Braking

Performance

Lateral Acceleration

Longitudinal Acceleration

Braking

Accelerating

Power Limited

Traction Limited

Speed, Distance & Time

Hill Climbing

Axle Load Distribution

From the previous figure (and assuming four identical tires), we can define the total friction force (F_f), the total rolling resistance (F_r) and the net vertical force (F_v):

F_f = F_ff + F_fr = μF_v

F_r = F_rf + F_rr = f_rF_v

F_v = mg − F_Lf − F_Lr = mg − F_L =

(

1 −

F_L

mg

)

mg =

(

1 −

0.5ρC_LAv²

mg

)

mg

Since practically all vehicles brake with both front and rear axles, the maximum force coming from the brake system will be the friction force from all tires (F_f).

Also, if we sum up the forces in the horizontal direction:

ma − F_D = F_f + F_r

(1a)

Stopping distance

From equation (1a) we get:(more)

a =

(

μ + f_r

)

g +

0.5ρ

m

[

C_DA −

(

μ + f_r

)

C_LA

]

v²

(1b)

This defines the maximum possible deceleration of the vehicle at any given speed v. To simplify future notations, we will write equation (1b) the following way: a = K_t + K_av².

Solving the equations of motion, we can find the stopping distance d necessary for decelerating from speed v to 0 with equation (2):

(more)

Because the velocity will vary throughout the deceleration process and that the acceleration depends on the velocity, we have to consider the instantaneous velocity and acceleration as the vehicle decelerates. Assuming s represents position, u the velocity, a the acceleration and t the time, then we can define velocity and acceleration in their derivative form:

u =

ds

dt

a =

du

dt

Rearranging both equations to eliminate dt leads to:

ds =

udu

a

Integrating both sides:

d

∫

s = 0

ds =

∫

u = v

udu

a

Solving the left-side equation:

d − 0 =

∫

u = v

udu

a

d =

∫

u = v

udu

a

To solve the right-side equation, we must integrate by substitution (see an example), as the acceleration depends on the velocity:

a = K_t + K_au²

da = 2K_audu

Rewriting the equation:

d =

2K_a

K_t + K_a(0)²

∫

a = K_t + K_a(v)²

da

a

d =

2K_a

(

K_t + K_av²

)

−

2K_a

(

K_t

)

Finally, by using the logarithmic identity:

d =

2K_a

(

1 +

K_a

K_t

v²

)

(2)

Where:

K_t =

(

μ + f_r

)

g

K_a =

0.5ρ

m

[

C_DA −

(

μ + f_r

)

C_LA

]

And:

`ρ`	= atmospheric air density (= 1.225 kg/m³)
`C_DA`	= Drag factor
`C_LA`	= Lift factor
`m`	= vehicle mass
`g`	= gravitational acceleration (= 9.80665 m/s²)
`μ`	= tire-road friction coefficient
`f_r`	= rolling resistance coefficient

This is the equation that is used to estimate the tire friction coefficient μ with a given stopping distance d in equation (2) of the TIRE COEFFICIENT section. Equation (2) was rewritten to isolate μ assuming C_LA = 0.

Axle load distribution

When braking, there will be a weight transfer from the rear axle to the front axle. It will be important to know the exact amount when we will design our brake system such that we can distribute adequately the brake force between the front and the rear axle, because the maximum brake forces at each axle will be:

F_ff = μW_f

F_fr = μW_r

(more)

we can find the portion of the vertical force on the rear axle:

W_r

F_v

= 1 −

W_f

F_v

(3a)

To find the portion of the vertical force on the front axle: (more)

W_f

F_v

l_r

L

(

μ + f_r

)

h

L

(3b)

Where:

`μ`	= tire-road friction coefficient
`f_r`	= rolling resistance coefficient

The first component of equation (3b), l_r/L, is the portion of the vehicle's weight on the front axle (at rest) and the other component of equation (3b) is the weight transferred from the rear axle to the front axle.

Note also that the portion of the vertical force on the front axle cannot exceed 100%. At this point, the vehicle is doing a «stoppie» as shown in the next figure.