Optimization of Kinematics of Inclined Swinging Pin

Stanislav Barton

### Abstract

The inclined swinging pin simply transforms the rotary motion into a rotational oscillation. It consists of three components, a rotating inclined tenon, a crossbeam, and a stirrup. The pitch angle of the inclined tenon relative to the x-axis of its rotation is decisive for the kinematics of this joint. Too small inclination angle will result in small amplitude of oscillation; too much inclination will lead to an impulse dynamic stress that can damage the pin. The optimal angle of inclination can be determined from the mathematical model, created in the Maple environment, which uses linear algebra resources to describe the behavior of the joint. Vectors of coordinates of the critical points are created for each joint component. Furthermore, transformation matrices are created which describe the behavior of the corresponding kinematic chain in relation to the inclination angle of the inclined tenon and time.

Keywords: mathematical modeling in Maple, transformation matrix, angular velocity, angular acceleration, optimization

#### 1. Introduction

The inclined swinging pin is used for the simple conversion of rotational motion into oscillation in many agricultural machines. For example, in combine harvesters, it is used to drive a mowing bar that cuts stems. It is composed of three basic components: an inclined tenon—red, a crossbeam—green, and a stirrup—blue, see Figure 1. We study the movement of individual components in a rectangular coordinate system, see Figure 1.

Similar mechanisms are described, for example, in the book [1] that is available in an electronic version [2]. Many animations can be found on the Web, if you use keywords rotation to oscillation mechanism, Google will offer you over 20,000 links.

#### 2. The basic components

#### 2.1 Tenon

The pin is a basic component. Its basic part is an oblique nose. Its axis of symmetry forms an angle ϕ with the axis of rotation of the whole pin, which is identical to the axis x of the coordinate system. The axis of symmetry of the nose

Figure 3. Crossbeam.

Optimization of Kinematics of Inclined Swinging Pin DOI: http://dx.doi.org/10.5772/intechopen.86589

Figure 4. Stirrup.

5

Figure 1. Joint-components and coordinate system.

passes through the beginning of the coordinate system. The spacer serves to secure the nose to the support shaft, see Figure 2.

#### 2.2 Crossbeam and stirrup

The crossbeam is the middle member of the kinematic chain. The tenon nose is pushed through its center hole. As the nose rotates, the crossbar turns. The stirrup is attached to its side pins, and therefore the axis of the side pins is constantly in the plane z = 0, see Figure 3.

Depending on how the crossbeam swings, the stirrup oscillates. The driven device connects to the vertical axis of the stirrup, see Figure 4.

Optimization of Kinematics of Inclined Swinging Pin DOI: http://dx.doi.org/10.5772/intechopen.86589

Figure 4. Stirrup.

passes through the beginning of the coordinate system. The spacer serves to secure

The crossbeam is the middle member of the kinematic chain. The tenon nose is pushed through its center hole. As the nose rotates, the crossbar turns. The stirrup is attached to its side pins, and therefore the axis of the side pins is constantly in the

Depending on how the crossbeam swings, the stirrup oscillates. The driven

device connects to the vertical axis of the stirrup, see Figure 4.

the nose to the support shaft, see Figure 2.

2.2 Crossbeam and stirrup

Figure 1.

Figure 2. Tenon.

4

Joint-components and coordinate system.

Kinematics - Analysis and Applications

plane z = 0, see Figure 3.

### 3. Deriving of transform matrices

We will derive the transformation matrices in the Maple13 program environment. The decisive element for which the transformation matrix will be derived is the crossbeam.

Allow us to specify axes of the base coordinate system in which tenon— Figures 1 and 2—are displayed as [x, y, z]. The axes of the moving coordinate system in which we plot the actual position of the crossbeam are denoted as [X, Y, Z]. Let us denote the transformation matrix from system [x, y, z] into system [X, Y, Z] as R\_xyz. The individual rows of the R\_xyz matrix are the x, y, z coordinates of the unit vectors in the X, Y, Z direction.

The direction of the X-axis is entered as the direction of the tenon's nose axis. The axis X, which is equal to the axis of the nose, creates the angle ϕ with the x-axis and generally the entire tenon is rotated by an angle ψ around the x-axis.

A detailed description of the properties of transformation matrices can be found, for example, in [3].

> restart; with(LinearAlgebra): with(plots): #Maple Start+Libraries > R\_z:=Matrix([[cos(phi),sin(phi),0],

[sin(phi),cos(phi),0],[0,0,1]]): #R\_z - Rotation around z


> R\_xz:=R\_x.R\_z;

$$\begin{aligned} R\_{-}\!\!\!\!\!\!\!\!\/) &= \begin{bmatrix} \cos\left(\not\!\/\rho\right) & \sin\left(\not\!\/\rho\right) & \mathbf{0} \\ -\cos\left(\not\!\/\rho\right)\sin\left(\not\!\/\rho\right) & \cos\left(\not\!\/\rho\right)\cos\left(\not\!\/\rho\right) & \sin\left(\not\!\/\rho\right) \\ \sin\left(\not\!\/\rho\right)\sin\left(\not\!\/\rho\right) & -\sin\left(\not\!\/\rho\right)\cos\left(\not\!\/\rho\right) & \cos\left(\not\!\/\rho\right) \end{bmatrix} \end{aligned} \tag{1}$$

> Z:=[Zx,Zy,Zz];

Movement of the unit vectors X, Y, Z.

Figure 5.

> Z:=subs(%,Z);

the angle of rotation ψ.

