HOW TO GET MAXIMUM ACCELERATION

**Copyright © 1998,
2013 by Mike Clements**

My interest in cars goes all the way back to my earliest memories, and ever since I took my first physics course as a young teenager I would constantly debate with my friends about torque, power, gearing and speed. Our standard topics of debate were what determined acceleration vs. what determined top speed, and at what RPM to shift for maximum acceleration. We all understood the basic physics about Force, Mass and Acceleration and their rotational analogs of Torque and Rotational Inertia. We also were all familiar with the standard phrase, "Torque is acceleration but power is top speed." But somehow this never seemed quite right (it isn't) and these ideas were never enough to resolve the debates.

It was not until many years later that I finally reached a decent understanding of these ideas. I've described them here as clearly as I can. I’ll never forget the first moment the fog cleared and I mentally “saw” what torque and power really were. It was an incredible “AHA” experience. I’ve written it all down for two reasons. First, to solidify my own understanding by going through the explanation in a methodical, logical way. Second, to help others to understand, and to get their input where my explanation is unclear. There's some math involved for people who like that sort of thing, but I try to explain the conclusions in plain English.

First, let's talk about torque and gearing. Given the number of people I've met in car and motorcycle racing, I'd say it is a common belief that torque is the single most important factor determining the acceleration of a car. And in fact, twisting force -- torque -- really does determine the acceleration of the car. But it is torque at the wheel of the car that matters, and this is where the confusion starts. You can get lots of torque at the wheel of the car without having much torque at the crankshaft of the engine (think motorcycle). And conversely, you can have lots of torque at the crankshaft yet still fail to get much at the wheel (think diesel).

Anyone who has ridden a 10 speed bicycle intuitively understands gearing: when you downshift it's easier to pedal but you go slower. It's easier to pedal because the gear ratio is a torque multiplier. You go slower because the cost of that torque multipler is reduction in rotational speed.

Suppose one gear is spinning at RPM1 with torque T1, and another gear is spinning twice as fast with half the torque. If the second gear is run through a 2:1 reduction gear the final output spins with torque T1 at speed RPM1. That is, the 2:1 reduction gear halves the speed and doubles the torque. Intuitively, gearing it down gives it more force, but sacrifices rotational speed.

In short, torque is not conserved through the drivetrain. One can always trade rotational speed for torque, and vice versa. This is precisely what gearing does.

In a car, the most fundamental gear ratio we are concerned with is the total ratio between the speed of the crankshaft and that of the wheel. If the crankshaft is spinning R times as fast as the wheel, the torque at the wheel is R times that at the crank. So if you can spin the crankshaft fast enough, you can get lots of torque at the wheel of the car, even if you don’t have much at the crankshaft.

Consider a Panoz Roadster powered by a '98 Mustang Cobra drivetrain. The engine produces 300 ft.lbs. of torque. First gear is 3.37:1 and the differential is 3.73:1, so the total gear ratio from the crankshaft to the wheels is 3.37 * 3.73 = 12.57. In 1st gear, the engine crankshaft is spinning 12.57 times as fast as the rear wheels, producing 300 * 12.57 = 3,771 ft.lbs. of torque at the rear wheels.

Consider a '95 Mazda RX-7 (twin turbo Wankel). The engine produces 250 ft.lbs. of torque. First gear is 3.483:1 and the differential is 4.1:1, so the total gear ratio from the crankshaft to the wheels is 3.483 * 4.1 = 14.28. In 1st gear, the engine crankshaft is spinning 14.28 times as fast as the rear wheels, producing 250 * 14.28 = 3,570 ft.lbs. of torque at the rear wheels.

Armed with this knowledge, here's a pop quiz:

Engine A generates a maximum torque of 100 ft.-lbs. at 3,000 RPM. Engine B also generates a maximum torque of 100 ft.-lbs., but it happens at 6,000 RPM. Which engine would you put in your race car?

Many people would take engine A, reasoning that its torque is more "accessible" since it doesn't have to rev as high. However, engine B is far better. It revs twice as high as A but gets the same torque. So engine B can use gear ratios that are twice as short without sacrificing speed. Since the wheel torque is equal to crankshaft torque multiplied by the gear ratio, engine B can produce twice as much torque at the wheel of the car. Yes, that means twice the acceleration.

Now I'm sure many of you are asking, "why not put those same short gears in engine A?" You can; but it limits your speed because engine A can't rev as high. You'll only be able to go half as fast in any given gear. And every time you shift up to the next gear, you're losing acceleration because you're getting less torque at the wheel of the car. For example, if you both start out in 1st gear you will be racing head-to-head. But soon, say around 20 mph, you have to shift to 2nd. But your competitor, using engine B, can stay in 1st gear until 40 mph (since you both have the same gear ratios but he can rev twice as high). So you begin to fall behind and your competitor blows by you. In fact, he's got twice the overall acceleration you have.

