In engineering, **simulations** play a critical part in the design phase of any system. Through simulation we can understand how a system works, how it behaves under predefined conditions and how the performance is affected by different parameters.

In this article we are going to use a simplified mathematical model of the longitudinal dynamics of a vehicle, in order to evaluate the **acceleration performance** of the vehicle (0-100 kph time) and determine the **maximum speed**.

To validate the accuracy of our mathematical model, we are going to compare the simulation result with the advertised parameters of the vehicle.

### Input data

The vehicle parameters are taken from a rear-wheel drive (RWD) 16MY Jaguar F-Type:

Engine | 3-litre V6 DOHC V6, aluminium-alloy cylinder block and heads | |

Maximum torque [Nm] | 450 | |

Engine speed @ maximum torque [rpm] | 3500 | |

Maximum power [HP] | 340 | |

Engine speed @ maximum power [rpm] | 6500 | |

Transmission type | automatic, ZF8HP, RWD | |

Gear ratio | 1^{st} | 4.71 |

2^{nd} | 3.14 | |

3^{rd} | 2.11 | |

4^{th} | 1.67 | |

5^{th} | 1.29 | |

6^{th} | 1.00 | |

7^{th} | 0.84 | |

8^{th} | 0.67 | |

Final drive (i_{0}) | 3.31 | |

Tire symbol | 295/30ZR-20 | |

Vehicle mass (curb) [kg] | 1741 | |

Aerodynamic drag, C_{d} [-] | 0.36 | |

Frontal area, A [m^{2}] | 2.42 | |

Maximum speed [kph] | 260 | |

Acceleration time 0-100 kph [s] | 5.3 |

From data available on the internet, we can also extract the static engine torque values at full load, function of engine speed:

Engine speed points (full load) [rpm] | 1000 | 2020 | 2990 | 3500 | 5000 | 6500 |

Engine static torque points (full load) [Nm] | 306 | 385 | 439 | 450 | 450 | 367 |

### Vehicle layout

The powertrain and drivetrain of a RWD vehicle consists of:

- engine
- torque converter (clutch)
- automatic (manual) transmission
- propeller shaft
- differential
- drive shafts
- wheels

For simplicity, for our simulation example we are going to make the following assumptions:

- the engine is only a source of torque, without any thermodynamics or inertia modeling
- the engine is running at full load all the time
- the effect of the torque converter is not considered
- the gear shifting is performed instantaneous disregarding any related slip or dynamics
- the effects of the propeller shaft and drive shafts are not considered
- the tires have constant radius and the effect of slip is not considered

The mathematical model is going to be implemented as a **block diagram in Xcos (Scilab)**, based on the following equations.

### Mathematical equations

The **vehicle movement** is described by the **longitudinal forces equation**:

where:

F_{t} [N] – traction force

F_{i} [N] – inertial force

F_{s} [N] – road slope force

F_{r} [N] – road load force

F_{a} [N] – aerodynamic drag force

The traction force can be regarded a a “positive” force, trying to move the vehicle forward. All the other forces, are resistant, “negative” forces which are opposing motion, trying to slow down the vehicle.

As long as the traction force will be higher than the resistances, the vehicle will **accelerate**. When the traction force is smaller compared with the sum of resistant forces, the vehicle will **decelerate** (slow down). When the traction force is equal with the sum of resistant forces, the vehicle will maintain a **constant speed**.

The **traction force** [N] depends on the engine torque, engaged transmission gear ratio, final drive ratio (differential) and wheel radius:

where:

T_{e} [Nm] – engine torque

i_{x} [-] – transmission gear ratio

i_{0} [-] – final drive ratio

η_{d} [-] – driveline efficiency

r_{wd} [m] – dynamic wheel radius

The **dynamic wheel radius** [m] is the radius of the wheel when the vehicle is in motion. It is smaller than the static wheel radius r_{ws} because the tire is slightly compressed during vehicle motion.

The **static wheel radius** [m] is calculated based on the tire symbol (295/30ZR-20). For a better understanding of the calculation method read the article How to calculate the wheel radius.

