---
title: "Roller chain drive: length, speed and factor of safety"
subject: "Machine Design I"
chapter: 8
key: "semester_6/machine_design_i:roller-chain-drive-selection"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/roller-chain-drive-selection
site: BCT Engineering
---

# Roller chain drive: length, speed and factor of safety

Machine Design I · Chapter 8: Design of belts

- Asked in: year not given, 6 marks
- Video: [Chain Drive Calculation - Step by Step](https://www.youtube.com/watch?v=yUpghqn68HU) — O Manual do Projetista
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/roller-chain-drive-selection
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-8

## Question

A roller chain drive of pitch 19.05 mm connects a 19-tooth driving sprocket at 1000 rpm to a 48-tooth driven sprocket. The power transmitted is 15 kW and the centre distance is about 40 pitches. Take the breaking load of the chain as 31.1 kN. Find the speed of the driven sprocket, the pitch diameters, the chain length in pitches (a whole even number), the exact centre distance, the chain velocity, the chain pull and the factor of safety.

## Short answer

Driven speed n₂=395.8 rpm. Pitch diameters: d₁=115.7 mm, d₂=291.3 mm. Chain length Lₚ=114 pitches (even). Exact centre distance C=762 mm. Velocity v=6.03 m/s, chain pull F=2.49 kN, factor of safety FS=31.1/2.49=12.5.

## Answer

Data: $p = 19.05$ mm, $z_1 = 19$, $z_2 = 48$, $n_1 = 1000$ rpm, $P = 15$ kW, $C \approx 40p$, breaking load $= 31.1$ kN.

#### Driven speed

$$
n_2 = n_1\,\frac{z_1}{z_2} = 1000 \times \frac{19}{48} = 395.8\ \text{rpm}
$$

#### Pitch diameters

$$
d_1 = \frac{p}{\sin(180^\circ/z_1)} = \frac{19.05}{\sin 9.474^\circ} = 115.7\ \text{mm}, \qquad d_2 = \frac{19.05}{\sin 3.75^\circ} = 291.3\ \text{mm}
$$

#### Chain length in pitches

$$
L_p = \frac{2C}{p} + \frac{z_1 + z_2}{2} + \left(\frac{z_2 - z_1}{2\pi}\right)^2 \frac{p}{C}
$$

$$
L_p = 80 + 33.5 + \left(\frac{29}{2\pi}\right)^2\frac{1}{40} = 80 + 33.5 + 0.53 = 114.03
$$

Take **114 pitches** (even number, so that no offset link is needed).

#### Exact centre distance

$$
\frac{C}{p} = \frac{1}{4}\left[\left(L_p - \frac{z_1 + z_2}{2}\right) + \sqrt{\left(L_p - \frac{z_1 + z_2}{2}\right)^2 - 8\left(\frac{z_2 - z_1}{2\pi}\right)^2}\right]
$$

$$
\frac{C}{p} = \frac{1}{4}\left[80.5 + \sqrt{80.5^2 - 8(21.30)}\right] = \frac{1}{4}[80.5 + 79.43] = 39.98
$$

$$
C = 39.98 \times 19.05 = 761.7\ \text{mm}
$$

#### Chain velocity and pull

$$
v = \frac{z_1 p n_1}{60\times 1000} = \frac{19 \times 19.05 \times 1000}{60\,000} = 6.03\ \text{m/s}
$$

$$
F = \frac{P}{v} = \frac{15\,000}{6.03} = 2487\ \text{N}
$$

#### Factor of safety

$$
FS = \frac{31\,100}{2487} = 12.5
$$

This is above the usual minimum (about 8 to 11 for ordinary speeds), so the chain is adequate. Chain drives give a constant average velocity ratio with no slip, but the velocity varies slightly with the chordal action, and they need lubrication.

**Answer: n2 = 395.8 rpm; d1 = 115.7 mm, d2 = 291.3 mm; 114 pitches; C = 762 mm; v = 6.03 m/s; F = 2.49 kN; FS = 12.5.**

## Check yourself

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

1. What is the driven sprocket speed?
   A. 1000 x 48/19 = 2526 rpm
   B. 1000 x 19/48 = 395.8 rpm
   C. 1000 x 19/(48 x 19.05) rpm
   D. 1000 rpm
   Answer: B. Speed is inversely proportional to teeth: n2 = n1 z1/z2 = 395.8 rpm. (Key point: Driven speed)

2. Which formula gives sprocket pitch diameter?
   A. d = p z
   B. d = p / sin(180 degrees / z)
   C. d = p / tan(z)
   D. d = p z / pi
   Answer: B. The chain pitch is a chord of the pitch circle, so d = p / sin(180/z): 115.7 mm and 291.3 mm. (Key point: Pitch diameter)

3. Why is the calculated 114.03 pitches taken as 114?
   A. It must be an even whole number of pitches to avoid an offset link
   B. It must be a multiple of 19
   C. It must be an odd number
   D. The chain must be shorter than the centre distance
   Answer: A. An even number of links closes the chain with a normal connecting link; an odd count would need an offset link. (Key point: Chain length)

4. How is the factor of safety found?
   A. Breaking load divided by chain pull F = P / v: 31100 / 2487 = 12.5
   B. Power divided by breaking load
   C. Chain pull divided by breaking load
   D. Breaking load divided by chain speed
   Answer: A. Chain pull F = P/v = 2487 N; FS = breaking load / pull = 12.5, above the usual minimum. (Key point: Factor of safety)

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