Some might argue that engine A would be quicker off the line, even though B would eventually pass it. This is based on the intuitive notion that both cars are initially standing still and engine A has more low end torque than engine B. But in this case intuition leads us astray with two mistaken assumptions. First, just because an engine achieves its peak torque at higher RPM does not mean it has less torque at lower RPMs. But more importantly, we are not concerned with torque at the engine crankshaft, we care only about torque at the wheel of the car. And since engine B has twice the gear ratio as engine A, it will get off the line just as quickly as engine A even if it has only half the low end torque at the engine crankshaft.

Regardless of the gear ratios you use with Engine A, your competitor using engine B will always blow you away because at any given speed, you've only got half the acceleration he does. Since we assumed that both cars have the same torque and the same weight, we just scrapped the common belief that torque determines acceleration.

Being able to rev the engine high is just as important as making good torque. Engine RPM multiples or leverages the torque at the crankshaft. But RPM alone doesn't make a car go fast, either. RPM and torque both are equally important, since it is the product of the two that makes the car go.

In a given gear, the acceleration of a car follows the torque curve. That is, acceleration directly corresponds to engine torque output. Many people, even experienced racers, believe this means they should shift gears just past the peak torque RPM before the engine's torque rolls off. While this seems like plain old common sense, it's wrong.

If you are screaming through 3rd gear with your foot on the floor as the engine is passing through its peak torque RPM, you may think you're getting the best acceleration you can. And you are, for 3rd gear. If you drop down to 2nd gear the engine will be spinning at a higher RPM and will be producing less torque. But if you think this means your acceleration would diminish in 2nd gear you just fell for a mistake that has fooled thousands before you -- mistaking crankshaft torque for wheel torque.

Second gear is a shorter ratio so the torque at the wheel of the car is a larger multiple of the torque at the engine crankshaft. In 2nd, the engine may produce more torque at the wheel of the car, even if it has less torque at the crankshaft. Remember it's the product of torque and gearing that pushes the car, so if you have a little less torque with a lot more gearing, you're going faster.

Here’s another way to look at this: remember that 3^{rd}
gear is a taller (numerically smaller) ratio than 2^{nd}, so it
provides less acceleration. Suppose the difference between the two ratios is
30% (any percentage will do, but this is typical). That means that when you are
accelerating in 2^{nd} gear, the moment you shift to 3^{rd} you
will have 30% less torque at the wheel of the car – you will lose 30% of your
acceleration. Thus, you should not shift to 3^{rd} gear until you have
passed the engine’s torque peak, and the torque has dropped at least 30% from
its peak value.

Many racers try to max out their acceleration by “feel”. And since the acceleration of the car follows the torque curve of the engine, you can feel the acceleration decreasing as the engine passes the peak torque RPM. And the moment you feel the car’s acceleration start to diminish, you instinctively grab for the shifter. But then you shift too soon. You’ve maximized the torque at the engine crankshaft, but you haven’t maximized the torque at the wheel of the car. As you rev the engine past the torque peak, you can feel your acceleration gradually diminishing. But if you were to shift, the taller gear ratio would reduce your acceleration even more. So you must resist the urge to shift and let the engine keep revving a bit higher than is intuitively obvious.

The bottom line of all this is the following equation:

P = __2π nt__

33000

**P ** engine power in horsepower

**π ** approx.
3.14159265358979323846264338327950288

**n ** engine speed in RPM

**t ** engine torque in ft. lbs.

Removing constant factors (2π / 33000), this equation becomes:

P = nt

** Which means power is the product of torque and RPM.**
Since RPM represents the potential gearing that can be used, power tells you how much
torque you can get at the wheel of the car at any given speed[1].

Incidentally, we mentioned above that gearing trades rotational speed (in this case, engine RPM) for torque and vice versa. Torque is not conserved through the drivetrain (neither is rotational speed). But their product, power, is conserved. Let's revisit the above examples to see how this works.

The Panoz Roadster had 300 ft.lbs. of torque at the engine crankshaft and 3,771 ft.lbs. at the rear wheels in 1st gear. What is the power? Its Ford Cobra V-8 makes 305 HP at the crankshaft. In first gear, the rear wheels are spinning 12.57 times slower than the engine crankshaft, with 12.57 times more torque. Since power is torque * RPM, power at the rear wheels is the same as at the crankshaft.

The Mazda RX-7 had 250 ft.lbs. of torque at the engine crankshaft and 3,570 ft.lbs. at the rear wheels in 1st gear. What is the power? It produces 275 HP at the crankshaft. In first gear, the rear wheels are spinning 14.28 times slower than the engine crankshaft, with 14.28 times more torque. Since power is torque * RPM, power at the rear wheels is the same as at the crankshaft.

The examples demonstrate that torque is not conserved, but power is. Power is the same at every point in the drivetrain. Gearing swaps rotational speed for torque, and the swap is always linear. Thus their product remains constant, and that product is power. In reality, drivetrain power efficiency is less than 100%, but it is very high, often 90% or better.

We can draw other conclusions from this mathematical relationship.

**First**, since power is the product of torque and RPM,
power cannot max out until torque is already decreasing. Thus, the peak power
RPM is always higher than the peak torque RPM.

**Second**, whether the peak torque RPM and the peak
power RPM are close together or far apart, depends on how rapidly the torque
rolls off after it peaks. If the torque rolls off quickly after it peaks, the
power peak will be close to the torque peak.

