Dynamic Computation of an Innovative Device for Reducing Reaction Torque

Abstract

As is well known, the torque produced by a rotating motor generates an opposite and equal reaction torque in the machine casing that must be transmitted to the base. In many applications, especially when the reaction moment has high values, it is necessary to apply some constructive solutions, which in certain cases are difficult to implement. In this context, the need to reduce the reaction moment from drive motors is a challenging topic which has not been completely exhausted. In this paper, the authors present an original concept of a device which uses the centrifugal force generated by some equidistantly placed weights on a chain drive for reducing the reaction torque of the motor used for driving a rotating tool. The proposed system is capable of producing a supplementary torque which can be added to the driving moment. Due to this fact, by using this system, the power of the driving motor can be decreased, with the consequence of reducing the reaction moment that must be absorbed by the base.

1. Introduction

According to Newton’s third law, there is an equal and opposite reaction to every action. This equation is most commonly applied to linear forces, but it also is valid for angular, or rotating, systems [1]. In this case, the torque is the angular equivalent of force. A torque can cause a mass to accelerate angularly in the same manner that a linear force can accelerate a mass linearly. Thus, a torque reaction is the equal and opposite response to a torque. Managing this reaction torque is fundamental in many technical applications, such as helicopter blade drives, motorcycle rear wheels, rotating tools operating under variable machining conditions, powered hand tools for fastening joints, and so on.

A helicopter creates lift by rotating a series of blades that deflect air downward. The upward force acting on the helicopter is the equal and opposite of this downward force acting on the air. There is also an equal and opposite reaction caused by the rotating blades. The remainder of the helicopter tends to spin in the opposite direction as the engine rotates the blades in one direction. Therefore, a tail rotor, consisting of a smaller set of blades oriented to blow air horizontally, is used to counter the torque reaction caused by the engine trying to spin the main blades. Thus, the stable flight of the helicopter can be achieved [2,3,4,5].

The torque reaction problem is solved in a different way by Boeing CH-47 Chinook helicopters [6]. They use two sets of blades, called tandem rotors, which spin in opposite directions. By spinning, each rotor produces a torque reaction of its own, with the two reactions canceling each other out. This principle is also in involved in other helicopters that are used to lift heavy loads [7,8,9,10].

The torque reaction is an important issue for terrestrial vehicles as well [11,12]. So, a motorcycle is operated by applying a torque to spin the rear wheel. If applying a low acceleration, the torque reaction is insufficient to counteract the motorcycle’s front weight. However, if the rider provides enough throttle, the reaction can cause the front of the motorcycle to lift off the ground or do a wheelie. Even without contact with the ground, the rest of a motorcycle tends to spin in the opposite direction of an accelerating rear wheel.

In the case of powered hand tools involved in manufacturing and assembly jobs, such as the tightening of screws, at the beginning of the fastening process, the operator holds and supports the tool. When the fastener joint begins to compress, the tool starts to build up torque and the corresponding torque reaction is transferred to the operator. When the tool torque overcomes the operator, the hand and arm exert a force to stabilize the tool in a direction opposite to the tool rotation. This may produce upper extremity musculoskeletal disorders [13,14].

In this study, the authors propose an original concept involving a special device for reducing the reaction torque of the motor used for driving a rotating tool. Similar devices have captivated the enthusiasm of many practicians, as well as that of many serious researchers of high academic background, including persons from the aerospace industry or from academia who have granted an important number of patents [15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35].

The interests of researchers on innovative driving systems vary. Some of them claimed to have discovered a means of subverting the rules of physics to accomplish action without reaction and thereby boost machine efficiency. Others have discovered a different method that saves energy by using vibrations instead of physical lifting to move big objects [36,37,38,39,40,41,42,43].

The present paper is structured as follows: in the next section, we present the principle of the proposed device, and we describe the main components. Further, in Section 3, we write the equations for the dynamic computation, and in Section 4, we present and discuss the calculus results of some numerical examples. Conclusions are given in Section 5.

2. Device Design and Principle of Operation

The principle of the device proposed by the authors for reducing the reaction torque of a rotating tool is shown in Figure 1. It consists of the following parts:

1-

Support frame;

2-

Chain wheel RC₁;

3-

Chain wheel RC₂;

4-

Chain;

5-

Mass weights;

6-

Drive motor;

7-

Tool clamping chuck;

8-

Tool;

9-

Belt wheel RB₁;

10-

Belt wheel RB₂;

11-

Synchronous belt;

12-

Fixing pin;

13-

Shaft.

