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In this paper, the Steiner area formula and the polar moments of inertia were expressed during one-parameter closed planar motions in complex plane. The Steiner points or Steiner normal concepts were described according to whe-ther rotation number was different from zero or equal to zero. The moving pole point was given with its components and its relation to the Steiner point or Steiner normal which was specified. The Steiner formula and the polar moments of inertia were expressed for the inverse motion. The fixed pole point was calculated for the inverse motion. The sagittal motion of a tele-scopic crane was considered as an example. This motion was described by a double hinge consisting of the fixed control panel of the telescopic crane and its moving arm. The results obtained in the first section of this study were applied to this motion.

Steiner explained some properties of the area of the path of a point for a geometrical object rolling on a line and making a complete turn [

We consider one parameter planar motion in complex plane between two reference systems: the fixed E ′ and the moving E , with their origins ( O ′ , O ) and orientations. Then, we take into account the motion relative to the fixed system (direct motion), and later the moving system (inverse motion). By taking the displacement vector O ′ O = u ′ and the total angle of rotation α ( t ) , the motion was defined by the transformation

x ′ ( t ) = x e i α ( t ) + u ′ ( t ) (1)

where x ′ ( t ) is the trajectory of the point x belonging to the moving system with the respect to the fixed system. If we replace u ′ = − u e i α in Equation (1), the motion can be written as

x ′ = ( x − u ) e i α . (2)

With the coordinates

x ′ = ( x ′ 1 x ′ 2 ) , x = ( x 1 x 2 ) , u = ( u 1 u 2 ) , (3)

the components of x ′ is obtained as

x ′ 1 = ( x 1 − u 1 ) cos α − ( x 2 − u 2 ) sin α x ′ 2 = ( x 1 − u 1 ) sin α + ( x 2 − u 2 ) cos α . (4)

Equation (4) can be written in matrix form

( x ′ 1 x ′ 2 ) = ( cos α − sin α sin α cos α ) ( x 1 − u 1 x 2 − u 2 ) . (5)

The trajectory area formula F of the point x ′ is given by

F = 1 2 ∮ x ′ 1 d x ′ 2 − x ′ 2 d x ′ 1 . (6)

If Equation (4) is differentiated,

d x ′ 1 = − ( x 1 − u 1 ) sin α d α − cos α d u 1 − ( x 2 − u 2 ) cos α d α + sin α d u 2 d x ′ 2 = ( x 1 − u 1 ) cos α d α − sin α d u 1 − ( x 2 − u 2 ) sin α d α − cos α d u 2 (7)

is found. If Equations (4) and (7) are placed in Equation (6),

2 F = ∮ ( x 1 2 + x 2 2 ) d α − 2 x 1 ∮ u 1 d α − 2 x 2 ∮ u 2 d α − x 1 ∮ d u 2 + x 2 ∮ d u 1 + ∮ ( u 1 2 + u 2 2 ) d α + ∮ u 1 d u 2 − u 2 d u 1 (8)

is obtained.

Since u 1 and u 2 are periodic functions, ∮ d u 1 = 0 and ∮ d u 2 = 0 . If x = 0 ( x 1 = 0 , x 2 = 0 ) is taken, then we have

2 F O = ∮ ( u 1 2 + u 2 2 ) d α + ∮ u 1 d u 2 − u 2 d u 1 (9)

for the trajectory area of the initial point. So we can rewrite Equation (8) as

2 ( F − F O ) = ( x 1 2 + x 2 2 ) ∮ d α − 2 x 1 ∮ u 1 d α − 2 x 2 ∮ u 2 d α . .

Using the abbreviations

a : = − 2 ∮ u 1 d α , (10)

b : = − 2 ∮ u 2 d α , (11)

we obtain

2 ( F − F O ) = ( x 1 2 + x 2 2 ) ∮ d α + x 1 a + x 2 b . (12)

F is the quadratic form of the coordinates x 1 and x 2 . The surface F ( x 1 , x 2 ) describes either a cone or a plane, so it follows that the sections of the constant area describe either concentric circles or parallel lines.