M\_XYZ ≔

4. Optimization

7

> M\_XYZ:=Matrix([X,Y,Z]);

> E1:=add(w,w=zip((u,v)->u\*v,X,Z));

Optimization of Kinematics of Inclined Swinging Pin DOI: http://dx.doi.org/10.5772/intechopen.86589

where %1 ¼

<sup>Z</sup> <sup>≔</sup> sin ð Þ <sup>ψ</sup> cosð Þ <sup>φ</sup> sin ð Þ <sup>φ</sup>

cosð Þ ψ sin ð Þ φ %1

� sin ð Þ <sup>ψ</sup> sin ð Þ <sup>φ</sup> cosð Þ <sup>φ</sup> %1

q

> E2:=add(w,w=zip((u,v)->u\*v,Z,Y)); E3:=Zx^2+Zy^2+Zz^2=1; > simplify(allvalues(solve({E1,E2,E3},{Zx,Zy,Zz}))[1],symbolic);

%1 ;

cosð Þ ψ

sin ð Þ ψ cosð Þ ψ sin ð Þ φ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

cosð Þ φ � cosð Þ ψ sin ð Þ φ sin ð Þ ψ sin ð Þ φ

cosð Þ φ %1

sin ð Þ ψ cosð Þ ψ sin ð Þ φ

%1

2 cosð Þ φ

%1 ; %1 " #,

<sup>2</sup> <sup>þ</sup> cosð Þ <sup>ψ</sup>

Finally, it is possible to create a graph showing the curves described by the endpoints of the X, Y, and Z vectors. For instance, if ϕ ¼ 2π=15 and ψ ∈h i 0; 2π are selected, then the resulting graph is Figure 5. The corresponding animation is stored

The orientation of the Y axis can be derived from the matrix M\_XYZ, see Eq. (5). The orientation of this axis indicates the angle of rotation of the stirrup about the z axis.

in the file An\_Axes.avi, available to download in the Videos section.

From the vectors X, Y, Z, it is possible to create a transformation matrix M\_XYZ, which will describe the orientation of the crossbeam depending on the angle ϕ and

2

2

:

0

%1

(4)

(5)

<sup>2</sup> <sup>þ</sup> cosð Þ <sup>φ</sup>

2

> X:=convert(R\_xz.<1,0,0>,list);

$$X \coloneqq [\cos(\varphi), -\cos(\psi)\sin(\varphi), \sin\left(\psi\right)\sin\left(\varphi\right)].\tag{2}$$

Equation (2) specifies the direction of the nose axis, thus X-axis vector in the coordinates [x, y, z]. Coordinates Y are calculated easily. We know that |Y| = 1 and Yz must always be equal to 0 and that X and Y must be perpendicular to each other, i.e., X�Y = 0, the scalar product X and Y must be equal to 0. Therefore, we choose the vector coordinates Y = [Yx, Yy, 0] and Yx and Yy are calculated from the following equations:

> Y:=[Yx,Yy,0]; > E1:=add(w,w=zip((u,v)->u\*v,X,Y)); E2:=Yx^2+Yy^2=1; > simplify(allvalues(solve({E1,E2},{Yx,Yy}))[2],symbolic); > Y:=subs(%,Y); 2

$$Y \coloneqq \left[ \frac{\cos\left(\wp\right)\sin\left(\wp\right)}{\sqrt{\cos\left(\wp\right)^2 - \cos\left(\wp\right)^2\cos\left(\wp\right)^2 + \cos\left(\wp\right)^2}}, \frac{\cos\left(\wp\right)}{\sqrt{\cos\left(\wp\right)^2 - \cos\left(\wp\right)^2\cos\left(\wp\right)^2 + \cos\left(\wp\right)^2}}, 0\right], \tag{3}$$

Equations E1 and E2 have two solutions. In order for the XYZ system to be oriented counterclockwise, a second solution has to be chosen. Coordinates Z are calculated from the conditions X�Z = 0, Y�Z = 0, and |Z| = 1.

Optimization of Kinematics of Inclined Swinging Pin DOI: http://dx.doi.org/10.5772/intechopen.86589

#### Figure 5.

3. Deriving of transform matrices

Kinematics - Analysis and Applications

coordinates of the unit vectors in the X, Y, Z direction.

> R\_z:=Matrix([[cos(phi),sin(phi),0],

> R\_x:=Matrix([[1,0,0],[0,cos(psi),sin(psi)],

the crossbeam.

for example, in [3].

> R\_xz:=R\_x.R\_z;

following equations: > Y:=[Yx,Yy,0];

> Y:=subs(%,Y);

2 6 4

6

R\_xz ≔

> X:=convert(R\_xz.<1,0,0>,list);

<sup>Y</sup> <sup>≔</sup> cosð Þ <sup>ψ</sup> sin ð Þ <sup>φ</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

cosð Þ <sup>ψ</sup> <sup>2</sup> � cosð Þ <sup>ψ</sup> <sup>2</sup> cosð Þ <sup>φ</sup> <sup>2</sup> <sup>þ</sup> cosð Þ <sup>ϕ</sup> <sup>2</sup>

We will derive the transformation matrices in the Maple13 program environment. The decisive element for which the transformation matrix will be derived is

The direction of the X-axis is entered as the direction of the tenon's nose axis. The axis X, which is equal to the axis of the nose, creates the angle ϕ with the x-axis

A detailed description of the properties of transformation matrices can be found,

cosð Þ φ sin ð Þ φ 0 � cosð Þ ψ sin ð Þ φ cosð Þ ψ cosð Þ φ sin ð Þ ψ sin ð Þ ψ sin ðÞ � φ sin ð Þ ψ cosð Þ φ cosð Þ ψ

X ≔ ½ � cosð Þ φ ; � cosð Þ ψ sin ð Þ φ ; sin ð Þ ψ sin ð Þ φ : (2)

Equation (2) specifies the direction of the nose axis, thus X-axis vector in the coordinates [x, y, z]. Coordinates Y are calculated easily. We know that |Y| = 1 and Yz must always be equal to 0 and that X and Y must be perpendicular to each other, i.e., X�Y = 0, the scalar product X and Y must be equal to 0. Therefore, we choose the vector coordinates Y = [Yx, Yy, 0] and Yx and Yy are calculated from the

<sup>q</sup> ; cosð Þ <sup>φ</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

Equations E1 and E2 have two solutions. In order for the XYZ system to be oriented counterclockwise, a second solution has to be chosen. Coordinates Z are

cosð Þ <sup>ψ</sup> <sup>2</sup> � cosð Þ <sup>ψ</sup> <sup>2</sup> cosð Þ <sup>φ</sup> <sup>2</sup> <sup>þ</sup> cosð Þ <sup>ϕ</sup> <sup>2</sup> <sup>q</sup> ; <sup>0</sup>

(1)

3 7 5:

(3)

> restart; with(LinearAlgebra): with(plots): #Maple Start+Libraries

and generally the entire tenon is rotated by an angle ψ around the x-axis.

