Play top

Fig. 1: Play top

The Spielkreisel is centrifugal theory a gyroscope which the gravitational field of the earth at any point held on a flat, level surface and this touching dances only one point, see picture. Of particular importance are the symmetrical gyroscopes shown , in which the center of mass lies on the figure axis .

Such gyroscopes can rotate around the vertical figure axis and perform regular precessions for a relatively long time, provided they are stable. Between the stable layers, gyroscopes aim at one of them in an unsteady sequence. With the fast gyroscope , the hanging vertical position is unstable, the upright stable and a regular precession between them is not possible. That is why this top stands up. The driving force is the friction at the point of contact. The description of the movements, existence and stability of their special forms, as well as the friction effects influencing them are the subject of the theory of the game top.

There are many, above all, mathematical treatises on the sometimes unexpected movements of gaming tops, most of which can hardly claim any technical significance. The interest shown in the top is not the last

"Because of the peculiar stimulus it exerts on our drive for knowledge."

- Richard Grammel (1920) :

history

The scientific treatment of tops began with JA Segner (1704 - 1777), who also correctly recognized the importance of friction for the erection of the top axis. Leonhard Euler published his gyro equations in 1750 and was also able to give solutions for the force-free Euler gyro . In 1788 Joseph-Louis Lagrange made an important contribution by solving the equations for the heavy symmetrical top with a base , which is closely related to the game top . Other special cases of peculiar movements on a plane of rolling bodies have been investigated by Jean-Baptiste le Rond d'Alembert (1761), Leonhard Euler (1765) and Siméon Denis Poisson (1811).

The first known attempts to explain the rising of the rate gyro were made by W. Thompson and Prof. Blackburn in the 19th century, in which JH Jellett (1872) also discovered the integral named after him . At the beginning of the 1950s, CM Braams, NM Hugenholtz and WM Plisken recognized that it is the frictional force at the point of contact that makes the spinning top stand up. In 1977, RJ Cohen formulated the now accepted model of the standing up top and was able to numerically simulate the standing up of the top . Between 1994 and 2005 the kinetics of the game top was further elucidated.

description

Realizations and mechanical model

Fig. 2: Gyro with radius of curvature ρ and center Z, center of mass G, contact point T, axial vector ê ₃ and vertical ê _z

Play tops come in many forms. A usually smooth surface and rotational symmetry as in Fig. 1 is not only aesthetically pleasing and easy to manufacture, but also keeps air resistance low. Hard and smooth materials for the top and the ground ensure low drilling, rolling and sliding friction .

The more blunt the lower tip of the top, the easier it is for the top to stand up. This effect is slightly reversed with a very pointed lower end of the top, so that the same cause lowers the top, see Fig. 4 . The more pointed the top is at the point of contact, the less the friction that occurs there slows down, so that the top dances longer. The actual shape forms a compromise between these conflicting aspects.

The turning gyro should turn from a "hanging" position, where the center of mass is below the center of curvature of the lower tip of the top (3 / 6a in Fig. 3), into an upright position, where the center of mass is above the center of curvature (3 / 6b in Fig. 3 ). Usual versions have a pin on which the gyro stands. When straightening up, it passes a critical point where the pin and the spherical part of the top touch the surface at the same time.

The inhomogeneous sphere with an eccentric center of mass as in Fig. 2 is a tried and tested model of a top that does not have a critical point. The center of curvature of the top of the top is then the geometric center of the sphere. With the hanging top, the center of mass is below the geometric center and with the upright top it is above.

Classification according to stability properties

Fig. 3: Stability diagram of game tops with a vertical figure axis

The stability of the game top with a vertical axis of the figure can be investigated using E. Routh's method of small oscillations and the result can be shown graphically in a stability diagram, see Fig. 3. The ratio of the main axial moment of inertia C to the main equatorial moments of inertia A is plotted on the abscissa . For a real symmetrical top, this ratio is in the range 0 <C / A <2. The height ratio is h = | GR | on the ordinate of the center of mass G above its base point R on the base and the radius of curvature ρ = | ZT | removed from the lower tip of the top at the contact point, see Fig. 2 . When the top is upright, h / ρ> 1, because with the distance s = | GZ | between the center of mass and the center of curvature is with a vertical figure axis h / ρ = (ρ + s) / ρ = 1 + α with dimensionless

α: = s / ρ.

In a vertically hanging top, the center of mass is below the center of curvature, h / ρ = (ρ - s) / ρ = 1 - α, and correspondingly h / ρ <1. The diagram is represented by the straight line h / ρ = 1 and the hyperbola

h / ρ = γ: = A / C

divided into four differently colored areas.

(a) γ ≥ 1 + α

In this area ( pink ) 1 <h / ρ ≤ A / C = γ, with the fast gyro regular precessions are only possible and stable with angles of inclination ϑ that satisfy cos (ϑ) <  , and rotations around vertical axes are always unstable . ${\displaystyle {\tfrac {\alpha }{\gamma -1}}}$

(b) γ ≤ 1 - α

In this area (olive green) γ = A / C ≤ h / ρ <1, with the fast gyroscope regular precessions in the cos (ϑ)> area are possible but unstable, and the rotations around vertical axes are always stable. This case is sometimes called indifferent , because the high-speed top can be accelerated as desired so that it strives for the upright or the hanging vertical position. ${\displaystyle {\tfrac {\alpha }{\gamma -1}}}$

(c) 1 - α <γ <1 + α

In the remaining areas (blue-gray and orange) a regular precession is impossible with the fast gyro, the vertically hanging gyro is only stable when turning slowly and the upright only when turning quickly. Rate gyros have this format.

The geometries 3a / b and 6a / b from area (c) are modeled on the gyro. If the top is set in rapid rotation in a vertically hanging position 3 / 6a, the top straightens up because its position is unstable. The upright positions 3 / 6b, on the other hand, are stable on the fast top. If the momentum is sufficient, the top can get over the critical moment when the spherical part and the pin touch the surface at the same time, it will reach the upright position. The top 6a cannot overcome the critical moment and is therefore unsuitable as a standing top top. The slowing down gyro 3b leaves the upright position as soon as this becomes unstable and strives for the now stable hanging position 3a.

With the not so fast gyroscope, the # stability of the stationary movements is not only of α and γ, but also of the momentum, expressed by #Jellett's integral J (the projection of the angular momentum onto the axis from the center of mass G to the contact point T in Fig. 2 ) and via the gyro's own parameter J _β  = ρ√ (Csmg) depending on its weight mg. The existence and stability of stationary motions is thus α only by the three properties, γ and J _β of the gyroscope and an initial condition J set.

Comparison with the Lagrange top

The game top has many parallels to the Lagrange top . Differences result from the fact that the Lagrange top rotates around the fixed contact point, but the play top rotates around the free center of mass . With the game top, the moments of inertia with respect to the center of mass are decisive, which are smaller than with a structurally identical Lagrange top.

In a vertical position, the game top is stable even at a lower speed and changes its inclination under the influence of friction significantly faster than a comparable Lagrange top. Like the Lagrange top, the play top can perform a regular precession with its own characteristics. Also pseudo regular Precessions are possible if the centrifugal rapidly rotates and the angular momentum is aligned close to the figure axis. The pseudoregular precessions are equally fast on both gyroscopes. However, both the center of mass and the point of contact with the game top fluctuate slightly, which is why we can no longer speak of a nutation cone. The orbit of the lower tip of the top around the axis of angular momentum also has a vertical component on the inclined top, with which the top pushes itself off the ground. If there is enough momentum, the top can take off and knock on the ground.

