Automotive Torque vs Horsepower: Auto Racing Power Showdown

[Stingray]

Actually, power and torque are seperate. Torque can be static. Take an electric motor and weld the output shaft to the casing and then turn the motor on. It will sit there and make its full start-up torque for as long as it is plugged in, but no work is being done (you just have a big heater). If you allow the torque to be used: The amount of work it will do in a given amount of time is horsepower.

That disabled motor will not be doing any mechanical work. Its power output is therefore zero. Of course it does require electrical power to run, but that all goes into heat. Internal combustion engines are always rotating when they're being useful, and their power output is equal to the torque multiplied by the (crankshaft's) angular velocity. Since your welded motor has no angular velocity, P = 0. At any nonzero rpm, dropping the power output by a certain percentage drops the torque by that same amount.

In a car, the effect of the flywheel depends on how quickly you try to spin it up. If you're not slipping the clutch, that's entirely determined by the car's acceleration together with the gear ratio between the wheels and flywheel. It follows that in an actual car, rotating driveline parts act like additional effective masses in terms of longitudinal acceleration. Those masses are dependent on the gear ratio. If you tried to say that the power changed instead, you'd find an effect which varied with gear ratio, rpm, and external load (e.g. drag). That's hardly a natural thing to do.

 

[rcgldr]

Flywheel versus acceleration

Is this is reference to race cars or to street cars? Race cars only launch at the start and out of the pits, so even first gear is pretty tall. Generally race cars are traction limited in 1st gear, so flywheel weight won't affect acceleration.

As I stated previously, the main advantage of a lighter flywheel is quicker shifts and less stress on the drive train. In order for an XTRAC like sequential shifter to shift in 30ms to 50ms (1/33 to 1/20 of a second), the flywheel has to be very light.

Some high end 4 cylinder motorcycle engines just rely on the clutch and counter-balancer for flywheel effect. Racing bikes remove the counter balancer.

 

[rbj]

Enzo Ferrari was once quoted as saying "Horsepower sells cars, torque wins races". Obviously Mr. Ferrari believed that torque was most important of the two provided that both were present in sufficient amounts. With lots of turns, acceleration, and braking, torque wins hands down IF there is sufficient horsepower. Without the horsepower, diesel trucks would win auto races. Without torque, jet-powered cars would win all the races. By the way, I believe a jet powered vehicle holds the land speed record... of course it just went straight and probably didn't handle well in the turns!

someone (maybe you) posted this before and i still don't get what the thing is. the issue with torque is gear ratio. providing you have sufficient HP in your engine, getting the torque you need to maximize accelleration is a matter of having the necessary gear ratio in the tranny. but getting the sufficient HP is a more fundamental problem of having an engine of sufficient capacity to convert chemical energy to mechanical energy at a sufficient rate. the latter affects the mass of the vehicle (which also affects acceleration) whereas the former does not to anywhere close to the same degree.

so i would say to Ferrari that, first make sure you have sufficient power output (which is the product of crackshaft angular speed and torque) for the total mass of vehicle to accomplish the performance goal needed, and then the rest is the design of the necessary gear ratios so that this power is transformed to the torque and speed values that suffice for the race.

 

[brewnog]

so i would say to Ferrari that, first make sure you have sufficient power output (which is the product of crackshaft angular speed and torque) for the total mass of vehicle to accomplish the performance goal needed, and then the rest is the design of the necessary gear ratios so that this power is transformed to the torque and speed values that suffice for the race.

Except that it's not nearly that simple, because the torque/power characteristics of an engine are not linear with engine speed.

 

[Stingray]

Is this is reference to race cars or to street cars? Race cars only launch at the start and out of the pits, so even first gear is pretty tall. Generally race cars are traction limited in 1st gear, so flywheel weight won't affect acceleration.

I was talking about street cars when I gave the 2-300 lb estimate.

In race cars, I agree that small flywheels are mainly to shift quickly. But their effective mass effect is also important. Although 1st gear is usually designed such that redline occurs at a relatively high speed, those redlines are usually much higher than in street cars. The overall gear ratio is therefore pretty big. The (static) weight of a race also tends to be small, and every little bit counts.

Just for fun, I've calculated what would happen if you gave an F1 car a flywheel from a roadgoing V8. Say that a redline of 20,000 rpm occurs at 70 mph, and that the rear tires are 22 inches in diameter. Then a 30 lb flywheel 11" in diameter would add an effective mass of about about 1,900 lb! It's obvious that nobody would ever try to race like that, but I think it shows that these things are still important.