The **inertial (resistant) force** [N] is given by the equation:

where:

m_{v} [kg] – total vehicle mass

a_{v} [m/s^{2}] – vehicle acceleration

The **total vehicle mass** [kg] consists of the curb vehicle mass, the driver’s mass and an additional mass factor. The mass factor takes into account the effect of the rotating components (crankshaft, gearbox shafts, propeller shaft, drive shafts, etc.) on the overall vehicle inertia.

where:

f_{m} [-] – mass factor

m_{cv} [kg] – curb vehicle mass

m_{d} [m] – driver mass

The **road slope (resistant) force** [N] is given by the equation:

where:

g [m/s^{2}] – gravitational acceleration

α_{s} [rad] – road slope angle

The **road load (resistant) force** [N] is given by the equation:

where:

c_{r} [-] – road load coefficient

The **aerodynamic drag (resistant) force** [N] is given by the equation:

where:

ρ [kg/m^{3}] – air density at 20 °C

c_{d} [-] – air drag coefficient

A [m^{2}] – vehicle frontal area

v [m/s] – vehicle speed

The traction force is limited by the wheel friction coefficient in the contact patch. The maximum **friction force** [N] that allows traction is:

where:

μ [-] – friction coefficient

c_{l} [-] – rear axle load coefficient

Replacing (4) in (1) and rearranging the terms, gives:

\[a_v = \frac{1}{m_v} \left [ F_t – \left (F_s + F_r + F_a \right ) \right ] \tag{10}\]By integrating equation (10) we obtain the vehicle speed [m/s]:

\[v = \frac{1}{m_v} \int \left [ F_t – \left (F_s + F_r + F_a \right ) \right ] dt \tag{11}\]The equations above will be used in the Xcos block diagrams. Before diving in the Xcos model, we need to prepare the input parameters for our simulation.

### Input parameters

The input parameters are mainly based on the advertised vehicle data and engineering assumptions. The most efficient way of defining them is using a Scilab script file (`*.sce`

).

// Engine eng.NmaxTq = 3500; // engine speed for maximum torque [rpm] eng.NmaxPwr = 6500; // engine speed for maximum power [rpm] eng.tqFullLoad = [306 385 439 450 450 367]; // engine torque curve at full load [Nm] eng.NtqFullLoad = [1000 2020 2990 3500 5000 6500]; // engine speed axis [rpm] eng.Nmax = 6500; // maximum engine speed [rpm] eng.Nmin = 1000; // minimum engine speed [rpm] // Transmission gbx.gearMin = 1; // lowest gear [-] gbx.gearMax = 8; // highest gear [-] gbx.gearRat = [4.71 3.14 2.11 1.67 1.29 1.00 0.84 0.67]; // gearbox gear ratios [-] gbx.i0 = 3.31; // final drive ratio (differential) [-] gbx.eff = 0.85; // driveline efficiency [-] // Tires: 295/30ZR-20 tire.W = 0.295; // tire width [m] tire.D = 20 * 0.0254; // rim diameter [m] tire.miua = 1.1; // wheel (tire) friction coefficient [-] tire.load = 0.65; // rear axle load coeeficient [-] // Vehicle veh.mass_curb = 1741; // [kg] veh.mass_driver = 80; // [kg] veh.mass_fm = 1.05; // mass factor [-] veh.g = 9.81; // gravitational acceleration [m/s2] veh.cd = 0.36; // drag coefficient [-] veh.fa = 2.42; // frontal area [m2] veh.ro = 1.202; // air density [kg/m3] road.slope = 0; // road slope angle [rad] road.cr = 0.011; // road load coefficient [-]

By running the script file, the input parameters are going to be loaded in the Scilab workspace. This step must be performed before the Xcos block diagram modeling, because the parameters are going to be loaded in some of the Xcos blocks.

### Parameters pre-processing

With the input parameters defined, we can calculate the remaining parameters used in the mathematical equations. The pre-processing instructions can be included in the same Scilab script file, after the input parameters definition.

// engine power curve at full load eng.pwrFullLoad = (1.36/1000)* eng.tqFullLoad .* (eng.NtqFullLoad * (%pi/30)); // [HP] // wheel dynamic radius tire.H = 30*tire.W/100; // tire height [m] tire.rws = tire.D/2 + tire.H; // wheel static radius [m] tire.rwd = 0.98 * tire.rws; // wheel dynamic radius [m] // vehicle total mass veh.mass = veh.mass_curb*veh.mass_fm + veh.mass_driver; // [kg] // simulation dT = 0.01; // sample (plot) time [s]

The **static engine power** at full load is calculated from the **static engine torque** at full load and **engine speed**:

where:

P_{e} [HP] – static engine power at full load

N_{e} [rpm] – engine speed

T_{e} [Nm] – static engine torque at full load

We can plot the static engine torque and power at full load, function of engine speed, using the Scilab instructions below.