As with the Lagrange top, Gaston Darboux's theorem on homologous tops applies to the friction-free game top . In both gyroscopes, the angular momentum L _z in the vertical and L ₃ in the axial direction and the total energy contained in the parameter k are constant. The gyro function of the game top is as long as it touches the surface, with the names of the Lagrange top:

${\displaystyle {\dot {u}}^{2}=U(u)={\frac {(k-L_{3}^{2}-2Ac_{0}u)(1-u^{2})-(L_{z}-L_{3}u)^{2}}{A^{2}{\color {blue}\left(1+{\frac {ms^{2}}{A}}(1-u^{2})\right)}}}}$

The top function of the Lagrange top (black) has an additional divisor (blue) on the top. The three Euler angles of rotation can be expressed with this gyro function with the same integrals as with the Lagrange gyro. For graphical representation of the movement its lower tip is used in Spielkreisel whose line in all respects to the Locuskurven the Lagrange's gyro corresponds. The analytically determined gyroscopic movements, however, differ significantly from the real ones due to the friction at the contact point.

Derivation of the gyro function
kinematics
If there is no friction, a spherical lower tip of the top has no influence and can and should be assumed to have an ideally pointed shape. The center of mass lies at a distance s from the point of contact on the figure axis, which has the inclination ϑ in relation to the vertical pointing upwards. The center of mass is at the top z = s cos (ϑ) = see below above the ground, a kinematic assumption that binds the top to the ground. Here u  =  cos (ϑ). The time derivative of this yields with sin² (ϑ) = 1 -  u ²: ${\displaystyle {\dot {u}}=-\sin(\vartheta ){\dot {\vartheta }}\quad \rightarrow \quad {\dot {\vartheta }}^{2}={\frac {{\dot {u}}^{2}}{1-u^{2}}}}$ The acceleration of the center of mass is the second time derivative . ${\displaystyle {\ddot {z}}=s{\ddot {u}}}$
Impulse rate
According to the law of impulses it results in the perpendicular direction ${\displaystyle N-mg=m{\ddot {z}}=ms{\ddot {u}}\rightarrow \;N=mg+ms{\ddot {u}}}$ where N is the normal force acting vertically upwards at the point of contact , m is the mass, g is the acceleration due to gravity and therefore mg is the weight force . The top does not take off as long as N> 0, and with N <0 the ground would pull the top down unrealistic. A solution found through time integration loses its validity if N <0 and must be checked for this.
Decomposition of the angular momentum
For all symmetrical gyroscopes , with the equatorial principal moment of inertia A and the axial C, the illustration applies ${\displaystyle {\vec {\omega }}={\frac {1}{A}}{\vec {L}}+\left({\frac {1}{C}}-{\frac {1}{A}}\right)L_{3}\,{\hat {e}}_{3}}$ the angular velocity due to the angular momentum and its component in the axial direction ê 3 . The node axis ${\displaystyle {\vec {\omega }}}$ ${\displaystyle {\vec {L}}}$ ${\displaystyle L_{3}:={\vec {L}}\cdot {\hat {e}}_{3}}$ sin (ϑ) ê N = ê z × ê 3 represents the direction of rotation of the angle of inclination ϑ of the figure axis, from which the angular momentum L N results in the direction of the node: ${\displaystyle {\dot {\vartheta }}={\vec {\omega }}\cdot {\hat {e}}_{N}={\frac {1}{A}}{\vec {L}}\cdot {\hat {e}}_{N}\;\rightarrow \;L_{N}:={\vec {L}}\cdot {\hat {e}}_{N}=A{\dot {\vartheta }}}$ In the precession plane, which is perpendicular to the node axis and spanned by the perpendicular ê z and the figure axis ê 3 , the vertical angular momentum is set according to ${\displaystyle L_{z}=L_{3}\cos(\vartheta )+L_{A}\sin(\vartheta )=L_{3}u+L_{A}{\sqrt {1-u^{2}}}}$ from the axial angular momentum L 3 and an equatorial L A perpendicular to it .
Swirl rate
With the torque M N  =  s  sin (ϑ)  N , which the supporting force N exerts at a horizontal distance s sin (ϑ) from the center of mass, results from the principle of twist around the center of mass ${\displaystyle {\begin{aligned}\qquad {\vec {M}}=&M_{N}\,{\hat {e}}_{N}=s\sin(\vartheta )N\,{\hat {e}}_{N}={\dot {\vec {L}}}\\\rightarrow {\vec {L}}\cdot {\dot {\vec {L}}}=&{\frac {1}{2}}{\frac {\mathrm {d} }{\mathrm {d} t}}({\vec {L}}\cdot {\vec {L}})=s\sin(\vartheta )NL_{N}=s\sin(\vartheta )(mg+ms{\ddot {u}})(A{\dot {\vartheta }})\\=&-A(mgs+ms^{2}{\ddot {u}}){\dot {u}}=-A{\frac {\mathrm {d} }{\mathrm {d} t}}\left(mgs\,u+{\frac {m}{2}}(s{\dot {u}})^{2}\right)\end{aligned}}}$ and after a time integration ${\displaystyle {\vec {L}}\cdot {\vec {L}}=k-2A\left({\frac {m}{2}}(s{\dot {u}})^{2}+c_{0}u\right)}$ where c 0 = mgs and k is a constant of integration . When the top is lifted, there is no moment and the angular momentum is constant, which is not taken into account at this point, see momentum set above. On the other hand, with the present angular momentum components ${\displaystyle {\vec {L}}\cdot {\vec {L}}=L_{N}^{2}+L_{A}^{2}+L_{3}^{2}=(A{\dot {\vartheta }})^{2}+{\frac {(L_{z}-L_{3}u)^{2}}{1-u^{2}}}+L_{3}^{2}}$ Combination of both equations for the angular momentum square leads with the kinematic relationships to the gyro function in the text. The constant also appears with the Lagrange top, but there the moments of inertia and angular momentum are decisive with regard to the contact point and the kinetic energy that contributes to the total energy E also has a translational component in the play top , see #Total energy of the play top . ${\displaystyle k=2AE+\left(1-{\tfrac {A}{C}}\right)L_{3}^{2}}$

Frictional play top

Erecting the turning top, even from a hanging position, is based on friction effects, which are therefore responsible for the essential forms of movement of the play top. With exclusive sliding friction, which is assumed in the gyroscopic theory, there is a conservation quantity with #Jellett's integral that is determined by the initial conditions . They can be used to read how the top will behave in the future. The gyroscope can hang or rotate upright in its vertical positions and assume the regular precession permanently. In between the movement is unsteady and the top approaches a stable one of these constant rotations and remains close to it. However, if they are not set from the beginning, stationary movements are only achieved asymptotically because of the speed dependence of the friction force .

Speed ​​dependence of the frictional force

The lower tip of the gyroscope generally moves with considerable slip on the ground . The sliding friction that occurs creates a frictional force R which is opposed to the speed v of the rising top. The frictional force is proportional to the normal force N and the coefficient of sliding friction μ, which is essentially determined by the material and the roughness of the lower tip of the top and the ground. Is at Coulomb friction law . In the case of the top of the game, the precession around the vertical superimposes a drilling friction, which changes the law of friction significantly: ${\displaystyle R=\mu N}$

${\displaystyle R=\mu Nv}$.

The frictional force is proportional to the sliding speed v. This viscous law of friction can be experienced in everyday life with a floor polishing machine that works with rotating brushes. When switched off, the machine is more difficult to move because of Coulomb friction than with viscous friction on rotating brushes.

In theory, the qualitative influence of friction is independent of the law of friction as long as the friction force as a function of the speed v is constant at the point v = 0 and disappears.

Influence of friction

The top cannot set up without friction for the following reasons. In the frictionless case, a translation of the center of mass has no influence and can be neglected. The center of mass only moves along the vertical and the total energy , consisting of positional energy and kinetic energy , is constant. In the vertical position, the kinetic energy is equal to the rotational energy , which decreases between the hanging and the upright position, because the top gains position energy. In a vertical position, the rotational energy is proportional to the square of the vertical angular momentum, which must therefore decrease when straightening up. According to the principle of twist , this requires a vertical torque that can only be applied by a horizontal force, and since friction is the only conceivable force that acts horizontally, it is indispensable for straightening.

