Simple Performance Model
Description of a very simple model to predict vehicle performance from power to weight ratio alone. The early stages need knowledge of A level maths to understand. Do not let this deter you, the resulting equations can be understood intuitively.
Two fundamental Physics equations to start with:
Power = Force x Speed P=Fv
Force = Mass x Acceleration F=ma
These can be rearranged and combined to give:
a=(P/M)/v
The equation shows that instantaneous acceleration is the power to weight ratio divided by the instantaneous speed. Hence maximum acceleration occurs at low speeds. Bit of a problem when v=0 because this then says that the acceleration is infinite.
This is true for an engine that puts out constant power, with no gear changes, transmission losses, drag losses or rolling resistance. Obviously all of these effects have an influence. A simple way round this is to introduce a scaling factor that converts from peak power to effective average power. This scaling factor is an experimental "bodge" that makes the results conform closer to reality. There's nothing wrong with that as long as it is understood. This number can be determined by fitting the equation to actual results and choosing it so the results sit nicely in the middle of real results. This constant will be called k, for konstant.
Hence:
a = k * (P/M)/v
Now, acceleration is the rate of change of speed, so the equation can be written as:
dv/dt =k * (P/M)/v
The solution of the equation is found by integrating both sides:
v dv = k * (P/M) dt
=> v^2 / 2 = k * (P/M) t
The equation for speed versus time is therefore:
v = (2k * (P/M))^(0.5) * t^(0.5)
This means that speed increases with the square root of time and is affected by the square root of the power to weigh ratio.
Equation for time to reach a speed
t = 1/(2k) * v^2 / (P/M)
Despite not knowing the value of k yet, this equation is very informative. It says that the time taken to reach a certain speed from rest goes up with the speed squared. So if it takes 7s to reach 60mph, then it will take 100 over 60 all squared times longer to reach 100mph. 7*(100/60)^2 = 19.5s. Sounds about right, a hot hatch capable of 7s to 60mph will take around 20s to reach 100mph.
A possible use of this is to estimate your 0-60time from the 1/8 mile time and speed on a 1/4 mile strip. Since time is proportional to speed squared:
t2 = t1 * (v1/v2)^2.
The fact that it is 1/8 mile is not used, but the speed and time can be used to scale. So if it took 9.8s to reach 73mph when I did my best 1/4 mile run, then the 0-60 time would be:
t(0-60)= 9.8 * (60/73)^2 = 6.6s. Sounds good. Bear in mind that the real time could well be more like 7s, or even the 7.2s I managed on a slightly gravelly surface. It's only approximate.
Notice that the time to reach a speed is also inversely proportional to power to weight ratio. So a 1% gain in power or a 1% loss in weight will improve the time by 1%. This will start to become unrealistic as the power increases to large amounts because traction issues will prevent the car from using all of that power.
Equation for time to travel a distance
The equation for speed versus time can subjected to a mathematical process called integration. This converts the equation from speed to distance travelled. The answer is:
x =2/3 * (2kP/M)^(1/2) * t^(3/2)
This really isn't that informative on its own. It's much useful to rearrange it to give the time taken to travel a required distance.
t = (x * 3/2)^(2/3) / (2kP/M)^(1/3)
This says that the time taken to travel a distance is proportional to one over the cubed root of the power to weight ratio. So each 1% increase in power or reduction in weight will reduce the 1/4 mile time by 0.33%.
Estimating 1/4 mile time changes
If you want to estimate the new 1/4 mile time based on your old 1/4 mile time use:
t2 = t1 * (P1/P2)^(1/3) * (M2/M1)^(1/3)
Alternatively, take your percentage change in power to weight ratio, divide by 3 and that will be the percentage change in 1/4 mile time. Increasing power or reducing weight will obviously reduce the time.
If I increased my power from 130 to 145BHP and my 1/4 mile time was 16.2s with 130BHP, my new 1/4 time would be:
t2 = 16.2 * (130/145)^(1/3) = 15.6s. Sounds about right. Note that the weight bit of the equation wasn't used since it would be equal to one.
If the car weight was reduced from 1050Kg to 950Kg then the 1/4 mile time would reduce to:
t2 = 16.2 * ( 130/145)^(1/3) * (950/1050)^(1/3) = 15.1s. Sounds about right.
Now, if the power was increased to 160BHP then the time would be reduced to:
t2 = 16.2 * (130/160)^(1/3) * (950/1050)^(1/3)=14.7s. Looking good..
Sadly thought, the roll cage added 30Kg, taking the vehicle mass with driver back up to 980Kg. The 1/4 mile would then be:
t2 = 16.2 * (130/160)^(1/3) * (980/1050)^(1/3)=14.8s. Still looking good.
Any of the above changes can be done by taking the percentage change in power to weight and then dividing that by 3 to give the percentage change in time.