The coefficient m

m : = ∮ d α = 2 π ν (13)

with the rotation number ν determines whether the lines with F = const describe circles or straight lines. If ν ≠ 0 , then we have circles. If ν = 0 , the circles are reduced to straight lines. If Equation (13) is replaced in Equation (12), then we have

2 ( F − F O ) = ( x 1 2 + x 2 2 ) m + x 1 a + x 2 b . (14)

In order to obtain the Steiner formula relative to the inverse motion, we begin by exchanging the fixed system with the moving system. If Equation (2) is solved according to x , we obtained the motion for the inverse motion as

x = x ′ e − i α + u . (15)

If u = − u ′ e − i α is replaced in Equation (15), we have

x = ( x ′ − u ′ ) e − i α . (16)

Moreover, for the components of x , from

x 1 + i x 2 = ( ( x ′ 1 − u ′ 1 ) + i ( x ′ 2 − u ′ 2 ) ) ( cos α − i sin α ) ,

we have

x 1 = ( x ′ 1 − u ′ 1 ) cos α + ( x ′ 2 − u ′ 2 ) sin α , (17)

x 2 = − ( x ′ 1 − u ′ 1 ) sin α + ( x ′ 2 − u ′ 2 ) cos α . (18)

Furthermore, if the coordinates is derived, we obtain

d x 1 = − ( x ′ 1 − u ′ 1 ) sin α d α − cos α d u ′ 1 + ( x ′ 2 − u ′ 2 ) cos α d α − sin α d u ′ 2 , (19)

d x 2 = − ( x ′ 1 − u ′ 1 ) cos α d α + sin α d u ′ 1 − ( x ′ 2 − u ′ 2 ) sin α d α − cos α d u ′ 2 . (20)

If Equations (17), (18), (19) and (20) are inserted into the formula for the area formula of inverse motion

F ′ = 1 2 ∮ x 1 d x 2 − x 2 d x 1 , (21)

we find

2 F ′ = − ( ( x ′ 1 ) 2 + ( x ′ 2 ) 2 ) ∮ d α + 2 x ′ 1 ∮ u ′ 1 d α + 2 x ′ 2 ∮ u ′ 2 d α − x ′ 1 ∮ d u ′ 2 + x ′ 2 ∮ d u ′ 1 − ∮ ( ( u ′ 1 ) 2 + ( u ′ 2 ) 2 ) d α + ∮ u ′ 1 d u ′ 2 − u ′ 2 d u ′ 1 . (22)

Since u ′ 1 and u ′ 2 are periodic functions, ∮ d u ′ 1 = 0 and ∮ d u ′ 2 = 0 . If x ′ = 0 ( x ′ 1 = 0 , x ′ 2 = 0 ) is taken, then we have

2 F ′ O = − ∮ ( ( u ′ 1 ) 2 + ( u ′ 2 ) 2 ) d α + ∮ u ′ 1 d u ′ 2 − u ′ 2 d u ′ 1 . (23)

Therefore, we can write Equation (22) as

2 ( F ′ − F ′ O ) = − ( ( x ′ 1 ) 2 + ( x ′ 2 ) 2 ) ∮ d α + 2 x ′ 1 ∮ u ′ 1 d α + 2 x ′ 2 ∮ u ′ 2 d α .

Using the abbreviations

a ′ : = ∮ u ′ 1 d α , (24)

b ′ : = ∮ u ′ 2 d α , (25)

m ′ : = − ∮ d α = − m , (26)

we have

2 ( F ′ − F ′ O ) = − ( ( x ′ 1 ) 2 + ( x ′ 2 ) 2 ) m ′ + x ′ 1 a ′ + x ′ 2 b ′ .. (27)

We begin by rewriting Equation (14) for the case m ≠ 0 ,

( x 1 2 + x 2 2 ) m + x 1 a + x 2 b = 2 ( F − F O ) . (28)

By dividing this equation by m and by completing the squares, we obtain the equation of a circle

( x 1 + a 2 m ) 2 + ( x 2 + b 2 m ) 2 = 2 ( F − F O ) m + a 2 4 m 2 + b 2 4 m 2 .. (29)

From Equation (29), we can relate the radius r of the circle with the area of the trajectories by

r 2 = 2 ( F − F O ) m + a 2 4 m 2 + b 2 4 m 2 . .