 

[rcgldr]

I underated the affect of flywheel weight versus acceleration in low gears, since I was thinking race cars, not street cars. Street cars use heavier flywheels to reduce vibration and for easier launches with a manual transmission, and a better idle. There is a peformance advantage in the lower gears with a lighter flywheel beyond faster shift times.

Some typical flywheel weights for street cars, stock and lightened.

Mitsubishi Starion 2.6 Turbo...35 lbs. stock / 21 lbs. lightenend
Datsun Z car 225mm...25 lbs. stock / 16 lbs. lightenend
Datsun Z car 240mm...28 lbs. stock / 18 lbs. lightenend
BMW 2002 215mm...18 lbs. stock / 11 lbs. lightenend
BMW 2002 228mm...24 lbs. stock / 16 lbs. lightenend
Datsun 240SX KA24DE 24-25 lbs. stock / 16-17 lbs lightened

The engines in these car only rev to 6500rpm or less, and the cars weigh 3000lbs or more. The cars probably redline at around 35mph in first gear.

I was previously thinking peformances cars, like a C6 Corvette Z06, redlines at 7100rpm at 61mph in 1st gear, or a motorcycle like a Hayabusa, which redlines at 11,000 rpm at 81mph in 1st gear. A F1 race car redlines at 18,000 to 20,000 rpm at around 100mph in 1st gear (note that 7th is geared around 185mph to 225mph, depending on wing setting for the track, a very close ratio tranny).

A F1 race engine has a tiny flywheel and can change rpms very quickly, as in the later part of this video:

f1 engine warm up

 

[IMP]

Maybe we are getting just a bit off topic here but: If you want to hear a Formula One engine sing, check this out:

 

[aliaze1]

Ive got a question, which is more powerful and better to have in auto racing, torque or horsepower? I've gotten mixed answers and wanted to ask some people who know their stuff.

you need a balance of the two, each has its purposes, i don't know the details exactly.

but if u notice, some of the amazing sports cars like the Subaru WRX STi have large numbers of both, in the case of the STi, 300hp and around 300ft-lb torque (i don't remember exactly)

 

[rcgldr]

The faster the shift time, and the more gears, the less important it is to have a wide powerband (shape of torque curve versus rpm). In Formula 1 racing, times have changed since Ferrari made that statement. Shift times are 50ms or less (electonic sequential no lift shifters, activated by paddles), and the cars use a 7 speed close ratio tranny. My guess is only the top 25% of the rpm range is used in these cars. In the older days of manual shifts and only 5 or 6 gears, the top 40% of the rpm range might have been used.

 

[Stingray]

Just to set the record straight, I'm pretty sure that the Ferrari quote is an urban myth. I've usually seen the statement associated with Carroll Shelby, but I doubt he said it either.

Also, F1 cars have had very close ratio gears for a long time. A typical racer from the 60's would rarely drop its revs more than 15% on an upshift (except for the 1->2 shift where you were pretty much traction-limited anyway).

 

[rcgldr]

For the 1967 season, the Lotus F1 race car (400 hp, 1400lb with driver and 20 gallons of fuel) had 1st gear set to redline around 100+mph, and 5th gear set for around 190mph to 200mph depending on the track. Gearing is track sensitive (choose gears to avoid extra shifting), but probably rpm drops around 30%, 20%, 15%, 12%, would be typical. Still to avoid an extra shift, a car might exit a turn in a higher gear, so having a power band for the upper 30% to 35% would help. A narrower powerband could be an issue if the jump in torque were sudden, making throttle inputs difficult to deal with in turns.

For a 6 speed tranny, the Suzuki Hayabusa motorcycle gears are set to redline at these speeds (actual top speed is around 190mph): 1st-80 mph, 2nd-108 mph, 3rd-137 mph, 4th-162 mph, 5th-183 mph, 6th-199 mph, for a rpm percentage drop of: 35%, 27%, 18%, 15%, 9%. 80% of peak torque is available from 3500 rpm to 10,500 rpm, about 2/3rds of the rpm range.

 

[u83rn00b]

Torque=force times distance, horsepower=power

 

[ME_262]

Well this is my first post here, I stumbled on to this thread with google looking for an answer to some questions I had.. I've learned quite a bit

Basically I've figured out that torque is better than HP every time... if.. and only if there was no such thing as a little thing called gearing.

Bring gearing into the picture and it's a whole new ball game. Even better bring in the ideal CVT and the inverse of the above is true.