// Engine static torque and power at full load subplot(1,2,1) plot(eng.NtqFullLoad,eng.tqFullLoad,"LineWidth", 3) ha=gca(); ha.data_bounds = [0 0;7000 500] xgrid() xlabel("Engine Speed [rpm]","FontSize",2) ylabel("Engine Torque [Nm]","FontSize",2) title("x-engineer.org") subplot(1,2,2) plot(eng.NtqFullLoad,eng.pwrFullLoad,"LineWidth", 3) ha=gca(); ha.data_bounds = [0 0;7000 400] xgrid() xlabel("Engine Speed [rpm]","FontSize",2) ylabel("Engine Power [HP]","FontSize",2) title("x-engineer.org")

which will output the following graphical window:

For simulation purposes, in the parameters file, a plot time constant is also defined `dT`

. This time constant is going to be used as sample time for the delay blocks and also for the plot interval.

### Xcos block diagram modeling

With all the parameters defined we can proceed with the Xcos block diagram modeling.

For an easier understanding and for a better readability, the Xcos model is split into 3 major subsystems (models): `Engine`

, `Transmission`

and `Vehicle`

.

**Top level model**

A `Clock`

block is used to define a global sampling time (`dT`

) for all the saved variables.

The `Engine`

model has the engine torque as output which is provided to the `Transmission`

model. Further, the transmission model is generating the wheel torque which is fed to the `Vehicle`

model.

The engine speed is calculates in the `Transmission`

model and the transmission speed in the `Vehicle`

model.

**Engine model**

The `Engine`

model is quite simplistic. The input (**current engine speed**) is filtered and saturated to a minimum and maximum values. The role of the filter is to simulated the mechanical inertia of the engine. The saturation block makes sure that the engine speed doesn’t go below idle speed (1000 rpm) and does not exceed the maximum value (6500 rpm).

The **current engine torque** is interpolated in the `Interp`

block, which has, as parameters, two vectors: the engine speed vector and the static engine torque at full load vector. Both vectors are defined in the Scilab parameter script file.

The **current engine power** is calculated using equation (12).

Using the sample time `dT`

, the current engine speed, torque and power are saved in the Scilab workspace.

**Transmission model**

The `Transmission`

model receives **two inputs**:

- the engine torque (coming from the
`Engine`

model) - the transmission speed (coming from the
`Vehicle`

model)

and calculates **two outputs**:

- the transmission torque (going to the
`Vehicle`

model) - the engine speed (going back to the
`Engine`

model)

The final drive gear ratio (of the differential) is included in the transmission model. Therefore the transmission torque is equal with the wheel torque and the transmission speed is equal with the wheel speed.

The **transmission (wheel) torque** [Nm] is calculated after the equation:

where:

T_{e} [Nm] – engine torque

i_{x} [-] – transmission gear ratio

i_{0} [-] – final drive ratio

η_{d} [-] – driveline efficiency

The final drive ratio and driveline efficiency are constant and defined in the parameter file. The transmission gear ratio depends on the engaged gear, which is determined by the `Shift Scheduler`

.

The gear index signal goes into an `Interp`

block which has as X axis the number of gears `[1:8]`

and as output Y, the gear ratio vector (define din the parameter file).

The **engine speed** [rpm] is calculated based on the equation:

where:

ω_{t} [rad/s] – is the transmission speed (equal with the wheel speed)

The **engaged gear** and **transmission (wheel) torque** are saved in the Scilab workspace.

**Shift Scheduler**

The shift scheduler is designed to keep the engine speed in the power band, between maximum torque and maximum power. This way the best acceleration performance is going to be obtained.

The shift scheduler is designed as IF ELSEIF conditions. The **(gear) upshift** is performed when the engine speed reached the engine speed value for maximum power and the maximum (8^{th}) gear is not engaged. The **(gear) downshift** is performed when the engine speed reached the engine speed value for maximum torque and the minimum (1^{st}) gear is not engaged. If none of the up/down shift conditions are true, the shift scheduler keeps the current gear engaged.

For this example, since we are interested in the acceleration of the vehicle, only the gear upshift condition will be active.

**Vehicle model**

The vehicle model receives the **wheel (transmission) torque** input from the `Transmission`

subsystem.

The friction force limit is calculated based on the equation (9).

The **wheel (traction) force** [N] is calculated based on the equation:

where:

T_{w} [Nm] – wheel torque (equal with transmission torque)

The resistant (rolling) forces are implemented based on the equations (6), (7) and (8).