For m ≠ 0 , the centre of the circles in the moving plane whose trajectories have the same area is called the Steiner point

s = − 1 2 m ( a b ) . (30)

In the case of m = 0 , Equation (28) can be written

2 ( F − F O ) = x 1 a + x 2 b . (31)

The Steiner circles are reduced to straight lines and then the Steiner point lies at infinity. The normal to the lines of the equal areas in Equation (31) is given by

n = ( a b ) (32)

which is called the Steiner normal [

The expressions for the inverse Steiner point and normal can be deduced in the same way as in the previous subsection. From the Equation (27), we can write the final results

s ′ = − 1 2 m ′ ( a ′ b ′ ) , (33)

n ′ = ( a ′ b ′ ) . (34)

The pole point is the point whose trajectories are instantaneously constant. For the motion

x ′ = ( x − u ) e i α ,

if we calculate the determining equation

d x ′ = 0 ,

we can obtain the moving pole point. The motion can be written in the matrix form as

( x ′ 1 x ′ 2 ) = ( cos α − sin α sin α cos α ) ( x 1 − u 1 x 2 − u 2 ) .

If we differentiate the matrix,

( d x ′ 1 d x ′ 2 ) = ( − sin α − cos α cos α − sin α ) d α ( x 1 − u 1 x 2 − u 2 ) + ( cos α − sin α sin α cos α ) ( − d u 1 − d u 2 ) = 0 (35)

is obtained. Then we have

( − sin α − cos α cos α − sin α ) ( x 1 − u 1 x 2 − u 2 ) d α = ( cos α − sin α sin α cos α ) ( d u 1 d u 2 ) . (36)

By following the necessary operations

( x 1 − u 1 x 2 − u 2 ) = 1 d α ( − sin α cos α − cos α − sin α ) ( cos α − sin α sin α cos α ) ( d u 1 d u 2 ) (37)

( x 1 − u 1 x 2 − u 2 ) = 1 d α ( 0 1 − 1 0 ) ( d u 1 d u 2 ) (38)

( x 1 − u 1 x 2 − u 2 ) = 1 d α ( d u 2 − d u 1 ) (39)

is obtained. From the Equation (39), we have the moving pole point

( p 1 p 2 ) = 1 d α ( d u 2 − d u 1 ) + ( u 1 u 2 ) . (40)

We can write the moving pole point

P M = ( p 1 p 2 ) = ( u 1 + d u 2 d α u 2 − d u 1 d α ) . (41)

If integral is taken over the total angle, we have

∮ P M d α = ∮ ( u 1 + d u 2 d α u 2 − d u 1 d α ) d α = ∮ ( u 1 d α + d u 2 u 2 d α − d u 1 ) = ∮ ( u 1 d α u 2 d α ) . (42)

For m ≠ 0 we arrive at the relation

∮ P M d α = s m = s ∮ d α

between the Steiner point and the pole point. Then we can write

s = ∮ P M d α ∮ d α . (43)

For m = 0 , we have

∮ P M d α = − 1 2 ( a b ) = − 1 2 n . (44)

We begin by differentiating the equation x = ( x ′ − u ′ ) e − i α . And then for d x = 0 , we can obtain the fixed pole point. The motion can be written in the matrix form as

( x 1 x 2 ) = ( cos α sin α − sin α cos α ) ( x ′ 1 − u ′ 1 x ′ 2 − u ′ 2 ) .

If we differentiate the matrix above,

( d x 1 d x 2 ) = ( − sin α cos α − cos α − sin α ) d α ( x ′ 1 − u ′ 1 x ′ 2 − u ′ 2 ) + ( − cos α − sin α sin α − cos α ) ( d u ′ 1 d u ′ 2 ) = 0 (45)

is obtained. Then we have

( − sin α cos α − cos α − sin α ) ( x ′ 1 − u ′ 2 x ′ 2 − u ′ 2 ) d α = ( cos α sin α − sin α cos α ) ( d u ′ 1 d u ′ 2 ) . (46)