[sin(phi),cos(phi),0],[0,0,1]]): #R\_z - Rotation around z

[0,-sin(psi),cos(psi)]]):]]): #R\_x - Rotation around x

> E1:=add(w,w=zip((u,v)->u\*v,X,Y)); E2:=Yx^2+Yy^2=1; > simplify(allvalues(solve({E1,E2},{Yx,Yy}))[2],symbolic);

calculated from the conditions X�Z = 0, Y�Z = 0, and |Z| = 1.

Allow us to specify axes of the base coordinate system in which tenon— Figures 1 and 2—are displayed as [x, y, z]. The axes of the moving coordinate system in which we plot the actual position of the crossbeam are denoted as [X, Y, Z]. Let us denote the transformation matrix from system [x, y, z] into system [X, Y, Z] as R\_xyz. The individual rows of the R\_xyz matrix are the x, y, z

Movement of the unit vectors X, Y, Z.

$$\succ \heartsuit := [\succ \heartsuit, \heartsuit, \heartsuit, \heartsuit] \colon$$


> Z:=subs(%,Z);

$$Z \coloneqq \left[ \frac{\sin \left( \wp \right) \cos \left( \wp \right) \sin \left( \wp \right)}{961}, \frac{\sin \left( \wp \right) \cos \left( \wp \right) \sin \left( \wp \right)^2}{961}, 961 \right],\tag{4}$$
 
$$\text{where}\ \mathfrak{W}\mathbf{1} = \sqrt{\cos \left( \wp \right)^2 + \cos \left( \wp \right)^2 \cos \left( \wp \right)^2 + \cos \left( \wp \right)^2}.$$

From the vectors X, Y, Z, it is possible to create a transformation matrix M\_XYZ, which will describe the orientation of the crossbeam depending on the angle ϕ and the angle of rotation ψ.

> M\_XYZ:=Matrix([X,Y,Z]);

$$M\_{-}XYZ \coloneqq \begin{bmatrix} \cos\left(\boldsymbol{\wp}\right) & -\cos\left(\boldsymbol{\wp}\right)\sin\left(\boldsymbol{\wp}\right) & \sin\left(\boldsymbol{\wp}\right)\sin\left(\boldsymbol{\wp}\right) \\ \frac{\cos\left(\boldsymbol{\wp}\right)\sin\left(\boldsymbol{\wp}\right)}{961} & \frac{\cos\left(\boldsymbol{\wp}\right)}{961} & \boldsymbol{0} \\ -\frac{\sin\left(\boldsymbol{\wp}\right)\sin\left(\boldsymbol{\wp}\right)\cos\left(\boldsymbol{\wp}\right)}{961} & \frac{\sin\left(\boldsymbol{\wp}\right)\cos\left(\boldsymbol{\wp}\right)\sin\left(\boldsymbol{\wp}\right)^{2}}{961} & \frac{\boldsymbol{\wp}}{961} \end{bmatrix} . \tag{5}$$

Finally, it is possible to create a graph showing the curves described by the endpoints of the X, Y, and Z vectors. For instance, if ϕ ¼ 2π=15 and ψ ∈h i 0; 2π are selected, then the resulting graph is Figure 5. The corresponding animation is stored in the file An\_Axes.avi, available to download in the Videos section.

#### 4. Optimization

The orientation of the Y axis can be derived from the matrix M\_XYZ, see Eq. (5). The orientation of this axis indicates the angle of rotation of the stirrup about the z axis. > Y:=evalm([0,1,0].M\_XYZ);

$$Y \coloneqq \left[ -\frac{\cos\left(\wp\right)\sin\left(\wp\right)}{\text{961}}, -\frac{\cos\left(\wp\right)}{\text{961}}, 0\right].\tag{6}$$

The angle of rotation of the stirrup is equal to the angle between the y and Y axes. Let us denote this angle as θ.

> theta:=arctan(rz[1]/rz[2]);

$$\theta \coloneqq \arctan\left(\frac{\cos\left(\wp\right)\sin\left(\wp\right)}{\cos\left(\wp\right)}\right).\tag{7}$$

The angle ψ indicates the angle of rotation of the pin about the x-axis. We can therefore substitute:

> su:=psi=2\*Pi\*tau:

$$
\mathfrak{su} := \mathfrak{w} = 2\,\pi\,\mathfrak{r}.\tag{8}
$$

value of ϕ further increases, then this local maximum becomes the local minimum. This means that the optimum value of the angle ϕ must be the one at which the transition from local maximum to local minimum occurs. From a mathematical

> � � � � <sup>τ</sup>¼<sup>1</sup> 2

The value of the angle ϕ can be easily determined from Eq. (12) using Maple.