**Third**, every engine has a range of RPM just above the
torque peak, where the torque is rolling off, but engine RPM is climbing faster
than torque is dropping. Since power is equal to the product of torque and RPM,
the engine’s power output is still increasing. Then, at some even higher RPM,
torque rolls off faster than RPMs climb. The peak power RPM is the transition
point where the rate of decreasing torque matches the rate of increasing RPM.
Past this point, torque is dropping rapidly – “like a rock” as some would say.

Many people like engines that get all their torque down low because "you don't have to rev it up to get to the power". While these engines are easy to drive, they are slower on a racetrack or autocross. Because they get their torque at low RPM, they cannot take advantage of short gearing to multiply that torque. Thus it is no mystery why some of the fastest cars in the world are the hardest to drive. They are designed to achieve their maximum torque at the highest possible RPM. The higher RPM at which they achieve their torque, the more gearing they can use to multiply the torque even higher. If achieving peak torque at 2,000 RPM allows you to use a 4:1 gear ratio, then getting that same torque at 6,000 RPM allows you to use a 12:1 ratio, which means three times the torque at the wheel of the car. That's three times the acceleration, which is the difference between doing 0 to 60 in nine seconds, or in three seconds[2].

Of course, developing strong torque at high RPM doesn’t necessarily mean the engine will have weak torque at low RPM. Some engines manage to develop peak torque at high RPM, but still get 90% of their peak torque throughout the entire RPM range. Features like variable intake geometry and valve timing help engines do this, and they are becoming common.

A couple of real world examples should demonstrate how these ideas apply equally well to two completely different engines. First, a 46ci Suzuki Katana motorcycle engine; next, the 454ci Chevrolet V8. In the charts shown below, the green/grey line is torque and the red/black line is power. In these examples, I assume that shifting up one gear lowers the engine RPM by 25% -- if you're at 3,000 RPM in a gear and you shift up, you will be at 2,250 RPM. While this is pretty close for most vehicles, the actual number used does not affect the conclusions (try using a different number and see for yourself).

Here is the torque/power curve for the Suzuki:

The torque peak is 48.5 ft.-lbs. at 10,290 RPM; the power peak is 100 hp. at
11,550 RPM. Following the torque curve, you'd shift up just past 10,290 RPM to
keep the engine near its peak torque output. Following the power curve, you'd
shift up just past 11,550 RPM to keep the engine near its peak power output.

Nobody would advocate shifting before the torque peak at 10,290 RPM. The question is how far past this point to go before shifting up to the next gear. In the chart below, each row tells you what happens before and after shifting gears at any point in time. For example, the first row says that if you are at 10,290 RPM in one gear and you upshift, you'll be at 7,718 RPM. The two shaded columns give you the final torque at the wheel of the bike before and after the shift -- in this case, 48.5 before the shift, vs. 34 after the shift. The actual wheel torque will depend on the final drive ratio, but that is irrelevant because it will be proportional to these numbers.

BEFORE SHIFT |
AFTER SHIFT |
|||||

RPM |
CRANK T |
RATIO |
RPM |
CRANK T |
RATIO |
FINAL T |

10290 |
48.5 |
1 |
7718 |
45 |
0.75 |
34 |

10500 |
48 |
1 |
7875 |
45 |
0.75 |
34 |

11000 |
46.5 |
1 |
8200 |
45 |
0.75 |
34 |

11500 |
45 |
1 |
8600 |
42 |
0.75 |
31.5 |

12000 |
40 |
1 |
9000 |
43 |
0.75 |
32 |

12500 |
38 |
1 |
9400 |
46 |
0.75 |
34.5 |

As the table above shows, the increased torque produced at lower RPM never overcomes the gearing disadvantage of the higher gear. The engine hits redline before reaching the optimal shift point. Thus, the optimal shift point is redline (12,500 RPM).

Here is the torque/power curve for the Chevy:

The torque peak is 440 ft.-lbs. at 3,500 RPM; the power peak is 333 hp. at
5,000 RPM.

Again, nobody would advocate shifting before the torque peak at 3,500 RPM. The question is how far past this point to go before shifting up to the next gear. This chart works just like the last one.

BEFORE SHIFT |
AFTER SHIFT |
|||||

RPM |
CRANK T |
RATIO |
RPM |
CRANK T |
RATIO |
FINAL T |

3500 |
440 |
1 |
2625 |
400 |
0.75 |
300 |

4000 |
407 |
1 |
3000 |
427 |
0.75 |
320 |

4500 |
376 |
1 |
3375 |
435 |
0.75 |
326 |

5000 |
351 |
1 |
3750 |
421 |
0.75 |
316 |

5500 |
306 |
1 |
4126 |
399 |
0.75 |
299 |

As the table above shows, the increased torque produced at lower RPM never overcomes the gearing disadvantage of the higher gear. As with the motorcycle, the optimal shift point is redline (5,500 RPM). This may comes as a surprise to some big V-8 owners!