In principle, we have a main transmission consisting of the electric motor (6), chuck (7), and tool (8). The second transmission consists of the synchronous belt (11) and the two identical belt wheels RB₁ (9) and RB₂ (10). The transmission ratio of the belt is i_B = 1.

The third transmission is the one that includes the chain (4) and the chain wheels RC₁ (2) and RC₂ (3). The diameters of the chain wheels are chosen in such a way that the transmission ratio is i_c = 8 and the chain wraps on the wheel RC₁ at an angle of 90° (π/2 radians) and on the wheel RC₂ at an angle of 270° (3π/2 radians).

The chain wheel RC₁ (2) is mounted on a shaft (13), which also supports the belt pulley RB₂ (10). The second chain wheel RC₂ (3) rotates on the pin (12). Both the pin (12) and the shaft (13) are fixed in the support frame (1). The support frame is connected to the tool (8) and rotates simultaneously with it.

The wheels that are part of the synchronous belt transmission are assembled in such a way that the belt pulley RB₂ (10) is mounted on the same shaft (13) as the chain wheel RC₁ (2), and the belt pulley RB₁ (9) is fixed on the driving shaft of the motor (6).

As in our previous research [44], equidistant mass weights (5) are mounted on the chain (4).

In operation, the drive motor (6) rotates the tool (8), the support frame (1), the belt transmission through the belt pulley RB₁ (9) and the chain transmission through the chain wheel RC₁ (2), all having the same direction of movement.

As a technical solution, for the system to be functional with the chain in any plane, guide discs profiled according to the shape of the chain with weights, mounted on the chain wheels, can be used.

When a moment of resistance appears in the tool (8), it is transmitted through the driving system to the motor (6). As a result, the drive motor together with the whole system tends to reduce its speed. Consequently, in the chain transmission, there is a moment of inertia amplified by the mass weights (5) placed on the chain (4) and which tends to mitigate the shock produced in the system. In this case, the masses have the role of reducing the moment of reaction of the chain transmission. This decrease is achieved by the fact that by rotating the chain around the tool axis, it produces an increase in the moment that appears at the tool. From here, we can deduce the fact that by resetting the moment that appears at the tool, in the initial phase, implicitly, we obtain a reduction in the reaction moment.

Due to the rotation around the motor axis, the acting centrifugal forces on the mass weights (5) can be calculated using the general equation of the centrifugal force F_c acting on a rotating mass:

$$F_c=M·{\omega }^2·r_c^O$$

(1)

where

• M—the mass of the rotating body;
• ω—the angular speed of the mass;
• $r_c^O$—the trajectory radius of the body’s center of gravity relative to the center of rotation O.

Next, the calculation of the centrifugal forces acting on the mass weights (5) of the induced reaction torques is presented.

3. Analytical Computation of System Dynamics

The calculation of the centrifugal forces acting on the mass weights (5) starts from the principle scheme shown in Figure 2, where O₁ and O₂ denote the centers of the chain wheels RC₁ and RC₂, while O is the center around which the support frame (1) rotates with the angular speed ω.

The radius of the small chain wheel RC₁ is denoted with r and, consequently, the radius of the big chain wheel RC₂ will be 8r. Further, the chain was divided into six sections, AB, BC, CD, DE, EF, and FA, of which two are linear and four are circular (in the form of an arc).

Point O is chosen at the intersection of the line defined by the straight chain section AB and a parallel line to the other straight chain section CD, which passes through the point F.

The mass of each section M_xy is calculated starting from a specific linear mass, assumed to be $\overline{m}$, and the length of the section L_xy:

$$M_{xy}=\overline{m}·L_{xy}$$

(2)

where $x,y \in \left\{A, B,C,D,E,F\right\}$.

Based on the notation from Figure 2, we obtain the following lengths of the sections:

$$L_{AB}=L_{CD}=7r$$

(3)

$$\widehat{L_{BC}}={\displaystyle \frac{\pi ·r}{2}}$$

(4)

$$\widehat{L_{DE}}=\widehat{L_{EF}}=\widehat{L_{FA}}=4·\pi ·r$$

(5)

Thus, based on Equations (1) and (2), the centrifugal forces acting on each of the upper mentioned sections become:

$$F_{cxy}=\overline{m}·{L_{xy}·\omega }^2·r_{cxy}^O$$

(6)

For the circular sections, the radii of the gravity centers are computed using the general relationship:

$$r_c^O=r·{\displaystyle \frac{sin\alpha }{\alpha }}$$

(7)

Based on the notation from Figure 3, the gravity center radius for the circular section BC may be calculated as follows:

$$r_{cBC}^{O_1}=r·{\displaystyle \frac{sin\frac{\pi }{4}}{\frac{\pi }{4}}}$$

(8)