<sup>6</sup> � 24 sin 2ð Þ πτ <sup>4</sup> tan ð Þ <sup>φ</sup> <sup>4</sup> <sup>þ</sup> 3 cos 2ð Þ πτ <sup>2</sup> tan ð Þ <sup>φ</sup>

<sup>4</sup> <sup>þ</sup> 5 tan ð Þ <sup>φ</sup>

<sup>2</sup> � �

1 þ tan ð Þ φ

ffiffi 5 p 5

Φ ≔ arctan

� �; �arctan

Sold ≔ ½ � 24:09484255 degrees; � 24:09484255 degrees; 0 degrees (16)

ffiffi 5 p 5

<sup>Þ</sup><sup>=</sup> <sup>1</sup> <sup>þ</sup> cos 2ð Þ πτ <sup>2</sup> tan ð Þ <sup>φ</sup> <sup>2</sup> � �<sup>4</sup>

¼ 0: (12)

<sup>2</sup> <sup>þ</sup> 9 cos 2ð Þ πτ <sup>4</sup> tan ð Þ <sup>φ</sup> <sup>4</sup>

2

(13)

<sup>6</sup> cos 2ð Þ πτ <sup>2</sup>

<sup>6</sup> <sup>þ</sup> 3 tan ð Þ <sup>φ</sup>

� �: (17)

<sup>2</sup> � �<sup>4</sup> (14)

ffiffi 5 p 5 � �; <sup>0</sup> � � (15)

<sup>d</sup><sup>2</sup> ε τð Þ ; <sup>φ</sup> dτ<sup>2</sup>

point of view, the following condition must be satisfied:

Angular acceleration for ϕ = 5, 10, 15, 20, 25, 30, 35, 40, and 45°.

Optimization of Kinematics of Inclined Swinging Pin DOI: http://dx.doi.org/10.5772/intechopen.86589

Ett <sup>≔</sup> � 16 cos 2ð Þ πτ <sup>π</sup><sup>4</sup> tan ð Þ ð� <sup>φ</sup> <sup>1</sup> <sup>þ</sup> 24 sin 2ð Þ πτ <sup>4</sup> tan ð Þ <sup>φ</sup>

<sup>16</sup>π<sup>4</sup> tan ðÞ� <sup>φ</sup> <sup>1</sup> <sup>þ</sup> 9 tan ð Þ <sup>φ</sup>

Sol ≔ arctan

> Sold:=map(u->evalf(convert(u,degrees)),Sol);

<sup>þ</sup> 8 sin 2ð Þ πτ <sup>2</sup> tan ð Þ <sup>φ</sup> <sup>4</sup> cos 2ð Þ πτ <sup>2</sup> � 20 sin 2ð Þ πτ <sup>2</sup> tan ð Þ <sup>φ</sup>

<sup>6</sup> cos 2ð Þ πτ <sup>4</sup>

> Ett:=normal(diff(epsilon,tau,tau));

<sup>þ</sup> 5 cos 2ð Þ πτ <sup>6</sup> tan ð Þ <sup>φ</sup>

<sup>þ</sup> 28 sin 2ð Þ πτ <sup>2</sup> tan ð Þ <sup>φ</sup>

> Ett2:=eval(subs(tau=1/2,Ett));

Ett2 ≔

Figure 6.

> Phi:=Sol[1];

9

> Sol:=[solve(Ett2,phi)];

> theta:=subs(su,sin(phi)=cos(phi)\*tan(phi),theta);

$$\theta \coloneqq \arctan\left(\cos\left(2\pi\tau\right)\tan\left(\rho\right)\right),\tag{9}$$

where τ = t/T = dimensionless time, t = time, and T = duration of one revolution = one period.

It is now possible to determine the angular velocity of the stirrup oscillations, ω ¼ dθ τð Þ=dτ, and the angular acceleration of the stirrup oscillations, <sup>ε</sup> <sup>¼</sup> <sup>d</sup><sup>2</sup> θ τð Þ=d<sup>τ</sup> 2 .

> omega:=diff(theta,tau);

$$\rho \coloneqq -\frac{2\sin\left(2\pi\tau\right)\pi\tan\left(\rho\right)}{T\left(1+\cos\left(2\pi\tau\right)^2\tan\left(\rho\right)^2\right)}\tag{10}$$

> epsilon:=normal(diff(omega,tau));

$$\varepsilon = -\frac{4\cos\left(2\pi t\right)\pi^2 \tan\left(\wp\right)\left(1+\cos\left(2\pi t\right)^2 \tan\left(\wp\right)^2 + 2\sin\left(2\pi t\right)^2 \tan\left(\wp\right)^2\right)}{\left(1+\cos\left(2\pi t\right)^2 \tan\left(\wp\right)^2\right)^2} \tag{11}$$

Now we plot the graph of ε(τ) for τ ∈h i 0; 1 : The values of ϕ will be chosen between 5 and 45° with a step of 5°, see Figure 6.

> EP:=plot([seq(epsilon,phi=expand(Pi/36\*[\$1..9]))], tau=0..1,color=red): EP;

It is quite clear from Figure 6 that with an increase in angle ϕ—that is, the angle between the pin nose and the x-axis—the course of the angular acceleration of the oscillations begins to deviate sharply from the sinusoidal shape. Thus, large variations in angular acceleration will occur for large ϕ values, resulting in rapid wear of the entire mechanism. This can lead to mechanical damage, because high angular acceleration values require high torque forces. Conversely, small values of the angle ϕ will lead only to a low oscillation amplitude, which may not be sufficient for the proper operation of the driven device. It is therefore necessary to find the optimum angle value ϕ.

After studying the graphs in Figure 6, the value at which the angular acceleration for τ = 1/2 reaches the local maximum possible value appears optimal. If the

Optimization of Kinematics of Inclined Swinging Pin DOI: http://dx.doi.org/10.5772/intechopen.86589

> Y:=evalm([0,1,0].M\_XYZ);

Kinematics - Analysis and Applications

Let us denote this angle as θ.

therefore substitute: > su:=psi=2\*Pi\*tau:

revolution = one period.