Before reading this thread I was under the misconception that 200Lbft at 3000RPM accelerated just as fast as 200Lbft at 6000RPM. This is true if gearing did not exist. But because it does you can trade the extra distance of the 6000RPM for more force and accelerate faster than the car at 3000RPM.

Basically it comes down to the amount of Watts or horsepower the car outputs, the higher this number the more it can be geared into torque at the rear wheels. And undisputedly in all cases, torque at the rear wheels is what will win you races.
Edit:
(Torque at the front or all wheels for some lol)

 

[punx0r1337i57]

I just had to put my 2¢ in on this one...

Torque is a pretty useless figure, but it can be a reflection of low RPM HP, and by extension, often reflect a wider power band.

HP is a much better figure, but the HP number you are given is peak HP, which in and of itself isn't too useful either in an automotive application unless the powerband is wide or you have a lot of gears.

Ultimately, the best data you could get would be a HP curve from idle to redline and all engine speeds in between.

See, the power a car can produce is inherently based on speed as much as force, which torque doesn't take into account. If I stood on a 1 foot long wrench pointed out at a 3:00 or 9:00 position, I would be putting 160 lb/ft of torque onto it. Which is more than my car. Does that mean I could pedal my car faster than the engine can pull it? Of course not, because in order to move the 3,000lb boat, I'd have to change my pedaling to a very slow speed. Even though I can produce that much torque, I could only get it around a few times a minute, resulting in less than 1 HP. Which is a better representation of how much power I can actually produce.

But before the torque elitists jump on me, peak HP doesn't mean too much either. Suppose a car can produce 200 HP at 6,500 RPM, but it produces only 50 HP at 6,000 RPM and redlines at 6,600. (Of course I'm using extreme examples for both of these, I'm well aware. I'm illustrating a point.) Unless you can shift really fast and have a few dozen gears (or a very effective CVT) you've got a dog-slow car.

Basically, what I'm saying is, what makes you fast is a good, wide curve of high HP. Torque is in and of itself completely useless, and peak HP is only worth marginally more. The value of these figures is in what they can reflect (but don't always, which is why both are imperfect figures.) Torque peaks are at lower RPMs, so they generally reflect how low down your powerband appears. HP peaks are higher and generally reflect how long an engine's powerband lasts at the high end.

So let's take it to a real-world situation. Let's take the most extreme examples of HP vs. Torque and compare them.

Honda Civic Si with the K20 engine, and a Chevrolet Camaro RS with the Throttle Body injected engine (This is late '80s, early '90s model).

The Honda comes in the lead in peak HP with 197 against the Camaro's 170, but the Camaro's mammoth 5.0 V8 has 255 lb/ft of peak torque, versus the Honda's measly 139. So which is better?

These are published 0-60 times:
Camaro: 7.4
Civic Si: 7.2

Of course, people have gotten better times on both with stock cars, and both can be a lot faster with modification, so don't flame me for conservative figures. Those are both official published figures.

So what do we have there? Not a whole lot of difference.

The Camaro will have much more power at the bottom. It's 170 HP comes at an astonishingly low 3800 RPM, courtesy of its huge displacement. The Civic, by comparison, feels its engine's tiny displacement, and by its torque peak of a far higher 6200 RPM is still falling behind at 164 HP (remember, HP is torque x RPM / 5252). Up until this point, advantage: Camaro.

However, shortly after that, the Camaro feels its archaic pushrod engine design and quickly starts to dump torque, and consequently HP to friction, inertia, and poor ventilation as RPMs increase, and then it chokes on its low 5,000 RPM redline. The Si, by comparison, breathes much better through 4 valves and doesn't have pushrod inertia and keeps increasing in power until a dizzying 7800 RPM, and then it keeps running at well over the Camaro's max output past its 8,000 RPM redline up until about 8,250 when the rev limiter cuts you off. Advantage: Si.

So which is ultimately better?

Well, that depends. The Honda is ultimately faster. But it is absolutely useless with an auto, which runs in the low RPM range, hence why all Si's are manuals. And you have to know what you are doing. It is much easier for a novice driver to run a 7.4 in a Camaro than a 7.2 in an Si. In the Si's defense, it also uses a lot less gas and weighs less, allowing the car to corner better.

