What are Braking Characteristics?
Braking characteristics describe how effectively and quickly a vehicle can be stopped once the brakes are applied. They determine the safe stopping distance, safe spacing between vehicles in a traffic stream, and the design sight distance for highways. The primary measure of braking performance on a road is the skid resistance (or coefficient of friction, μ) of the pavement surface.
A braking test is conducted by driving a vehicle at a known speed and applying brakes fully. From the test results, the skid resistance of the pavement can be calculated. At least two of three parameters are needed:
- Initial velocity (v or u) — speed at start of braking
- Braking distance (L) — distance from brake application to complete stop
- Duration of brake application (t) — time from brake application to stop
Three Cases for Calculating Skid Resistance
Case 1: When Braking Distance (L) and Initial Velocity (u) are Known
By equating work done against frictional force to the change in kinetic energy of the vehicle:
fmgL = ½mu² → μ = u²/(2gL) = V²/(2gL)
Where u = initial velocity in m/s, L = braking distance in metres, g = 9.81 m/s².
Case 2: When Initial Velocity (u) and Brake Duration (t) are Known
Since the vehicle comes to rest (final velocity = 0): a = u/t (deceleration). Equating frictional force to retardation force: mgf = ma → f = a/g = u/(gt).
μ = u/(gt) = V/(gt₀)
Case 3: When Braking Distance (L) and Duration (t) are Known
From kinematics: L = ut/2 (since average velocity = u/2 when decelerating uniformly from u to 0). Combining with a = fg:
μ = 2L/(gt²)
Also: u = fgt and L = fgt²/2
Brake Efficiency Formula
Real brakes are never 100% efficient. The brake efficiency is the ratio of the friction actually achieved in the braking test to the maximum known friction of the pavement:
η (Brake Efficiency) = (f_obtained from braking test / f_max known) × 100%
For highway design purposes, IRC assumes 50% brake efficiency, meaning the effective friction used in SSD calculations is 50% of the actual skid resistance.
Solved Examples
Example 1
Problem: During a braking test, a vehicle travelling at 35 km/h was stopped within (i) 2 seconds after brake application, (ii) skid marks 5.8 m long. Find average skid resistance for both cases.
Solution:
V = 35 km/h → v = 0.278 × 35 = 9.73 m/s
(i) Using Case 2: μ = V/(gt₀) = (0.278 × 35)/(9.81 × 2) = 9.73/19.62 = 0.496
(ii) Using Case 1: μ = V²/(2gL) = (0.278×35)²/(2×9.81×5.8) = 94.7/113.8 = 0.832
Example 2 — Brake Efficiency
Problem: V = 65 km/h, L = 25.5 m, f_max = 0.70. Find brake efficiency.
μ_obtained = V²/(2gL) = (0.278×65)²/(2×9.81×25.5) = (18.07)²/500 = 326.5/500 = 0.653
Brake efficiency = (0.653/0.70) × 100 = 93.2%
Key Summary
- Three braking parameters: v (velocity), L (distance), t (duration) — need any 2
- Case 1 (L,v known): μ = v²/2gL | Case 2 (v,t known): μ = v/gt | Case 3 (L,t known): μ = 2L/gt²
- Brake efficiency = (f_obtained/f_max) × 100%
- IRC assumes 50% brake efficiency in highway design
- PCU value: passenger car = 1.0 | bus/truck = 3.0 | motorcycle = 0.5