General concepts
Driving mechanics: science of movements and forces on the vehicle
Driving resistances: forces agains the main movement of the vehicle

Residual braking torque
The wheel brakes are usually disc brakes with pads. Some hydraulic cylinders press the pads against the disc to create friction and convert the energy into heat. The gap between pads and disc should be really small in order to make the braking action almost instantaneous (specially in cases of emergency brakes). Sometimes, the gap between pads and disc are so small that after braking there are still some friction, creating this residual braking torque

Engine loss

• Combustion engines: very low efficiency, being 40% in the best case, and usually 20%. A lot of heat is produced.
• Electric motors: efficiency is around 95%, it depends on the torque and speed.

Battery/Fuel cell loss
In case of electrical vehicles. Losses due to internal resistances. The internal resistance is very low, making the currents very high, originating losses that can go up to 20%

Accessories loss
Air condition, steering assistant systems, electronic control units (specially autonomous driving). Usually min. 500 W.

Suspension resistance
Not very relevant. It is related with the shock absorber when the vehicle is driving through uneven roads to keep the vehicle body quiet. It converts the oscillations into heat. Usually the loss is not higher than 1kW. Neglectacble in smooth roads

F[RPT] — Power train resistance
proportional to P[Eng] - around 0.1 - 0.2 P[Eng])

F[RIn] — Climbing resistance
Can be a high source of losses in case of the vehicle going up a very steep hills/road. The benefit of it is that when the car is on the top of the hill, it can go down without any energy. The mass of the vehicle act like a storage of potential energy

F[RIn] = F[G] ⋅ sin(α)

• F[G] — gravitational force (weight)
• α — inclination
  
  

⚠ The inclination percentage q = tan(α). This means an inclination of 100% corresponds to 45º (a = b)

Wheel standing

• F[zW,i] — comes from the weight of the car acting on that individual wheel.
• Tire deflection due to its elasticity, creating a contact patch. On the contact patch there is a pressure distribution. If this distribution of pressure is integrated over the area of the contact patch, it generates a force of the same magnitude of F[zW,i] but opposite
  
  

Wheel rotating

• To model the internal losses of the rubber material of the tire we can use springs and viscous dampers
• The moment a point of the tire starts touching the road (beginning of contact patch) the spring is compressed (energy is necessary). The moment that same point is about to leave the contact patch zone the spring will be decompressed, making the tire go back to its original shape. This processes do not causes losses
• The damping creates losses. This losses creates a higher pressure in the leading edge of the contact patch and lower pressure at the rear edge of the contact patch. This creates a non-symmetrical pressure distribution, leading to a F[zW,i] with a offset e from the center of the wheel.
• To overcome the driving resistance torque generated by the F[zW,i] with an offset, we need another force — F[RFlex, i] — applied on the center of the wheel pointing to the driving direction.
  

P[st] — Driving Power on Street
P[st] = P[Eng] - P[RPT] = sum of all external driving resistances

F[D] — Driving force
The force necessary to overcome all the resistances. Created by the Driving Train/Motor

F[RFl, i] — Flood resistance
If the vehicle is driving on a road with a water film on it, the wheel needs to push away the water in order to get in direct contact with the surface of road. The direct contact is necessary to create force transfer between tire and road. There are three zones in this phenomena:

• Approaching zone
• Transition zone
• Contact zone
  
  

Note: the grooves on the tire are created for the water to flow away, guaranteeing the contact between tire and road. It's important to say that the faster the car goes, the smaller is the contact area is. If the grooves are too small, the vehicle is going too fast or the there is a lot of water, the tire won't be in contact with the road anymore, becoming unstable. This instability is caused because the force is now transmitted by viscous friction, causing aquaplaning/hydroplaning

F[RFl, i] = ƒ(Volume of water/time) = ƒ(h⋅l⋅b / t)

Power train resistance
Specially on the transmission (around 2%). Viscous friction due to lubrication also causes losses and it depends on the transmission ratios. the higher it is, the higher the losses

F[Flex, i] — Flexing resistance
ω[i] = 2π ⋅ƒ[i]
ω[i] = 2π ⋅V[ω]/C[dyn]
C[dyn] = 2π ⋅V[ω]/ω[i]

therefore,
r[dyn] = V[ω]/ω[i] = C[dyn]/2π

By calculating the power for both deformed and undeformed tires, it will be clear that in reality r[dyn] = r[1]