Ultimately, it comes down to the driver. Proving one to be better wasn't my intention (the comparison is ridiculous anyway... a 15 year old car with a 5L V8 to a new car with a 2L I4) and I don't drive either, so I don't much care. The point is, neither is a perfect figure, and are in fact, in and of themselves, pretty useless. Their value is in reflecting points of an overall HP curve.
However, I do have a bone to pick with Briggs & Stratton for using torque instead of HP on their new engines. Peak horsepower really *is* the best figure on small engines that run at a constant RPM, and I think they are just using torque (which is, to be honest, a pretty useless figure on a generally fixed-speed engine) to jack up their performance figures since they are increasingly unable to compete in HP per CC against Honda and Robin engines with OHC valvetrains and are confusing the whole small engine comparison process along the way.

Alright, that's enough for now.

 

[rcgldr]

Camaro vs Si 0 to 60 times

There's more going on here than just torque and power. 0 to 60 times can be affected by how much momentum is built up by the rotating mass of the engine and flywheel. The Camaro with rear wheel drive, and a lot of engine and flywheel mass, is going to get a lot of help with it's initial launch, regardless of the torque and power it makes. 1/4 miles would be a better indicator of power.

 

[ME_262]

The only way I could see torque peak being a valid number, is if it was a device that ran at a constant speed and ran at a 1 to 1 ratio with the engine. In a 1 to 1 ratio with none of the RPM being geared down for extra torque, the torque peak is the point where the device would output the most power.

In addition to this, this is because the overall output of any engine is indeed torque, it is changed from the torque value in the engine by gearing, but giving the output torque would also be valid to me.

 

[rcgldr]

In a 1 to 1 ratio with none of the RPM being geared down for extra torque, the torque peak is the point where the device would output the most power.

The most power always occurs at the power peak. In a fuel buring engine, the power peak is alway at some faster rate of revolution than the torque peak (assuing that torque versus rpm doesn't fall off vertically as some magic point, or due to a rev-limiting device set at the torque peak).

In 1 to 1 applications, such as aircraft the prop pitch and diameter are choosen to work with the power range of the motor, so in effect, there's still the equivalent of gearing. In the case of a car with a 1 to 1 gear the diameter of the wheel could be adjusted.

Most electrical motors are more interesting, in that they are almost constant power motors across a wide range of rpm. Torque decreases with increases in rpm, and power remains approximately constant, and it's the load and the input voltage that determines the rpm. Current demands will vary to match the load and rpms. Although output power is nearly constant across a wide range of rpms, the associated input power will vary, and there is a specific rpm where the motor runs most efficiently. So once again, gearing or the equivalent is still used to allow the motor to run at an efficient range of rpm.

To eliminate the need for gearing, "outboard" motors are becoming popular with radio control models. (The housing rotates while the shaft remains fixed). These motors have peak effeciencies at much lower rpms than conventional motors, eliminating the need for gearing. Again for rc aircraft the prop size and pitch still have to be matched with the motor.

 

[punx0r1337i57]

There's more going on here than just torque and power. 0 to 60 times can be affected by how much momentum is built up by the rotating mass of the engine and flywheel. The Camaro with rear wheel drive, and a lot of engine and flywheel mass, is going to get a lot of help with it's initial launch, regardless of the torque and power it makes. 1/4 miles would be a better indicator of power.

This is true, but the Civic is also lighter. It certainly isn't a perfect comparison, but it is the best I could think of. The extra traction of large rear drive wheels probably gives the Camaro an unfair advantage. Plus the fact that that figure is actually for the 190 HP Camaro rather than the throttle body 170 HP model, since I couldn't find a figure for it. Maybe an S2000 would have been better for comparison? But an S2000 really blows the Camaro away, and I didn't want to be accused of being anti-American biased.

Of course, more recent Camaros are much faster than the old ones, but that is largely because they have a much better HP figure, so they aren't so relevant to the discussion.

 

[ME_262]

So basically... I'm trying to unite your thoughts with mine on this issue...

What it comes down to is the horsepower output of the engine which is converted to a torque value at the wheels by gearing. So the higher the HP the faster you are going to accelerate. So for example two cars going down the road both with a 1 to 1 ratio of gearing all factors including tires the same and both making 25HP would be accelerating at the same rate at that instant.

 

[rcgldr]

Well yes, identical cars will have identical acceleration rates (assuming identical driver inputs).

One missing issue here is "drivablity". It's one thing to setup a car for racing where the engine is always kept at high rpms. It's another thing to setup a car for everyday driving where the engine is mostly running at lower rpms. For a street car, having a wider power band is more important than max power, and most street cars are setup that way.

 

[danielrm26]

Greetings all,

I've been grappling with this question for over a year now and I just now found this forum. Excellent content.

