Tension ratio of a flat belt drive

A belt transmits power by friction between the belt and the pulley. The tight side tension $T_1$ is greater than the slack side tension $T_2$, and their difference is the effective pull.

Derivation

Consider a small element of belt subtending angle $\delta\theta$ at the centre of the pulley (angle of contact $\theta$). Let the belt mass per metre be $m$ and speed $v$. The forces on the element are:

• tensions $T$ and $T + \delta T$ at its ends,
• normal reaction $R$ from the pulley,
• friction $\mu R$,
• centrifugal force $m v^2\,\delta\theta$ (equal to $T_c\,\delta\theta$, where $T_c = mv^2$).

Resolving radially:

$$R + T_c\,\delta\theta = (T + \delta T)\sin\frac{\delta\theta}{2} + T\sin\frac{\delta\theta}{2} \approx T\,\delta\theta$$

so $R = (T - T_c)\,\delta\theta$.

Resolving tangentially:

$$\mu R = (T + \delta T)\cos\frac{\delta\theta}{2} - T\cos\frac{\delta\theta}{2} = \delta T$$

Substituting $R$:

$$\mu (T - T_c)\,\delta\theta = \delta T \quad\Rightarrow\quad \frac{\delta T}{T - T_c} = \mu\,\delta\theta$$

Integrating from $T_2$ to $T_1$ over the angle $0$ to $\theta$:

$$\ln\frac{T_1 - T_c}{T_2 - T_c} = \mu\theta \quad\Rightarrow\quad \frac{T_1 - T_c}{T_2 - T_c} = e^{\mu\theta}$$

If the centrifugal tension is neglected (low speed, $T_c = 0$), this becomes $\dfrac{T_1}{T_2} = e^{\mu\theta}$, where $\theta$ is in radians and is taken for the smaller pulley (or the smaller $\mu\theta$ product).

Condition for maximum power

Power $P = (T_1 - T_2)\,v$. Put $T_1 = T_{max}$ (the maximum permissible tension). From the result above, $T_2 - T_c = (T_1 - T_c)\,e^{-\mu\theta}$, so $T_1 - T_2 = (T_1 - T_c)\left(1 - e^{-\mu\theta}\right)$ and

$$P = \left(T_{max} - m v^2\right)\left(1 - e^{-\mu\theta}\right) v$$

For maximum power, $dP/dv = 0$:

$$\frac{d}{dv}\left(T_{max}v - m v^3\right) = T_{max} - 3 m v^2 = 0$$ $$v = \sqrt{\frac{T_{max}}{3m}}$$

At this speed the centrifugal tension is $T_c = mv^2 = T_{max}/3$, i.e. one third of the maximum permissible tension.