2 . > omega:=diff(theta,tau);

> epsilon:=normal(diff(omega,tau));

between 5 and 45° with a step of 5°, see Figure 6.

tau=0..1,color=red): EP;

4 cos 2ð Þ π τ <sup>π</sup><sup>2</sup> tan ð Þ <sup>φ</sup> <sup>1</sup> <sup>þ</sup> cos 2ð Þ <sup>π</sup> <sup>t</sup> <sup>2</sup> tan ð Þ <sup>φ</sup>

> EP:=plot([seq(epsilon,phi=expand(Pi/36\*[\$1..9]))],

<sup>ε</sup> <sup>¼</sup> <sup>d</sup><sup>2</sup> θ τð Þ=d<sup>τ</sup>

ε ≔ �

8

> theta:=arctan(rz[1]/rz[2]);

<sup>Y</sup> <sup>≔</sup> � cosð Þ <sup>ψ</sup> sin ð Þ <sup>φ</sup>

θ ≔ arctan

> theta:=subs(su,sin(phi)=cos(phi)\*tan(phi),theta);

%1 ; � cosð Þ <sup>φ</sup>

cosð Þ ψ sin ð Þ φ cosð Þ φ 

The angle of rotation of the stirrup is equal to the angle between the y and Y axes.

The angle ψ indicates the angle of rotation of the pin about the x-axis. We can

where τ = t/T = dimensionless time, t = time, and T = duration of one

ω ¼ dθ τð Þ=dτ, and the angular acceleration of the stirrup oscillations,

It is now possible to determine the angular velocity of the stirrup oscillations,

<sup>ω</sup> <sup>≔</sup> � 2 sin 2ð Þ π τ <sup>π</sup> tan ð Þ <sup>φ</sup>

<sup>1</sup> <sup>þ</sup> cos 2ð Þ <sup>π</sup> <sup>t</sup> <sup>2</sup> tan ð Þ <sup>φ</sup>

Now we plot the graph of ε(τ) for τ ∈h i 0; 1 : The values of ϕ will be chosen

It is quite clear from Figure 6 that with an increase in angle ϕ—that is, the angle between the pin nose and the x-axis—the course of the angular acceleration of the oscillations begins to deviate sharply from the sinusoidal shape. Thus, large variations in angular acceleration will occur for large ϕ values, resulting in rapid wear of the entire mechanism. This can lead to mechanical damage, because high angular acceleration values require high torque forces. Conversely, small values of the angle ϕ will lead only to a low oscillation amplitude, which may not be sufficient for the proper operation of

After studying the graphs in Figure 6, the value at which the angular acceleration for τ = 1/2 reaches the local maximum possible value appears optimal. If the

the driven device. It is therefore necessary to find the optimum angle value ϕ.

<sup>T</sup> <sup>1</sup> <sup>þ</sup> cos 2ð Þ π τ <sup>2</sup> tan ð Þ <sup>φ</sup>

%1 ; <sup>0</sup>

su ≔ ψ ¼ 2π τ: (8)

<sup>2</sup> (10)

<sup>2</sup>

<sup>2</sup> <sup>2</sup> (11)

<sup>2</sup> <sup>þ</sup> 2 sin 2ð Þ <sup>π</sup> <sup>t</sup> <sup>2</sup> tan ð Þ <sup>φ</sup>

θ ≔ arctan cos 2 ð Þ ð Þ πτ tan ð Þ φ , (9)

: (6)

: (7)

Figure 6. Angular acceleration for ϕ = 5, 10, 15, 20, 25, 30, 35, 40, and 45°.

value of ϕ further increases, then this local maximum becomes the local minimum. This means that the optimum value of the angle ϕ must be the one at which the transition from local maximum to local minimum occurs. From a mathematical point of view, the following condition must be satisfied:

$$\left. \frac{d^2 \varepsilon(\tau, \rho)}{d\tau^2} \right|\_{\tau = \frac{1}{2}} = 0. \tag{12}$$

The value of the angle ϕ can be easily determined from Eq. (12) using Maple. > Ett:=normal(diff(epsilon,tau,tau));

$$\begin{aligned} \text{Et} & \coloneqq -16 \cos \left( 2\pi \tau \right) \pi^4 \tan \left( \wp \right) \left( -1 + 24 \sin \left( 2\pi \tau \right)^4 \tan \left( \wp \right)^6 \cos \left( 2\pi \tau \right)^2 \right. \\ & \left. + 8 \sin \left( 2\pi \tau \right)^2 \tan \left( \wp \right)^4 \cos \left( 2\pi \tau \right)^2 - 20 \sin \left( 2\pi \tau \right)^2 \tan \left( \wp \right)^2 + 9 \cos \left( 2\pi \tau \right)^4 \tan \left( \wp \right)^4 \right. \\ & \left. + 5 \cos \left( 2\pi \tau \right)^6 \tan \left( \wp \right)^6 - 24 \sin \left( 2\pi \tau \right)^4 \tan \left( \wp \right)^4 + 3 \cos \left( 2\pi \tau \right)^2 \tan \left( \wp \right)^2 \\ & \left. + 28 \sin \left( 2\pi \tau \right)^2 \tan \left( \wp \right)^6 \cos \left( 2\pi \tau \right)^4 \right) / \left( 1 + \cos \left( 2\pi \tau \right)^2 \tan \left( \wp \right)^2 \right)^4 \end{aligned} \tag{13}$$

> Ett2:=eval(subs(tau=1/2,Ett));

$$\text{Ett2} \coloneqq \frac{\mathbf{16}\,\pi^4 \tan\left(\varphi\right) \left(-\mathbf{1} + \mathbf{9}\tan\left(\varphi\right)^4 + \mathbf{5}\tan\left(\varphi\right)^6 + \mathbf{3}\tan\left(\varphi\right)^2\right)}{\left(\mathbf{1} + \tan\left(\varphi\right)^2\right)^4} \tag{14}$$