This whole discussion becomes much simpler if we take a different perspective.
Think about this from the perspective of Energy.
Kinetic energy depends on the mass and speed of the car.
At rest, a car has zero kinetic energy.
Suppose it has kinetic energy ** E** at 60 mph.
Energy

From this perspective, it becomes obvious that Torque and RPM alone are each totally irrelevant. If you want to accelerate a certain car to 60 mph, it takes a fixed amount of work. If you want to do it in a specific time, say 5 seconds, it takes a fixed amount of power. ANY engine that produces that power can do the job. It doesn't matter how much torque the engine generates, or how high it revs. 1 ft.lb. at 1 Million RPM is the same as 1 Million ft.lbs. at 1 RPM. Either way, it's 190 HP.

Finally, we're ready to answer the original questions from the first paragraph.

First, what
is more important for acceleration, power or torque? The answer should by now be obvious: **POWER**. The above
discussion should make it clear that neither torque nor RPM is alone sufficient
to determine acceleration. Since one is force and the other is leverage, it is
their product that determines acceleration. And their product is power. The
single most important factor determining acceleration is a car's power to
weight ratio. For good acceleration you want a light car with lots of power.

Second, what determines the top speed of a car? The answer is much simpler than it seems:
once again, **POWER**. The whole question of acceleration vs. top speed is a
red herring. Both depend on the amount of torque at the wheels of the car. And
power at the engine crankshaft determines how much torque you can get to the
wheels of the car. So the single most important factor determining top speed is
the ratio of a car's power to its coefficient of wind resistance. For high top
speeds you want an aerodynamic car with lots of power.

Third, when should you shift gears for maximum
acceleration? The answer is that you
shift on the **POWER** curve. That means your shift points should surround the peak power RPM.
More precisely, the optimum shift point for any engine and transmission is *always*
between the peak power RPM and redline. If you're not inclined to get
into the physics of it, just split the difference and shift halfway between the
two -- you'll always be very close.

To be precise, the optimum shift RPM is the point at which
the engine’s power output is the same before and after the shift. Ride
the power curve to the peak and over just far enough so that when you shift,
you’re at the same power output but you’re climbing the curve again. Thus, how
far past the peak power RPM you go, depends on how close together your gear
ratios are spaced. Since most cars have the lower gears spaced farther apart than the
higher gears, the optimum shift point is usually redline in 1^{st}
gear, dropping by a hundred RPM or so in each successive gear. But since the
peak power RPM and redline are usually close together, splitting the difference
and using the same point for all gears is a good strategy.

Here are the key points which the above discussion and examples have demonstrated:

1. Power is conserved but Torque is not -- do not mistake crankshaft torque with torque at the wheel of the car.

2. Engine RPM and Torque are both equally important for acceleration.

3. Power alone tells you what the engine can do – torque and RPM alone each mean nothing.

4. For maximum acceleration, always shift between the engine’s peak power RPM and redline.

Armed with this knowledge, let's tackle some common misperceptions.

**You can't tow a load with a high power, low torque engine.
To tow a load, you need torque!**

There is a thread of truth to this, but only a thread.
The people who say it typically don't understand that it is POWER that tows the load.
That should be clear already, but if it's not, consider this:
Suppose you need to tow a boat up a hill.
** Any** engine can do the job. Any engine at all.
Your lawnmower, or the electric motor that rolls your window up and down.
The only problem is it's going to take a long time.

Thus the question is not whether the engine can tow the load.
Any engine, no matter small, can tow any load up any hill, given enough time.
The question is how **quickly** the engine can tow the load.

How quickly: the time component is absolutely essential and critical. Any work that is ever actually done, is done over a certain amount of time. Power is work over time. It tells you how quickly the engine performs work.

Consider a truck with a 250 HP diesel that makes 500 ft.lbs. of torque. At full throttle, it tows a certain boat up a certain hill at 35 mph. Since POWER is what moves the load, you can pull that same boat up the same hill at the same speed with ANY engine that makes 250 HP. Consider a motorcycle engine that makes 250 HP with only 95 ft.lbs. of torque. Why does it sound ridiculous for this engine to tow the boat up the hill?

It's all about how well the engine can sustain its power output. Since power = torque * RPM, given any 2 of them you can compute the third. We were given power and torque for these engines so we can compute the RPM at which the power is generated. That diesel makes its power at about 2,600 RPM. It's just loafing along while producing full power. The motorcycle makes its power at about 14,000 RPM. How long do you think that motorcycle engine will last spinning at 14,000 RPM, at full throttle, going 35 mph?

The reason high torque diesels are good at towing loads is only indirectly because they have a lot of torque. Torque doesn't pull the load - power does. But since power is torque * RPM, high torque means they operate at low RPM. Making the same power at lower RPM has some benefits:

Because they operate at low RPM, they can produce their maximum rated power continuously even at slow speeds, and are more efficient when producing their max rated power. Most high power, low torque engines cannot produce their maximum rated power continuously at low speeds, and are not efficient when producing their max rated power. They run hotter and require more airflow and cooling.