By following the necessary operations

( x ′ 1 − u ′ 1 x ′ 2 − u ′ 2 ) = 1 d α ( − sin α − cos α cos α − sin α ) ( cos α sin α − sin α cos α ) ( d u ′ 1 d u ′ 2 ) (47)

( x ′ 1 − u ′ 1 x ′ 2 − u ′ 2 ) = 1 d α ( 0 − 1 1 0 ) ( d u ′ 1 d u ′ 1 ) (48)

and finally

( x ′ 1 − u ′ 1 x ′ 2 − u ′ 2 ) = 1 d α ( − d u ′ 2 d u ′ 1 ) (49)

is found. From the Equation (50), we have the fixed pole point

( p ′ 1 p ′ 2 ) = 1 d α ( − d u ′ 2 d u ′ 1 ) + ( u ′ 1 u ′ 2 ) . (50)

We can write the fixed pole point

P F = ( p ′ 1 p ′ 2 ) = ( u ′ 1 − d u ′ 2 d α u ′ 2 + d u ′ 1 d α ) . (51)

For m ′ ≠ 0 , we arrive at the relation

∮ P F d α = s ′ m ′ = s ′ ∮ d α (52)

between the Steiner point and the fixed pole point.

For m ′ = 0 , we have

∮ P F d α = − 1 2 ( a ′ b ′ ) = − 1 2 n ′ . (53)

Blaschke and Müller gave a relation between the Steiner formula and the polar moment of inertia around the pole for a moment [

In this section we find a formula for the polar moment of inertia and we arrive at the relation between the polar moments of inertia “ T ” and the formula of area “ F ”.

If we use α as a parameter, we need to calculate

T = ∮ ( ( x ′ 1 ) 2 + ( x ′ 2 ) 2 ) d α . (54)

Then by using Equation (4) in Equation (54),

T = ∮ [ ( ( x 1 − u 1 ) cos α − ( x 2 − u 2 ) sin α ) 2 + ( ( x 1 − u 1 ) sin α + ( x 2 − u 2 ) cos α ) 2 ] d α = ∮ ( ( x 1 − u 1 ) 2 ( cos 2 α + sin 2 α ) + ( x 2 − u 2 ) 2 ( cos 2 α + sin 2 α ) ) d α = ∮ ( ( x 1 − u 1 ) 2 + ( x 2 − u 2 ) 2 ) d α = ( x 1 2 + x 2 2 ) ∮ d α − 2 x 1 ∮ u 1 d α − 2 x 2 ∮ u 2 d α + ∮ ( u 1 2 + u 2 2 ) d α (55)

is found.

If we calculate the polar moments of inertia for the origin of the moving system, for T o = T ( x 1 = 0 , x 2 = 0 ) , we have

T O = ∮ ( u 1 2 + u 2 2 ) d α . (56)

If Equation (56) is replaced in Equation (55),

T − T O = ( x 1 2 + x 2 2 ) ∮ d α − 2 x 1 ∮ u 1 d α − 2 x 2 ∮ u 2 d α (57)

is obtained. If Equations (10), (11) and (13) is replaced in Equation (57),

T − T O = ( x 1 2 + x 2 2 ) m + x 1 a + x 2 b (58)

can be written.

As a result, we arrive at the relation between the polar moments of inertia and the formula for the area,

T − T O = 2 ( F − F O ) . (59)

If we use α as a parameter, we need to calculate

T ′ = ∮ ( x 1 2 + x 2 2 ) d α . (60)

Then by using Equations (17) and (18) in Equation (60),

T ′ = ∮ [ ( ( x ′ 1 − u ′ 1 ) cos α + ( x ′ 2 − u ′ 2 ) sin α ) 2 + ( − ( x ′ 1 − u ′ 1 ) sin α + ( x ′ 2 − u ′ 2 ) cos α ) 2 ] d α = ∮ [ ( x ′ 1 − u ′ 1 ) 2 ( cos 2 α + sin 2 α ) + ( x ′ 2 − u ′ 2 ) 2 ( cos 2 α + sin 2 α ) ] d α = ∮ [ ( x ′ 1 − u ′ 1 ) 2 + ( x ′ 2 − u ′ 2 ) 2 ] d α = − ( ( x ′ 1 ) 2 + ( x ′ 2 ) 2 ) ∮ d α + 2 x ′ 1 ∮ u ′ 1 d α + 2 x ′ 2 ∮ u ′ 2 d α + ∮ ( ( u ′ 1 ) 2 + ( u ′ 2 ) 2 ) d α (61)

is found.