The dissipative character of the friction becomes more and more apparent when the top is slowing down:

1. With a pseudoregular precession, the tremors of the figure axis disappear over time.
2. Depending on the initial condition and the friction situation at the lower tip of the gyro, the gyro lowers or straightens up and strives for a stable stationary movement in a vertical or hanging position or in the form of a stable, regular precession in between.
3. In such a position of equilibrium, only the drilling, rolling and air friction are effective . If these are low, which is usually the case and is assumed here in theory, the top will remain in the assumed state for a relatively long time.
4. But it is only a matter of time before the slowing top descends and finally hangs vertically or hits the surface on the way there.
5. Finally, the top is in a rest position.

Only in the case of smooth unrolling or rotation with a vertical, stationary axis of the figure, there is no frictional force and the top can theoretically dance permanently, see #Dissipation of energy through friction . When the top slides, however, this creates a frictional force that is opposed to the sliding speed, which results from its own rotation around the figure axis and its precession around the vertical.

The frictional force acts in the support plane in two ways. Firstly, according to the law “ force equals mass times acceleration ”, it accelerates the center of mass also perpendicular to the figure axis, which is already precessing around the vertical due to the torque of the weight force. The center of mass in the floor plan is currently directed into a circular path, see Fig. 4.

In addition, the frictional force creates a moment of friction, the component of which in the plane of precession, perpendicular to the figure axis, raises or lowers the top. Only that part of the frictional force which is perpendicular to the figure axis and which is essentially determined by the circumferential speed v _{t of} the lowest gyro point T contributes to this moment .

Fig. 4: Play top (yellow) with decreasing radius of curvature ρ of the lower top of the top in parts a, b and c.

Fig. 4 shows the outline of the movement of three tops with different radii of curvature of the lower tip of the top. If nothing else is changed, the circumferential speed v _{t of} the standing point T decreases with the radii . The gravity-driven precession speed and its contribution v _P to the resulting speed v of T remain almost unaffected. The frictional force R is opposed to the resulting speed v, which is oriented reversely in Fig. 4a and c and is zero in 4b. In Fig. 4a the frictional torque erects the top and in Fig. 4c the top is lowered by it.

Lagrange function of the game top

The Lagrange function of the top is:

${\displaystyle {\begin{aligned}{\mathcal {L}}=&{\frac {m}{2}}({\dot {x}}^{2}+{\dot {y}}^{2})+{\frac {1}{2}}[ms^{2}\sin ^{2}(\vartheta )+A]{\dot {\vartheta }}^{2}+{\frac {A}{2}}\sin ^{2}(\vartheta ){\dot {\psi }}^{2}\\&+{\frac {C}{2}}[{\dot {\varphi }}+{\dot {\psi }}\cos(\vartheta )]^{2}-mg[\rho +s\cos(\vartheta )]\end{aligned}}}$

In it means

• m is the mass of the top,
• ${\displaystyle {\dot {x}},{\dot {y}}}$the velocities of the center of mass in the horizontal xy-plane ,
• s is the distance of the center of mass from the center of curvature of the spherical lower top of the top,
• ρ is the radius of curvature of this top,
• {ψ, ϑ, φ} the Euler angles , and
• A, C are the equatorial and the axial main moment of inertia of the top around its center of mass.

The Lagrange equations

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{i}}}\right)-{\frac {\partial {\mathcal {L}}}{\partial q_{i}}}=Q_{i}}$

result with the generalized coordinates q _{1,…, 5}  =  x, y, ψ, ϑ, φ and the generalized force

${\displaystyle Q=\sum _{i=1}^{5}Q_{i}\,\mathrm {d} q_{i}={\vec {R}}\cdot ({\hat {e}}_{x}\,\mathrm {d} x+{\hat {e}}_{y}\,\mathrm {d} y)+({\vec {a}}\times {\vec {R}})\cdot (\,{\hat {e}}_{z}\,\mathrm {d} \psi +{\hat {e}}_{N}\,\mathrm {d} \vartheta +{\hat {e}}_{3}\,\mathrm {d} \varphi )}$

As with #Jellett's integral points from the center of mass to the point of contact and sin (ϑ) ê _N  = ê _z  × ê ₃ in the direction of the node. The frictional force acts on the base, which is represented by the xy plane . ${\displaystyle {\vec {a}}}$ ${\displaystyle {\vec {R}}}$

Equations of motion

The equations of motion of the game top result from the principle of momentum " force equals change in momentum ", the principle of twist "moment equals change in angular momentum" and the kinematics :

${\displaystyle {\begin{aligned}m{\ddot {\vec {G}}}=&(N_{z}-mg)\,{\hat {e}}_{z}+{\vec {R}}\\{\dot {\vec {L}}}=&(-s\,{\hat {e}}_{3}-\rho \,{\hat {e}}_{z})\times (N_{z}\,{\hat {e}}_{z}+{\vec {R}})\\{\dot {\hat {e}}}_{3}=&{\frac {1}{A}}{\vec {L}}\times {\hat {e}}_{3}\\0\leq &w,N_{z},\;wN_{z}=0\end{aligned}}}$

In it mean

• m the mass,
• ${\displaystyle {\vec {G}}}$the center of mass (G in Fig. 2 ),
• N _z ≥ 0 the existing only in contact in the contact point, never negative, the weight force opposing normal force ,
• g is the acceleration due to gravity ,
• ê _z the z-direction pointing upwards antiparallel to the weight force,
• ${\displaystyle {\vec {R}}}$the frictional force acting in the plane on contact at the contact point ,
• ${\displaystyle {\vec {L}}}$ the angular momentum with respect to the center of mass,
• s is the axial distance of the center of mass from the center of curvature (GZ in Fig. 2 ),
• ê ₃ the axial vector parallel to the figure axis (from Z to G in Fig. 2 ),
• ρ is the radius of curvature of the hemispherical lower tip of the top (ZT in Fig. 2 ),
• A is the equatorial principal moment of inertia ,
• ${\displaystyle w:=({\vec {G}}-s\,{\hat {e}}_{3}-\rho \,{\hat {e}}_{z})\cdot {\hat {e}}_{z}\geq 0}$ the never negative height of the lowest point T above the ground,
• the supreme point the time derivative , "×" the cross product and "·" the scalar product .

The equations of motion are a system of autonomous differential equations in nine variables with a secondary condition. ${\displaystyle {\vec {G}},\,{\vec {L}},\,{\hat {e}}_{3}}$

In the third equation of motion, the relationship between angular velocity and angular momentum for a symmetrical top was used :

${\displaystyle {\begin{aligned}\quad {\vec {\omega }}=&{\frac {1}{A}}{\vec {L}}+{\frac {A-C}{AC}}L_{3}\,{\hat {e}}_{3}\\\rightarrow \;{\dot {\hat {e}}}_{3}=&{\vec {\omega }}\times {\hat {e}}_{3}={\frac {1}{A}}{\vec {L}}\times {\hat {e}}_{3}\end{aligned}}}$

Therein is L ₃ , the axial angular momentum and C the axial principal moment of inertia.