• ƒ[i] — rotational frequency
• C[dyn] — Rolling circumference of the tire
• r[dyn] — dynamic radius of the wheel
• r[1] — undeformed radius of the wheel
• r[2] — deformed radius of the wheel from its center to the central point of the contact patch
  
  

To calculate F[Flex, i]
Write the moment equilibrium equation equals to zero for the wheel rotating diagram. From this same equation we can take the flexing resistance coefficient ƒ[RFlex,i]

F[Air,i] — Air resistance

• It depends on the speed of the tire. The tire kinda works like a fan in and there is a lot of air going around it

F[Rα, i] — Side slip resistance
When there is a steering angle between the direction of the velocity and the orientation of the tire, it creates a deformation on the tire. There is some adhesion and friction. This creates a side force due to the tire elasticity. When the adhesion friction is exceeded, it starts to slide. Sliding also transmit forces, but lower, creating some relaxation.

F[Rα, i] = F[yW, i] ⋅ sin(α)
If α is really small, sin(α) ~ α, therefore
F[Rα, i] = F[yW, i] ⋅ α

The relation between F[yW, i] and α for very small α's are linear. The coefficient of proportionality for this relation is called Cornering Stiffness — C[α]
F[yW, i] = C[α] ⋅ α

Therefore,
F[Rα, i] ~= C[α] ⋅ α^2

Road surface

• Smooth road: it creates little deformation on the surface of the tire
• Rough road (e.g. with large stones): it creates big deformation on the tire surface, leading to additional losses

Inflation pressure
The higher the inflation pressure, the smaller the contact patch, therefore the lower the rolling resistance

Standing waves
Waves generated behind the contact patch due to oscillation. The tire is deformed due to the wheel load. When a portion of the tire surface comes to the end of the contact patch, it will return to its original force. That happens for two reasons: (1)The elasticity of the tire like springs decompressing, (2)The centrifugal forces. The centrifugal forces are so strong that it pulls the tire too much (pulling the springs). at a certain point it pulls it so much that its elastic force overcome the centrifugal forces and it creates an oscillation similar to a pendular system.

Road deformation
Not solid roads like sand, snow, etc. the road will deform.

• Resistance through ground compression
• Resistance through bulldozing — accumulating material in front of contact patch
• Resistance through creation of wheel ruts — friction around the wheel because it is "inside" of the road (like mud for example)

Shapes, Pressure forces and Friction forces

• Horizontal plate — F[P] = 0.0 ; F[fric] = 1.0
• Wing-shape —F[P] = 0.1 ; F[fric] = 0.9
• Cylinder — F[P] = 0.9 ; F[fric] = 0.1
• Vertical plate — F[P] = 1.0 ; F[fric] = 0.0

Other factors and phenomena to take into consideration
Induced flow — negative pressure on the top of the vehicle and high pressure below the vehicle, making the air from below going towards the top

Total driving resistance F[R, tot] - velocity graph

• In this graph the distance between xOy and the 0% curve represents the roller resistance F[RR]
• If we draw a parallel line to the xx axis passing by the point where 0% crosses the yy axis, the distance between this line and the 0% curve for any velocity v gives the air resistance F[RAir]
• for any given velocity v, the distance between the 0% curve and any other curve represents the climbing resistance F[RIn]

Tractive effort diagram

• In this diagram we can see the torque - rotation curves representing each gear. the higher the curve is, the lower is the gear
• The total driving resistances are also represented in the diagram
• The hyperbola represents the tractive force for a constant power. Power is equal to force times velocity (P = F⋅V). This means each gear has its own hyperbola that is tangent to its torque curve. If the hyperbola plotted is related to the third gear, the curves of first and second gears would not touch it, the curve from third gear would be tangent to it, and the curves from forth and fifth gears would cross it.
• To accelerate the fastest way possible, we should always go to the maximum rotation from each gear and only then switch to the next gear. This is not the most energy efficient way, though.

Torque - Rotation graph

• Combustion engines — the curve can't touch the yy axis because combustion motors cannot provide torque without rotation below idle speed, therefore it requires a starter drive clutch to overcome the gap in the graph.

Engine characteristics map

• Thick line represents the maximum torque the engine can deliver
• Topological thin lines indicates points with equal fuel consumption for 1kW of mechanical energy, which means lines with equal efficiency.
• Most efficient way to drive a internal combustion engine: accelerate fast, using high torque and not going too far away from optimal speed