What is Braking Distance? Braking Power and Engine Power
The braking distance is the distance from the moment the driver presses the brake pedal to the moment the vehicle stops.
The stopping distance is the time taken until the driver presses the brake pedal and the response time of the brake system and the distance traveled during the stopping time of the vehicle. In this case, the stopping distance will be greater than the braking distance.
Tire characteristics and road conditions affect the braking (and stopping) distance of the vehicle; worn poor quality tires, wet-icy road conditions, etc. Regarding our subject, let's talk about the mass and speed of the vehicle that affects the braking distance. As the mass of the vehicle increases, the braking distance increases at the same rate.
Increasing the speed of the vehicle increases the braking distance much more. In other words, if the speed of the vehicle increases by 2 times, the braking distance increases by 4 times.
For example, if a vehicle traveling at 30 km/h stops 10 meters later,
The same vehicle can stop after 40 meters when the brake is pressed while traveling at 60 km/h. The speed has doubled, but the stopping distance has increased 4 times.
The reason for this is hidden in the formula of the motion energy of the vehicle.
The energy of motion is found by the formula K=1/2.m.v2. Here m is the mass, V is the velocity, as you can see the energy of motion is equal to half the mass times the velocity squared. If the speed is doubled, the energy is quadrupled.
Relationship Between Vehicle Speed and Braking Distance
Sample:
Calculation of kinetic energy (calculation of energy of motion)
Let's find the energy of a 1000kg vehicle traveling at 50km/h. (we take the speed in m/s, 50km/h= 13.8m/s)
K= (1/2).m.v2 -- > K= (1/2).1000kg.(13.8m/s)2
There is an energy of K = 95,220 jolues.
Again, let's find the energy of a 1000kg vehicle when it doubles its speed this time and goes at 100km/h. (We convert the speed to m/s and get 100km/h = 27.7m/s)
K= (1/2).m.v2 -- > K= (1/2).1000kg.(27.7m/s)2
There is an energy of K = 383,645 joules.
As seen in the example, when the speed increased by 2 times, the energy increased 4 times. Since this motion energy is converted into heat energy when braking, when the energy increases 4 times, the braking distance increases 4 times. For example, instead of stopping at 10 meters, it stops at 40 meters.
Do Wide Tires Increase Braking Force?
Looking at the braking force formula (Fs= μ . FN), installing tires with wider surfaces on the same vehicle does not increase the braking force.
Fs = μ . FN
Fs: Friction (braking) force (force that slows down the vehicle)
μ: Friction coefficient (between the road and the tire) (read as mu) (it is the most important factor, the friction coefficient depends on the structure and type of the friction surfaces (tyre-asphalt), that is, it depends on the asphalt, gravel, wet, snowy roads and the quality of the tire and does not depend on width.
FN : Vertical force acting on the tire. (Normal force), this force is the weight corresponding to each wheel of the vehicle.
The braking distance of the same vehicle with wide tires and normal tires is the same. The braking force is transferred to the road by the contact surface of the tire. Large-surface tires have more contact with surfaces, but this has no effect on the coefficient of friction or friction (braking force).
Wide tires have more contact surfaces with the road, wider contact surface reduces the amount of force per unit area, that is, the pressure exerted by the tire on the road. In this case, it can be said that less wear occurs on wide tires.
Main Topic (See: Brake System Types and Parts)
Braking Force Calculation and Friction Coefficient
Calculation of Brake Force and Types of Braking Forces
There are two different calculation methods in calculating the braking force,
one is “theoretical braking force” and the other is
The “transmitted (actual) brake force” is the brake force.
*Theoretical Braking Force; It is the braking force in which the speed-acceleration and mass variables of the vehicle are used and the braking force required for the vehicle to come to a stop in a certain time is calculated. This force is found by the formula F=m.a.
*Transmitted (actual) Brake Force; this force is actually the braking force between the tire and the road, this actually occurring braking force also gives the maximum braking force that can be achieved, and this braking force has nothing to do with speed.
A detailed description of these two braking forces is given below.
THEORETICAL BRAKING FORCE
The formula for braking force is the same as Newton's formula for force.
F=m.a
F= Force, unit: Newton (N) (we call it braking force here)
m = Mass, unit kilogram (kg)
a= Acceleration, unit (m/s2) meters per second squared
(1 Newton is the force that accelerates or slows a 1kg mass with an acceleration of 1 m/s2.)
Here;
* F braking force,
* m is the mass of the vehicle, for example a passenger car is about 1300kg.
* a acceleration is the deceleration of the vehicle. By definition, acceleration is the change in velocity per unit time.
The unit of acceleration is m/s2. There is also acceleration or deceleration.
F = m. According to the formula a, Braking force depends on the mass and acceleration (deceleration) of the vehicle. The greater the mass of the vehicle or the greater the deceleration (negative acceleration), the greater the braking force required to bring the vehicle to a stop.
The braking force (F) required to stop a truck traveling at 100 km/h (10,000kg) is also greater than the braking force (F) of a car traveling at 100 km/h (1500kg).
Or, if two vehicles of the same weight are traveling at 100 km/h and one stops in 5 seconds and the other in 10 seconds, the braking force of the vehicle that stops faster in 5 seconds is greater.
