Why do we need gears

The vast majority of vehicles sold today are powered by internal combustion engines (ICE). The most common types of ICE are spark-ignition (gasoline / petrol) and compression ignition (diesel) engines. From the traction point of view, regardless of the type of fuel used, the internal combustion engines has some fundamental disadvantages:

• it is not capable to produce torque from rest (0 rpm)
• the maximum engine power is produced only at a certain engine speed
• fuel consumption is directly linked to the operating point of the engine (speed and torque)
• it can only rotate one way
• it has a minimum operating speed (idle speed)

From the traction point of view, the maximum vehicle acceleration is obtained when we constantly apply at the wheel the maximum engine power.

Power is the product between force and speed:

\[P_{max}= F_{t} \cdot v \tag{1}\]

where:

P_max [W] – maximum power at the wheel
F_t [N] – traction force at the wheel
v [m/s] – vehicle speed (wheel tangential speed)

From (1) we can extract the formula of the traction force at the wheel, function of maximum power and vehicle speed:

\[F_{t} = P_{max} \cdot \frac{1}{v} \tag{2}\]

Let’s suppose that the maximum power at the wheel is a constant 80 kW. If the vehicle speed varies between 10 and 90 kph, applying equation (2), the traction force at the wheel will look like:

Thus, if the maximum engine power could be deployed at the wheel through the whole speed range of the vehicle, the traction hyperbola depicted in the above figure will result. This is called ideal traction hyperbola (ITH). In real applications, at low speeds, the traction force is limited by the friction limit of the wheels and at high speed by the road and aerodynamic vehicle losses.

Assuming that there are no gears (no gearbox, transmission and differential) on the vehicle, the engine torque can be translated into wheel force (traction force) by dividing it with the wheel radius.

\[F_{e} = \frac{T_{e}}{r_{w}} \tag{3}\]

where:

F_e [N] – engine force (at the wheel)
T_e [Nm] – engine torque
r_w [m] – wheel radius

With equation (3) we can take the engine’s maximum torque line and convert it into engine available traction line at the wheel. If we overlap the traction line of the engine over the ideal traction hyperbola, we get something like this:

As you can see, the engine’s traction line can only cover a small area of the required (ideal) traction performance (blue area). Without a way of converting the traction output of the engine, the vehicle will not be able to cover the blue area, beneath the ideal traction hyperbola. This means that the vehicle will not be able to pull-away and it will have very poor acceleration performance.

Here is when the gearbox / transmission comes into play. The multi-step gearbox is a torque and speed converter which adapts the traction output of the internal combustion engine to match the traction requirement of the vehicle. Multi-step gearbox means that the gearbox has at least two or more forward gears and one reverse gear.

Given a transmission with 4 gears and a differential, the engine traction force at the wheel will be:

\[F_{e_{x}} = i_{0} \cdot i_{x} \cdot \frac{T_{e}}{r_{w}} \tag{4}\]

where:

i_x [-] – is the gear ratio (x = 1, 2, 3, 4)
i₀ [-] – is the final drive (differential) ratio

From (4) we can see that, thanks to the gearbox, we now have 4 traction lines of the engine, one for each gear. Overlapping the engine’s traction lines (for each gear) with the required (ideal) traction hyperbola of the vehicle, gives:

As we can see, using a multi-step gearbox, the traction output of the engine approximates better the ideal traction hyperbola. The higher the numbers of gears, the better the approximation.

A battery electric vehicle, which is using electric motors for traction, doesn’t need a multi-step gearbox because the traction output of the electric motors matches the ideal traction hyperbola and the electric motor can also output torque rotating in opposite direction. Also, a vehicle equipped with an internal combustion engine and continuously variable transmission (CVT) will also have an ideal traction hyperbola.