The last of the equations of motion connects like a complementarity condition the never negative height w of the lowest gyro point above the base with the vertical force component N _z , which only occurs on contact . It results from the consistency condition to ${\displaystyle {\ddot {w}}=0}$

${\displaystyle N_{z}={\frac {mA\left\{g+{\frac {s}{A^{2}}}{\big [}L_{3}L_{z}-({\vec {L}}\cdot {\vec {L}})u{\big ]}\right\}}{A+ms^{2}(1-u^{2})-ms(su+\rho )\,{\hat {e}}_{3}\cdot {\vec {R}}_{1}}}}$

We used and ( μ , N _z , v), which is the frictional force due to a unit force . ${\displaystyle u={\hat {e}}_{3}\cdot {\hat {e}}_{z}=\cos(\vartheta ),\,L_{3}={\vec {L}}\cdot {\hat {e}}_{3},\,L_{z}={\vec {L}}\cdot {\hat {e}}_{z}}$ ${\displaystyle {\vec {R}}_{1}={\tfrac {1}{N_{z}}}{\vec {R}}}$

Derivation of the normal force N z

With the angular velocity ${\displaystyle {\vec {\omega }}={\frac {1}{A}}{\vec {L}}+{\frac {A-C}{AC}}L_{3}\,{\hat {e}}_{3}}$ the time derivatives of the figure axis are provided using the twist rate and the Graßmann identity : ${\displaystyle {\begin{aligned}{\dot {\hat {e}}}_{3}=&{\frac {1}{A}}{\vec {L}}\times {\hat {e}}_{3}\\{\ddot {\hat {e}}}_{3}=&{\frac {1}{A}}{\dot {\vec {L}}}\times {\hat {e}}_{3}+{\frac {1}{A}}{\vec {L}}\times {\dot {\hat {e}}}_{3}\\=&{\frac {1}{A}}[(-s\,{\hat {e}}_{3}-\rho \,{\hat {e}}_{z})\times (N_{z}\,{\hat {e}}_{z}+{\vec {R}})]\times {\hat {e}}_{3}+{\frac {1}{A^{2}}}{\vec {L}}\times ({\vec {L}}\times {\hat {e}}_{3})\\=&-{\frac {N_{z}}{A}}[({\vec {R}}_{1}\cdot {\hat {e}}_{3}+u){\vec {a}}+(s+\rho u)({\vec {R}}_{1}+{\hat {e}}_{z})]+{\frac {1}{A^{2}}}[L_{3}{\vec {L}}-({\vec {L}}\cdot {\vec {L}})\,{\hat {e}}_{3}]\end{aligned}}}$ It was set here. The acceleration of the geometric center (Z in Fig. 2 ), which has the constant distance ρ to the lowest gyro point, is calculated using the momentum law : ${\displaystyle {\vec {R}}=N_{z}{\vec {R}}_{1}}$ ${\displaystyle {\vec {Z}}:={\vec {G}}-s\,{\hat {e}}_{3}}$ ${\displaystyle {\begin{aligned}{\ddot {\vec {Z}}}=&{\ddot {\vec {G}}}-s{\ddot {\hat {e}}}_{3}\\=&{\frac {1}{m}}[(N_{z}-mg)\,{\hat {e}}_{z}+{\vec {R}}]-{\frac {s}{A^{2}}}[L_{3}{\vec {L}}-({\vec {L}}\cdot {\vec {L}})\,{\hat {e}}_{3}]\\&+{\frac {sN_{z}}{A}}[({\vec {R}}_{1}\cdot {\hat {e}}_{3}+u){\vec {a}}+(s+\rho u)({\vec {R}}_{1}+{\hat {e}}_{z})]\end{aligned}}}$ If the top is not lifted off, the z-component of it is 0, which is provided by the equation for N z : ${\displaystyle {\begin{aligned}{\ddot {\vec {Z}}}\cdot {\hat {e}}_{z}=&{\frac {1}{m}}(N_{z}-mg)-{\frac {s}{A^{2}}}[L_{3}L_{z}-({\vec {L}}\cdot {\vec {L}})u]\\&+{\frac {sN_{z}}{A}}[-({\vec {R}}_{1}\cdot {\hat {e}}_{3}+u)(\rho +su)+s+\rho u]{\stackrel {!}{=}}0\\\rightarrow N_{z}=&{\frac {g+{\frac {s}{A^{2}}}[L_{3}L_{z}-({\vec {L}}\cdot {\vec {L}})u]}{{\frac {1}{m}}+{\frac {s}{A}}[s+\rho u-({\vec {R}}_{1}\cdot {\hat {e}}_{3}+u)(\rho +su)]}}\end{aligned}}}$ A few transformations lead to the expression in the text.

Jellett's integral

Fig. 5: Symmetrical game top according to Jellett (1872), p. 181.

When the game top moves under exclusive sliding friction, there is a constant of motion found by JH Jellett and generally proven as such by E. Routh

${\displaystyle J:={\vec {L}}\cdot {\vec {a}}}$

It is the projection of the angular momentum onto the connecting line GT = from the center of mass G to the contact point T on the spherical lower tip of the top, see Fig. 5 ( s  = | GC | and ρ = | CT |.) ${\displaystyle {\vec {L}}}$ ${\displaystyle {\vec {a}}=-s\,{\hat {e}}_{3}-\rho \,{\hat {e}}_{z}}$

Jellett's integral is independent of the frictional force and thus of the surface properties of the subsoil and the top. The drilling, rolling and air friction , which are neglected in the theory, realistically consume the Jellett constant over time. The significance of Jellett's integral is based on the fact that the other forms of friction in the normal game top from Fig. 1 are much less pronounced than the sliding friction.

During the regular precession of the game top, the center of mass G rests and the top rolls smoothly on the ground. Therefore, the point of contact T is also at rest at the moment, otherwise there would be slippage. Even with the permanent rotation around the vertical figure axis, the center of mass and the point of contact rest. With these uniform forms of movement, GT = the current axis of rotation. For the analysis of the game top, the following decomposition of the angular momentum and the rotational energy is useful: ${\displaystyle {\vec {a}}}$

${\displaystyle {\begin{aligned}{\vec {L}}=:&{\frac {J}{|{\vec {a}}|^{2}\Theta _{a}}}\mathbf {\Theta } \cdot {\vec {a}}+{\vec {L}}_{\bot },\quad {\vec {L}}_{\bot }\cdot {\vec {a}}=0\\E_{\text{rot}}=&{\frac {J^{2}}{2|{\vec {a}}|^{2}\Theta _{a}}}+{\frac {1}{2}}{\vec {L}}_{\bot }\cdot \mathbf {\Theta } ^{-1}\cdot {\vec {L}}_{\bot }\end{aligned}}}$

Here, Θ is the inertia tensor and Θ _{a is} the mass moment of inertia that acts when rotating around GT . At the transition of the gyro to a uniform rotation around GT goes to zero and the angle of inclination, the distance | GT | and the mass moment of inertia Θ _a become constant. ${\displaystyle {\vec {L}}_{\bot }}$

Constancy of Jellett's integral

The time derivative of the Jellett constant disappears with exclusive sliding friction, because the moment of the contact force converts the axial angular momentum into vertical one without loss and vice versa. Because with a symmetrical top, the angular momentum, the angular velocity and the figure axis are coplanar:, see Symmetrical top # Angular velocity and angular momentum . The time derivative of body-fixed vectors is calculated from the cross product with the angular velocity, which has the consequence in particular . ${\displaystyle {\vec {L}}\cdot ({\vec {\omega }}\times {\hat {e}}_{3})=0}$ ${\displaystyle {\vec {L}}\cdot {\dot {\hat {e}}}_{3}={\vec {L}}\cdot ({\vec {\omega }}\times {\hat {e}}_{3})=0}$ According to the principle of twist , the time derivative of the angular momentum around the center of mass is equal to the moment that is applied in the center of mass by the force acting at the point of contact . The force does not have to act perpendicular to the base and expressly includes any frictional forces. Now is according to the product rule ${\displaystyle {\vec {M}}={\dot {\vec {L}}}}$ ${\displaystyle {\dot {\vec {L}}}}$ ${\displaystyle {\vec {M}}={\vec {a}}\times {\vec {N}}}$ ${\displaystyle {\vec {N}}}$ ${\displaystyle {\begin{aligned}({\vec {L}}\cdot {\hat {e}}_{3}{\dot {)\;}}=&{\dot {\vec {L}}}\cdot {\hat {e}}_{3}+{\vec {L}}\cdot {\dot {\hat {e}}}_{3}={\vec {M}}\cdot {\hat {e}}_{3}\\=&{\big (}(-s\,{\hat {e}}_{3}-\rho \,{\hat {e}}_{z})\times {\vec {N}}{\big )}\cdot {\hat {e}}_{3}=\rho ({\hat {e}}_{z}\times {\hat {e}}_{3})\cdot {\vec {N}}\end{aligned}}}$ where it was exploited that the factors in the late product may be interchanged cyclically. In the same way it shows ${\displaystyle {\begin{aligned}({\vec {L}}\cdot {\hat {e}}_{z}{\dot {)\;}}=&{\dot {\vec {L}}}\cdot {\hat {e}}_{z}={\vec {M}}\cdot {\hat {e}}_{z}\\=&{\big (}(-s\,{\hat {e}}_{3}-\rho \,{\hat {e}}_{z})\times {\vec {N}}{\big )}\cdot {\hat {e}}_{z}=-s({\hat {e}}_{z}\times {\hat {e}}_{3})\cdot {\vec {N}}\end{aligned}}}$ That can be combined to ${\displaystyle 0=-s({\vec {L}}\cdot {\hat {e}}_{3}{\dot {)\;}}-\rho ({\vec {L}}\cdot {\hat {e}}_{z}{\dot {)\;}}=[{\vec {L}}\cdot (-s\,{\hat {e}}_{3}-\rho \,{\hat {e}}_{z}){\dot {]\;}}=({\vec {L}}\cdot {\vec {a}}{\dot {)\;}}}$ which is why Jellett's integral delivers a constant independent of the contact force and its friction component. ${\displaystyle {\vec {N}}}$