Estimating speed after traveling a given distance
The time taken to travel a distance is given by:
t = (x * 3/2)^(2/3) / (2kP/M)^(1/3)
Speed at a known time is given by:
v = (2kP/M)^(0.5) * t^(0.5)
Substituting the time from the first equation into the second and doing some simplification gives:
v = (2kP/M)^(1/3) * (x * 3/2)^(1/3)
Hence the trap speed will increase by the cubed root of the power to weight ratio. So a 1% increase in power or 1% reduction in weight will increase the trap speed by 0.33%.
Estimating 1/4 mile trap speed changes
If you want to estimate the new trap speed then use:
v2 = v1 * (P2/P1)^(1/3) * (M1/M2)^(1/3)
Or alternatively take the percentage change in power to weight ratio and divide by 3 to give the percentage change in trap speed. Increasing power or reducing weight will increase the trap speed.
In going from 130BHP and a trap speed of 88mph, I moved up to 145BHP and a trap speed of:
v2 = 88*(145/130)^(1/3)=91mph.
Or 145/130 is a 9% increase in power, so there will be a 3% increase in trap speed. 1.03*88 is 91mph.
The equation for time to reach a speed is:
t = 1/(2k) * v^2 / (P/M)
The constant can be rewritten as:
t = C * v^2 / (P/M)
A ready source of data was the 0-60 times listed in the back of Evo magazine (car_stats.xls). The plot below shows the data and curve fits for 0-60mph and 0-100mph times.
Look at the power of x in the curve fits. For 0-60mph it is -0.86 rather than the -1 expected. For the 0-100mph time it is -1.13 rather than the -1 expected. This means that over the 0-60mph increment cars with high power to weight ratios don't do as well as expected, whilst those with low power to weight ratios do better than expected. On the 0-100mph increment cars with high power to weight ratios do better than expected. On the 0-60mph increment very high power levels don't produce the expected improvement because of traction issues. On the 0-100mph increment then cars with higher power to weigh will also tend to be those with higher power levels and hence will be better at overcoming air drag forces. These forces were lumped together into a single loss constant in the model. Bear in mind that the range of weights in the input data is actually quite narrow, probably 800-1500Kg.
The average of the indices two is -0.99, which is conveniently close to the expected single value of -1.
By calculating (t/v^2 * P/M) for all the cars listed for both 0-60mph and 0-100mph increments, the average value for C was found to be 0.29. The standard deviation was 0.026, so about 99% of the population (3 standard deviations) would be contained within 0.29 +/- 0.08. ie: about +/-27%. This was for units of mph, BHP and tonnes. If it was in the SI units of m/s, W and Kg then the constant would be 1.07. This means that k would be about 0.47. Hence the acceleration of the vehicle is actually about half of what it would be if it made peak power over all the operating rev range, there was no drag or rolling resistance, or traction issues, or gear changes or transmission losses.
The graph below shows the predicted times and the real data. The overall curve power is a pretty good agreement for such a simple model.
Prediction equation for acceleration time:
t = C1 * v^2 / (P/M)
Also
v = (t/C1 * P/M)^(1/2)
| v Speed units | P Power units | M Mass units | Value of C1 | Unit Multiplier Used |
| m/s | W | Kg | 1.07 | 1 |
| mph | BHP | Tonnes | 0.29 | (1/60^2*1000/0.6214)^2 / 0.7355 |
| km/h | KW | Tonnes | 0.08 | (1/60^2*1000)^2 |
Remember that this is only approximately true and takes no account of traction issues, gearing, vehicle drag differences and power b and.
Examples:
0-60mph time of car with 130BHP and 1020Kg.
t = 0.29 * 60^2 / (130/1.02) = 8.2s
If power increased to 145BHP, t=7.3s. If weight dropped to 950Kg, t=6.8s, if power increased to 160BHP and weight increased to 980Kg, t=6.4s.
Prediction equation for time and speed at a distance
t = (x * 3/2)^(2/3) / (2kP/M)^(1/3)
The constants can be grouped together into a new constant.
t = C2 * x^(2/3) / (P/M)^(1/3)
where C2 = (3/2)^(2/3) * C1^(1/3)
Since C1 has already been evaluated as 1.05 for SI units, then C2 = 1.33.
| x Distance units | P Power units | M Mass units | Value of C2 | Multiplier |
| m/s | W | Kg | 1.34 | 1 |
| miles | BHP | Tonnes | 204 | (1000/0.6214)^(2/3) / 0.7355^(1/3) |
| km | KW | Tonnes | 134 | 1000^(2/3) |
| feet | BHP | Tonnes | 0.67 | (25.4* 12/1000)^(2/3) / 0.7355^(1/3) |
The trap speed can be used by using the time to cover the distance in:
v = (t/C1 * P/M)^(1/2)
Comparison of this prediction with real data showed that it generally estimated 1/4 mile time quite well, but the trap speeds were generally too slow. This shows up the limitations of the model. This detract from the usefulness of seeing the relationships between the various variables, but it does mean that it isn't a full enough model to allow accurate predictions to be made over the 1/4 mile strip.
I've coded the equations on this page into an Excel spreadsheet. Simple_Performance_Model.xls.
Remember that this is a very simple model based on recorded 0-60mph times and cannot be expected to work in all cases, especially when speeds are high or power is very high.