If we calculate the polar moments of inertia for the origin of the moving system, for T o = T ( x 1 = 0 , x 2 = 0 ) , we have

T ′ O = ∮ ( ( u ′ 1 ) 2 + ( u ′ 2 ) 2 ) d α . (62)

If Equation (62) is replaced in Equation (61)

T ′ − T ′ O = − ( ( x ′ 1 ) 2 + ( x ′ 2 ) 2 ) ∮ d α + 2 x ′ 1 ∮ u ′ 1 d α + 2 x ′ 2 ∮ u ′ 2 d α (63)

is obtained. If Equations (24), (25) and (26) is replaced in Equation (63)

T ′ − T ′ O = − ( ( x ′ 1 ) 2 + ( x ′ 2 ) 2 ) m ′ − x ′ 1 a ′ − x ′ 2 b ′ (64)

can be written.

Finally, we arrive at the relation between the polar moments of inertia and the formula for the area,

T ′ − T ′ O = − 2 ( F ′ − F ′ O ) . (65)

In the previous sections geometrical objects as the Steiner point or the Steiner normal, the pole point and the polar moments of inertia for closed motions are emphasized in a complex plane. In this section, we want to visualize the experimentally measured motion with these objects.

We choose the sagittal part of the movement of the telescopic crane as an example. The motion of the telescopic crane has a double hinge. The double hinge means that it has two systems, a fixed arm and a moving arm (

By taking

R ( t ) = ( cos ( l ( t ) − k ( t ) ) − sin ( l ( t ) − k ( t ) ) sin ( l ( t ) − k ( t ) ) cos ( l ( t ) − k ( t ) ) ) , U ′ ( t ) = ( L cos ( l ( t ) ) L sin ( l ( t ) ) ) , (66)

we have Equation (1) namely,

X ′ ( t ) = R ( t ) X + U ′ ( t ) .

Also we have U ′ = − R U and

U = ( u 1 u 2 ) = ( − L cos ( k ( t ) ) − L sin ( k ( t ) ) ) . (67)

Then the double hinge motion can be written as

x ′ 1 ( t ) = cos ( l ( t ) − k ( t ) ) ( x 1 + L cos ( k ) ) − sin ( l ( t ) − k ( t ) ) ( x 2 + L sin ( k ) ) x ′ 2 ( t ) = sin ( l ( t ) − k ( t ) ) ( x 1 + L cos ( k ) ) + cos ( l ( t ) − k ( t ) ) ( x 2 + L sin ( k ) ) (68)

where α = l − k is the resulting total angle.

If Equation (68) is derived, we obtain the velocities x ′ 1 • ( t ) and x ′ 2 • ( t ) and so we find

x ′ 1 x ′ 2 • − x ′ 2 x ′ 1 • = ( x 1 2 + x 2 2 ) ( l • ( t ) − k • ( t ) ) + x 1 ( 2 L cos ( k ( t ) ) ( l • ( t ) − k • ( t ) ) + L cos ( k ( t ) ) k • ( t ) ) + x 2 ( 2 L sin ( k ( t ) ) ( l • ( t ) − k • ( t ) ) + L sin ( k ( t ) ) k • ( t ) ) + L 2 l • ( t ) . (69)

We now integrate the previous equation using periodic boundary conditions while assuming that the integrands are periodic functions. The periodicity of f implies that the integrals of the following types vanish ∮ d f = ∫ 1 F f • d t = f | 1 F = 0 .

