Wheel Size and System Efficiency
How wheel size acts as mechanical gearing in direct-drive systems, and why this has a strong impact on efficiency and performance.
Key insight: In direct-drive systems such as hub motors, wheel size is the only form of mechanical gearing. A smaller wheel, with faster rotation, allows a motor with fewer winding turns and thicker wire, resulting in lower winding resistance, lower copper losses, and more mechanical output at the same electrical input.
Introduction
In direct-drive electric motor systems, such as hub motors for e-bikes and electric mopeds, a traditional gearbox is absent. This means that wheel size becomes the only form of mechanical gearing between the motor and the road.
The central thesis of this article is that a smaller wheel, by acting as a higher gear ratio, enables the use of a motor with a higher Kv value. This higher-Kv motor, in turn, has lower winding resistance, which results in lower copper losses and higher efficiency at the same mechanical output power.
Theoretical foundation
Core motor constants: Kv and Kt
For a brushless DC motor (BLDC/PMSM), there is a fundamental and unavoidable relationship between the speed constant (Kv) and the torque constant (Kt). This relationship is a direct consequence of energy conservation and electromagnetic induction:
Kt = 60 / (2π × Kv) ≈ 9.55 / Kv [Nm/A]
Here, Kv is expressed in RPM per volt (no-load), and Kt in newton-meters per ampere. This relationship is physically constrained by motor design and cannot be bypassed.
Physical background
At no-load, the motor rotates at a speed where the induced back-EMF balances the supply voltage. This speed defines Kv. At stall (zero RPM), the motor produces maximum torque proportional to current, defined by Kt.
Winding design and resistance
A motor's Kv value is primarily determined by the number of winding turns per phase:
- High-Kv motor: Fewer winding turns with thicker wire
- Low-Kv motor: More winding turns with thinner wire
Conductor resistance is given by R = ρ × L / A, where ρ is resistivity, L is conductor length, and A is cross-sectional area. For a motor with twice as many turns (half Kv), conductor length doubles and cross-sectional area is halved. This gives:
R ∝ 1/Kv²
Resistance scales with the square of 1/Kv
This is the key insight: A motor with twice the Kt (half the Kv) has four times higher resistance.
Resistance and Kt vs Kv value
The chart shows how resistance (relative to Kv=10) and torque constant Kt change with Kv. Note the quadratic increase in resistance at lower Kv.
At Kv=5, resistance is about 4x higher than at Kv=10. At Kv=15, resistance is about 44% of Kv=10.
Wheel size as mechanical gearing
In a direct-drive system, the wheel acts as a lever arm. Tractive force at the ground relates to motor torque as:
F = M / r
Tractive force = Motor torque / Wheel radius
A smaller wheel therefore provides higher tractive force for the same motor torque. Conversely, less motor torque is required to achieve the same tractive force with a smaller wheel.
Vehicle speed relates to motor speed as v = ω × r. A smaller wheel means the motor must rotate faster for the same vehicle speed. This is equivalent to a higher gear ratio in a traditional drivetrain.
Key principle
By choosing a smaller wheel, you can use a motor with higher Kv. This higher-Kv motor has lower resistance, which reduces copper losses at the same tractive force.
Mathematical derivation of the loss difference
Let us compare two systems designed for the same top speed and voltage, but with different wheel sizes:
| Parameter | System A (small wheel) | System B (large wheel) |
|---|---|---|
| Wheel radius | rA | rB = k × rA (k > 1) |
| Kv for the same vmax | KvA | KvB = KvA / k |
| Kt | KtA | KtB = k × KtA |
| Resistance | RA | RB = k² × RA |
To achieve the same tractive force F at the wheel, different motor torques are required, but interestingly the currents become identical:
MA = F × rA MB = F × rB = k × MA
IA = MA / KtA IB = (k × MA) / (k × KtA) = IA
But copper losses differ dramatically:
PCu,B = k² × PCu,A
The large wheel has k² times higher copper losses!
If k = 1.34 (the 29"/20" ratio), losses become 1.34² = 1.80 times higher, that is 80% more.
