How to Calculate Engine Torque from Wheel Torque

Title: How to Calculate Engine Torque from Wheel Torque Category: physics

You can estimate engine torque from wheel torque by reversing the torque multiplication produced by the transmission and final drive. The result depends on the selected gear, drivetrain losses, and whether the wheel-torque figure refers to one wheel or all driven wheels.

The basic relationship

Let $i_g$ be the transmission gear ratio and $i_f$ the final-drive ratio. Use the common reduction convention: each ratio is input speed divided by output speed. A reduction gear therefore has a ratio greater than 1. The overall reduction is

$$i_{\text{total}}=i_g i_f.$$

Ignoring losses, the drivetrain reduces rotational speed and increases torque by the same factor. Including drivetrain efficiency $\eta$, the relationship is approximately

$$T_{\text{wheel}}=T_{\text{engine}}i_g i_f\eta.$$

Rearrange it to calculate engine torque:

$$T_{\text{engine}}=\frac{T_{\text{wheel}}}{i_g i_f\eta}.$$

Here, $\eta$ is the fraction of input power that reaches the wheels. For example, $\eta=0.85$ means an estimated 85% of the engine’s power makes it through the drivetrain under the conditions being considered. If you want an ideal, loss-free calculation, set $\eta=1$.

Calculate it step by step

1. Identify what the wheel-torque number represents. Check whether it is the torque at one wheel, the sum across the driven wheels, or a value derived from a dynamometer.
2. Use the ratios for the operating gear. Multiply the transmission ratio by the final-drive ratio. Do not use the ratio for a different gear.
3. Choose an efficiency value. Use a measured or specified value if available. Otherwise, state your assumption; there is no single exact efficiency that applies to every vehicle and operating condition.
4. Divide wheel torque by the ratio and efficiency. Keep torque units consistent, such as newton-metres (N·m) or pound-feet (lb-ft).

If a gear ratio is supplied as output speed divided by input speed—for example, 0.25 rather than 4:1—take its reciprocal before using it as a reduction ratio in this formula.

Worked example

Suppose a measurement gives a total of 3,600 N·m at the driven wheels. The transmission is in a gear with a ratio of 3.6:1, the final-drive ratio is 3.9:1, and estimated drivetrain efficiency is 85%, or $0.85$.

First calculate the overall reduction:

$$i_{\text{total}}=3.6\times3.9=14.04.$$

Then include efficiency and solve for engine torque:

$$\begin{aligned} T_{\text{engine}}&=\frac{3{,}600}{3.6\times3.9\times0.85}\\ &=\frac{3{,}600}{11.934}\\ &\approx301.7\text{ N·m}. \end{aligned}$$

The estimated engine torque is about 302 N·m at the operating point represented by that measurement. If you ignored drivetrain losses, the estimate would be $3{,}600/14.04\approx256.4$ N·m. With losses, the engine must supply more torque to produce the same torque at the wheels.

Check whether the torque is per wheel or total

The formula works directly when the wheel-torque figure is the total torque delivered to the driven wheels. If it is torque at just one wheel, first establish how torque is divided between the wheels. With a symmetric, open differential and comparable conditions, each wheel receives roughly half the axle’s total torque. In that case the total is approximately twice the per-wheel reading. A limited-slip differential, unequal traction, or a measurement taken at only one wheel can change that distribution.

You can also estimate wheel torque from tractive force. If $F$ is the total forward force at the tire contact patches and $r$ is the tire’s effective rolling radius, then

$$T_{\text{wheel,total}}\approx F r.$$

For example, a total tractive force of 10,000 N acting through an effective radius of 0.32 m corresponds to about $10{,}000\times0.32=3{,}200$ N·m across the driven wheels. This assumes the force and radius refer to the same wheels and ignores details such as tire deformation. A dynamometer may report torque at its rollers rather than at the wheel hub, so check what its displayed value represents before using it.

Common mistakes

• Using only one ratio: The transmission ratio alone is not enough; include the final drive too.
• Multiplying instead of dividing: To work backward from wheel torque to engine torque, divide by the total reduction and efficiency.
• Mixing up ratio conventions: A 4:1 reduction multiplies torque by about four; a speed ratio of 0.25 means the same reduction but must be inverted for this calculation.
• Confusing per-wheel and axle torque: A single-wheel value is not interchangeable with the total at all driven wheels.
• Assuming efficiency is exactly known: Losses depend on the vehicle, gear, speed, load, temperature, and measurement method. Treat an assumed efficiency as an estimate, not a universal constant.

When the simple formula needs another factor

The equation assumes a fixed mechanical ratio between engine and wheels, as in a manual gearbox with the clutch engaged or an automatic transmission with its torque converter locked. An unlocked torque converter can multiply torque by a changing amount, particularly at low speeds. Let $M$ be the ratio of torque at the converter turbine to torque at the engine, and let $s$ be the ratio of turbine speed to engine speed. If $\eta_g$ is the efficiency of the gears and final drive downstream of the converter, then

$$T_{\text{wheel}}\approx T_{\text{engine}}M i_g i_f\eta_g.$$

To recover engine torque from this relation, divide by $M i_g i_f\eta_g$. The overall engine-to-wheel power efficiency in this unlocked-converter case is $\eta_{\text{total}}=M s\eta_g$, so it is not generally the same as $\eta_g$. If you know $\eta_{\text{total}}$ instead, use

$$T_{\text{wheel}}\approx T_{\text{engine}}i_g i_f\frac{\eta_{\text{total}}}{s}.$$

You need the converter’s operating condition or the relevant multiplication factor and speed ratio to make that correction.

A useful consistency check is speed: in a locked drivetrain, engine speed is approximately wheel speed multiplied by $i_g i_f$, with wheel speed measured at the driven wheels. Torque rises as speed falls through a reduction gear, while power is approximately conserved apart from losses. If the torque, ratios, and speeds imply a large power gain, recheck the units, ratio convention, and measurement point.