Interactive Lecture
Torque Distribution, Speed Kinematics, and Cornering Behaviour in Automotive Drivetrains
by Dr Arnaldo Delli Carri
When a car turns a corner, the outer wheel traces a longer arc than the inner wheel. If both driven wheels were locked to the same shaft, one would be forced to slip — wasting energy and causing tyre wear. The differential solves this by allowing the two output shafts to rotate at different speeds while still transmitting engine torque to both wheels.
The open differential uses a set of bevel gears arranged so that the speed of the two output shafts averages to the speed of the input (ring gear). The planet gears — the small bevel gears that sit on the spider — are the key: they are free to rotate about their own axes, which lets one output speed up as the other slows down, without disturbing the sum.
The fundamental rule of the open differential: the average output speed equals the input speed. Torque is always split equally between the two outputs, regardless of the speed difference. This is both its strength and its limitation.
Label the ring gear (carrier) angular velocity \(\omega_c\), and the two output shafts (left and right axles) \(\omega_L\) and \(\omega_R\). The differential imposes the constraint:
\[\omega_c = \frac{\omega_L + \omega_R}{2}\]
Where:
Rearranging, the speed difference between the outputs is:
\[\omega_R - \omega_L = 2(\omega_R - \omega_c)\]
When the vehicle travels straight, \(\omega_L = \omega_R = \omega_c\) and the planet gears do not spin about their own axes. In a corner, they begin to rotate, allowing the speed difference to develop.
The planet gears mesh with the side (output) gears at the pitch-line, so in the carrier's rotating frame the two gears' surface speeds must match at that mesh. Writing the planet's pitch radius as \(r_p\) and the side gear's as \(r_s\) gives the planet's own spin rate:
\[\omega_p = \frac{\omega_R - \omega_L}{2} \cdot \frac{r_s}{r_p}\]
Where:
Driving straight gives \(\omega_R = \omega_L\), so \(\omega_p = 0\) — the planets orbit with the carrier but do not spin on their own axes. Cornering makes them spin at a rate set by the speed difference and the gear ratio \(r_s / r_p\).
For a vehicle of track width \(2w\) (distance between tyre contact patches) turning with radius \(R\) measured to the vehicle centreline, the required speed ratio is:
\[\frac{\omega_R}{\omega_L} = \frac{R + w}{R - w}\]
Where:
The carrier runs at the average: \(\omega_c = \frac{\omega_L + \omega_R}{2}\).
In an open differential, the planet gears are in static equilibrium on the spider. Summing torques about the spider axis shows that the driving torque \(T_c\) delivered to the carrier splits equally:
\[T_L = T_R = \frac{T_c}{2}\]
This equal split is the defining property of the open differential. It is independent of the speed difference between the outputs. The torque available at each wheel is limited by whichever side has the least traction — the Achilles' heel of the open design.
A limited-slip differential (LSD) adds a mechanism — friction clutch packs, viscous coupling, or Torsen worm gears — that introduces a torque bias ratio \(k\):
\[T_{\text{high}} = k \cdot T_{\text{low}}\]
Where:
The left panel shows a top-down view of the driveshaft, differential, and two driven wheels. The right panel shows the differential internals — ring gear, planet bevel gears on the spider, and side gears. Press the buttons to steer; adjust the turning radius to see how the inner/outer speed ratio depends on the corner geometry. The sim runs at a fixed display speed, so try \(R = 8\) m to reproduce Problem 1's turning radius and 1.222 speed ratio — not its 200 rpm carrier speed; try \(R = 3\) m for a tight roundabout.
Select a road scenario. The width of the torque arrow on each axle reflects how much drive reaches that wheel. A wheel spinning freely (red) is receiving torque it cannot use. Notice how the LSD scenario recovers drive that the open differential wastes. Wheel speeds animate at a fixed schematic rate for legibility, not to any real vehicle speed — the rpm and kW readouts show how power (torque × angular speed) splits between useful drive and slip losses, not an actual drivetrain's numbers.
| Concept | Equation | Description |
|---|---|---|
| Differential constraint | \(\omega_c = \dfrac{\omega_L + \omega_R}{2}\) | Carrier speed equals the average of the two output speeds |
| Speed difference | \(\Delta\omega = \omega_R - \omega_L\) | Proportional to how fast planet gears spin on their own axes |
| Cornering speed ratio | \(\dfrac{\omega_R}{\omega_L} = \dfrac{R + w}{R - w}\) | Required ratio for smooth rolling; outer wheel faster than inner |
| Open diff torque split | \(T_L = T_R = \dfrac{T_c}{2}\) | Equal torque to both outputs regardless of speed difference |
| LSD torque bias | \(T_{\text{high}} = k \cdot T_{\text{low}}\) | Higher-traction side receives \(k\) times more torque than the slipping side |
| Total input torque (LSD at bias) | \(T_c = T_{\text{low}}(1 + k)\) | Only when the low-traction side has reached its grip limit and the LSD is at its bias ratio; below that, the diff acts as open |
A rear-wheel-drive car has a track width of 1.6 m (so \(w = 0.8\) m). The rear axle carrier rotates at 200 rpm. The car turns a corner with a turning radius of 8 m measured to the vehicle centreline.
Find: (a) the required speed ratio \(\omega_R / \omega_L\), and (b) the angular speeds of the inner and outer rear wheels.
