Engine & Tuning Homework Section
I'm in the process of completely rewriting this section. The original pages contained some useful information, but were half finished and some of the information was not up to date with my latest thoughts after spending six years thinking about this stuff all of the time.
The combustion cycle allows a continuous stream of explosions to occur in the engine. Each one of these pushes down on the pistons. These transfer the load into the crank shaft via the connecting rods. The end result is that the crank is pushed round.
How hard the crank is twisted is referred to as torque. Torque is defined as a load acting some distance from a pivot point. Think of a bike. The rider pushes down using their weight onto the pedals. This is a linear pushing force. Because the pedal is some distance from the pivot of the bottom bracket then the force is turned into torque. In the UK, the normal units of torque are lb-ft. One pound foot (or foot pound. It's the same thing) means one pound of weight pushing one foot away from the pivot. The same effect can be produced by two pounds pushing down half a foot from the pivot. Or half a pound pushing down two feet away. This is why you use a long breaker bar on a tight nut. More torque from the same force.
Another common unit for torque is the Newton - Metre (Nm). One Newton metre can be made by applying one Newton of turning force (aka 0.1Kg) one metre away from the pivot. There is a simple conversion factor: 1 lb-ft = 1.356 Nm
Power is calculated by multiplying the torque by the speed of the crank. The answer has to be multiplied by a constant to turn it into the required units. Forgetting about the constant for a moment, the same power can be made from various combinations of speed and torque. So a small engine making low torque but running at very high rpm can make the same power as a large engine running at low rpm.
To calculate power from torque use:
P = k x T x N (thousands of rpm)
| Value of K for various unit combinations. | ||
| Power | Torque | |
| Lb Ft | Nm | |
| BHP | 0.193 | 0.142 |
| KW | 0.142 | 0.105 |
Difference between torque and power
At any engine speed then it is twisting the crank with a certain torque. From the speed and the torque you can work out how much power it is making at that point in the rev range. Both torque and power are measures of the same physical thing. Torque is how hard the crank is being twisted. Power is how hard it's being twisted multiplied by how fast it's going. When you make torque, power results.
The twisting torque is transmitted through the gearbox (where it is multiplied by the gears). This ends up as a torque at the wheels. Dividing this by the radius of the wheels gives a force acting on the tarmac. The equal and opposite to this pushes on the car. At speed there are constant forces to overcome, like rolling resistance and air drag. Any excess force can be used to accelerate the vehicle. The more torque from the engine then the more acceleration possible.
The above paragraph can also be written in terms of power. The engine output power is transmitted through the gearbox to the wheels. There is some loss in the gearbox (around 15-20% in a fwd car). The power at the wheels is used to overcome rolling resistance and drag power losses. Any excess can be used to accelerate the car. Power is force multiplied by speed. So dividing the excess power at the wheels by the road speed gives the force available to accelerate the car.
Both torque and power are measures of the same thing. When comparing engines though, peak power is a more useful figure. This is because power already includes the engine speed at which a particular torque is made. A high rpm engine will need more gearing to get down to the same road speed as a low rpm engine. This will multiply the torque more. Hence if both engines make the same torque, the high rpm engine will have better acceleration.
ie: If two engines made the same torque figure, but one made at it twice the rpm, then it would need twice as much gearing for the same road speed, so would have twice as much torque available at the wheels for the same road speed. It would have twice as much acceleration if there were no losses. Much simpler to think of it as having twice as much power.
Common phrases about torque and power
People often say things like "mid range torque is very important", or "power is great but torque wins races".
"mid range torque is very important" This is only true if you understand the full picture. Ultimately, all that matters is the engine power output over the operating rev range. If a car was geared to always operate between 6000 and 7000rpm and had a perfect computer controlled gearbox then it wouldn't matter at all what the torque/power was below 6000rpm. In fact, it wouldn't matter at all what the torque figure was anyway, all that matters is the power output over the operating rev range. It doesn't matter if the engine makes less torque than an electric drill, providing the rpm is high enough to make decent power. However, gear changes take time, and having an engine that can pull at lower rpm all help in the real world where drivers make mistakes. On a road car, running at high rpm is noisy and leads to increased engine wear rates. So what people mean is that a peaky engine that doesn't go well outside a narrow rev range can actually be slower than a more tolerant engine that pulls better over a wider rpm range. On a road car, you can drive unknown roads and pull round unexpected corners in 4th gear at 2500rpm without trouble.
"power is great but torque wins races". Once again, only true if you understand the full picture. In strict terms, it's wrong; power is all that matters. What this expression means is don't just think about peak power, but consider power over the operating rev range.
An engine is essentially an air pump. For every unit of air volume then a certain amount of chemical energy can be released. The power output depends on two things: how much energy is released for each volume of air; and how much air is moved through the engine in a set time. For standard normally aspirated engines then the amount of energy per unit volume of air is pretty much constant. The energy per unit volume is directly related to the peak torque potential of an engine. Hence, most normally aspirated engines make a very similar amount of torque per litre. Power is related to rate of air flow. A large engine turning slowly can pump the same amount of air as a small engine turning quickly. Hence a small engine at high rpm can make the same sort of power as a large engine at low rpm.
I analysed some typical engine data. Click.
