The model has been extensively rewritten. It's still in progress. All this description will be updated soon to match. Download the new model here. Save as then run in Excel stand alone. It won't work in an internet browser window.

This page describes a computer based mathematical model used to estimate and predict the acceleration statistics for any vehicle. It has gone through a number of iterations to make it produce more realistic results. Since some of the loss factors can have a large effect on the results it is best to adjust the model to match the manufacturer's figures before forging ahead with predictions for a car.

Download the old acceleration model.

The model estimates the acceleration performance of a vehicle. It produces acceleration-time, speed-time and distance-time graphs. It uses the torque curve, gear ratios, wheel diameter, power train losses, tyre losses,  mass of car and drag factor.

 The model is quite simple to understand. If the car is travelling at a particular speed in a known gear then the engine speed can be worked out from a knowledge of the wheel diameter and the gear ratios. If the wheel diameter is known, then the circumference is Pi*D (where Pi is 3.14...). If the car speed is known in m/s then the wheel revs/second will be the speed divided by the wheel circumference. The engine speed will be faster than this, being multiplied first by the final drive ratio and then the gear ratio. To convert from revs/sec to revs/min multiply by 60. Therefore

n (rpm) =	60 x v x gr x fdr
	Pi x D
n is engine speed in rpm	v is car speed in m/s
gr is gear ratio	fdr is final drive ratio
Pi is 3.14...	D is wheel rolling diameter in m

If the engine speed is known then the torque available can be looked up from the data table of input values. Next the torque at the wheels can be calculated. The gearbox acts to reduce the speed and also multiplies the torque. It's the same as a low gear on a bike providing more "leverage" in exchange for more pedal speed. The actual torque at the wheels will be a bit less than expected because of transmission losses and tyre friction. If the losses are ignored, then

t_wheels  =	T_engine * gr * fdr
t_wheels is torque at wheels in Nm	T_engine is flywheel torque in Nm
gr is gear ratio	fdr is final drive ratio

A major difficulty is in coming up with realistic models for the transmission torque loss. The early model used a curve fit through some rolling road data. This was replaced by a simple percentage loss on later models. Much more realistic results were produced if two loss factors were used: one a simple percentage loss and another loss that was proportional to wheel speed. This was discovered by trial and error, though the physical process could be regarded as tyre losses.

 

t_wheels =	(T_engine * gr * fdr)ktlf - v2 kwlf
t_wheels is torque at wheels in Nm	T_engine is flywheel torque in Nm
gr is gear ratio	fdr is final drive ratio
ktlf is the torque loss factor	kwlf is the tyre loss factor

This now gives  a simple way of estimating the torque at the wheels. Dividing this torque by the wheel radius (half of the diameter) gives the propulsive force acting on the car. Next the drag losses need to be subtracted. These grow with speed cubed. That means if the speed is doubled then the drag losses increase by 2*2*2=8 times. In order to estimate the drag losses I looked in a text book that I own and rearranged things to get the handy formula...

If the drag force is subtracted then the total force acting to accelerate the car is known. Dividing this by car mass in Kg gives the acceleration in m/s/s.

This number expresses how many m/s of speed the car will increase by in 1 second. If the acceleration is 5m/s/s then in one second the car speed would increase by 5m/s. If it started at 0m/s then it would get to 5m/s. If it started at 200 m/s it would get to 205 m/s. The above allows an estimate of the the acceleration at a particular instant if the speed is known at that instant. But the trouble is, if a  1 second period of time is taken and the speed is known at the start of that instant then the speed will change during that second and hence the acceleration will change as well. The way round this is to use hundreds or thousands of much smaller time steps. It the speed is known at the start of a time step then the acceleration can be calculated. The speed at the end of the time step is approximately the acceleration multiplied by that time step. Eg: speed is 10m/s, acceleration is 3m/s/s, use time step of 0.001s, speed will end up as 10.003 at end of time step and time will have increase by 0.001s. ie: v(new)=v(old)+acceleration*time_step.

The same logic can be applied to distance travelled.

x(new)=x(old)+speed*time_step

The program starts at time = 0 seconds with v (speed) = 0 m/s and x (distance) = 0 m.. The car is in first gear. The acceleration is calculated. The new speed is calculated. The new position is calculated. This is repeated hundreds of times over small time steps. As the speed increases then eventually the engine speed reaches the rev limit or the user selected gear change rpm. When this happens the model changes the gear ratio and disengages the drive for a preset length of time. This simulates a gear change. When the engine reaches the rev limit in top gear then the speed is limited to that value. In most standard cars the top speed is limited by the drag forces so that the redline is not reached in top gear.

After fiddling around with the drag and loss constants, the car mass and the gear change time I started to get realistic looking results. The selection of realistic loss factors is important. In fact, it doesn't actually matter what physical process each factor represents as long as they provide a good fit to experimental data. Essentially the model says that there is a speed proportional loss and a speed squared loss. There is no reason why some sort of constant couldn't be added as well. Essentially, there is some sort of loss curve that.

When I came across a series of articles reprinted from magazine tests in the late 80's I was in a position to fine tune the model some more. Once again after fiddling around with the grip limit, change time, car mass, drag factor and change rpm I managed to get results that looked plausible.

The model has been used by several other people, most of whom find that the results agree with their quarter mile times and trap speeds remarkably closely. Those that don't get good matches do so because they don't properly tune the loss factors for the standard car. Garbage in, garbage out as they say. The model is realistic because the physics it uses is fundamental. Only the loss factors could be considered as a gross simplification.

Effect of engine and wheel inertia.

Firstly a model that didn't include transmission or drag losses was used. If engine and wheel inertia are ignored then the acceleration of the vehicle is given simply by:

If engine and wheel inertia are included then several equations can be written. These can be combined to produce:

Where: x'' is vehicle acceleration in m/s2, T is engine torque, gr is gear ratio, fdr is final drive ratio, r is wheel rolling radius, m is mass of vehicle, I_f is the flywheel inertia and I_w is the inertia of a single wheel.

The effective mass of the vehicle has now been increased, with a term depending on the engine inertia and a term due to the wheel inertia. Some typical numbers might be: car mass 800-1200Kg, first gear 3.42, final drive 3.55, engine inertia 0.04Kgm2, wheel rolling diameter 0.58m, radius 0.29m, inertia of a wheel 0.8 kgm2. This means that in first gear the engine inertia appears as an extra 55Kg of weight. This falls off in the higher gears. The wheel inertia appears as approximately an extra 35Kg of mass.