In a similar manner, for the other three circular sections, we obtain:

$$r_{cDE}^{O_2}=r_{cEF}^{O_2}=r_{cFA}^{O_2}=8r·{\displaystyle \frac{sin\frac{\pi }{4}}{\frac{\pi }{4}}}$$

(9)

Since the entire chain transmission rotates with the angular velocity ω around point O, we will have to calculate the gravity radii centers relative to O. Using the annotation from Figure 4, the following vector relations may be written:

$${\overrightarrow{r}}_{cAB}^O=\left\{\begin{array}{c}0\\ OA+{\displaystyle \frac{AB}{2}}\end{array}\right\}=\left\{\begin{array}{c}0\\ 8·r+{\displaystyle \frac{7·r}{2}}\end{array}\right\}=\left\{\begin{array}{c}0\\ 11.5·r\end{array}\right\}$$

(10)

$$\begin{gathered} {\overrightarrow{r}}_{cBC}^O=\overrightarrow{O{O}_1}+{\overrightarrow{r}}_{cBC}^{O_1}=\left\{\begin{array}{c}{O}_{1}B\\ 0B\end{array}\right\}+\left\{\begin{array}{c}{-r}_{cBC}^{O_1}·cos{\displaystyle \frac{\pi }{4}}\\ r_{cBC}^{O_1}·sin{\displaystyle \frac{\pi }{4}}\end{array}\right\}=\left\{\begin{array}{c}r-r·{\displaystyle \frac{sin\frac{\pi }{4}·cos\frac{\pi }{4}}{\frac{\pi }{4}}}\\ 15·r+r·{\displaystyle \frac{{sin}^2\frac{\pi }{4}}{\frac{\pi }{4}}}\end{array}\right\} \\ =\left\{\begin{array}{c}0.363·r\\ 15.363·r\end{array}\right\} \end{gathered}$$

(11)

$${\overrightarrow{r}}_{cCD}^O=\left\{\begin{array}{c}{O}_{1}B+{\displaystyle \frac{DC}{2}}\\ 16·r\end{array}\right\}=\left\{\begin{array}{c}r+{\displaystyle \frac{7·r}{2}}\\ 16·r\end{array}\right\}=\left\{\begin{array}{c}4.5·r\\ 16·r\end{array}\right\}$$

(12)

$$\begin{gathered} {\overrightarrow{r}}_{cDE}^O=\overrightarrow{O{O}_2}+{\overrightarrow{r}}_{cDE}^{O_2}=\left\{\begin{array}{c}OF\\ OA\end{array}\right\}+\left\{\begin{array}{c}{r}_{cDE}^{O_2}·cos{\displaystyle \frac{\pi }{4}}\\ r_{cBC}^{O_2}·sin{\displaystyle \frac{\pi }{4}}\end{array}\right\}=\left\{\begin{array}{c}8·r+8·r·{\displaystyle \frac{sin\frac{\pi }{4}·cos\frac{\pi }{4}}{\frac{\pi }{4}}}\\ 8·r+8·r·{\displaystyle \frac{{sin}^2\frac{\pi }{4}}{\frac{\pi }{4}}}\end{array}\right\} \\ =\left\{\begin{array}{c}13.093·r\\ 13.093·r\end{array}\right\} \end{gathered}$$

(13)

$$\begin{gathered} {\overrightarrow{r}}_{cEF}^O=\overrightarrow{O{O}_2}+{\overrightarrow{r}}_{cEF}^{O_2}=\left\{\begin{array}{c}OF\\ OA\end{array}\right\}+\left\{\begin{array}{c}{r}_{cEF}^{O_2}·cos{\displaystyle \frac{\pi }{4}}\\ {-r}_{cEF}^{O_2}·sin{\displaystyle \frac{\pi }{4}}\end{array}\right\}=\left\{\begin{array}{c}8·r+8·r·{\displaystyle \frac{sin\frac{\pi }{4}·cos\frac{\pi }{4}}{\frac{\pi }{4}}}\\ 8·r-8·r·{\displaystyle \frac{{sin}^2\frac{\pi }{4}}{\frac{\pi }{4}}}\end{array}\right\} \\ =\left\{\begin{array}{c}13.093·r\\ 2.907·r\end{array}\right\} \end{gathered}$$