Practical example: 20" vs 29" wheels
We compare two direct-drive hub-motor systems with the following shared specifications:
| Parameter | Value |
|---|---|
| System voltage | 72 V |
| Maximum battery current | 80 A |
| Maximum battery power | 5 760 W |
| Target top speed | 70 km/h |
| Total mass | 100 kg |
Wheel dimensions
| Parameter | 20" wheel | 29" wheel | Ratio k |
|---|---|---|---|
| Total diameter | 560 mm | 750 mm | 1.34 |
| Radius (r) | 0.280 m | 0.375 m | 1.34 |
| Circumference (C) | 1.76 m | 2.36 m | 1.34 |
Motor design for the same top speed
To reach 70 km/h at 72 V, different Kv values are required:
| Parameter | 20" motor | 29" motor |
|---|---|---|
| Kv | 7.89 RPM/V | 5.89 RPM/V |
| Kt | 1.21 Nm/A | 1.62 Nm/A |
| Resistance (R) | 0.025 Ω | 0.045 Ω (+80%) |
Calculated results at different speeds
| Speed | Parameter | 20" wheel | 29" wheel | 20" advantage |
|---|---|---|---|---|
| 20 km/h | Copper loss | 980 W | 1 458 W | -33% |
| Efficiency | 83.0% | 74.7% | +8.3% | |
| 30 km/h | Copper loss | 533 W | 845 W | -37% |
| Efficiency | 90.7% | 85.3% | +5.4% | |
| 45 km/h | Copper loss | 260 W | 441 W | -41% |
| Efficiency | 95.5% | 92.3% | +3.2% |
Efficiency at different wheel sizes
Efficiency comparison for 20", 26", and 29" wheels at constant battery power (5760W). All systems are designed for the same top speed (70 km/h) at 72V.
Acceleration: 0-45 km/h
| Parameter | 20" wheel | 29" wheel | Difference |
|---|---|---|---|
| Average Pmech | ~4 900 W | ~4 400 W | +11% |
| Time 0-45 km/h | 1.59 s | 1.78 s | 12% faster |
| Energy from battery | 9.2 kJ | 10.2 kJ | 10% less |
Acceleration simulation
The graphs below show simulated acceleration from standstill using the parameters from the example: 72V system voltage, 80A battery current (5760W), 100 kg total mass, and all systems designed for a 70 km/h top speed.
Because the smaller wheel enables a motor with higher Kv and therefore lower resistance, more electrical power becomes available as mechanical power. This produces faster acceleration, even though all systems receive exactly the same electrical input power.
Calculated Kv values for 70 km/h top speed at 72V
| Wheel size | Diameter | Circumference | RPM at 70 km/h | Kv |
|---|---|---|---|---|
| 20" | 506 mm | 1590 mm | 734 RPM | 10.19 RPM/V |
| 26" | 665 mm | 2089 mm | 558 RPM | 7.76 RPM/V |
| 29" | 732 mm | 2300 mm | 507 RPM | 7.05 RPM/V |
20" wheel
Kv = 10.19 RPM/V
26" wheel
Kv = 7.76 RPM/V
29" wheel
Kv = 7.05 RPM/V
How to read the charts
Power lines (left Y-axis)
- Red solid (Mech. power): Mechanical output to the wheel. 20" is highest and the difference is "free" power from higher efficiency.
- Orange dashed (Elec. power): Electrical input from the battery. The difference between electric and mechanical power is the losses (I2R).
- Gray line (Load): Power needed to overcome resistance (air + rolling + grade). Identical for all wheel sizes at the same speed.
Torque and reference
- Blue line (Torque, right Y-axis): Wheel torque. 29" is highest torque due to lower Kv (higher Kt), but that does not help.
- Green vertical line (Equilibrium): Marks equilibrium speed where motor power equals load power. The line follows the pointer when hovering and shows speed.
Tip: Hover the chart to see detailed data, including time to speed. The green vertical line marks the current position.
Why does the 20" wheel win?
The counterintuitive result: The 29" wheel has higher torque but worse acceleration. How is that possible?