As a result, only the integrals appearing in the second row of Equation (69) do not become equal to zero and we finally obtain a simplified expression for the area namely,

F = 1 2 ∫ t 1 t 2 ( x ′ 1 x ′ 2 • − x ′ 2 x ′ 1 • ) d t

2 F = x 1 ( ∫ t 1 t 2 2 L cos k l • d t ) + x 2 ( ∫ t 1 t 2 2 L sin k l • d t ) . (70)

In the last equation, by taking

∫ t 1 t 2 2 L cos k l • d t = a (71)

∫ t 1 t 2 2 L sin k l • d t = b , (72)

we can write

2 F = a x 1 + b x 2 . (73)

In this case, we have the steiner normal

n = ( a b ) . (74)

If Equation (67) is replaced in Equation (41), we obtain the pole point P = ( p 1 p 2 ) with the components

p 1 = − L cos k l • l • − k • (75)

p 2 = − L sin k l • l • − k • (76)

is obtained. If integral is taken over the total angle, we have

∮ P d α = ∮ P ( l • − k • ) d t = − L ∫ ( cos k sin k ) l • d t .

If we consider Equations (71) and (72) in the last equation, we obtain

∮ P d α = − 1 2 ( a b ) . (77)

If we consider Equations (54) and (68), then Equation (67) is replaced in Equation (55)

T = x 1 ∮ 2 L cos k ( l • − k • ) d t + x 2 ∮ 2 L sin k ( l • − k • ) d t

and finally

T = x 1 ∮ 2 L cos k l • d t + x 2 ∮ 2 L sin k l • d t (78)

is obtained. If we consider Equations (71), (72) and (73) together, we arrive at the relation between the polar moments of inertia and the formula for the area namely,

T = 2 F . (79)

By taking

R ( t ) = ( cos ( l ( t ) − k ( t ) ) − sin ( l ( t ) − k ( t ) ) sin ( l ( t ) − k ( t ) ) cos ( l ( t ) − k ( t ) ) ) , U ( t ) = ( u 1 ( t ) u 2 ( t ) ) = ( − L cos ( k ( t ) ) − L sin ( k ( t ) ) ) , (80)

we have

X ( t ) = ( R ( t ) ) T ( X ′ − U ′ ( t ) ) .

Also we have U ′ = − R U and

U ′ ( t ) = ( u ′ 1 ( t ) u ′ 2 ( t ) ) = ( L cos ( l ( t ) ) L sin ( l ( t ) ) ) . (81)

So the double hinge may be written as

x 1 ( t ) = cos ( l ( t ) − k ( t ) ) ( x ′ 1 − L cos ( l ) ) + sin ( l ( t ) − k ( t ) ) ( x ′ 2 − L sin ( l ) ) x 2 ( t ) = − sin ( l ( t ) − k ( t ) ) ( x ′ 1 − L cos ( l ) ) + cos ( l ( t ) − k ( t ) ) ( x ′ 2 − L sin ( l ) ) (82)

where α = l − k is the resulting total angle.

By following the same operations similar to the direct motion, we finally obtain the Steiner formula for the inverse motion

2 F ′ = a ′ x ′ 1 + b ′ x ′ 2 (83)

where

− ∫ t 1 t 2 2 L cos l k • d t = a ′ (84)

− ∫ t 1 t 2 2 L sin l k • d t = b ′ . (85)

In this case, we have the Steiner normal

n ′ = ( a ′ b ′ ) . (86)

If Equation (81) is replaced in Equation (51), we obtain the pole point P ′ = ( p ′ 1 p ′ 2 ) with the components

p ′ 1 = − L cos l k • l • − k • p ′ 2 = − L sin l k • l • − k • . (87)

If we consider Equations (60) and (82), then Equation (81) is replaced in Equation (78)

T ′ = x ′ 1 ∮ ( 2 L cos l k • ) d t + x ′ 2 ∮ ( 2 L sin l k • ) d t (88)

is obtained. If we consider Equations (83), (84), (85) and (88) together, we arrive at the relation between the polar moments of inertia and the formula for the area below:

T ′ = − 2 F ′ . (89)

This study is supported by the University of Ondokuz Mayis Project Management Office.

Sener, O. and Tutar, A. (2017) The Steiner Formula and the Polar Moment of Inertia for the Closed Planar Motions in Complex Plane. Advances in Pure Mathematics, 7, 277-289. https://doi.org/10.4236/apm.2017.73014