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\end{document}