The answer is efficiency. To reach the same top speed at the same voltage, the 29" wheel needs a lower-Kv motor (more winding turns). More turns mean higher resistance (R proportional to 1/Kv^2), which means larger copper losses (I2R).
Even though all systems get the same electrical input (72V x 80A = 5760W), the 29" system loses more as heat in the windings. The 20" system gets "free" mechanical output from higher efficiency.
Results summary
Compare the values in the panels above. Notice how they connect:
Key insight: Acceleration is determined by power, not torque. The formula a = P/(m x v) shows that higher mechanical power gives higher acceleration at the same speed. The smaller wheel wins through higher efficiency, not more torque.
When the difference is largest
The efficiency gain from higher gearing (smaller wheel) is greatest in:
- Low speeds and high forces: During launch and acceleration, copper losses dominate
- Hill climbing: High force at low speed requires high current
- Stop-and-go driving: Frequent accelerations from low speed
The difference decreases at high speeds, where back-EMF approaches the supply voltage and current, and therefore I²R losses, naturally decreases.
Copper losses (I2R) at different wheel sizes
Shows how copper losses in motor windings vary with speed. At low speeds, the difference is largest: the 29" wheel loses nearly twice as much power as the 20" wheel.
At 20 km/h: 20" loses about 900W, 29" loses about 1600W, a 700W difference (78% more).
Practical trade-offs
Smaller wheels do not provide only benefits. The following factors must also be considered:
✓ Advantages of smaller wheels
- Higher efficiency during acceleration
- More mechanical output per electrical watt
- Better hill-climbing capability
- Lower energy consumption at varying speeds
✗ Disadvantages of smaller wheels
- Lower comfort, does not roll over obstacles as well
- Generally higher rolling resistance
- Less gyroscopic stability at high speed
- Limits maximum motor size for hub motors
Optimal solution: For performance and efficiency during acceleration, smaller wheels are advantageous. For long-distance comfort at constant speed, larger wheels may be preferable. 26 inches often represents a good compromise for general use.
Application to electric cars
Although electric motors can deliver full torque from standstill, virtually all electric cars use a single-speed gearbox (reduction gear) between the motor and the wheels. The reason is exactly the same as discussed above.
| Parameter | Typical value |
|---|---|
| Maximum power | 150-300 kW |
| Maximum motor RPM | 12 000-18 000 RPM |
| Maximum torque (at motor) | 300-500 Nm |
| Reduction gear | 8:1 to 12:1 |
| Wheel torque | 2 400-6 000 Nm |
Porsche Taycan: Two-speed transmission
Porsche Taycan is notable as one of the few electric cars with a two-speed transmission on the rear axle. The reason illustrates the principles in this article perfectly:
| Gear | Gear ratio | Use | Advantage |
|---|---|---|---|
| 1st gear | ~15:1 | 0-100 km/h acceleration | Maximum force at low motor current |
| 2nd gear | ~8:1 | High-speed driving | Efficient at high speed |
The result is up to 10-15% better efficiency at highway speeds and improved acceleration, without compromising top speed (260 km/h for Taycan Turbo S).
Conclusions
Summary of the efficiency gain
The fundamental mechanism can be summarized in the following chain:
- Smaller wheel = higher "gearing": the motor must spin faster for the same vehicle speed
- Higher gearing enables higher Kv: to reach the same top speed
- Higher Kv = lower resistance: thicker wire, fewer turns
- Lower resistance = lower I²R losses: at the same tractive force
- Torque multiplication from the smaller wheel compensates for lower Kt
Core principle: Let mechanical gearing provide torque multiplication instead of relying on a "strong" motor with high Kt and high resistance. This shifts losses from electrical (I²R) to mechanical, where they are much lower.
In the practical example, the 20" wheel delivered higher efficiency and 32% faster acceleration compared with the 29" wheel, despite the same input power.
For direct-drive systems where no traditional gearbox is available, such as direct-drive hub motors for e-bikes, wheel size becomes the only available gearing mechanism. Choosing a smaller wheel is therefore an effective way to improve system performance and efficiency, especially during acceleration and low-speed riding.