The `Integrator`

block integrates equation (11) and outputs the vehicle speed v [m/s]. The integrator output (vehicle speed) is saturated to 0 m/s (minimum value).

The **wheel speed** [rad/s] is calculated based on the equation:

where:

v [m/s] – vehicle speed

The friction (force) limit, vehicle acceleration, vehicle speed, wheel (traction) force and the sum of the resistant forces are saved in the Scilab workspace.

### Simulation results post-processing

The simulation is run for `60 s`

. Several Scilab instructions are used to plot the saved simulation variable and analyze the behavior of the vehicle.

**Engine speed**

// Engine speed in time plot(sEngSpd_rpm.time,sEngSpd_rpm.values,"LineWidth", 3) ha=gca(); ha.data_bounds = [0 0;60 7000] xgrid() xlabel("Time [s]","FontSize",2) ylabel("Engine Speed [rpm]","FontSize",2) title("x-engineer.org")

The engine starts to accelerate from idle speed (1000 rpm) up to the engine speed value for maximum power (6500 rpm). Notice that there are 6 gear shift performed, the last engaged gear being 7^{th}.

In the lower gears, because the gear ratio is higher, the wheel torque is higher thus the vehicle (engine) accelerates more rapidly. As we approach the top gear (8th), which means smaller gear ratios, the acceleration of the vehicle (engine) is slower.

**Engine torque**

// Engine torque in time plot(sEngTq_Nm.time,sEngTq_Nm.values,"LineWidth", 3) ha=gca(); ha.data_bounds = [0 0;60 500] xgrid() xlabel("Time [s]","FontSize",2) ylabel("Engine Torque [Nm]","FontSize",2) title("x-engineer.org")

The minimum engine torque starts at 306 Nm, which is the full load torque at idle speed. It ramps up, towards the maximum value (450 Nm) and ramps down since the engine speed keeps increasing. At every gear shift, the engine torque goes up, to maximum value and then starts decreasing.

**Engine power**

// Engine power in time plot(sEngPwr_HP.time,sEngPwr_HP.values,"LineWidth", 3) ha=gca(); ha.data_bounds = [0 0;60 400] xgrid() xlabel("Time [s]","FontSize",2) ylabel("Engine Power [HP]","FontSize",2) title("x-engineer.org")

The engine power has the same behavior as the engine torque. Notice that it never exceeds the maximum engine power value (340 HP).

**Engaged gear**

// Engaged gear in time plot(sGear.time,sGear.values,"LineWidth", 3) ha=gca(); ha.data_bounds = [0 0;60 9] xgrid() xlabel("Time [s]","FontSize",2) ylabel("Engaged Gear [-]","FontSize",2) title("x-engineer.org")

This plot shows how the timing of the gear shift during vehicle acceleration. As expected, lower gears are active for a shorter duration. The reason is the higher torque wheel given by the higher gear ratio allows the engine to accelerate faster.

Notice that 1^{st} gear is engaged for a longer time than 2^{nd} gear. This happens because in first gear the engine has to accelerate from idle speed (1000 rpm) to maximum power engine speed (6500 rpm).

**Wheel torque**

// Wheel torque in time plot(sTrnTq_Nm.time,sTrnTq_Nm.values,"LineWidth", 3) ha=gca(); ha.data_bounds = [0 0;60 7000] xgrid() xlabel("Time [s]","FontSize",2) ylabel("Wheel Torque [Nm]","FontSize",2) title("x-engineer.org")

The wheel torque is actually the sum between the left and right wheel torques. Notice that the value of the torque which can be applied to the wheels is much higher than the engine torque. The torque amplification is done in the transmission with the current gear ratio and final drive ratio.

**Wheel traction force**

// Wheel traction force in time plot(sWhlTotTrc_N.time,sWhlTotTrc_N.values,"LineWidth", 3) ha=gca(); ha.data_bounds = [0 0;60 20000] plot(sWhlFricLim_N.time,sWhlFricLim_N.values,"g","LineWidth", 1) plot(sWhlRolRes_N.time,sWhlRolRes_N.values,"r","LineWidth", 1) xgrid() xlabel("Time [s]","FontSize",2) ylabel("Wheel Force [N]","FontSize",2) title("x-engineer.org") legend("Total Traction","Friction Limit","Rolling losses")

This plot is very interesting because it highlights some critical aspects of the vehicle dynamics.

Notice that the **total force available at the wheels for traction** is obtained in 1^{st} gear and it’s 17766 N. Even if the engine and transmission are capable to deliver this high amount of force, the wheel can not deploy it on the road. The maximum traction force which can be deployed by the wheels is limited by the **friction force**, which is 13383 N.