(14)

$$\begin{gathered} {\overrightarrow{r}}_{cFA}^O=\overrightarrow{O{O}_2}+{\overrightarrow{r}}_{cFA}^{O_2}=\left\{\begin{array}{c}OF\\ OA\end{array}\right\}+\left\{\begin{array}{c}-r_{cFA}^{O_2}·cos{\displaystyle \frac{\pi }{4}}\\ {-r}_{cFA}^{O_2}·sin{\displaystyle \frac{\pi }{4}}\end{array}\right\}=\left\{\begin{array}{c}8·r-8·r·{\displaystyle \frac{sin\frac{\pi }{4}·cos\frac{\pi }{4}}{\frac{\pi }{4}}}\\ 8·r-8·r·{\displaystyle \frac{{sin}^2\frac{\pi }{4}}{\frac{\pi }{4}}}\end{array}\right\} \\ =\left\{\begin{array}{c}2.907·r\\ 2.907·r\end{array}\right\} \end{gathered}$$

(15)

The moduli of the vectors (their length) expressed by Equations (10)–(15) are:

$$r_{cAB}^O=11.5·r$$

(16)

$$r_{cBC}^O=15.367·r$$

(17)

$$r_{cCD}^O=16.621·r$$

(18)

$$r_{cDE}^O=18.516·r$$

(19)

$$r_{cEF}^O=13.412·r$$

(20)

$$r_{cFA}^O=4.111·r$$

(21)

Next, using Equation (6), the section lengths computed with Equations (3)–(5), and the radii of the gravity centers deduced with Equations (16)–(21), we obtain the following centrifugal forces acting on the six sections of the chain:

$${F_c}_{AB}=\overline{m}·7·r·{\omega }^2·11.5·r$$

(22)

Using the notation:

$$\overline{m}·{\omega }^2·r^2=K$$

(23)

Equation (3) can be written as:

$${F_c}_{AB}=80.5·K$$

(24)

In the same way, the centrifugal forces acting on the other 5 sections are:

$${F_c}_{BC}=24.139·K$$

(25)

$${F_c}_{CD}=116.347·K$$

(26)

$${F_c}_{DE}=232.679·K$$

(27)

$${F_c}_{EF}=168.540·K$$

(28)

$${F_c}_{FA}=51.660·K$$

(29)

Figure 5 shows the arrangement of the centrifugal forces calculated above. These forces, applied on the six sections of the chain, produce additional torques at the rotation points O₁ and O₂ of the chain wheels (RC₁ and RC₂) according to the centralization presented in

Table 1. Additional torques produced by the centrifugal forces.

Centrifugal Force	Produced Torque at the Rotation Point
	O1	O2
${{\mathrm{F}}_{\mathrm{c}}}_{\mathrm{A}\mathrm{B}}$	0	${{\mathrm{M}}_{\mathrm{A}\mathrm{B}}}_{O_2}$
${{\mathrm{F}}_{\mathrm{c}}}_{\mathrm{B}\mathrm{C}}$	${{\mathrm{M}}_{\mathrm{B}\mathrm{C}}}_{O_1}$	0
${{\mathrm{F}}_{\mathrm{c}}}_{\mathrm{C}{\mathrm{D}}_x}$	${{\mathrm{M}}_{\mathrm{C}\mathrm{D}}}_{O_1}$	0
${{\mathrm{F}}_{\mathrm{c}}}_{\mathrm{D}\mathrm{E}}$	0	0
${{\mathrm{F}}_{\mathrm{c}}}_{\mathrm{E}\mathrm{F}}$	0	${{\mathrm{M}}_{\mathrm{E}\mathrm{F}}}_{O_2}$
${{\mathrm{F}}_{\mathrm{c}}}_{\mathrm{F}\mathrm{A}}$	0	0

.

It must be mentioned that there is a torque only around the rotation point at which the centrifugal force “pulls” the chain. Furthermore, for the section CD, the centrifugal force can be decomposed along the x and y directions. Only the component along the x-axis produces a torque, as the y-component tends to bend the chain segment. Obviously, on the sections where the direction of the centrifugal force passes through the point of rotation, there is no torque, since the distance from the force to the center of rotation is zero.

The computation of torques produced by centrifugal forces starts from the well-known relationship:

$$M_{{xy}_{O_{1\left(2\right)}}}=F_{cxy}·d({O_{1\left(2\right)}, F}_{cxy})$$

(30)

where

• $x,y \in \left\{A, B,C,D,E,F\right\}$ are the characteristic points of the chain sections;
• O₁₍₂₎ are the rotation points of the chain wheels;
• F_cxy are the centrifugal forces acting on the six sections of the chain;
• $d({O_{1\left(2\right)},F}_{cxy})$ is the distance from the centrifugal force to the rotation points O₁₍₂₎ of the chain wheels.