If it were possible to build a high RPM low torque engine that could operate continuously and efficiently at its maximum rated power output, even at slow speeds, without overheating, that engine would tow the boat up the hill just as well as the high torque diesel. The point is, for towing you don't necessarily need high torque. What you need is an engine that has enough power to tow the load, and can produce that power output continuously and efficiently. High torque engines are just one way to meet those goals and have proven to be practical. Also, diesel engines have higher compression than gasoline engines, which makes diesels more fuel efficient. We'll talk more about compression and efficiency below...

The power an engine can produce depends on air density. If the air sucked into the intake has greater density (and more fuel is added to keep the air-fuel ratio the same), it will produce higher pressure when it ignites, which will push the crankshaft harder, which means the engine develops more torque at the same RPM, which is more power. Alternately, one could say that if the A/F ratio is held constant, more air means more fuel. Thus the engine is burning more fuel with each revolution. If efficiency is constant, burning more fuel in the same amount of time means doing more work, which is more power.

There is a common misperception that an engine’s
compression ratio is a measurement of this firing pressure. But the compression
ratio is just an abstract geometric number - it does not indicate firing pressure.
Compression ratio compares the
geometric volume of the cylinder chamber when the piston is at the bottom of the
stroke (large), to when it is at the top of the stroke (small). This ratio influences,
but it does not determine, the firing pressure. For example, imagine adding a
turbocharger or supercharger to an engine. This considerably increases the
firing pressure but it does not change the compression ratio at all.
In fact, most turbocharged and supercharged cars have ** lower** compression ratio
than normally aspirated cars.

Firing pressure is expressed as BMEP which stands for Brake Mean Effective Pressure. The torque an engine produces depends on its BMEP and its total displacement. Essentially, this means torque is the average firing pressure multiplied by the volume in which the pressure is exerted. BMEP is expressed in PSI, but in unitless form it is easy to calculate: BMEP is simply torque divided by displacement.

BMEP = __Torque__

Displacement

The firing pressure is determined by how full the intake can fill the cylinder during the intake stroke. For example, if the valves are small and don’t open very long, or if the intake manifold restricts the air flow, the engine will not suck in very much mixture and will produce low firing pressures. An engine with a well designed intake manifold, with high flow valves and optimized cam timing, will fill up the cylinder quickly and efficiently, thus producing a high BMEP.

The ultimate goal is to produce an engine that generates a high BMEP and is able to sustain that BMEP even when spinning at high RPM. As RPM increase, this is hard to do because there is less time to fill up the cylinder. That is why high revving engines use 4 or more valves per cylinder, variable intake runners, and other technologies to get the engine to “breathe” at high RPM.

Most normally aspirated gasoline engines produce roughly the same BMEP, which means an engine’s peak torque is more or less linearly related to displacement. If you compute the BMEP for just about any gasoline engine from Hondas to Ferraris to Fords to motorcycles, you'll find them suprisingly close.

For example, consider the Katana 750 engine versus the 454ci Chevy above. The Katana 750 has 48.5 / 46 ci = 1.05 ft.lbs. per cubic inch. The Chevy 454 has 440 / 454 = 0.97 ft.lbs. per cubic inch. It's hard to image two more different engines, and one is 25 years older than the other, yet their BMEP is almost the same (only 8% different).

Engines that rev higher usually have higher geometric compression ratios even though their BMEP is the same as lower revving engines. The higher geometric compression ratio is used to help increase the firing pressure which would otherwise drop because there is less time to fill the combustion chamber at high RPM. Conversely, increasing the geometric compression ratio of an engine without making any other changes, will increase the BMEP.

If on a hot afternoon the air is less dense than the cold morning air, then the engine’s BMEP will drop. A drop in BMEP means less torque at the same RPM, which is less power. Since the ambient air density typically varies 5% to 10% above or below “normal”, this has a significant effect on the power an engine produces. Thus, when measuring the power output of an engine it is important to correct the measurement to “standard” air density. This correction may be upward or downward, depending on what we define as “standard” and what were the conditions during the test.

What we are talking about is known as RAD, which stands for Relative Air Density. RAD is expressed as a percentage, which at ground level under normal conditions is usually somewhere between 90% and 110%. Since air density is determined by temperature, pressure and humidity, there are many different conditions that all equal 100% RAD. For example:

60 degrees F, 30” Hg, 0% humidity = 100% RAD

62 degrees F, 30.3” Hg, 40% humidity = 100% RAD

70 degrees F, 30.5” Hg, 0% humidity = 100% RAD

Thus, the optimum conditions for maximum power are low temperature, low humidity and high pressure.

There are two standards commonly used for correcting power: STD and SAE.

STD is defined as: 59 F, 29.92” Hg, 0% humidity. It is the same thing as 100% RAD.

SAE is defined as: 77 F, 29.23” Hg, 0% humidity. It is the same thing as 94.4% RAD.

Suppose you dyno test your engine and measure 100 HP. You determine that the atmospheric conditions during the test were equivalent to 105% RAD. That means the air was 5% denser than “normal”, thus your engine would produce 5% less power under “normal” circumstances. Your STD corrected power is 95 HP, and your SAE corrected power is 90 HP.