I have created my own primer on this topic and I'm wondering if I can get some input from you guys. I'm not a physicist, but I am a hobbyist and am somewhat familiar with the main concepts.

Anyway, my hobby is to try and simplify things into relatively concise explanations; and this primer is one example. As with all my primers of this type, the main goal is to understand the concept myself and then have a reference to come back to. I'd very much like to see what you guys think about it. Is it totally botched? Am I on the right track? What am I missing?

http://dmiessler.com/study/horsepower/

Thanks much!

 

[rcgldr]

Hang on a second, I'm blowing off the dust of this old thread.

Look at this post I made:

post #54

Horse power never "equals" torque at any rpm, they aren't the same units. For specific units, their magnitudes are the same at 5252.113122... rpms, in English units and are the same at 9549.2966... rpms in metric units. If you measure angular rate in radians per second, the magnitudes are the same at 33,000 radians per second for English units, and are the same at 60,000 radians per second for metric units. So are power and torque the same at 60,000 radians per second? A person has 32 teeth, so is it more efficient for that person to generate 32 lbs or 32 Newtons of force because both numbers are 32?

Most chassis dynamometers measure power, not torque. Either a variable load is used, or the rate of angular acceleration of a heavy drum is used. In either case, there are two inputs, the measured force, and the speed at which the force is measured over a range of speeds. In semi English units, horsepower = force (lbs) times speed (mph) divided by 375 (conversion factor). In metric units, watts = force (Newtons) time speed (meters / second). Without additional input, such as engine rpm, chassis dynamometers can't determine the angular speed of the engine, so they can't calculate the effective torque without the additional input.

I did some dyno runs on my old Caterham, and because of issues sensing the spark line, they could only plot the power and not the torque, because they couldn't determine engine rpms. Since a chassis dyno simulates the real world, this should help make it clear that it's power and not torque that determines the performance of a car.

As another example, imagine a small linear inductive motored train powered by plutonium buttons and thermal couples. No internal moving parts (except at the atomic level), nothing is rotating, so there is no torque, but there is power, and the rate of acceleration of the train will be determined by the power to mass ratio (and aerodynamic drag).

A follow up to my previous post:


Some typical flywheel weights for street cars, stock and lightened.

Mitsubishi Starion 2.6 Turbo...35 lbs. stock / 21 lbs. lightenend
Datsun Z car 225mm...25 lbs. stock / 16 lbs. lightenend
Datsun Z car 240mm...28 lbs. stock / 18 lbs. lightenend
BMW 2002 215mm...18 lbs. stock / 11 lbs. lightenend
BMW 2002 228mm...24 lbs. stock / 16 lbs. lightenend
Datsun 240SX KA24DE 24-25 lbs. stock / 16-17 lbs lightened


Flywheel weight of a Formula 1 race car - 4 lbs

 

[46weaver]

Hi All, I am trying to translate the theory into real world situation.
Hope you guys can help me out. Here's my long list of questions -
** note - the hp and torque figures quoted are made up numbers**

1. An engine produces peak 200 lb ft at 300rpm, and peak 220 hp at 5500rpm.
In the same gear, is the acceleration force stronger at 3000 rpm compared to at 5500rpm?

2. A Ferrari engine with peak 300 lb ft at 5000 rpm and peak 400hp at 7000rpm vs a BMW 7 series engine with peak 400lb ft at 3000 rpm and peak 340hp at 5200 rpm
Both engines used on a 1500kg car and same gear ratios (using Ferrari's closely spaced ratio).
Which one accelerates faster?

3. A Ferrari engine with peak 300 lb ft at 5000 rpm and peak 400hp at 7000rpm vs a BMW 7 series engine with peak 400lb ft at 3000 rpm and peak 340hp at 5200 rpm.
Both engines used on a 1900kg car and same gear ratios. (using a BMW wider spaced ratios)
Which one accelerates faster?

4. A Ferrari engine with peak 300 lb ft at 5000 rpm and peak 400hp at 7000rpm vs a BMW 7 series engine with peak 400lb ft at 3000 rpm and peak 340hp at 5200 rpm
Both engines used on a 1500kg car and same gear ratios. (using a CVT transmission)
Which one accelerates faster?

5. A Ferrari engine with peak 300 lb ft at 5000 rpm and peak 400hp at 7000rpm vs a BMW 7 series engine with peak 400lb ft at 3000 rpm and peak 340hp at 5200rpm
Both engines used on a 1900kg car and same gear ratios. (using a CVT transmission)
Which one accelerates faster?