Example:
To give an example of braking force, let a 1500kg vehicle travel at 100km/h, apply the brake and stop after 5 seconds.
The deceleration acceleration of this vehicle: Acceleration is calculated by the formula (speed change / time change).
a= ΔV/ Δt -- > a=(100km/h)/(5s) Here we need to write the unit of speed, which is km/h, in meters/second. 100km/h = 27.7m/h.
In this case, acceleration is found as a= (27.7m/s) / (5 s) -- > a=5.54 m/s2.
Braking force:
F= m.a -- > F= 1500kg . 5.54 m/s2 -- > F= 8.310N (in Newton).
(Another unit of Newton is kgm/s2.)
It takes about 10 seconds for an ordinary middle class vehicle to accelerate from 0 to 100 km/h. But it takes less than 5 seconds to reach 0 from 100km/h, that is, to stop.
From this we can deduce that the braking force of each vehicle is at least 2 times greater than the engine drive force. This difference can be up to 8 times.
TRANSMITTES (ACTUAL) BRAKE FORCE
The braking effect of the vehicle, that is, the effect that slows down or stops the vehicle, depends on the friction force between the tire and the road. We can also call the friction force, the braking force transferred from the tire to the road. The braking force is created by the friction of the lining to the disc, but it works when it is transferred from the tire to the road, the important thing here is that the braking force created in the brake system can be transmitted to the road at maximum. The greater the frictional force between the tire and the road, the greater the braking force actually achieved and the easier it is to brake the vehicle.
Braking force is obtained from the friction of the pad and disc in the brake system, but ultimately, the braking force that works for us is the "transmitted braking force" between the tire and the road. This is the effect that stops the vehicle.
Friction force (Fs) and braking force are used interchangeably. In fact, the friction force between the tire and the road is the force that provides the road holding, at the same time, this force makes the vehicle go and stop.
The braking force is found by multiplying the coefficient of friction and the vertical force exerted on the tire resulting from the mass of the vehicle.
Fs = μ . FN
Fs: Friction (brake) force
μ: coefficient of friction (between road and tire) (pronounced mu)
FN : Vertical force acting on the tire. (Normal force)
What is Vertical Force (Normal Force)?
As it can be understood from the formula, if the weight (mass) of the vehicle increases or the friction coefficient increases, the braking force of the vehicle also increases. The braking force of a loaded vehicle's tire is greater than that of an empty vehicle. Of course, the brake force required to be created in the brake system (pad-disc) must also be more (sufficient). If there is not enough braking force on the pad-disc, the braking force (friction force) between the tire and the road has practically no meaning.
You may have seen that sandbags are placed in the trunks of the vehicles in the winter, or people jumping on the wheels of a skidding vehicle, the logic and purpose of all this is the same; It is to increase the mass of the vehicle on the wheels, that is, to increase the vertical force acting on the tire and thus to increase the friction force (road grip) between the tire and the road. A pickup truck driver, who puts a bag of sand in his car in winter, increases the grip and braking force of the vehicle tires on snow.
What is Coefficient of Friction? (μ)
The coefficient of friction is the most important variable in obtaining the braking force, the coefficient of friction is related to the structure of the road and the tire, that is, it depends on the material properties of the two friction surfaces. For example, on the snowy-icy-wet road surface, the friction coefficient value decreases, which prolongs the braking distance.
The friction coefficient on the dry asphalt road is the best value of 0.8, on the wet road the coefficient of friction is about 0.5 - on the snowy icy road the coefficient of friction is about 0.2.
Summer tires harden in the cold in winter and the coefficient of friction decreases, winter tires do not harden in the cold in winter and their friction coefficients do not decrease.
Calculation of the braking force generated in a tire.
Example: Let's calculate the maximum braking force that can be obtained from the tire of a 1200kg vehicle.
Fs = μ . FN
FN : 300kg is equivalent to a single wheel, if we convert kilograms to Newtons (1kg=10N)
FN: becomes 3000N
μ : 0.7
FS:?
Fs = μ . FN -- > Fs= 0.7 . 3000N
Fs= 2100N (Braking force)
The total braking force of the vehicle is found by adding the braking force of the 4 wheels.
Fs= 0.7 for 4 wheels. 12000N = 8400N.
When the same vehicle is driving on an icy road, the maximum braking force will be much less.
μ : 0.2 (coefficient of friction between tire and road on icy road)
Fs = μ . FN -- > Fs= 0.2 . 3000N
Fs= 2100N (Braking force)
Brake Performance – Brake Test
The brake performance of the vehicles is made by the brake measuring devices at the vehicle inspection stations. Brake performance of vehicles: It is found by dividing the brake force of the vehicle by the weight of the vehicle and is expressed as a %. Usually this ratio should be at least 50%.
The brake calculation formula is as follows:
Z = (F/G) . one hundred%
Z: braking rate
F : Brake force
G: The weight of the vehicle.
In the example above, if the braking force of the vehicle on the dry road is 8400N, and the mass of the vehicle is 1200kg = 12000N;
Z = (F/G) . 100% -- > Z= (8400/1200) . It is found as 100% -- > Z= 70/100 = 70%.
Main Topic (See: Brake System Types and Parts)