More about the decomposition of the angular momentum

The angular momentum is according to ${\displaystyle {\vec {L}}={\frac {J}{{\vec {a}}\cdot \mathbf {\Theta } \cdot {\vec {a}}}}\mathbf {\Theta } \cdot {\vec {a}}+{\vec {L}}_{\bot }=:{\vec {L}}_{\parallel }+{\vec {L}}_{\bot }}$ disassembled into a component that owes ${\displaystyle {\vec {L}}_{\bot }}$ ${\displaystyle {\vec {a}}\cdot {\vec {L}}_{\bot }={\vec {a}}\cdot {\vec {L}}-{\frac {J}{{\vec {a}}\cdot \mathbf {\Theta } \cdot {\vec {a}}}}{\vec {a}}\cdot \mathbf {\Theta } \cdot {\vec {a}}=0}$ is perpendicular to GT, and a component which it turns out to be in the plane of precession. For each axial vector is determined by the inertia tensor with only the factor C and each equatorial only by the factor A stretched . Thus with = GT also lies in the precession plane generated by ê 3 and ê z . Generally is . The angular velocity is ${\displaystyle {\vec {L}}_{\parallel }}$ ${\displaystyle {\vec {a}}=-s\,{\hat {e}}_{3}-\rho \,{\hat {e}}_{z}}$ ${\displaystyle \mathbf {\Theta } \cdot {\vec {a}}}$ ${\displaystyle {\vec {L}}_{\parallel }\cdot {\vec {L}}_{\bot }\neq 0}$ ${\displaystyle {\vec {\omega }}=\mathbf {\Theta } ^{-1}\cdot {\vec {L}}={\frac {J}{{\vec {a}}\cdot \mathbf {\Theta } \cdot {\vec {a}}}}{\vec {a}}+\mathbf {\Theta } ^{-1}\cdot {\vec {L}}_{\bot }}$ with which the rotational energy around the center of mass is calculated to ${\displaystyle {\begin{aligned}E_{\text{rot}}:=&{\frac {1}{2}}{\vec {\omega }}\cdot {\vec {L}}={\frac {J}{2{\vec {a}}\cdot \mathbf {\Theta } \cdot {\vec {a}}}}\underbrace {\overbrace {{\vec {\omega }}\cdot \mathbf {\Theta } } ^{\vec {L}}\cdot {\vec {a}}} _{J}+{\frac {1}{2}}\left({\frac {J}{{\vec {a}}\cdot \mathbf {\Theta } \cdot {\vec {a}}}}{\vec {a}}+\mathbf {\Theta } ^{-1}\cdot {\vec {L}}_{\bot }\right)\cdot {\vec {L}}_{\bot }\\=&{\frac {J^{2}}{2|{\vec {a}}|^{2}\Theta _{a}}}+{\frac {1}{2}}{\vec {L}}_{\bot }\cdot \mathbf {\Theta } ^{-1}\cdot {\vec {L}}_{\bot }\end{aligned}}}$ Furthermore, through algebraic transformations ${\displaystyle {\begin{aligned}|{\vec {a}}|=&\rho {\sqrt {(\alpha +u)^{2}+1-u^{2}}}\\\Theta _{a}:=&{\frac {{\vec {a}}\cdot \mathbf {\Theta } \cdot {\vec {a}}}{|{\vec {a}}|^{2}}}={\frac {(\alpha +u)^{2}+\gamma (1-u^{2})}{(\alpha +u)^{2}+1-u^{2}}}C\end{aligned}}}$ be detected. Here, α = s / ρ, γ = A / C and u  = cos (ϑ) is the cosine of the angle of inclination ϑ of the top.

Regular precession

See also : Regular precession

In regular precession, the center of mass is at rest and the angular velocities around the plumb line and figure axis as well as the inclination of the figure axis are constant. In the case of a game top that is subject to friction , there is only no dissipation of energy through friction when the lower tip of the top rolls smoothly on the ground. To do this, the precession speed, self-rotation and inclination must be coordinated with one another. The precession  velocity results from the condition on the angular momentum in the regular precession L _f  L _v = Amgs

${\displaystyle {\dot {\psi }}={\sqrt {\frac {mgs}{C[\alpha +(1-\gamma )u]}}}}$

For a regular precession to take place, the expression in the square brackets in the denominator must be positive. Because of the roll condition, unlike in the frictionless case, there is at most one precession speed for each angle of inclination.

The amount of #Jellett's integral takes the form

${\displaystyle J_{R}(u)=J_{\beta }{\frac {(\alpha +u)^{2}+\gamma (1-u^{2})}{\sqrt {\alpha +(1-\gamma )u}}}}$

where J _β  = ρ√ ( Cmgs ) is a gyroscopic own constant with the dimension of the Jellett integral.

Derivation of the precession velocity
The angular velocities p, q, r = ω 1,2,3 and principal axes ê 1,2,3 , see gyroscopic theory # reference systems and Euler angles , with a stationary contact point ( see below) , are expressed with the Euler angles ψ, ϑ and φ  : ${\displaystyle {\dot {\varphi }}=\alpha {\dot {\psi }}}$ ${\displaystyle {\begin{aligned}L_{z}:=&{\vec {L}}\cdot {\hat {e}}_{z}=Ap\sin(\vartheta )\sin(\varphi )+Aq\sin(\vartheta )\cos(\varphi )+Cr\cos(\vartheta )\\=&{\big (}A\sin ^{2}(\vartheta )+C\cos ^{2}(\vartheta ){\big )}{\dot {\psi }}+C\cos(\vartheta ){\dot {\varphi }}\\L_{3}:=&{\vec {L}}\cdot {\hat {e}}_{3}=Cr=C{\big (}{\dot {\psi }}\cos(\vartheta )+{\dot {\varphi }}{\big )}\\{\vec {L}}=&L_{v}{\hat {e}}_{z}+L_{f}{\hat {e}}_{3}\\L_{v}=&{\frac {L_{z}-L_{3}\cos(\vartheta )}{\sin ^{2}(\vartheta )}}=A{\dot {\psi }}\\L_{f}=&{\frac {L_{3}-L_{z}\cos(\vartheta )}{\sin ^{2}(\vartheta )}}=C[\alpha +(1-\gamma )\cos(\vartheta )]{\dot {\psi }}\end{aligned}}}$ The condition on the angular momentum in the regular precession L f  L v  =  Amgs finally provides the precession velocity given in the text ${\displaystyle {\dot {\psi }}={\sqrt {\frac {mgs}{C[\alpha +(1-\gamma )u]}}}}$
Rest condition of the point of uprising
The center of mass G is identified with the vector and the lowest point T with the top, with the vector from G to T being solid. The speed of T is therefore ${\displaystyle {\vec {G}}}$ ${\displaystyle {\vec {T}}={\vec {G}}+{\vec {a}}}$ ${\displaystyle {\vec {a}}=-s\,{\hat {e}}_{3}-\rho \,{\hat {e}}_{z}}$ ${\displaystyle {\dot {\vec {T}}}={\dot {\vec {G}}}+{\vec {\omega }}\times {\vec {a}}={\dot {\vec {G}}}-{\vec {\omega }}\times (s\,{\hat {e}}_{3}+\rho \,{\hat {e}}_{z})}$ In regular precession, the angular velocity is parallel to GT ( ) and T is at rest, which is why the center of mass stands still. ${\displaystyle {\vec {\omega }}\times {\vec {a}}={\vec {0}}}$ The angular velocity can be expressed with the Euler angles ψ, ϑ and φ in the Euler base system: where the vector points in the direction of the node, see gyro theory # reference systems and Euler angles . The speed results from α = s / ρ ${\displaystyle {\vec {\omega }}={\dot {\psi }}\,{\hat {e}}_{z}+{\dot {\vartheta }}\,{\hat {e}}_{N}+{\dot {\varphi }}\,{\hat {e}}_{3}}$ ${\displaystyle \sin(\vartheta )\,{\hat {e}}_{N}={\hat {e}}_{z}\times {\hat {e}}_{3}}$ ${\displaystyle {\dot {\vec {T}}}={\dot {\vec {G}}}_{h}+[{\dot {G}}_{z}+s{\dot {\vartheta }}\sin(\vartheta )]\,{\hat {e}}_{z}+({\vec {G}}\cdot {\hat {e}}_{z}){\dot {\vartheta }}\,{\hat {e}}_{z}\times {\hat {e}}_{N}+\rho ({\dot {\varphi }}-\alpha {\dot {\psi }})\,{\hat {e}}_{z}\times {\hat {e}}_{3}}$ There is no horizontal speed and no vertical speed and the inclination angle ϑ is constant. If the contact point T is stationary, it follows from the last summand . ${\displaystyle {\dot {\vec {G}}}_{h}}$ ${\displaystyle {\dot {G}}_{z}}$ ${\displaystyle {\dot {\varphi }}=\alpha {\dot {\psi }}}$

