FCHEV Longitudinal Dynamics Model: From Mechanics to Electrics

In our previous discussions on Dynamic Programming and Fuel Cell Structure, we explored the control algorithms and the power source itself. However, to execute these algorithms effectively, we require the most critical input variable: $P_{demand}$ (Power Demand).

The Energy Management System (EMS) does not inherently “know” if the car is climbing a steep hill or facing a strong headwind; it simply asks: “How many kilowatts of electricity does the electric machine (E-machine) need right now?”

Therefore, this article will construct a fundamental Longitudinal Dynamics Model for a vehicle to convert Mechanical Power (such as velocity, acceleration, and road grade) into Electrical Power ($P_{elec}$).

Step 1: Calculating Resistive Forces

When a vehicle moves, it must overcome four fundamental resistive forces. The sum of these forces is known as the total tractive force at the wheels ($F_{total}$).

1.1. Rolling Resistance Force

This is the friction between the tires and the road surface.

1.2. Aerodynamic Drag Force

This is the air resistance acting against the vehicle. It increases with the square of the velocity (the faster the car moves, the greater the drag).

1.3. Road Gravity Force (Gradient Resistance)

This is the component of gravity pulling the vehicle back when climbing a slope (or assisting when descending).

1.4. Acceleration Force (Inertial Force)

The force required to change the vehicle’s speed, derived from Newton’s Second Law.

Nomenclature Table:

Step 2: From Force to Torque & Mechanical Power

Once we have the total resistive force $F_{total}$, we can calculate the required Torque at the wheels ($T_{wheel}$) and the Mechanical Power Demand ($P_{demand}$).

2.1. Power Demand at Wheels

This represents the instantaneous mechanical power required at the tire-road interface to maintain the vehicle’s current trajectory. It is worth noting that if the vehicle is stationary ($v=0$), the power demand is zero, regardless of the magnitude of resistive forces (e.g., holding position on a steep gradient).

2.2. Wheel Torque

This is the rotational moment required at the wheel axle to overcome the linear resistive force $F_{total}$. Based on the principles of mechanics, the wheel acts as a lever arm with a length equal to the tire radius ($r_{tire}$). This relationship highlights the impact of wheel geometry. For a constant resistive force, a larger tire radius ($r_{tire}$) necessitates a higher torque output from the powertrain to propel the vehicle.

($r_{tire}$: Tire radius in meters)

Step 3: Drivetrain Transmission

The Electric Machine (Motor) is not connected directly to the wheels because the rotational speed and torque of the motor are significantly higher and lower, respectively, than those required at the wheels. Therefore, the motor transmits power through a reduction gear (Transmission/Differential) to multiply torque.

3.1. Torque at the Motor Shaft

After calculating the required torque at the wheels ($T_{wheel}$), we must convert it back to the motor shaft. Why?

Because there is always a Transmission/Gearbox between the wheels and the motor. As mentioned, electric motors typically spin very fast but produce relatively low torque. The gearbox amplifies this torque to enable the vehicle to move. Thus, we calculate $T_{motor}$ to determine how much effort the motor must exert.

Where:

• $N_g$: Gear Ratio. For example, $N_g = 10$ means the motor spins 10 times for every 1 revolution of the wheel. This multiplies the torque at the wheel by a factor of 10 (mechanical advantage).
• $\eta_g$: Transmission Efficiency. Gears create friction and heat, causing energy loss (typically $\eta_g \approx 0.95 - 0.97$).
• Note: Since $N_g$ is in the denominator and usually greater than 1, $T_{motor}$ will be significantly smaller than $T_{wheel}$.

3.2. Motor Angular Velocity

The efficiency of an electric motor ($\eta_{motor}$) varies greatly depending on its rotational speed (RPM). To look up the Efficiency Map in Step 4, we must know exactly how fast the motor is spinning.

First, convert the vehicle’s linear velocity ($v$) to the wheel’s angular velocity ($\omega_{wheel}$):

$\omega_{wheel} = \frac{v}{r_{tire}}$

Then, convert this to the motor shaft via the gear ratio:

$\omega_{motor} = \omega_{wheel} \cdot N_g$

(Standard Unit: $rad/s$. To convert to RPM, multiply by $60/2\pi$).

• Contrast to Torque: Due to the gear ratio $N_g$, the motor must spin much faster than the wheels.

3.3. Mechanical Power at the Motor Shaft

This is the actual physical power at the motor’s rotating shaft. It is also the power that the drivetrain must transmit from the motor to the wheels. It is the product of torque ($T$) and angular velocity ($\omega$).

Note: If we ignore transmission losses ($\eta_g = 1$), then $P_{mech}$ would be exactly equal to $P_{demand}$ at the wheels (Law of Conservation of Energy: Power remains constant through gears; only Torque and Speed change inversely).

Step 4: Electric Motor Model

This is the final step for the Energy Management System to determine how much electrical power the vehicle requires. We must convert Mechanical Power ($P_{mech}$) into Electrical Power ($P_{elec}$) that the Battery or Fuel Cell must supply. However, this process incurs losses due to motor efficiency ($\eta_{motor}$).

Motor efficiency is not constant; it is a function of Torque and Speed: $\eta_{motor} = f(T_{motor}, \omega_{motor})$.

Calculation Logic for $P_{elec}$

Case 1: Motoring Mode ($P_{mech} > 0$) The vehicle consumes electricity to move.

Case 2: Regenerative Braking Mode ($P_{mech} < 0$) The vehicle generates electricity to recharge the battery.

Case Study: Toyota Mirai Calculation 🚗

Let’s calculate the electrical power required for a Toyota Mirai (Gen 2) cruising on a highway.

Assumed Specifications:

• Mass ($m$): $1900 \, kg$
• Aerodynamic Drag Coefficient ($C_d$): $0.29$
• Frontal Area ($A$): $2.4 \, m^2$
• Rolling Resistance Coefficient ($C_r$): $0.01$
• Tire Radius ($r$): $0.35 \, m$
• Gear Ratio ($N_g$): $10$
• Transmission Efficiency ($\eta_g$): $0.97$
• Average Motor Efficiency ($\eta_m$): $0.92$

Scenario: Constant velocity of $100 \, km/h$ ($27.78 \, m/s$) on a flat road ($\alpha = 0, a = 0$).

1. Calculate Resistive Forces

• Rolling Resistance: $F_{roll} = 1900 \cdot 9.81 \cdot 0.01 = 186.39 \, N$
• Aerodynamic Drag: $F_{aero} = 0.5 \cdot 1.225 \cdot 2.4 \cdot 0.29 \cdot (27.78)^2 \approx 328.76 \, N$
• Total Tractive Force: $F_{total} = 186.39 + 328.76 = 515.15 \, N$

2. Calculate Mechanical Power at Wheels

3. Calculate Power at Motor Shaft (Including Gear Losses)

4. Calculate Electrical Power Demand ($P_{load}$)

This is the final figure fed into the EMS.

Conclusion: To maintain a speed of 100 km/h, the powertrain system (Fuel Cell + Battery) must continuously supply 16.03 kW. The EMS controller will then decide how to split this 16.03 kW between Hydrogen and the Battery.

I hope this post clarifies the “origin” of the $P_{load}$ or $P_{demand}$ variable often seen in simulation models.