> Sol:=[solve(Ett2,phi)];

$$Sol := \left[ \arctan\left(\frac{\sqrt{5}}{5}\right), -\arctan\left(\frac{\sqrt{5}}{5}\right), 0 \right] \tag{15}$$

> Sold:=map(u->evalf(convert(u,degrees)),Sol);

Sold ≔ ½ � 24:09484255 degrees; � 24:09484255 degrees; 0 degrees (16)

> Phi:=Sol[1];

$$\Phi \coloneqq \arctan\left(\frac{\sqrt{5}}{5}\right). \tag{17}$$

> Epsilon:=simplify(subs(phi=Phi,epsilon),symbolic);

$$\mathbf{E}' \coloneqq \frac{4\pi^2\sqrt{5}\cos\left(2\pi\tau\right)\left(-7+\cos\left(2\pi\tau\right)^2\right)}{25+10\cos\left(2\pi\tau\right)^2+\cos\left(2\pi\tau\right)^4}.\tag{18}$$

It is now possible again to plot curves describing the angular acceleration dependence on τ and ϕ, and highlight the curve for ϕ = Φ, see Figure 7.

> EO:=plot(subs(phi=Phi,epsilon),tau=0..1,color=blue,thickness=3):

> display({EP,EO});

To make sure that the angle Φ is optimal, we will plot the angular velocity depending on τ, see Figure 8.

> Omega:=simplify(subs(phi=Phi,T=1,omega),symbolic);

$$\Omega \coloneqq -\frac{2\pi\sqrt{5}\sin\left(2\pi\tau\right)}{5+\cos\left(2\pi\tau\right)^2}.\tag{19}$$

> OO:=plot(Omega,tau=0..1,color=blue,thickness=3):

Θ ≔ arctan

one kinematic variable is plotted against another kinematic variable.

angular velocity of the oscillations—ε(ω(τ)), see Figure 12.

> TW:=plot([seq([theta,omega,tau=0..1],

> TE:=plot([seq([theta,epsilon,tau=0..1],

> OE:=plot([seq([omega,epsilon,tau=0..1],

> TP:=plot([seq(theta,phi=expand(Pi/36\*[\$1..9]))],

> TO:=plot(Theta,tau=0..1,color=blue,thickness=3):

5. Plotting of the dependencies in phase spaces

> Theta:=simplify(subs(phi=Phi,theta));

Optimization of Kinematics of Inclined Swinging Pin DOI: http://dx.doi.org/10.5772/intechopen.86589

tau=0..1,color=red):

And yet it is possible to plot the dependence of the angle θ on τ, see Figure 9.

ffiffi 5 p 5

Further evidence of the choice of the optimum angle value Φ, that is, the choice of a suitable compromise between the oscillation amplitude and smooth oscillation over time, can be given by plotting the phase diagrams, which are graphs in which

In this case, we will present three phase diagrams, where we will plot the dependence of the oscillation angular velocity on the current value of the oscillation angle—ω(θ(τ)), see Figure 10, then the course of the angular acceleration of the oscillations against the current value of the oscillation angle—ε(θ(τ)), see Figure 11 and finally the course of the angular acceleration of the oscillations versus the

phi=expand(Pi/36\*[\$1..9]))],color=red,numpoints=500):

phi=expand(Pi/36\*[\$1..9]))],color=red,numpoints=500): > TEO:=plot([Theta,Epsilon,tau=0..1],color=blue,thickness=3):

> TWO:=plot([Theta,Omega,tau=0..1],color=blue,thickness=3):

cos 2ð Þ π τ � �: (20)

> display({OP,OO});

Dependence of angle θ on τ.

Figure 9.

> display({TP,TO});

> display({TW,TWO});

> display({TE,TEO});

11

> OP:=plot([seq(omega,phi=expand(Pi/36\*[\$1..9]))], tau=0..1,color=red):

Figure 7. Angular acceleration and angular acceleration for the optimized ϕ.

Figure 8. The angular velocity of the oscillations.

Figure 9. Dependence of angle θ on τ.

> Epsilon:=simplify(subs(phi=Phi,epsilon),symbolic);

<sup>5</sup> <sup>p</sup> cos 2ð Þ� π τ <sup>7</sup> <sup>þ</sup> cos 2ð Þ π τ <sup>2</sup> � �

<sup>5</sup> <sup>p</sup> sin 2ð Þ π τ

It is now possible again to plot curves describing the angular acceleration

To make sure that the angle Φ is optimal, we will plot the angular velocity

dependence on τ and ϕ, and highlight the curve for ϕ = Φ, see Figure 7. > EO:=plot(subs(phi=Phi,epsilon),tau=0..1,color=blue,thickness=3):

<sup>Ω</sup> <sup>≔</sup> � <sup>2</sup><sup>π</sup> ffiffi

> Omega:=simplify(subs(phi=Phi,T=1,omega),symbolic);

> OP:=plot([seq(omega,phi=expand(Pi/36\*[\$1..9]))],

tau=0..1,color=red):

Angular acceleration and angular acceleration for the optimized ϕ.

<sup>25</sup> <sup>þ</sup> 10 cos 2ð Þ π τ <sup>2</sup> <sup>þ</sup> cos 2ð Þ π τ <sup>4</sup> : (18)

<sup>5</sup> <sup>þ</sup> cos 2ð Þ π τ <sup>2</sup> : (19)

4π<sup>2</sup> ffiffi

Ε´ ≔

Kinematics - Analysis and Applications

> display({EP,EO});

Figure 7.

Figure 8.

10

The angular velocity of the oscillations.

depending on τ, see Figure 8.

> OO:=plot(Omega,tau=0..1,color=blue,thickness=3):

> display({OP,OO});

And yet it is possible to plot the dependence of the angle θ on τ, see Figure 9. > Theta:=simplify(subs(phi=Phi,theta));

$$\Theta \coloneqq \arctan\left(\frac{\sqrt{5}}{5}\cos\left(2\pi\tau\right)\right). \tag{20}$$


### 5. Plotting of the dependencies in phase spaces

Further evidence of the choice of the optimum angle value Φ, that is, the choice of a suitable compromise between the oscillation amplitude and smooth oscillation over time, can be given by plotting the phase diagrams, which are graphs in which one kinematic variable is plotted against another kinematic variable.