Derivation of the function J R ( u )

The  angular momentum arises from the intermediate results from the derivation of the precession velocity and the abbreviation u = cos (entsteht) ${\displaystyle {\vec {L}}=L_{v}\,{\hat {e}}_{z}+L_{f}\,{\hat {e}}_{3}={\dot {\psi }}\{A\,{\hat {e}}_{z}+C[\alpha +(1-\gamma )u]\,{\hat {e}}_{3}\}}$ and with the precession speed ${\displaystyle {\dot {\psi }}={\sqrt {\frac {mgs}{C[\alpha +(1-\gamma )u]}}}={\frac {J_{\beta }}{\rho C{\sqrt {\alpha +(1-\gamma )u}}}}}$ the amount of the Jellett integral ${\displaystyle J_{R}(u):=|{\vec {L}}\cdot {\vec {a}}|={\vec {L}}\cdot (s{\hat {e}}_{3}+\rho {\hat {e}}_{z})=J_{\beta }{\frac {(\alpha +u)^{2}+\gamma (1-u^{2})}{\sqrt {\alpha +(1-\gamma )u}}}}$ Here J β  = ρ√ ( Cmgs ) is a constant of its own.

If ❘ J ❘ =  J _R ( u ), the stability of the stationary movements is given, because then the total energy of the top as a function of the angle of inclination in u is stationary. Below are some characteristics of J _R ( u ).

The marginal values

${\displaystyle {\begin{aligned}J_{R1}:=&J_{R}(1)={\frac {J_{\beta }(1+\alpha )^{2}}{\sqrt {\alpha -\gamma +1}}}\\J_{R-1}:=&J_{R}(-1)={\frac {J_{\beta }(1-\alpha )^{2}}{\sqrt {\alpha +\gamma -1}}}\end{aligned}}}$

tops are constants of their own. If J _R ( u ) in the interval u  ∈ ⅅ ⊆ [-1: 1] is a monotonic function of u , then for every Jellett integral between the boundary values ​​in ⅅ there is exactly one possible inclination of the top with which the top is regular can precess. Otherwise if anywhere in ⅅ the derivative

${\displaystyle {J_{R}}'(u):={\frac {\mathrm {d} J_{R}}{\mathrm {d} u}}=J_{\beta }{\frac {3[\alpha +(1-\gamma )u]^{2}-\gamma (1-\alpha ^{2}-\gamma )}{2[\alpha +(1-\gamma )u]^{\frac {3}{2}}}}}$

is zero, there are extreme values in the vicinity of which there are two possible angles of inclination for a value of the integral. In physically attainable zeros J _R '( u _m ) = 0 applies

${\displaystyle {\begin{aligned}u_{m}=&{\frac {-\alpha +{\sqrt {{\frac {\gamma }{3}}(1-\alpha ^{2}-\gamma )}}}{1-\gamma }}\\J_{Rm}:=&J_{R}(u_{m})={\frac {4J_{\beta }}{1-\gamma }}\left({\frac {\gamma }{3}}(1-\alpha ^{2}-\gamma )\right)^{\frac {3}{4}}\end{aligned}}}$

what can only happen with real movements if

½ <γ = A / C <1 - α²

i.e. with a flattened top with a blunt top α = s / ρ <1 / √2 ≈ 0.7.

Total energy of the top

The total energy E  =  E _tra  +  E _rot  +  E _{pot of} the top is made up of its

• Translational energy , ${\displaystyle E_{\text{tra}}={\tfrac {m}{2}}{\dot {\vec {G}}}\cdot {\dot {\vec {G}}}}$
• Rotational energy with respect to the center of mass and ${\displaystyle E_{\text{rot}}={\tfrac {1}{2}}{\vec {\omega }}\cdot {\vec {L}}={\tfrac {1}{2A}}{\vec {L}}\cdot {\vec {L}}+{\tfrac {A-C}{2AC}}L_{3}^{2}}$
• Position energy ${\displaystyle E_{\text{pot}}=mg{\vec {G}}\cdot {\hat {e}}_{z}}$

With the relationships given in the section #Motion equations, in particular

${\displaystyle {\vec {\omega }}={\frac {1}{A}}{\vec {L}}+{\frac {A-C}{AC}}({\vec {L}}\cdot {\hat {e}}_{3})\,{\hat {e}}_{3}}$

shows when the top is standing up:

${\displaystyle {\begin{aligned}E=&{\frac {J^{2}}{2|{\vec {a}}|^{2}\Theta _{a}}}+mg\rho (1+\alpha u)\\&+{\frac {1}{2}}{\vec {L}}_{\bot }\cdot \mathbf {\Theta } ^{-1}\cdot {\vec {L}}_{\bot }+{\frac {m}{2}}\left[{\dot {G}}_{h}^{2}+\left({\frac {s}{A}}{\vec {L}}_{\bot }\cdot ({\hat {e}}_{z}\times {\hat {e}}_{3})\right)^{2}\right]\end{aligned}}}$

Here an angular momentum appears perpendicular to and #Jellett's integral J , which were defined there. The horizontal velocity of the center of gravity and the angular momentum approach zero when approaching a stationary movement, so that only the first two summands remain in such a movement. ${\displaystyle {\vec {L}}_{\bot }}$ ${\displaystyle {\vec {a}}}$ ${\displaystyle {\dot {G}}_{h}}$ ${\displaystyle {\vec {L}}_{\bot }}$

Dissipation of energy through friction

Without contact with the substrate, the total energy remains under the restriction made here to sliding friction. Upon contact of the lowermost point gyroscope has no velocity perpendicular to the base, ie . With this and with the #movement equations, the time derivative of the total energy of the rising top is calculated ${\displaystyle ({\dot {\vec {G}}}+{\vec {\omega }}\times {\vec {a}})\cdot {\hat {e}}_{z}=0}$

${\displaystyle {\dot {E}}=\left({\dot {\vec {G}}}+{\vec {\omega }}\times {\vec {a}}\right)\cdot {\vec {R}}}$

The friction force is always opposite to the speed of the contact point in the brackets, which is why the scalar product is never positive and the total energy decreases monotonically over time . ${\displaystyle {\vec {R}}}$

In the case of a stationary movement on the surface, the lowest gyro point must be at rest at all times, which is the case with rotation with a vertical figure axis and # regular precession with a stationary center of mass.

