Read Car Physics - part 2
So far we have seen the torque generated by the engine. But how does this translate to motion? In this article we will materialize the torque into motion. How does the torque influence the kinetic state of the car? How does the accelerator pedal interfere with all that?
We are starting with the relation between the wheel and the surface of the road. The wheel is either rolling or slipping. When torque is applied to it it's the friction that keeps it from slipping and actually pushes the car ahead. The friction is determined by the slip factor and the vertical force between the two surfaces. In our case the vertical force is the fraction of weight of the car that goes on the drive wheels. This friction translates to a torque on the axis of the wheel. This is the maximum amount of torque that can be applied on the wheel without leading to wheel spin. if the driver depresses the accelerator pedal more the wheels will start spinning and although this is impressive it is a waste of power and leads to slower acceleration.
First let's see how the engine torque and the friction work together to accelerate the car. Suppose that the engine is revving at 3000 rpm. The maximum torque at this speed is 130 Nm. If we are in second gear, at the wheels this is 1036.28 Nm. Half of this goes to the left and half to the right wheel. The tyres of the car have the following dimensions: 185-60-14. Knowing this, it is fairly easy to calculate the radius of the wheel. Let me explain these numbers and you will see. The first (185) is the width of the tyre in mm. The second (60) is the ratio of the height to the width in this case 60%. The third (14) is the diameter of the rim in inches. It really has it all but do not panic. The radius is the height of the tyre plus the radius of the rim.
wr = (tw*hr/100) + (rd*25.4/2) (1)
where wr is the wheel radius, tw is the tyre width, hr is the height ratio, rd is the rim diameter, 25.4 is the number of mm in an inch and divide by 2 for radius.
The second step to calculating the maximum torque the wheel can take is to find the friction between the wheel and the road. This is simple. The friction force is the product of the vertical force pressing the surfaces together, in our case the weight of the car on the drive wheels, multiplied by a friction factor indicating the relation between the surfaces. For the needs of a drive simulator we can use the generally acceptable value of 1.0. Thus we have
F = cw * ff / 4.0 (2)
where F is the generated force cw is the car weight, ff is the friction factor and the product is divided by 4.0 since half the weight is applied to the drive wheels and we have 2 drive wheels (aka 1/4 of the weight is applied on one wheel).
The torque diagram shows the maximum torque the engine produces at a given speed. The actual amount of torque depends on the driver and how much he presses on the accelerator pedal. If we push the pedal all the way down and fully open the throttle this is the torque we will get. If the torque (aka force) produced is greater than the resisting forces the car accelerates, else it decelerates. There is always a balance point between the resistance and the engine torque. If the engine torque is greater than the torque the wheels can handle they will start spinning. Wheel spin is undesirable for two reasons. First it is a waste of energy. The car is not accelerating any faster. Second it decreases the friction factor, the grip of the tyres is less and this leads to even more loss of power and slower acceleration.
Enough with all the driving theory. This is not a driving school. Let's put it all together and see what we will get.
Assume we are cruising at 3000 rpm in 4th gear. The maximum engine torque is 130 Nm. The drive torque at the wheels is 535.3 Nm. If we put the pedal own to 80% then the drive torque is 428.24 Nm or 214.12 Nm per wheel. The wheel diameter for 185-60-14 is 289 mm. The reference car is 1200 Kg and with driver and fuel 1300 kg which multiplied with gravity acceleration (9.81) is 12753 N.
The friction force is from equation (2)
F = 12753 * 1 / 4 = 3188.25 N
The drive force is (convert the torque to force)
Fa = (535.3 / 2) / 0.289 = 926.12 N
The driving force is less than the friction so there will be no wheel spin. But is the torque enough to accelerate the vehicle? In the first article we introduced the forces applied on the moving car. These are the aerodynamic drag and the rolling friction of the wheels. Both of these forces increase with speed so first we must calculate the speed. At 3000 rpm engine speed the wheel spins at 728.6 rpm or 12.14 revolutions per second. This determines the speed of the car. Actually the car speed is the linear speed of a point on the tread. All we have to do is multiply the wheel rotational speed by the length of its circumference. The circumference of the circle is
circ = radius * 3.14 * 2
and in our case radius = 0.289 m, hence the circumference is 1.81m. This gives us a speed of 21.97 m/s, which is approximately 79 kph or 49 mph.
Aerodynamic drag is
Fd = Cd * v * v = 0.43 * 21.97 * 21.97 = 207.55 N
Rolling friction is
Fr - Cr * v = 12.9 * 21.97 = 283.4 N
And the total resistance is
Fres = Fd + Fr = 207.55 + 283.4 = 490.95 N
This is less than the 926.12N of the driving force and the car will accelerate. The accelerating force is
Ft = Fa - Fres = 435.17N
The acceleration generated by that force is
a = Ft / m = 435.17 / 1300 = 0.33 m / sec2
As the car accelerates the drag is getting higher, the engine revs faster the torque changes accordingly, and eventually we come to an equilibrium point where the acceleration stops and the car is moving in constant speed.