# Efficiency Estimation of Roller Chain Power Transmission System
## 2 Chain Drive Efficiency
### Sliding Friction Force  
Figure 2 shows the interaction between the chain and the sprocket, where the roller is engaged with the sprocket tooth, resulting in normal contact force \(N_{1}\) and friction force \(\mathrm{F_{1}}\) between the pin and the bush. Articulation from the chain span to the seated sprocket turns a maximum angle \(\theta\). Normal contact force \(N_{1}\) between the pin and the bush is generated, where \(T_{c}\) represents the link tension with the chain span, \(\theta_{N}\) is the angle of normal contact force; \(R_{p}\) and \(R_{bi}\) are the radius of the pin and the bush inner surface, respectively; \(\theta\) is the articulation angle of the pin between axes of the chain span and the seated link in the sprocket; and t is the width of the bush. [PERSON] [5] and [PERSON] [6] determined that sliding frictional losses during pin articulation are dependent on the pin sliding against the bush. Static or slip motion between the pin and the bush depends on the magnitude of the friction angle of \(\theta_{N}\) compared with the articulation angle \(\theta\). The frictional force is sufficiently large enough to ensure pin stiction under the condition of tan \(\theta<\) tan \(\theta_{N}\). In the case of the pin slipping, it should be satisfied with the condition of tan \((\theta)\geq\)  
Figure 2: Schematic of contact for pin articulation. \(\mu_{1}\) and \(\mu_{1}=F_{1}/N_{1}=\tan(\theta_{N})\), where \(\mu_{1}\) is the friction coefficient between the pin and the bush. The relationship of contact force between the pin and the bush can be written as follows:  
\[N_{1}=T_{c}\cdot\cos(\theta_{N})=T_{c}\cdot\frac{1}{\sqrt{1+\tan^{2}(\theta_{ N})}}=T_{c}\cdot\frac{1}{\sqrt{1+{\mu_{1}}^{2}}} \tag{1}\]  
Normally, a normal force can be found by integrating the contact pressure over the area. The link tension is then related to the surface contact pressure and area. [PERSON] [3] calculated pressure related with contact surface between the pin and the bush. It was assumed that the contact pressure is a constant, and the contact angle between the pin and the bush is also a constant. Here, the normal contact force N\({}_{1}\) between the pin and the bush is considered to be a normal force. The relationship of contact pressure and normal force is shown as Equation (2), where it is assumed that contact pressure P\(\cdot\cos(\theta)\) acts radially and is a constant:  
\[N_{1}=R_{p}\cdot t\cdot\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}P\cdot \cos(\theta)d\theta \tag{2}\]  
Then, friction force between the pin and the bush can be given by Equation (3):  
\[F_{1}=\mu_{1}\cdot(P\cdot dA) \tag{3}\]  
The tension in the chain links owing to centripetal acceleration is considered to have little effect on friction force. The energy losses W\({}_{\text{pin}}\) of pin articulation over the contact area between the pin and the bush are given by:  
\[W_{pin}=\frac{\pi}{2}\cdot\left(\frac{1}{\sqrt{1+{\mu_{1}}^{2}}}\cdot T_{c} \right)\cdot\mu_{1}\cdot R_{bi}\cdot\theta \tag{4}\]  
Figure 3 shows that the chain is driven under lateral offset owing to the derailleur system. The contact force between the pin and the bush with lateral offset angle \(\gamma\) is generated. This analysis differs from that in [PERSON] [1] and [PERSON] [3], in that their analysis did not consider the contact of chain links. In fact, contact between the pin and the bush is not generated along the vertical direction of the inner bush surface, but rather occurs at the boundary of the bush plate, shown as normal force in one side \(N_{0}^{1}\) and other side \(N_{0}^{2}\) due to lateral offset. When the link engages the sprocket, the sliding friction force of pin articulation \(F_{0}^{1}\) and \(F_{0}^{2}\) are generated.