Stability of stationary movements

A stationary movement of the gyroscope is stable exactly when its total energy is in a local minimum. The table summarizes the conditions for this ( u  = cos (ϑ)).

symbol	Form of movement	Tilt	Stability criterion
▼	Rotation around a vertical upright figure axis	u = 1	γ <1 + α and ❘ J ❘>  J R1
▲	Rotation with vertically hanging figure axis	u = -1	γ <1 - α or ❘ J ❘ <  J R-1
●	Regular precession	-1 <u <1	γ> 1 - α² or u> u m

Under the opposite conditions, the total energy is at a local maximum and the movement is unstable.

symbol	Form of movement	Tilt	Criterion for instability
▽	Rotation around a vertical upright figure axis	u = 1	γ ≥ 1 + α or ❘ J ❘ ≤  J R1
△	Rotation with vertically hanging figure axis	u = -1	γ ≥ 1 - α and ❘ J ❘ ≥  J R-1
○	Regular precession	-1 <u <1	γ ≤ 1 - α² and u ≤ u m

The size comparison with J _R1 , J _R-1 or u _m is only used for real values.

Total energy of the stationary movements

The total energy not only remains constant when the top is lifted, but also when the bottom top point is stationary on the surface, i.e. it is not sliding. Then the top rotates with a vertical figure axis or the lower tip of the top rolls smoothly on the surface during regular precession. In these equilibrium solutions the center of mass is at rest , the energy in a local extremum and given Jellett's integral J is a function of the angle of inclination alone: ${\displaystyle {\vec {L}}_{\bot }={\vec {0}}}$ ${\displaystyle {\begin{aligned}E=&{\frac {2J_{\beta }^{2}(1+\alpha u)[(\alpha +u)^{2}+\gamma (1-u^{2})]+\alpha J^{2}}{2\rho ^{2}C\alpha [(\alpha +u)^{2}+\gamma (1-u^{2})]}}\\E_{,u}:=&{\frac {\mathrm {d} E}{\mathrm {d} u}}={\frac {J_{\beta }^{2}[(\alpha +u)^{2}+\gamma (1-u^{2})]^{2}-J^{2}[\alpha +(1-\gamma )u]}{\rho ^{2}C[(\alpha +u)^{2}+\gamma (1-u^{2})]^{2}}}\\E_{,\vartheta }:=&{\frac {\mathrm {d} E}{\mathrm {d} \vartheta }}={\frac {\mathrm {d} E}{\mathrm {d} u}}{\frac {\mathrm {d} u}{\mathrm {d} \vartheta }}=-\sin(\vartheta )E_{,u}\end{aligned}}}$ The constant J β  = ρ√ ( CMGs ) has the dimension of Jellett integral J . The total energy is stationary when E , ϑ  = 0 and thus sin (ϑ) = 0, i.e. the figure axis is perpendicular, or when E , u  = 0. The numerator in E , u can be converted into a product with the third binomial formula , which vanishes if ❘ J ❘ =  J R ( u ), an identity that satisfies the # regular precession . The total energy is stationary when rotating with a vertical figure axis and with regular precession.

Minima of energy

In a minimum of the total energy the movement is stable and the second derivative of the total energy according to the angle of inclination is positive. It is calculated: ${\displaystyle {\begin{aligned}E_{,uu}:=&{\frac {\mathrm {d} ^{2}E}{\mathrm {d} u^{2}}}={\frac {J^{2}}{\rho ^{2}C}}{\frac {3[\alpha +(1-\gamma )u]^{2}+\gamma (\alpha ^{2}+\gamma -1)}{[(\alpha +u)^{2}+\gamma (1-u^{2})]^{3}}}\\E_{,\vartheta \vartheta }:=&{\frac {\mathrm {d} ^{2}E}{\mathrm {d} \vartheta ^{2}}}=-\cos(\vartheta )E_{,u}+\sin ^{2}(\vartheta )E_{,uu}=-uE_{,u}+(1-u^{2})E_{,uu}\end{aligned}}}$ If the figure axis is perpendicular ( u  = ± 1), E , ϑϑ has the sign of - u E , u . When the top is upright, u  = 1 and the energy is at the minimum, if E , u  <0 ${\displaystyle {\begin{aligned}&(\alpha -\gamma +1)J^{2}>J_{\beta }^{2}(1+\alpha )^{4}\\\rightarrow \quad &\gamma <1+\alpha \quad {\text{und}}\quad |J|>{\frac {J_{\beta }(1+\alpha )^{2}}{\sqrt {\alpha -\gamma +1}}}=J_{R1}\end{aligned}}}$ For the vertically hanging top, u  = -1 and the energy is at the minimum, if E , u  > 0 ${\displaystyle {\begin{aligned}&(\alpha +\gamma -1)J^{2}<J_{\beta }^{2}(1-\alpha )^{4}\\\rightarrow \quad &\gamma <1-\alpha \quad {\text{oder}}\quad |J|<{\frac {J_{\beta }(1-\alpha )^{2}}{\sqrt {\alpha +\gamma -1}}}=J_{R-1}\end{aligned}}}$ In a regular precession the figure axis is not perpendicular, -1 <  u  <1, α + (1 - γ) u  > 0 and E , u  = 0. Then E , ϑϑ has the sign of E , uu , which is because of ${\displaystyle E_{,uu}>0\quad \leftrightarrow \quad 3[\alpha +(1-\gamma )u]^{2}>\gamma (1-\alpha ^{2}-\gamma )}$ under the conditions ${\displaystyle \gamma >1-\alpha ^{2}\quad {\text{oder}}\quad u>{\frac {-\alpha +{\sqrt {{\frac {\gamma }{3}}(1-\alpha ^{2}-\gamma )}}}{1-\gamma }}=u_{m}}$ is positive.

Division of the α-γ parameter space

The limits shown in Fig. 6 result from the following conditions that are relevant for regular precessions: ${\displaystyle {\begin{aligned}\left.\alpha +(1-\gamma )u\right|_{u=1}=0\;\rightarrow &\;\;\gamma =1+\alpha \\\left.\alpha +(1-\gamma )u\right|_{u=-1}=0\;\rightarrow &\;\;\gamma =1-\alpha \\\left.\alpha +(1-\gamma )u\right|_{u=u_{m}}=0\;\rightarrow &\;\;\gamma =1-\alpha ^{2}\\u_{m}=1\;\rightarrow &\;\;\gamma _{1,2}={\frac {1}{8}}(1+\alpha )\left(7-\alpha \pm {\sqrt {(7-\alpha )^{2}-48}}\right)\\u_{m}=-1\;\rightarrow &\;\;\gamma ={\frac {1}{8}}(1-\alpha )\left(7+\alpha +{\sqrt {(7+\alpha )^{2}-48}}\right)\\J_{R-1}=J_{R1}\;\rightarrow &\;\;\gamma ={\frac {(1-\alpha ^{2})(1+3\alpha ^{2})}{1+6\alpha ^{2}+\alpha ^{4}}}\end{aligned}}}$ The first three conditions border areas in which the precession speed can increase beyond all limits and regular precessions are not possible with every inclination (Ib, IIb). The two following conditions border areas in which J R ( u ) has a local minimum J Rm  =  J R ( u m ) in the physically reachable area, i.e. the top at a given ❘ J ❘ ∈ [  J Rm : min ( J R -1 ,  J R1 )] can perform regular precessions with two different angles of inclination. The sixth and last condition separates the areas ④ and ⑤, in which there is at a point u k  ∈] -1: 1 [a regular precession with ❘ J ❘ =  J R-1 or ❘ J ❘ =  J R1 : In the range ④, J R-1  >  J R1 , J R ( u ) in the interval [-1:  u k [decreasing monotonically and for every ❘ J ❘ ∈] J Rm :  J R1 ] there are two regular precessions with u 1 , 2  ∈ [ u k : 1]. In the domain ⑤, on the other hand, J R-1  <  J R1 , for every ❘ J ❘ ∈] J Rm :  J R-1 ] there are two regular precessions with u 1,2  ∈ [-1:  u k ] and is J R ( u ) in the interval] u k : 1] monotonically increasing.