In this case, we will present three phase diagrams, where we will plot the dependence of the oscillation angular velocity on the current value of the oscillation angle—ω(θ(τ)), see Figure 10, then the course of the angular acceleration of the oscillations against the current value of the oscillation angle—ε(θ(τ)), see Figure 11 and finally the course of the angular acceleration of the oscillations versus the angular velocity of the oscillations—ε(ω(τ)), see Figure 12.

> TW:=plot([seq([theta,omega,tau=0..1],

	- phi=expand(Pi/36\*[\$1..9]))],color=red,numpoints=500):

phi=expand(Pi/36\*[\$1..9]))],color=red,numpoints=500): > OEO:=plot([Omega,Epsilon,tau=0..1],color=blue,thickness=3): > display({OE,OEO});

6. Results and conclusions

Optimization of Kinematics of Inclined Swinging Pin DOI: http://dx.doi.org/10.5772/intechopen.86589

Figure 13.

13

amplitude and the mechanical stress of the entire joint.

Positions of components for rotation angles ψ = 0, 90, 180, and 270°.

All of the presented calculations and graphs show that the optimal angle of the nose inclination is <sup>ϕ</sup> = arctan(5�1/2) = 0.4205343352 rad = 24° 5<sup>0</sup> 41.43″. If the angle ϕ is less than the optimum value, then the amplitude of the resulting oscillations will be low and gears will be required. This will lead to a significant increase in the price of the whole pin. Otherwise, when the value of the angle ϕ is higher than the optimum value, there is a sharp increase in the instantaneous angular acceleration of the oscillations. High angular acceleration values mean that the entire pin must transmit high torque values to which the entire pin must be dimensioned. This will mean either increasing the size of the pin or using materials with satisfactory mechanical properties. This solution again leads to an increase in the price of the whole pin. If the pin is not appropriately sized for the transmitted torque, it may be mechanically damaged or blocked. The condition for the optimum angle value ϕ in Eq. (12), thus, appears to be the optimal compromise between the oscillation

Of course, the optimum shape of the oscillations should have a sinusoidal waveform, i.e., θ(t) = a sin(2πt/T), where θ = oscillation angle, t = time, a = oscillation amplitude, and T = oscillation period. The corresponding dependencies of angular velocity and angular acceleration over time are also in the form of sine or cosine

#### Figure 10. The phase diagram ω(θ(τ)).

Figure 11. The phase diagram of ε(θ(τ)).

Figure 12. The phase diagram of ε(ω(τ)).

### 6. Results and conclusions

phi=expand(Pi/36\*[\$1..9]))],color=red,numpoints=500): > OEO:=plot([Omega,Epsilon,tau=0..1],color=blue,thickness=3):

> display({OE,OEO});

Kinematics - Analysis and Applications

Figure 11.

Figure 10.

The phase diagram ω(θ(τ)).

Figure 12.

12

The phase diagram of ε(ω(τ)).

The phase diagram of ε(θ(τ)).

All of the presented calculations and graphs show that the optimal angle of the nose inclination is <sup>ϕ</sup> = arctan(5�1/2) = 0.4205343352 rad = 24° 5<sup>0</sup> 41.43″. If the angle ϕ is less than the optimum value, then the amplitude of the resulting oscillations will be low and gears will be required. This will lead to a significant increase in the price of the whole pin. Otherwise, when the value of the angle ϕ is higher than the optimum value, there is a sharp increase in the instantaneous angular acceleration of the oscillations. High angular acceleration values mean that the entire pin must transmit high torque values to which the entire pin must be dimensioned. This will mean either increasing the size of the pin or using materials with satisfactory mechanical properties. This solution again leads to an increase in the price of the whole pin. If the pin is not appropriately sized for the transmitted torque, it may be mechanically damaged or blocked. The condition for the optimum angle value ϕ in Eq. (12), thus, appears to be the optimal compromise between the oscillation amplitude and the mechanical stress of the entire joint.

Of course, the optimum shape of the oscillations should have a sinusoidal waveform, i.e., θ(t) = a sin(2πt/T), where θ = oscillation angle, t = time, a = oscillation amplitude, and T = oscillation period. The corresponding dependencies of angular velocity and angular acceleration over time are also in the form of sine or cosine

Figure 13. Positions of components for rotation angles ψ = 0, 90, 180, and 270°.

functions. If we draw phase diagrams for these oscillations, then the graph of the dependencies [θ(τ), ω(τ)] = ω(θ(τ)) and [ω(τ), ε(τ)] = ε(ω(τ)) is the shape of an ellipse and the dependency [θ(τ), ε(τ)] = ε(θ(τ)) is the shape of a line segment. The amplitude of the oscillations is then equal to the length of the main half-axis of the ellipse, which displays ε(θ(τ)).

References

9780831110321

2019]

15

[1] Horton HL, Newell JA. Ingenious Mechanisms. New York: Industrial Press

Optimization of Kinematics of Inclined Swinging Pin DOI: http://dx.doi.org/10.5772/intechopen.86589

[2] Horton H. Ingenious Mechanisms. Volume IV [Internet]. 2017. Available from: https://books.industrialpress.c om/ingenious-mechanisms-vol-iv-eb ook-only.html [Accessed: March 15,

[3] Leon S. Linear transformations. In: Leon S, editor. Linear Algebra with Applications. 8th ed. New Jersey: Pearson Prentice Hall; 2010.

pp. 166-198. ISBN: 139780136009290

Inc.; 1967;4:245-265. ISBN:

If we examine the graphs in Figures 10–12, we find that the highlighted curve that corresponds to the optimum angle ϕ are an excellent compromise between the shape of the curve (ellipse or line segment) and the dimensions of the curve that correspond to the oscillation amplitude. Similarly, the graphs of functions θ(τ), ω(τ), and ε(τ), which correspond to the optimal value of the angle ϕ, are the best compromise between the desired function shape and the oscillation amplitude, see the graphs in Figures 7–9.