If, for the centrifugal forces ${F_c}_{AB}$ and ${F_c}_{C{D}_x}$, the distances to the points of rotation are easy to deduce ($d({O_2, F}_{cAB})=8·r$ and $d\left({O_1, F}_{c{CD}_x}\right)=r$), for the forces ${F_c}_{BC}$ and ${F_c}_{EF}$, the following general relationship applies:

$$d(M,L)={\displaystyle \frac{\left|a·x_M+b·y_M+c\right|}{\sqrt{a^2+b^2}}}$$

(31)

where

• d(M, L)—the distance of the point M to the straight line L;
• x_M, y_M—the coordinates of point M;
• a·x + b·y + c = 0—the equation of line L.

In the case of a straight line passing through two points P($x_{MP}$, $y_P$) and R($x_R$, $y_R$), its equation has the form:

$${\displaystyle \frac{x-x_P}{x_R-x_P}}={\displaystyle \frac{y-y_P}{y_R-y_P}}$$

(32)

Consequently, based on Equations (11) and (14), the support lines of the forces ${F_c}_{BC}$ and ${F_c}_{EF}$ can be identified with the following equations:

$$\mathrm{L}({F_c}_{BC}):{\displaystyle \frac{x-x_O}{0.363·r-x_0}}={\displaystyle \frac{y-y_O}{15.363·r-y_0}}$$

(33)

and:

$$\mathrm{L}({F_c}_{EF}):{\displaystyle \frac{x-x_O}{13.093·r-x_0}}={\displaystyle \frac{y-y_O}{2.907·r-y_0}}$$

(34)

That is:

$$\mathrm{L}\left({F_c}_{BC}\right):15.363·r·x-0.363·r·y=0$$

(35)

and:

$$\mathrm{L}\left({F_c}_{EF}\right):2.907·r·x-13.093·r·y=0$$

(36)

Accordingly, using the coordinates O₁(r, 15·r) and O₂(8·r, 8·r), together with applying Equations (31), (35) and (36), we can obtain the following distances:

$$\mathrm{d}\left({{O_1,F}_c}_{BC}\right)={\displaystyle \frac{\left|15.363·r·r-0.363·r·7·r\right|}{\sqrt{{\left(15.363·r\right)}^2+{\left(0.363·r\right)}^2}}}=0.834·r$$

(37)

$$\mathrm{d}\left({{O_2,F}_c}_{EF}\right)={\displaystyle \frac{\left|2.907·r·8·r-13.093·r·8·r\right|}{\sqrt{{\left(2.907·r\right)}^2+{\left(13.093·r\right)}^2}}}=6.076·r$$

(38)

As a result, the torques produced by the centrifugal forces are:

$$M_{{AB}_{O_2}}={{-F}_c}_{AB}·d\left({O_2, F}_{cAB}\right)=-80.5·K·8·r=-644·K·r$$

(39)

$$M_{{BC}_{O_1}}={{-F}_c}_{BC}·d\left({O_1, F}_{cBC}\right)=-24.139·K·\mathrm{0,834}·r=-20.132·K·r$$

(40)

$$\begin{gathered} M_{{CD}_{O_1}}={{-F}_c}_{{CD}_x}·d\left({O_1, F}_{cD{C}_x}\right)={{-F}_c}_{CD}·\sin \left[arctan\left({\displaystyle \frac{\mathrm{4,5}·r}{16·r}}\right)\right]·r \\ =31.5·K·r \end{gathered}$$

(41)

$$M_{{DE}_{O_2}}=0$$

(42)

$$M_{{EF}_{O_2}}{{=F}_c}_{EF}·d\left({O_2, F}_{cEF}\right)=168.54·K·6.076·r=1024.049·K·r$$

(43)

$$M_{{FA}_{O_2}}=0$$

(44)

The “+” or “−“ signs for the torques were established according to the right hand rule.

Considering the transmission ratio of the chain drive, the above computed torques can be reduced at the center of rotation O₁ as follows:

$$\begin{gathered} {\mathrm{M}}_{{\mathrm{r}\mathrm{e}\mathrm{d}}_{O_1}}=\sum M_{O_1}+{\displaystyle \frac{\sum M_{O_2}}{i_c}}=M_{{BC}_{O_1}}+M_{{CD}_{O_1}}+{\displaystyle \frac{M_{{AB}_{O_2}}+M_{{EF}_{O_2}}}{8}} \\ =-\mathrm{20,132}·K·r+31.5·K·r+{\displaystyle \frac{-644·K·r+1024.049·K·r}{8}}=58.874·K·r \end{gathered}$$

(45)

This supplementary torque is transmitted through the belt transmission (with the ratio i_B = 1) to the tool (8).