Show Solution
Solution:
Given:
(a) Speed ratio:
\[\frac{\omega_R}{\omega_L} = \frac{R + w}{R - w} = \frac{8 + 0.8}{8 - 0.8} = \frac{8.8}{7.2} = 1.222\]
(b) Individual speeds:
Using the differential constraint \(\omega_c = \frac{\omega_L + \omega_R}{2}\), and the ratio \(\omega_R = 1.222\,\omega_L\):
\[200 = \frac{\omega_L + 1.222\,\omega_L}{2} = \frac{2.222\,\omega_L}{2} = 1.111\,\omega_L\]
\[\omega_L = \frac{200}{1.111} = 180.0 \text{ rpm (inner)}\]
\[\omega_R = 1.222 \times 180.0 = 220.0 \text{ rpm (outer)}\]
Check: \(\frac{180 + 220}{2} = 200\) rpm ✓
Answer: Speed ratio = 1.222; inner wheel = 180 rpm, outer wheel = 220 rpm.
A vehicle's rear differential receives 400 N·m of input torque. The right rear wheel sits on ice and can transmit a maximum of 30 N·m before spinning. The vehicle has an open differential.
Find: (a) the torque delivered to each wheel, and (b) the total forward thrust force available, given a tyre rolling radius of 0.32 m.
Show Solution
Solution:
Given:
(a) Torque at each wheel:
An open differential always delivers equal torque to both outputs. The torque available is limited by the wheel that can transmit less:
\[T_L = T_R = \min\!\left(\frac{T_c}{2},\; T_{\text{traction limit}}\right)\]
The available half-torque is \(\frac{400}{2} = 200\) N·m, but the ice side can only absorb 30 N·m. Therefore:
\[T_L = T_R = 30 \text{ N·m}\]
The wheel on the high-grip surface also receives only 30 N·m — the remaining 170 N·m is wasted spinning the ice-side wheel.
(b) Total forward thrust:
\[F = \frac{T_L + T_R}{r} = \frac{30 + 30}{0.32} = \frac{60}{0.32} = 187.5 \text{ N}\]
Answer: Each wheel receives 30 N·m; total thrust = 187.5 N.
The same vehicle from Problem 2 (400 N·m input, ice-side traction limit 30 N·m, \(r = 0.32\) m) is now fitted with an LSD having a torque bias ratio of \(k = 3\).
Find: (a) the torque at each wheel, and (b) the total forward thrust, and compare with the open differential result.
Show Solution
Solution:
Given:
(a) Torque at each wheel:
With an LSD, the high-grip side receives \(k\) times the torque of the low-grip side:
\[T_{\text{low}} = 30 \text{ N·m (ice side, limited by traction)}\]
\[T_{\text{high}} = k \cdot T_{\text{low}} = 3 \times 30 = 90 \text{ N·m}\]
Check that the sum does not exceed the input torque:
\[T_{\text{low}} + T_{\text{high}} = 30 + 90 = 120 \text{ N·m} \leq 400 \text{ N·m} \checkmark\]
The differential can transmit only 120 N·m through the wheels under these conditions. If the engine continues to demand 400 N·m, the carrier (and the engine) will simply spin up faster — the engine cannot force more torque through the diff than the road allows.
(b) Total forward thrust:
\[F = \frac{T_{\text{low}} + T_{\text{high}}}{r} = \frac{30 + 90}{0.32} = \frac{120}{0.32} = 375 \text{ N}\]
Comparison: The LSD delivers 375 N versus 187.5 N for the open differential — exactly double the thrust on this surface.
Answer: Ice side = 30 N·m; grip side = 90 N·m; total thrust = 375 N.
In a bevel differential, the ring gear (carrier) rotates at \(\omega_c = 120\) rpm. The left output shaft rotates at \(\omega_L = 90\) rpm. The bevel planet gears have a pitch circle radius of \(r_p = 40\) mm, and the side gear (output bevel) has a pitch circle radius of \(r_s = 80\) mm.
Find: (a) the right output shaft speed, and (b) the angular velocity of a planet gear about its own axis relative to the carrier.
Show Solution
Solution:
Given:
(a) Right output speed:
From the differential constraint:
\[\omega_c = \frac{\omega_L + \omega_R}{2} \implies \omega_R = 2\omega_c - \omega_L\]
\[\omega_R = 2 \times 120 - 90 = 150 \text{ rpm}\]
(b) Planet gear angular velocity (relative to carrier):
Work in the carrier's rotating reference frame (subtract \(\omega_c\) from all components). In this frame:
Convert the left side gear's relative speed to rad/s so the pitch-line velocity comes out in m/s:
\[(\omega_L - \omega_c) = -30\,\text{rpm} \times \frac{2\pi}{60} = -3.142\,\text{rad/s}\]
The planet gear meshes with the left side gear. The pitch-line velocity at the mesh must match:
\[v = (\omega_L - \omega_c) \cdot r_s = (-3.142\,\text{rad/s}) \times 0.080\,\text{m} = -0.2513\,\text{m/s}\]
The planet gear rotates about its own axis to produce this pitch-line velocity at its own pitch circle, giving \(\omega_p\) in rad/s:
\[\omega_p = \frac{v}{r_p} = \frac{-0.2513}{0.040} = -6.283\,\text{rad/s}\]
Converting back to rpm:
\[\omega_p = -6.283 \times \frac{60}{2\pi} = -60 \text{ rpm (relative to carrier)}\]
The magnitude is 60 rpm; the negative sign means it rotates in the direction that accelerates the right side and decelerates the left side — consistent with \(\omega_R > \omega_L\).
Answer: Right output = 150 rpm; planet gear spins at 60 rpm relative to the carrier.