The maximum power of an engine can be estimated by applying some maths to the following logic...
Mathematically:
| P = (Vol effic) x (Ind effic) x (Mech effic) x | ( | 1 | ) x ( | v | ) x ( | n | ) x ( | air density | ) | x (fuel calorific value) |
| 2 | 1000 | 60 | air fuel ratio |
| Symbol | Property | Typical value | Units |
| P | Engine power | 40BHP to 200BHP | W |
| Vol effic | Volumetric efficiency | 0.8 (80%) | - |
| Ind effic | Indicated efficiency (fuel conversion efficiency) | 0.4 (40%)-0.45 (45%) | - |
| Mech effic | Mechanical efficiency | 0.75 (75%) - 0.8 (80%) | - |
| v | Engine capacity | 0.8 to 3 | litres |
| n | Engine speed | 5500 | rpm |
| air density | 1.2 at 20deg C and standard atmosphere | Kg/m3 | |
| air fuel ratio | 14 (14:1) | - | |
| fuel calorific value | 43 | MJ/Kg | |
Plugging the typical numbers into the equation gives the useful formulae:
|
P(KW) = 7.4 x v (litres) x n (thousands of rpm) |
P(BHP) = 10 x v (litres) x n (thousands of rpm) |
The graph shows the predicted power versus engine size for two values of n. The results are
in pretty good agreement with the Vauxhall 8v engine data. The larger engines
are better than the graph line predicts. This is because the compression ratio is
larger for these models and also because they tend to be tuned more for power,
with multi-point fuel injection etc..
| P = (Vol effic) x (Ind effic) x (Mech effic) x | ( | 1 | ) x ( | v | ) x ( | n | ) x ( | air density | ) | x (fuel calorific value) |
| 2 | 1000 | 60 | air fuel ratio |
The equation also reveals the methods for increasing power. These are
The acceleration performance of a car at low speeds depends primarily on power to weight ratio. This is because air drag forces are low. At higher speeds then aerodynamic drag forces rise very quickly, so outright power and low drag is important.
In overall performance terms, power to weight ratio will instantly allow an estimate of acceleration times. So what is faster: a 1.6 making 160BHP or a 3L V6 making 160 BHP? If they were both in the same car with the same total weight? Answer: If they both used continuously variable transmission so that the both put out exactly 160.0000BHP all the time then they would accelerate identically. However, most cars use fixed gear ratios. So the 3L engine would have an advantage, because would make more power at the lower end of its rpm operating range. Hence the average power over the operating rev range would be higher. When I say operating rev range, I mean the rev range used when changing gear at high rpm. I don't mean power at idle speed. The 3L engine would also allow a wider rev range to be used, which would make driving easier, especially if the gear ratios are not optimum for the track/road in use.
There's quite a lot to this. It is true, more torque does mean better acceleration. But more power means more torque and better acceleration. So what's more important? Well peak power is the priority. BUT, not at the expense of average power available over the operating rev range. People often call this "maximum area under the torque graph". That's the same thing. It means don't push peak power up but lose power lower down the operating rev range. Once again, operating rev range means the rev range the engine will be operated over when being driven. Not the whole rev range available.
I developed a very simple model to predict acceleration performance from power to weight ratio. Click
I spent a long time developing a much more detailed acceleration model. It's partially written up here. Click.
The maximum speed of a vehicle is limited by the power required to overcome rolling resistance and air drag. Of these two it is air drag that is the more important. The graph below shows the power requirement in BHP as a function of speed. The power requirement scales with speed cubed (speed^3), frontal area, air density and the drag coefficient. Because of the cubic relationship, a 1% increase in speed requires a 3% increase in power. For this reason a 15% power increase will yield only about a 5% speed increase. To convert from power increase to speed increase divide the percentage gain by 3. Strictly speeking you should use speed_increase%=(1+power_increase%/100)^(1/3). The number^(1/3) means find the cubed root of the number. This only differs from speed_increase%=power_increase%/3 by a fraction of 1% for increases less than 50%. That's good enough.
What this means is that you need a hell of lot of power to go a bit faster.
To convert from engine rpm to mph think through the following. You know the engine speed in rpm, that's shaft revolutions per minute. When the drive goes through the gearbox the speed is reduced first by the gear ratio and then by the final drive ratio. So if the gear ratio is 2 and the final drive ratio is 4 then each rev from the engine will be divided by 2 and then divided by 4 to come up with 1/8th of a rev at the wheels. This will still be in revs per minute. Multiply this by sixty and you will get the number of wheel revolutions per hour. For each wheel revolution the car will travel a distance equal to the circumference of the wheel. So you need to know the wheel effective diameter in metres, multiply this by Pi (3.14) and you will get the wheel circumference, ie: the car distance travelled per revolution. Multiply this by the number of revs per hour and you will get the distance travelled in metres per hour. Divide by a 1000 to convert to Km, then divide by 1.609 to get it in mph.
So...
| V(mph) = | n(rpm) x 60 x 3.14 x D(m) |
| gr x fdr x 1609 |
The formula can be arranged by combining the constants and putting n=1000rpm in as a constant. You get
| V(mph)per 1000rpm = 117.09 | D(m) |
| gr x fdr |