Fig. 6: α-γ parameter space of the game top

The gyroscopes can be assigned to eight areas by their parameters α = s / ρ and γ = A / C, in which they have the same stability behavior, see Fig. 6. The eight areas can in turn be divided into four cases Ia / b, IIa / b group.

Gyro type I: 1 - α² ≤ γ

The function J _R (u) for the # regular precession is, where it is real in the physically reachable area, monotonic and for every J there is at most one regular precession, which is then also stable (●).

The rotation around the hanging vertical figure axis with u = -1 is stable if ❘ J ❘ <  J _R-1 (▲) and unstable if ❘ J ❘ ≥  J _R-1 (△).

Gyro type Ia: 1 - α² ≤ γ <1 + α

Regular precessions are possible and stable in the entire interval -1 <u <1 (●).

Rotations around the vertical upright figure axis with u = 1 is stable if ❘ J ❘>  J _R1 (▼) and unstable if ❘ J ❘ ≤  J _R1 (▽).

Gyro type Ib: 1 + α ≤ γ

Regular precessions are only possible in the interval -1 <u <  and then also stable (●). ${\displaystyle {\tfrac {\alpha }{\gamma -1}}}$

The rotation around the vertical upright figure axis with u = 1 is unstable (▽).

Gyro type II: γ ≤ 1 - α²

Because γ> ½, α <1 / √2 ≈ 0.7 for these gyroscopes.

In the zones ③, ⑥ and ⑧, the function J _R (u), where it is real in the physically reachable area, is monotonic, whereas in the other areas, ④, ⑤ and ⑦, a local minimum J _Rm  =  J _R (u _m ) occurs.

The rotation with a vertical figure axis (u = 1) is stable if ❘ J ❘>  J _R1 (▼), and otherwise unstable (▽).

Gyro type IIa: 1 - α <γ ≤ 1 - α²

The top can execute a regular precession with every inclination u ∈ [-1: 1], because the function J _R (u) is defined in the whole interval.

• In ③ u _m  > 1> u and the associated regular precession is unstable (○).
• In ④ and ⑤ there is a minimum J _Rm at u _m  ∈ [-1: 1]. The regular precessions with u <u _m are unstable (○) and the others are stable (●). For every ❘ J ❘ ∈]  J _Rm : min ( J _R-1 ,  J _R1 )] there are two regular precessions with different inclinations, the less inclined being correspondingly stable (●) and the more inclined being unstable (○).
• In ⑥ u _m  <-1 <u and the associated regular precession is stable (●).

The rotation with the figure axis hanging vertically (u = -1) is stable if ❘ J ❘ <  J _R-1 (▲), and otherwise unstable (△).

Gyro type IIb: ½ <γ ≤ 1 - α

Regular Precessions can only in the interval and on amount arbitrarily large J occur. In the domain ⑧, u _m  > 1 and for every ❘ J ❘>  J _{R1 there is} exactly one regular precession and this is unstable (○). In zone ⑦, J _R (u), coming from infinity, drops to the minimum J _Rm and then increases to J _R1 . In the falling branch] ∞:  J _R1 ] there is a regular precession for every J and this is unstable (○). Between J _Rm and J _R1 there are two regular precessions for every J , of which the less inclined is stable (●) and the other is unstable (○). ${\displaystyle -{\tfrac {\alpha }{1-\gamma }}<u<1}$

The vertically hanging top is stable (▲).

Web links

Wiktionary: Spielkreisel  - explanations of meanings, word origins, synonyms, translations

Commons : Play Top  - Collection of images, videos and audio files

Individual evidence

1. ↑ Magnus (1971), p. 266, Grammel (1920), pp. 111 and 123, see literature.
2. ↑ ^{a b} Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 352.
3. ↑ Magnus (1971), p. 266.
4. ↑ Grammel (1920), p. 111.
5. ↑ Felix Klein , Conr. Müller: Encyclopedia of the Mathematical Sciences with inclusion of its applications . Mechanics. Ed .: Academies of Sciences in Göttingen , Leipzig, Munich and Vienna. Fourth volume, 1st part volume. BG Teubner, 1908, ISBN 978-3-663-16021-2 , p. 546 , doi : 10.1007 / 978-3-663-16021-2 ( limited preview in the Google book search [accessed on January 24, 2020] see also wikisource }).
6. ↑ Clifford Truesdell : The Development of the Swirl Theorem . In: Society for Applied Mathematics and Mechanics (ed.): Journal for Applied Mathematics and Mechanics (=  Issue 4/5 ). tape 44 , April 1964, p. 154 , doi : 10.1002 / zamm.19640440402 ( wiley.com ).
7. ^ Joseph-Louis Lagrange : Mécanique Analytique . Tome Second. Corucier, Paris 1815, p. 265 f . (French, archive.org [accessed August 20, 2017]). or Joseph-Louis Lagrange: Analytical Mechanics . Vandenhoeck and Ruprecht, Göttingen 1797 ( archive.org [accessed on August 20, 2017] German translation by Friedrich Murhard).
8. ↑ Magnus (1971), p. 266.
9. ↑ ^{a b c} JH Jellett : A treatise on the theory of friction . Macmillan Publishers , London 1872, pp. 185 ( archive.org [accessed December 15, 2018]). . In gyro theory, constants of motion are called integral because their time derivative vanishes and, conversely, the time integral provides a constant.
10. ↑ see Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 333 and the references there.
11. ↑ see Rauch-Wojciechowski, Sköldstam and Glad (2005)
12. ↑ Edward Routh : A TREATISE ON THE STABILITY OF A GIVEN STATE OF MOTION . PARTICULARLY STEADY MOTION. Macmillan Publishers , London 1877 ( archive.org - This essay was awarded the Adams Prize at Cambridge University in 1877. ).
13. ↑ Magnus (1971), pp. 271 + 273, Kuypers and Ucke (1994), p. 215.
14. ↑ smoke Wojciechowski, Sköldstam and Glad (2005), S. 361st
15. ↑ Kuypers and Ucke (1994), p. 215.
16. ↑ ^{a b} Grammel (1920), p. 113.
17. ↑ ^{a b} Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 339.
18. ↑ ^{a b c} Grammel (1920), p. 116.
19. ↑ Magnus (1971), p. 267, Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 336 or Ciocci and Langerock (2007).
20. ↑ ^{a b} Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 337.
21. ↑ Kuypers and Ucke (1994), p. 214.
22. ↑ Grammel (1920), p. 124 ff.
23. ↑ Grammel (1920), p. 125 f.
24. ↑ Grammel (1920), p. 127.
25. ↑ MC Ciocci, B. Long Rock: Dynamics of the Top Touch via Routhian Reduction . In: Regular and Chaotic Dynamics . tape 12 , no. 6 . Springer Nature, 2007, ISSN  1468-4845 , p. 602–614 , doi : 10.1134 / S1560354707060032 , arxiv : 0704.1221 (English, The Euler angles φ and ψ have interchanged meanings.).
26. ↑ Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 336.
27. ↑ Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 335.
28. ↑ Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 338.
29. ↑ Edward Routh : The Dynamics of Rigid Body Systems . The higher dynamic. second volume. BG Teubner, Leipzig 1898, p. 192 ( archive.org ).
30. ↑ ^{a b} Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 340.
31. ↑ Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 343.
32. ↑ Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 347.
33. ↑ ^{a b} Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 353 ff. There cos (ϑ) = - u .
34. ↑ Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 356. There η ₃ = - u , λ = J and β = J _β . The symbols are defined in the section # Regular precession .
35. ↑ Rauch-Wojciechowski, Sköldstam and Glad (2005), p. 358 ff. There η ₃ = - u , λ = J and β = J _β .

literature

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