If the traction force generated by the engine and transmission exceeds the friction limit, the wheels will start to slip. This is how a tire “**burnout**” happens. In the vehicles equipped with Electronic Stability Program (ESP) systems, the wheel slip is detected and the engine torque (wheel force) is decreased below the friction limit.

The difference between the traction force and the road (rolling) resistant forces (also called rolling losses) is called **reserve force**. The higher the value of this force, the higher the vehicle acceleration.

As the vehicle speeds up, the resistant forces increase (mainly due to aerodynamic drag) and the traction force decrease. The point where these **forces are equal** (lines overlap or cross each other) is equivalent with the **maximum speed of the vehicle**. At this point the reserve force is zero, therefore the vehicle can not accelerate anymore and has a constant maximum speed.

**Vehicle acceleration**

// Vehicle acceleration in time plot(sVehAcc_mps2.time,sVehAcc_mps2.values,"LineWidth", 3) ha=gca(); ha.data_bounds = [0 0;60 7] xgrid() xlabel("Time [s]","FontSize",2) ylabel("Vehicle acceleration [m/s2]","FontSize",2) title("x-engineer.org")

The maximum vehicle acceleration (6.9 m/s^{2}) is obtained in 1st gear. This is happening because the reserve force is at its maximum. Notice that the vehicle acceleration in first gear is kept nearly constant for a short time. This happens due to the friction force limitation (`MIN`

block in the Xcos diagram).

We can also calculate the maximum G-force obtained during acceleration, by dividing the maximum acceleration of the vehicle with the gravitational acceleration.

`-->maxGforce_G = max(sVehAcc_mps2.values)/veh.g`

`maxGforce_G =`

` 0.7035715`

The vehicle (in simulation) is capable of reaching 0.7 G in 1^{st} gear, during full load acceleration.

**Vehicle speed**

// Vehicle speed in time plot(sVehSpd_kph.time,sVehSpd_kph.values,"LineWidth", 3) ha=gca(); ha.data_bounds = [0 0;60 300] xgrid() xlabel("Time [s]","FontSize",2) ylabel("Vehicle speed [kph]","FontSize",2) title("x-engineer.org")

From the vehicle speed plot we can extract the time to reach 100 kph and the maximum speed. These values can be obtained by running the following Scilab instructions:

// maximum vehicle speed [kph] maxVehSpd_kph = max(sVehSpd_kph.values); // 0 - 100 kph time [s] idx=find(sVehSpd_kph.values>100); t_0_100_s = sVehSpd_kph.time(idx(1));

For a 60 s simulation, the maximum vehicle speed is 253 kph. For a 100 s simulation, the maximum vehicle speed is **258 kph**.

The time to reach 100 kph from standstill is **5.00 s**.

### Conclusions

The model gives a very good match compared with the advertised vehicle parameters.

Advertised | Simulation | |

Maximum vehicle speed [kph] | 260 | 258 |

0 – 100 kph acceleration time [s] | 5.3 | 5.0 |

Even if the simulation model is relatively simple, by running it, we can extract a lot of information about the vehicle dynamics:

- what is the maximum speed of a given vehicle
- what is the 0 – 100 kph acceleration time of a given vehicle
- what is the impact of different parameters (mass, gear ratios, shift scheduler, etc.) on the vehicle performance
- what is the impact of different engine torque curves (at full load) on the vehicle performance

This simulation model is also a very good source of understanding the basic phenomena of vehicle dynamics, which can be valuable to engineering students, engineers or vehicle enthusiasts.

If you like this tutorial and want more for the future, you can support me on Patreon.

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## FStein

Not sure if my last comment went through, so here it is again:

in the engine model you implemented a mechanical inertia filter. In this formula, what does the variable s represent? Thanks!

## Anthony Stark

The “s” is the output variable from a Laplace transform of a first order differential equation of a low-pass filter…. The idea is that a transfer function was used to simulate the inertia of the engine.

To understand transfer functions, read:

https://x-engineer.org/graduate-engineering/signals-systems/control-systems/find-transfer-function-system/

## F

In the engine model you implemented a mechanical inertia filter. What is the value “s” in this formula?

## Fred

Is the Scilab file available anywhere? I’ve never used Scilab and am having a hard time recreating the block diagrams.

## Anthony

Where exactly are you stuck? The best way of learning how to use Xcos is by making a block diagram.