In addition to the effect of generating torques in the chain wheel rotation centers, these centrifugal forces (which occur in the six portions of the transmission) also induce parallel loads at the rotation centers O₁ and O₂. These forces, denoted with $R_{xy}^{O_{1\left(2\right)}}$, are shown in Figure 6 in red, being equal and having the same sense as the centrifugal forces by which they were generated.

In the case of the horizontal chain segment CD, because the centrifugal force acts in the middle of the segment, its vertical component (${F_c}_{C{D}_y}$) generates forces in both centers of rotation (O₁ and O₂), equal to half of this component. The situation of the load generated by the horizontal component (${F_c}_{C{D}_x}$) is completely different. The induced force acts only at the point O₁, because, in the direction of the point O₂, the chain is “pushed”, without achieving a load at this point.

Taking these into account and considering Equations (24)–(29), we obtain:

$$R_{AB}^{O_2}=F_{c_{AB}}=80.5·K$$

(46)

$$R_{BC}^{O_1}={F_c}_{BC}=24.139·K$$

(47)

$$R_{C{D}_x}^{O_1}={{\mathrm{F}}_{\mathrm{c}}}_{\mathrm{C}{\mathrm{D}}_x}={F_c}_{CD}·\sin \left[arctan\left({\displaystyle \frac{4.5·r}{16·r}}\right)\right]=31.5·K$$

(48)

$$R_{C{D}_y}^{O_1}={{0.5·\mathrm{F}}_{\mathrm{c}}}_{\mathrm{C}{\mathrm{D}}_y}=0.5·{F_c}_{CD}·\cos \left[arctan\left({\displaystyle \frac{4.5·r}{16·r}}\right)\right]=56.001·K$$

(49)

$$R_{C{D}_y}^{O_2}={{0.5·\mathrm{F}}_{\mathrm{c}}}_{\mathrm{C}{\mathrm{D}}_y}=0.5·{F_c}_{CD}·\cos \left[arctan\left({\displaystyle \frac{4.5·r}{16·r}}\right)\right]=56.001·K$$

(50)

$$R_{DE}^{O_2}={F_c}_{DE}=232.679·K$$

(51)

$${{R_{EF}^{O_2}=F}_c}_{EF}=168.540·K$$

(52)

$${{R_{FA}^{O_2}=F}_c}_{FA}=51.660·K$$

(53)

These forces, acting in the rotation centers of the two chain wheels, generate additional torques in the rotational point O, which can be computed as:

$$M_{{xy}_O}=R_{xy}^{O_{1\left(2\right)}}·d(O,R_{xy}^{O_{1\left(2\right)}})$$

(54)

where

• $x,y \in \left\{A, B,C,D,E,F\right\}$ are the characteristic points of the chain sections;
• O is the rotation point of the tool;
• $R_{xy}^{O_{1\left(2\right)}}$ are the loads induced at the rotation centers O₁ and/or O₂ by the centrifugal forces acting on the six sections of the chain;
• $d(O,R_{xy}^{O_{1\left(2\right)}})$ is the distance from the force $R_{xy}^{O_{1\left(2\right)}}$ to the rotation point O.

The distances $d(O,R_{xy}^{O_{1\left(2\right)}})$ are:

$$d\left(O,R_{AB}^{O_2}\right)=O_{2}A=8·r$$

(55)

$$d\left(O,R_{BC}^{O_1}\right)=\mathrm{d}\left({{O_1,F}_c}_{BC}\right)=0.834·r$$

(56)

$$d\left({O,R}_{C{D}_x}^{O_1}\right)=OB=7·r$$

(57)

$$d\left({O,R}_{C{D}_y}^{O_1}\right)=O_{1}B=r$$

(58)

$$d\left({O,R}_{C{D}_y}^{O_2}\right)=O_{2}A=8·r$$

(59)

$$d\left(O,R_{DE}^{O_2}\right)=0$$

(60)

$$d\left({O,R}_{EF}^{O_2}\right)=\mathrm{d}\left({{O_2,F}_c}_{EF}\right)=6.076·r$$

(61)

$$d\left(O,R_{FA}^{O_2}\right)=0$$

(62)

Thus, the torques defined by Equation (54) are:

$$M_{{AB}_O}=R_{AB}^{O_2}·d\left(O,R_{AB}^{O_2}\right)=80.5·K·8·r=644·K·r$$

(63)

$$M_{{BC}_O}=R_{BC}^{O_1}·d\left(O,R_{BC}^{O_1}\right)=24.139·K·0.834·r=20.132·K·r$$