The STD or SAE correction is *not* correcting for
drivetrain losses or friction. It is merely correcting atmospheric conditions.
For example, one could have an SAE corrected rear wheel horsepower, or an SAE
corrected brake horsepower.

Forced induction refers to using a compressor on the intake of an engine. The compressor dramatically increases the density of the incoming air, producing significantly higher BMEP, torque and power. Superchargers and turbochargers are the two primary methods of forced induction. The difference between them is what powers the compressor.

A supercharger is powered directly from the crankshaft of the engine. Normally, it is powered by installing an extra pulley in the serpentine belt. Some of the power the engine produces, goes into to the supercharger instead of going to the wheels of the car. But the supercharger adds more power than it draws. That is, the additional power gained by increasing the BMEP, is greater than the power required to drive the compressor.

A turbocharger is powered by inserting a fan in the exhaust. As the exhaust gases flow past the blades of the fan, they cause the fan to spin. The fan is connected to another fan on the intake, which is the compressor. Thus, the intake compressor is powered by the moving exhaust gases. As the engine revs higher, the exhaust gases have increased energy and velocity, which provides greater power to the compressor, which increases the intake compression, which increases BMEP. The turbo fans do impede the exhaust gases, but the impediment they introduce is much smaller than the drag a supercharger draws from the engine crankshaft. Without the turbo, the energy of the exhaust gases would have been unused, so the turbo is powered by harnessing otherwise unused energy. Because of this, turbochargers are usually more efficient than a supercharger. Compared to a supercharger, the same level of boost from a turbo provides greater additional power because less of the power it generates is lost driving the compressor.

The drawback to turbos is design complexity. It is harder to engineer a turbo because one must place an extremely light, low friction, efficient fan directly into the incredibly hot exhaust of the engine. It is difficult to overcome the engineering difficulties. Superchargers are less efficient, but easier to design and build.

One interesting advantage of forced induction is that it can make the engine totally impervious to changes in atmospheric conditions. If the relative air density is low, a computer controlled boost monitor can dial up more boost, essentially driving the compressor harder so the density of the intake charge is the same. This is especially effective on airplanes. As the plane gains altitude, the engine computer dials in more boost to compensate. The use of superchargers was pioneered during WWII on aircraft engines such as the Rolls Royce Merlin.

All of the examples so far ignored the shape of the torque curve. This is essential to simplify the problem and highlight the physical effects at work. But now it is time to drop this assumption for a closer look at reality.

While it is true that power determines acceleration, and
that the optimum shift point is always between the peak power RPM and redline,
there is more to the story. Since F = ma, and F is determined by power, one
might expect to calculate the acceleration of any car given only its power to
weight ratio. For example, a stock 3rd gen RX-7 can accelerate at 1.09 G (3099
lbs. thrust / 2850 lbs. weight) in 1st gear. From this one might assume that it
can go 0 - 60 in 2.5 seconds – which is half its actual 0 - 60 time of about
4.9 seconds. Why is it off by a factor of 2? Our estimate would have been
accurate only if the car did 0-60 entirely in 1^{st} gear, and if the
engine had a perfectly flat torque curve. But the car spends some of its 0-60
run in 2^{nd} gear, where it has less acceleration. In addition, even
in 1^{st} gear it accelerates that fast only as it passes through its
peak torque RPM. Throughout the rest of the RPM range, it is producing less
torque and thus accelerating slower.

The power to weight ratio provides a good "first guess" rough approximation of acceleration. Indeed, one rule of thumb is to divide the curb weight by the horsepower, then take half this number as the 0-60 time in seconds. But actual results can vary from this calculation. The problem is that no engine has a perfectly flat torque curve, so it's only producing its peak torque and power throughout a fraction of its RPM range. This fraction can be relatively small or large, depending on the engine. Big V8s tend to have relatively flat torque curves, while small, high revving engines tend to have increasing torque with RPM. But these are only general tendencies and there are plenty of exceptions. Indeed, the reason I chose the Suzuki GSX 750cc engine in the previous example is that it is one of very few high power, high RPM engines that has a relatively flat torque curve.

An engine of any size can be designed with a flat torque curve, or it can be designed to develop greater high RPM peak power by sacrificing power at lower RPM, or it can be designed in the opposite manner -- to develop strong low end and midrange torque at the expense of high RPM power. Each design optimizes the engine for different purposes. Which is best depends on how the engine will be used.

For all-out maximum drag racing acceleration, one needs the best possible power to weight ratio. We might like a monster V8 or larger engine with tremendous power output, but it's big and heavy, which impairs the power to weight ratio. If we can get that same power from a smaller, lighter engine, then the reduced weight will give us even greater acceleration (in addition to a better balanced car, better cornering and better stopping). We can do this if we are willing to engineer it carefully and make some sacrifices. In the comparison of the two engines above, observe that the Chevy is about 10 times larger than the Suzuki, but has only about 3 times more power, and the Suzuki’s torque curve is almost as flat as the Chevy’s. Of course, this is partially the effect of 25 years of engine design improvement (1969 to 1994).