Figure 1 shows the entire inclined swinging pin, while Figure 2 shows the tenon plotted for this angle. Several bottom views of the entire joint during one quarter of the working period are shown in Figure 13.

The main advantages of an inclined swinging pin compared to similar mechanisms are:


For these reasons, it is used in a wide variety of agricultural machines. An animation (An\_Joint.avi) that displays the entire device's activity can be downloaded from the Videos section.

### 7. Videos

All videos referenced in this chapter are available from: https://bit.ly/2vZVGXR

### Author details

Stanislav Barton Faculty of Electrical Engineering, Automatics and Informatics, Department of Informatics, Technical University of Opole, Poland

\*Address all correspondence to: s.barton@po.opole.pl

© 2019 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/ by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Optimization of Kinematics of Inclined Swinging Pin DOI: http://dx.doi.org/10.5772/intechopen.86589

## References

functions. If we draw phase diagrams for these oscillations, then the graph of the dependencies [θ(τ), ω(τ)] = ω(θ(τ)) and [ω(τ), ε(τ)] = ε(ω(τ)) is the shape of an ellipse and the dependency [θ(τ), ε(τ)] = ε(θ(τ)) is the shape of a line segment. The amplitude of the oscillations is then equal to the length of the main half-axis of the

If we examine the graphs in Figures 10–12, we find that the highlighted curve that corresponds to the optimum angle ϕ are an excellent compromise between the shape of the curve (ellipse or line segment) and the dimensions of the curve that correspond to the oscillation amplitude. Similarly, the graphs of functions θ(τ), ω(τ), and ε(τ), which correspond to the optimal value of the angle ϕ, are the best compromise between the desired function shape and the oscillation amplitude, see

Figure 1 shows the entire inclined swinging pin, while Figure 2 shows the tenon plotted for this angle. Several bottom views of the entire joint during one quarter of

The main advantages of an inclined swinging pin compared to similar mecha-

1. Simplicity: The whole pin consists of only three main components that can be

3.Durability: All parts of the mechanism are subjected to approximately the

All videos referenced in this chapter are available from: https://bit.ly/2vZVGXR

Faculty of Electrical Engineering, Automatics and Informatics, Department of

© 2019 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/ by/3.0), which permits unrestricted use, distribution, and reproduction in any medium,

Informatics, Technical University of Opole, Poland

provided the original work is properly cited.

\*Address all correspondence to: s.barton@po.opole.pl

same forces. No component will receive more wear than others.

For these reasons, it is used in a wide variety of agricultural machines. An animation (An\_Joint.avi) that displays the entire device's activity can be

2. Compactness: All components are similar in dimensions.

ellipse, which displays ε(θ(τ)).

Kinematics - Analysis and Applications

the graphs in Figures 7–9.

nisms are:

7. Videos

Author details

Stanislav Barton

14

forged.

the working period are shown in Figure 13.

downloaded from the Videos section.

[1] Horton HL, Newell JA. Ingenious Mechanisms. New York: Industrial Press Inc.; 1967;4:245-265. ISBN: 9780831110321

[2] Horton H. Ingenious Mechanisms. Volume IV [Internet]. 2017. Available from: https://books.industrialpress.c om/ingenious-mechanisms-vol-iv-eb ook-only.html [Accessed: March 15, 2019]

[3] Leon S. Linear transformations. In: Leon S, editor. Linear Algebra with Applications. 8th ed. New Jersey: Pearson Prentice Hall; 2010. pp. 166-198. ISBN: 139780136009290

Chapter 2

Jiří Ondrášek

contact mechanics.

1. Introduction

17

Abstract

The General Kinematic Pair

At present, there are still increasing demands on the performance parameters of machinery equipment as well as cam mechanisms that belong to it. For this reason, the operating speeds and hence inertial effects of moving bodies, which limit the utilizable working frequency of machines, are increasing. These facts are the cause of higher wear and a decrease of the overall lifetime and reliability of machines. The force ratios in the general kinematic pair created by contact between the cam and the follower cause the contact stress. The generated stresses are transient and have a pulse shape. Fatigue damage of the cam working surface or the follower working surface may occur after exceeding a certain limit value of these stresses during the cam mechanisms running. This damage is in the form of cavities (pitting), which develop from cracks on the working surface. The chapter aim is to outline the issues of the dynamic stress of a general kinematic pair of a cam mechanism. One of the possible methods of the complex solution of the stress of the general kinematic pair is to use the possibilities of the finite element method in combination with the knowledge and conclusions of the

Keywords: cam mechanism, cam, follower, general kinematic pair, contact stress

Cam mechanisms are one of the basic objects in the design of production machines and equipment, whose characteristic feature is a high degree of automation and optimization of production and work processes. These mechanical systems are characterized by the transmission of large load possibility at high speed and positional accuracy of the working member of the relevant machinery. Their application is mainly connected with the so-called hard automation, which is characterized by unchangeable or difficult to change operations of the given technical equipment. Their widespread use is known in manufacturing and handling

machines of the manufacturing industry, with their dynamic effects and properties greatly affecting the overall behavior, operation, and efficiency of such machinery. At present, increasing demands are placed on the performance parameters of such machinery. Therefore, the operating speeds and thus the inertial effects of the moving bodies are increased, thereby reducing the usable operating frequency of the machines. These facts cause greater wear and reduce overall machine lifetime and reliability and must be taken into account when designing and developing

of a Cam Mechanism

### Chapter 2