(64)

$$\begin{gathered} M_{{CD}_O}={-R}_{C{D}_x}^{O_1}· d\left({O,R}_{C{D}_x}^{O_1}\right)+R_{C{D}_y}^{O_1}·d\left({O,R}_{C{D}_y}^{O_1}\right)+R_{C{D}_y}^{O_2}·d\left({O,R}_{C{D}_y}^{O_2}\right)= \\ -31.5·K·7·r+56.001·K·r+56.001·K·8·r=283.509·K·r \end{gathered}$$

(65)

$$M_{{DE}_O}=R_{DE}^{O_2}· d\left(O,R_{DE}^{O_2}\right)=0$$

(66)

$$M_{{EF}_O}=-R_{EF}^{O_2}· d\left(O,R_{EF}^{O_2}\right)=-168.540·K·6.076·r=-1024.049·K·r$$

(67)

$$M_{{FA}_O}=R_{FA}^{O_2}· d\left(O,R_{FA}^{O_2}\right)=0$$

(68)

and their sum is:

$$\sum M_{{xy}_O}=M_{{AB}_O}+M_{{BC}_O}+M_{{CD}_O}+M_{{DE}_O}+M_{{EF}_O}+M_{{FA}_O}=-76.409·K·r$$

(69)

The total additional torque produced by the device in point O is:

$$M_{{tot}_O}=\sum M_{{xy}_O}+{\mathrm{M}}_{{\mathrm{r}\mathrm{e}\mathrm{d}}_O}=-76.409·K·r+58.874·K·r=-17.535·K·r$$

(70)

The negative sign from Equation (70) indicates that the produced torque is in a counterclockwise sense and coincides with the tool rotation sense, denoted by ω in Figure 2, Figure 4 and Figure 5. Thus, the additional torque generated by the system through the centrifugal forces acting on the masses added to the chain is accumulated to the working torque of the tool.

4. Numerical Examples and Discussion

First, let us further consider a device such as the one described in Chapter 2, which has the technical data given in

Table 2. Technical data of the proposed device.

Technical Data
Toll driving power	P = 500 [W]
Driving speed	n = 600 [min−1]
Specific mass of the weights	$\overline{m}=0.5 [\mathrm{k}\mathrm{g}/\mathrm{m}]$
Radius of the small chain wheel	r = 0.05 [m]

.

For the above indicated technical data, we have a driving torque:

$$M_O=9.55·{\displaystyle \frac{P}{n}}=7.958 \left[\mathrm{N}·\mathrm{m}\right],$$

(71)

an angular velocity:

$$\omega ={\displaystyle \frac{\pi ·n}{30}}=62.83 \left[\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\right],$$

(72)

a device-specific force, calculated according to Equation (23):

$$K=0.5·{62.83}^2·{0.05}^2=4.93 \left[\mathrm{N}\right],$$

(73)

and an additional torque produced by the device in point O, computed with Equation (70):

$$M_{{tot}_O}=17.535·K·r=17.535·4.93·0.05=4.32 \left[\mathrm{N}·\mathrm{m}\right].$$

(74)

Thus, it follows that the additional torque produced by the system is, for this specific example, 54.3% from the nominal driving torque of the motor. This means that the specific power absorbed from the supply network of the electric motor can be reduced by implementing such a device. In addition, if the torque required to drive the tool increases suddenly, the reaction torque that occurs at the motor is mitigated by this device, making the handling of the entire system easier.

As the additional torque produced by the system is independent of the drive power, the system is even more efficient the lower the motor power. However, there is a limitation as the moment is considered necessary to overcome the inertia of the system during the starting process.

It should also be mentioned that, for reasons related to the complexity of the dynamic model, the losses due to friction and transmissions efficiency were not taken into consideration. Normally, for the proposed system, these losses do not exceed 10%.

Further, we studied the consequences of varying different parameters (driving speed, specific mass, and radius of small chain wheel) on the additional torque produced by the system.

4.1. Effect of Driving Speed

To investigate this influence, we considered four driving speed steps: 600, 750, 900, and 1200 [min⁻¹], while keeping the other data constant. Performing the calculation as described above, we obtained the total torques produced by the device as shown in

Table 3. Influence of driving speed on the additional torque.

Technical Data	Values
Toll driving power P [W]	500
Driving speed n [min−1]	600	750	900	1200
Specific mass of the weights $\overline{m} [\mathrm{k}\mathrm{g}/\mathrm{m}]$	0.5
Radius of the small chain wheel r [m]	0.05
Driving torque M0 [N·m]	7.96	6.37	5.31	3.98
Additional torque MtotO [N·m]	4.32	6.75	9.73	17.30
Torque increase [%]	54.3	106.0	183.4	434.8

.