To get a certain amount of power, we need a combination of torque and RPM. It is harder for a smaller engine to generate the same amount of torque, but it is easier for it to rev higher. So if we can design a smaller, lighter engine that revs higher, we can get the same amount of power as a larger, heavier engine. The trick is to get our torque at high RPM. This is not easy, because in a piston engine we are fighting the inertial losses of a piston moving back and forth, which grow exponentially as RPM increases linearly. To minimize the inertial losses of the pistons, we can use a shorter stroke and larger bore to get the same overall displacement. The shorter stroke means that at any given RPM the piston is moving slower, which reduces the inertial losses at high RPM. But shortening the stroke gives the downward force of the piston less leverage on the crankshaft, which gives the engine less torque. However, with more RPM and less torque we are likely to have the same power, and we've obtained it from a lighter, smaller engine. The larger bore gives us room to use larger valves, which lets the engine “breathe” better and get more mixture into each compression stroke. In addition, we can use overhead cams instead of pushrods to reduce inertia in the valve train, and we can tune the valve timing and the exhaust to resonate at high RPM. All of these together will enable the engine to continue developing torque at high RPM, which gives us more power. But they also can reduce the low end torque, which means the engine lugs at low RPM. This narrows the operating RPM range. Engines like this are often called "peaky", "pipey" or “binary” since they are fairly weak at low RPM, and come on strong at high RPM.

To compensate for this, we would want to use gear ratios spaced close together so that when you shift to the next higher gear, the engine RPM drops only a little bit, staying in the high RPM range where the engine has good power. You end up with a car that is harder to drive, because the driver has to pay close attention to his gear and speed to keep his RPM up. But it will accelerate faster because smaller, high revving engines have inherently greater power to weight ratios.

One way to view this is that RPM and torque are equally important for producing power, and RPM is lighter than torque.

However, drag racing is not the only form of racing, perhaps not even the most exciting. If the race involves turning, the driver will have to slow down and speed up while turning and braking. Slowing down without downshifting could make our pipey engine fall off the power curve. But we'd like to avoid constantly having to shift gears. In this case, peak power becomes less meaningful, especially if it is achieved only in a narrow RPM range. The shape of the torque curve becomes more important.

An engine with a flat torque curve is consistent and predictable. Put your foot down and it accelerates through each gear at a constant rate. Contrast this with an engine like the turbocharged Wankel in the 3rd generation RX-7, which doesn't do much until it hits 4500 RPM, at which point the high speed turbo kicks in full boost at the same time the Wankel's natural torque curve is increasing. Such an engine produces more peak power, but is less consistent and less predictable because it sacrifices low RPM power. When you put your foot down, the engine responds in a significantly different way, depending on the gear and RPM.

When driving through curves, it is easier to control the car if the engine has a flat torque curve. With a flat torque curve, you can brake for the turn and when you hit the gas again you'll always have the same acceleration. You can't fall off a flat torque curve!

Most drag racers spend a lot of time thinking about the amount of time the engine spends in the power band; i.e. the distance between the shift points. If your gears are spaced too far apart, then your engine drops out of power band after each shift, and has to climb back up the power curve. People want to get the gears spaced closer together so the engine stays in the power band. But there is a downside: if the engine is already staying in the power band with the current gears, then putting the ratios closer together slows you down because you’re not getting any extra power, but you are shifting gears more often and losing time. There is a happy optimum gear ratio spacing for every engine, based on the shape of the engine's torque curve.

Many drag racers use ridiculously short (high ratio) differential ratios in their cars, some nearly twice the stock gearing. Many of them think this moves their shift points closer together, or enables the engine to spend “more time in the power band”. It also makes the car feel faster, since it is pulling harder in every gear (though at lower speeds).

But this issue is a giant red herring.

You might wonder why we never considered the differential
ratio in any of the examples so far. Changing the differential ratio has no
effect on the spacing of the gear ratios. For example, if with a stock 3.27
differential ratio, a shift from 2nd gear at 6500 RPM starts 3rd gear at 4750
RPM, this also will be true with 4.1, 4.3 or any other differential gear ratio
you choose. If the percentage difference in the gear ratio between two gears,
say, 2^{nd} and 3^{rd}, is X %, then the percentage drop in RPM
when shifting from one gear to the next is also X % -- no matter what your differential
ratio is. Using a shorter gear ratio in the differential does not move the
shift points closer together. So if you think the gears are spaced too far
apart, you better replace your transmission. It's the tranny gearing that keeps
the engine in the power curve, not differential gearing.

You might be wondering what the differential ratio does do.
It affects two key areas: getting off the line and shift points. The first is
the most important. You want a differential ratio that is big enough (short
enough) to make your 1^{st} gear acceleration traction limited. To be
more precise, you want to use the lowest (tallest) differential ratio, that is
still high (short) enough to make your 1^{st} gear acceleration limited
only by tire traction.

Why is this optimal? If you gear taller than this, you will not be getting all the acceleration you can and you'll take longer to climb the power curve. If you gear shorter than this, you will not get off the line any quicker (because you were already traction limited), but you will hit your shift points sooner and spend more time shifting gears.