We can summarize the role of a transmission (gearbox) on a vehicle with internal combustion engine as:

• adapts the traction output torque of the engine to match (approximate) the required (ideal) traction hyperbola of the vehicle (through the forward gears)
• allows the vehicle to move in reverse (backwards) with the engine rotating in the same direction (through the reverse gear)
• allows the engine to be disconnected from the wheels (through the clutch / torque converter or putting the gearbox in Neutral)

Example. Plot the traction lines of a vehicle with the following input parameters:

Maximum power, Pmax [kW]		250
Gear ratios	i1	4.71
	i2	3.14
	i3	2.11
	i4	1.67
	i5	1.29
	i6	1.00
	i7	0.84
	i8	0.67
	Final drive (i0)	3.31
Wheel radius, wr [m]		0.33565
Maximum speed, Vmax [kph]		260

The engine traction force at the wheel will be calculated from the maximum static engine torque line.

Engine speed points (full load), Ne [rpm]	1000	2020	2990	3500	5000	6500
Engine static torque points (full load), Te [Nm]	306	385	439	450	450	367

First, we need to calculate the minimum vehicle speed, which is achieved with the engine at idle and the first gear engaged. The idle speed of the engine is considered to be 1000 rpm.

\[V_{min} \text{ [kph]} = \frac{1000 \cdot \frac{\pi}{30} \cdot r_{w} \cdot 3.6}{i_{1} \cdot i_{0}} \tag{5} \]

With equation (2) we’ll calculate the ideal traction hyperbola (ITH). Using equations (4) we can calculate the engine wheel force for each gear. For an easier implementation, we are going to use a Scilab script.

// Engine and vehicle data
Pmax = 250; // [kW]
Te = [306 385 439 450 450 367]; // [Nm]
Ne = [1000 2020 2990 3500 5000 6500]; // [rpm]
ix = [4.71 3.14 2.11 1.67 1.29 1.00 0.84 0.67]; // [-]
i0 = 3.31; // [-]
rw = 0.33565; // [m]

// Calculate vehicle speed window
vmin = (Ne(1)*(%pi/30)*rw*3.6)/(ix(1)*i0); // [kph]
vmax = 260; // [kph]
v = linspace(vmin,vmax,100); // [kph]

// Ideal traction force at wheel
Ft = Pmax*1e3 ./ (v/3.6); // [N]

// Plot ideal traction hyperbola (ITH)
plot(v,Ft)

// Calculate and plot engine traction lines
for i=1:length(ix)
    for j = 1:length(Ne)
        Fex(i,j) = Te(j)*ix(i)*i0/rw; // [N]
        vx(i,j) = (Ne(j)*(%pi/30)*rw*3.6)./(ix(i)*i0); // [kph]
    end
    plot(vx(i,:),Fex(i,:),'r');
end

// Plot formatting
hf=gcf();
hf.background=8;
hf.children.font_size=2;
ha=gca();
ha.data_bounds = [0 0;vmax 22e3];
xgrid();
xlabel("Vehicle (wheel) speed [kph]","FontSize",2)
ylabel("Traction force (wheel) [N]","FontSize",2)
title("x-engineer.org","FontSize",2)

Running the Scilab script will output the following graphical window:

Having 8 forward gears brings the engine traction output very close to the ideal traction hyperbola. From the traction plot we can also notice that the maximum speed of the vehicle is being obtained in 7^th gear (where the ideal traction hyperbola touches the 7^th engine traction line). The 8^th gear is only used for cruising at high vehicle speeds in order to reduce fuel consumption.

For a vehicle with several gears (multi-step gearbox), one of the gears has the gear ratio around 1.00. In our example is the 6^th gear. This means that the input speed and torque are not converted at all by the 6^th gear. If there was no final drive (differential) ratio, driving in 6^th gear would mean that the vehicle had no gearbox.

In order to experiment how the vehicle would behave without a gearbox, although the differential can not be bypassed, try launching and driving your vehicle in the gear which has the gear ratio close to 1.00. This is almost impossible since there is not enough torque at the wheel to launch the vehicle. Further, trying to launch the vehicle in a high gear will put severe stress on the clutch (slip) which can cause burn-out (not recommended).

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