---
title: "Force components on a helical gear tooth"
subject: "Theory of Mechanism and Machine I"
chapter: 8
key: "semester_6/theory_of_mechanism_and_machine_i:helical-gear-force-components-axial-thrust"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/helical-gear-force-components-axial-thrust
site: BCT Engineering
---

# Force components on a helical gear tooth

Theory of Mechanism and Machine I · Chapter 8: Force Analysis of Mechanisms

- Asked in: year not given, 6 marks
- Video: [Gear Forces of HELICAL and BEVEL GEARS in Just Over 10 Minutes!](https://www.youtube.com/watch?v=921ToiOynMA) — Less Boring Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/helical-gear-force-components-axial-thrust
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-8

## Question

Resolve the normal tooth force in a helical gear into tangential, radial and axial components in terms of the normal pressure angle and helix angle. Show that the axial thrust equals W_t tan(helix angle), and explain how double helical (herringbone) gears remove the axial thrust.

## Short answer

Normal force W_n resolved: tangential W_t = W_n cosφ_n cosψ, radial W_r = W_n sinφ_n, axial W_a = W_n cosφ_n sinψ = W_t tanψ. Double helical gears use two opposite-hand helices to cancel axial thrust, allowing larger helix angles with no thrust bearing.

## Answer

#### Geometry of the tooth force
In a helical gear the tooth force $W_n$ (normal to the tooth surface) acts in the **normal plane**, which is inclined to the transverse plane. $\phi_n$ is the normal pressure angle, $\psi$ the helix angle.

```text
 Normal plane:        Pitch plane:
      W_n             axial ^   /
      /|                    |  / helix
     / | W_r                | /  (psi)
    /phi_n                  |/
   +-----               ----+------> tangential
   W_n cos(phi_n)
```

#### Step 1: Components in the normal plane
- Radial component: $W_r = W_n\sin\phi_n$.
- Component in the pitch plane, perpendicular to the tooth (helix) direction: $W_n\cos\phi_n$.

#### Step 2: Resolve the pitch-plane component
The helix line makes angle $\psi$ with the axis. Along the transverse tangent (tangential) and the axis:
- Tangential component: $W_t = W_n\cos\phi_n\cos\psi$
- Axial component: $W_a = W_n\cos\phi_n\sin\psi$

#### Results
$$
W_n = \frac{W_t}{\cos\phi_n\cos\psi}
$$

$$
W_r = W_n\sin\phi_n = \frac{W_t\tan\phi_n}{\cos\psi} = W_t\tan\phi_t
$$

$$
W_a = W_n\cos\phi_n\sin\psi = W_t\tan\psi
$$

where $W_t = 2T/d$ is found from the torque, and $\tan\phi_t = \tan\phi_n/\cos\psi$.

#### Axial thrust
$W_a = W_t\tan\psi$ increases with helix angle. For $\psi = 15^\circ$, $W_a = 0.268W_t$; for $\psi = 30^\circ$, $W_a = 0.577W_t$. Thrust bearings (angular contact or taper roller) must take it. The direction of the thrust depends on the hand of the helix and direction of rotation, and the gear and pinion have equal and opposite axial forces.

#### Double helical (herringbone) gears
A double helical gear is two helical gears of opposite hand cut side by side on one blank. Each half produces an axial thrust of $W_t\tan\psi/2$, and the two thrusts are equal and opposite, so the net axial force on the shaft is zero. This allows larger helix angles (up to 45 degrees), giving smoother and more powerful drives with no thrust bearing needed. Applications: heavy reduction gearboxes, turbines, rolling mills.

## Check yourself

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

1. In a helical gear, the radial component of the normal tooth force W_n is given by:
   A. W_r = W_n cos(phi_n)
   B. W_r = W_n sin(phi_n)
   C. W_r = W_n sin(psi)
   D. W_r = W_n cos(phi_n) sin(psi)
   Answer: B. In the normal plane W_n splits into W_n sin(phi_n) radially and W_n cos(phi_n) in the pitch plane. (Key point: Radial component)

2. The pitch-plane component W_n cos(phi_n) is resolved into tangential and axial parts. Which pair is correct?
   A. W_t = W_n cos(phi_n) sin(psi), W_a = W_n cos(phi_n) cos(psi)
   B. W_t = W_n sin(phi_n) cos(psi), W_a = W_n sin(phi_n) sin(psi)
   C. W_t = W_n cos(phi_n) cos(psi), W_a = W_n cos(phi_n) sin(psi)
   D. W_t = W_n cos(psi), W_a = W_n sin(psi)
   Answer: C. The helix line makes angle psi with the axis, so the tangential part carries cos(psi) and the axial part sin(psi), both times W_n cos(phi_n). (Key point: Tangential and axial split)

3. Dividing W_a by W_t gives the axial thrust in terms of the tangential load. The result is:
   A. W_a = W_t tan(psi)
   B. W_a = W_t sin(psi)
   C. W_a = W_t tan(phi_n)
   D. W_a = W_t / cos(psi)
   Answer: A. W_a / W_t = [W_n cos(phi_n) sin(psi)] / [W_n cos(phi_n) cos(psi)] = tan(psi). (Key point: Axial thrust formula)

4. Why does a double helical (herringbone) gear need no thrust bearing?
   A. Its helix angle is zero
   B. The two halves have opposite hands, so their axial thrusts are equal and opposite and cancel
   C. The two halves have the same hand, so the thrusts add up and are taken by the blank
   D. The radial force is increased to balance the axial force
   Answer: B. Two helical gears of opposite hand on one blank each give W_t tan(psi)/2 axially in opposite directions, so the net thrust on the shaft is zero. (Key point: Herringbone removes thrust)

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