The second (less important) factor in choosing a differential ratio is shift points. You want to be near -- but not at -- redline when you drive through the finish line. For most cars, this usually means you want 3 tall gears, or 4 short gears. If you're bumping the rev limiter just before the end of the run, it's better to use a slightly taller gear to finish in that gear without hitting the limiter, thus avoiding time lost in the extra gear shift. On the other hand, if you're well below redline when you pass the finish line, you could use a shorter ratio and get more acceleration, without requiring an extra shift.

But the first factor – launching – is usually dominant. So
even if you are well below redline at the finish line, using a shorter gear
ratio will make you faster *only* if you have the traction to handle it.
If not, it will slow you down.

Also remember that you can fine tune your differential gearing by your selection of rear tire sizes. Of course this applies only if you have a rear wheel drive car. But we already knew you had rear wheel drive, else you'd be a ricer and you would not be reading this.

In summary, the primary points regarding differential gearing are the following:

1. Differential gear ratios have no effect on the optimal gear shift RPM or on the amount of time the engine spends in the power band.

2. If you are already traction limited in 1st gear, then bigger differential ratios will slow you down.

3.
Bigger differential ratios will make you faster only if *both*
of the following are true: You have enough traction to handle it, and the
improvement is enough to compensate for any extra shifts required by the
shorter gear ratio.

Most 4 stroke piston engines run in what is known as
**Otto Cycle**:

Some modern engines claim to operate in
**Atkinson Cycle**
but this is a misnomer.
Atkinson cycle was invented in the 1800s as an engine with an articulated piston rod
giving a short intake stroke with a long power stroke to maximize efficiency.
The system that modern engines use, calling it by the same name,
is quite different yet serves the same goal with a simpler design.
That goal is to increase efficiency.

As explained in this article, for maximum power,
a 4-cycle engine crams the cylinder as full as possible during the intake stroke.
In Atkinson cycle, the cylinder is only partially filled during the intake stroke.
This improves efficiency in two key ways:

Put conversely, there are 2 drawbacks to cramming the cylinder as full as possible:

As you might guess, the increased efficiency that Atkinson cycle provides comes at a price: lower power output. This makes it non-useful to racing, but it's a clever idea, and an engine can be designed to operate in either Otto or Atkinson as needed.

In Otto Cycle the intake valve closes at the bottom of the intake stroke. In Atkinson Cycle, the intake valve remains open longer - it closes late, after the cylinder has risen part way up the compression stroke. Some of the mixture drawn in during the intake stroke is pushed back out the intake valve before it closes. Why would anyone do this? It seems counterproductive! It means less mixture captured in the cylinder, which lowers BMEP, torque and power. But this smaller amount of mixture is easier to compress, so there is less pumping resistance. And when ignited, it can fully expand during the power stroke. Ideally, at the end of the power stroke the mixture is fully expanded and the pressure in the cylinder equals atmospheric. This means no energy is wasted.

Because modern engines implement Atkinson cycle with valve timing, they can shift between Otto and Atkinson cycle automatically on demand. At low RPM, part throttle, you aren't demanding lots of power. The engine shifts intake valve timing to run in Atkinson Cycle for maximum efficiency. As the throttle is opened, or as the engine revs higher, you are demanding power. The engine shifts intake valve timing to Otto Cycle for maximum power.

Earlier, we presented the equation showing the mathematical relationship between power and torque. Here is where that equation comes from.

Torque is the rotational version of Force, expressed in units of Newton Meters or Pound Feet. Revolutions are the rotational version of distance, usually expressed in Radians. Work is the transfer of energy from one object or system to another, expressed as the product of force and distance. The rotational version of work is expressed as the product of torque and radians. Power is the rate at which work is performed, expressed in any system as work divided by time, otherwise known as work per second.

In the metric system, 1 Joule is the same as 1 Newton Meter. Thus, with Power expressed in Watts, we would have:

P = __Newton * Meter * radians__

second

But since radians are unitless, we have

P = __Newton * Meter__

second

But the equation presented in section 3 of this document was expressed in British units. So we must convert the units.

1 Horsepower = 746 Watts

1 Joule = 0.738 Ft. Lbs.

1 rad = 1 / 2π revolutions

1 second = 1 / 60 minute

1 foot = 0.305 meter

Now remember the original equation:

P = __2π nt__

33000

Thus, the 33,000 is an approximation that from the expression (746 * 0.738 * 60), which is actually about 33033. And the 2π factor comes from converting radians to revolutions.

[1] If you combine the constant factors from the above equation, you have 2π/33000 which is approximately 1/5,252. This means that for all engines -- whether a Ferrari V12, a Kawasaki 4, a Cummins diesel or a Mazda rotary -- for every engine ever made and any that will ever be made, at 5,252 RPM power in horsepower is equal to torque in foot pounds. You don't believe it? Take a look at the graphs that follow. Still not convinced? Take a look at the torque/power curve of any other engine you like.

[2] You can
always use shorter gear ratios in any engine, but without RPM, you must
sacrifice speed. The key advantage of high RPM torque is that you can use the
shorter ratios without sacrificing speed.