---
title: "Tension ratio of a flat belt drive"
subject: "Machine Design I"
chapter: 8
key: "semester_6/machine_design_i:belt-tension-ratio-derivation"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/belt-tension-ratio-derivation
site: BCT Engineering
---

# Tension ratio of a flat belt drive

Machine Design I · Chapter 8: Design of belts

- Asked in: year not given, 6 marks
- Video: [Tension ratio, Belt drives, Derive tension Ratio, Ratio of tensions, #KTM #TOM #Belt #TensionRatio](https://www.youtube.com/watch?v=R6ZKkczl8iQ) — Mechanical Engineering Management
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/belt-tension-ratio-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-8

## Question

Derive the relation T1/T2 = e^(mu theta) for a flat belt drive. Include the effect of centrifugal tension and show that the power transmitted is maximum when the belt speed is v = sqrt(Tmax/(3m)), where m is the mass per metre length of the belt.

## Short answer

Resolving forces on a belt element including centrifugal tension Tc=mv² gives (T−Tc) dθ = dT/μ. Integrating yields (T₁−Tc)/(T₂−Tc)=e^(μθ); with Tc=0, T₁/T₂=e^(μθ). For max power, differentiate P=(Tmax−mv²)(1−e^(−μθ))v, set dP/dv=0 → v=√(Tmax/(3m)).

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does the centrifugal force on a belt element of angle d(theta) equal, and how does it affect the radial balance?
   A. m v^2 d(theta); it reduces the normal reaction so R = (T - Tc) d(theta)
   B. m v d(theta); it increases R
   C. m v^2 d(theta); it has no effect on R
   D. T d(theta); it equals the friction
   Answer: A. The centrifugal force Tc d(theta) with Tc = m v^2 acts outward and reduces the pulley reaction: R = (T - Tc) d(theta). (Key point: Forces on element)

2. With centrifugal tension included, which relation results from the derivation?
   A. T1 / T2 = e^(mu theta) always
   B. (T1 - Tc) / (T2 - Tc) = e^(mu theta)
   C. (T1 + Tc) / (T2 + Tc) = e^(mu theta)
   D. T1 - T2 = e^(mu theta)
   Answer: B. Integrating d(T)/(T - Tc) = mu d(theta) gives ln[(T1 - Tc)/(T2 - Tc)] = mu theta. T1/T2 = e^(mu theta) holds only when Tc is neglected. (Key point: Tension ratio result)

3. For maximum power, T1 is set to Tmax and P is written as which expression in v?
   A. P = (Tmax - m v^2)(1 - e^(-mu theta)) v
   B. P = Tmax (1 - e^(mu theta)) v
   C. P = (Tmax + m v^2)(1 - e^(-mu theta)) v
   D. P = Tmax m v^2
   Answer: A. T1 - T2 = (T1 - Tc)(1 - e^(-mu theta)) with T1 = Tmax and Tc = m v^2, then multiplied by v. (Key point: Power expression)

4. At the belt speed for maximum power, what is the centrifugal tension?
   A. Tmax / 2
   B. Tmax
   C. Tmax / 3
   D. Zero
   Answer: C. d/dv (Tmax v - m v^3) = 0 gives Tmax = 3 m v^2, so Tc = m v^2 = Tmax / 3. (Key point: Optimum speed)

## Related in Machine Design I

- [Open flat belt power and cross belt length](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aopen-flat-belt-power-and-length) — Machine Design I, chapter 8
- [V-belt compared with flat belt](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Avbelt-versus-flat-belt) — Machine Design I, chapter 8
- [Roller chain drive: length, speed and factor of safety](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aroller-chain-drive-selection) — Machine Design I, chapter 8

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