For a better visualization of this influence, Figure 7 shows in graphic form the effect of the driving speed variation on the additional torque produced by the system.

As can be observed, the variation in the speed influences both the driving torque M₀ and the additional produced torque M_totO. Thus, when maintaining the same driving power and increasing the speed, the driving moment decreases and the additional torque increases. Consequently, by doubling the driving speed, the system can generate an additional torque of 434.8%. This exponential increase, however, implies some challenges related to the dynamic stability of the system, due to the eccentric masses in rotational motion.

4.2. Effect of the Specific Mass of the Weights

To observe this impact, we investigated the increase in the specific masses, as presented in

Table 4. Influence of the specific mass of the weights.

Technical Data	Values
Toll driving power P [W]	500
Driving speed n [min−1]	600
Specific mass of the weights $\overline{m} [\mathrm{k}\mathrm{g}/\mathrm{m}]$	0.5	0.6	0.7	0.8
Radius of the small chain wheel r [m]	0.05
Driving torque M0 [N·m]	7.96
Additional torque MtotO [N·m]	4.32	5.18	6.04	6.91
Torque increase [%]	54.3	5.1	75.9	86.8

, where the computed total torques are also shown.

To obtain a better image of this influence, Figure 8 depicts the variation in the produced (additional) torque by increasing the specific mass.

By observing the above Figure, it can be noted that the increase in the specific mass $\overline{m}$ produces a directly proportional raise of the torque.

4.3. Effect of Small Chain Wheel Radius

To examine this influence, we considered four different radii of the small chain wheel 0.05, 0.06, 0.07, and 0.08 [m] by maintaining the ratio of the chain wheel transmissions and keeping the other data constant. By performing the calculus of the total torques produced by the device, the results shown in

Table 5. Influence of small chain wheel radius on the additional torque.

Technical Data	Values
Toll driving power P [W]	500
Driving speed n [min−1]	600
Specific mass of the weights $\overline{m} [\mathrm{k}\mathrm{g}/\mathrm{m}]$	0.5
Radius of the small chain wheel r [m]	0.05	0.06	0.07	0.08
Driving torque M0 [N·m]	7.96
Additional torque MtotO [N·m]	4.32	7.46	11.85	17.69
Torque increase [%]	54.3	93.7	148.9	222.3

were obtained.

Figure 9 presents the influence of the small chain wheel radius on the additional torque produced by the system in graphic form.

As one can observe, the torque produced by the system can be raised by increasing the radii of the chain wheels. Compared to the influence of the driving speed, where by doubling it, we generate an additional torque of 17.30 Nm, a similar value (17.69 Nm) is obtained by increasing the small chain wheel radius by only 60% (from 0.05 to 0.08 mm).

5. Conclusions

The present study presents a novel device capable of producing, by means of centrifugal forces generated by masses placed on a chain drive, a supplementary torque for driving a rotating tool. In addition, this paper details the dynamics of the system and exemplifies its benefits for a concrete example. Furthermore, several numerical examples showing the influence of various parameters (driving speed, specific mass, and radius of small chain wheel) on the additional torque produced by the system were presented.

The proposed device could be incorporated in powered hand tools such as screw-tightening devices used in extreme working conditions (when assembling wind turbine blades, for example) or in motor drills for digging holes. In these applications, the implementation of the concept would help reduce the risk of injury. As the system is able to reduce the reaction torque, the possibility that the operator could suffer musculo-skeletal disorders is minimized.

The main conclusions which can be drawn are as follows:

• The torque developed by the device is constant if the polygonal effect of the chain is neglected.
• The additional torque generated by the device is independent of the drive power, with the system showing a higher efficiency as the nominal motor power decreases.
• The torque produced by the system can be increased by raising the masses which are added to the chain by increasing the radii of the chain wheels and by raising the driving speed.
• If, during the operation, an increase in the resisting torque suddenly appears at the working tool, the reaction moment that occurs at the motor is mitigated by this device, making the handling of the entire system easier.

The main disadvantages and limitations of the proposed system lie in its relatively large size and the need to take special protective measures, because the support frame (1) and the other associated elements rotate together with the tool (8). Furthermore, the system only properly works in one direction of rotation. In case of a change in the direction of rotation, the centrifugal forces acting on the masses attached to the chain lower the driving torque of the tool.

As future research directions, the authors propose the following:

• Optimize the design so as to minimize the size of the device;
• Build a prototype of the device and carry out test trials;
• Investigate how the additional moment, generated by the masses attached to the chain, is influenced, if the transmission ratio of the synchronous belt is different from 1.