 

[Lsos]

Even if I calculated and answered all 5 questions...or 10, or 50...would you be any closer to understanding what hp and torque is?

If you don't understand something, ask it. When you finally understand it, it's very simple to figure out all the answers yourself.

 

[rcgldr]

1. - Maximum force at the tires in a specific gear occurs at the speed of peak torque. Acceleration would depend on the total opposing force, rolling resistance and aerodynamic drag.

2. 3. - If the distance to accelerate is long enough for cars to get near top speed, then the higher powered car will accelerate faster, although torque versus gearing may help the higher torque lower powered car in 1st gear, ingoring the fact that a clutch can be used in first gear to get around this.

4. 5. - With a CVT transmission, the higher the power, the more torque that is delivered to the driven tires, at any speed.

 

[Fahlin Racing]

Very interesting thread on HP & TQ.

If 'torque gets you moving, horsepower keeps you moving'. This definition fits the best from what I have observed. Then the comments on one or the other being useless perhaps should be discarded. Simply because HP is calculated from torque production. In a piston type engine the torsional rotational we need is just a kinetic version. Maybe if we were to speak about jet engines where we do not use a drivetrain, then torque can very well be removed from 'stat advertisements' then.

If torque were not important, then we wouldn't increase the cubic inch displacement values of piston engines by extending the crankpin throw for more leverage and cubes. Smokey's theory on connecting rod length as fit the longest con-rod in the engine that is suitable, reason was to utilize the piston dwell time ATDC for in-cylinder pressure to build up and effectively push the piston/con-rod. Now if smaller chambers create more rapid pressure rises, I think the long rod theory would still have a slight advantage even today, but with the quality cylinder heads of today short and long rod engines win races. If somebody may use a factory head of 90s and earlier american V8s not-worked, the longer rod could produce a stronger suction on the induction port because of the somewhat slower movement of the piston ATDC. You could say it allows the induction charge to flow with the piston decending down the bore to a point since the long rod is faster at BDC.

Where I grew I heard a little story from someone, that a factory 80s IROC Camaro took a 1971 Nova with a blower in a race. Now I am not sure the story is absolutely true, but I wouldn't doubt the possibility. It all comes down to the 'Tuner's skills'. That includes chassis, engine, drivetrain, tire selection and surface.

From how I believe things get done is 'torque gets you moving, horsepower keeps you moving'. Now building HP & TQ is a whole world in itself.

 

[xxChrisxx]

Oh no! Not this thread again.

Sweeping statements about torque vs horsepower are all wrong, just to varying degrees. (Which is the only sweeping statement that is true). The fact they are so intrinsically linked it depends on the question and what you want to know. Catchy phrases don't cut the mustard.

GEARBOX >> POWER vs TORQUE
The gearing tends to render the very basic questions about torque vs horsepower pointless.
EDIT: Grrr, thread is millions of years old.

 

[jack action]

I never understood why this question was so difficult to answer for most. Let's clear this up.

Looking at the vehicle:

Acceleration_{car}=\frac{Force_{car}}{Mass_{car}}

Acceleration_{car}=\frac{Power_{car}/Velocity_{car}}{Mass_{car}}

There you have it: for a given car velocity, acceleration is proportional to power. It doesn't matter if the power comes from an internal combustion engine (ICE), an electric motor or a set of pedal for that matter: No body cares about the torque, power is all you need to know.

And before someone says:

Acceleration_{car}=\frac{Torque_{wheels}/Radius_{wheels}}{Mass_{car}}

This is not a valid relationship because the torque and radius are properties of the wheels and not of the car. It is impossible to measure the "torque of the car" or the "radius of the car". Furthermore, reducing the wheel radius by a factor of 2 will increase the acceleration of the car by a factor of 2, but what the equation doesn't say is that it also reduces the speed of the car by 2. So, two car-related variables changes with wheel radius: whenever you are gaining in acceleration, you loose in speed and vice versa.

But wait, there is more.

What if power is varying? For example, if we are using an ICE with a know RPM range (say 3000-6000 rpm) where, obviously, the power increases with the RPM.

Then what we care about is the AVERAGE power throughout the RPM range. No matter what is the extent of the RPM range.

Race cars (of any type) will try to maintain a very small RPM range across the peak power of the engine. Therefore, the average power will be very close to the peak power.

But if you take a more useful vehicle and you want to accelerate from standstill, dropping the clutch at 5000 rpm is not recommended such that you can get the peak power. So in such a case the POWER at low RPM becomes very important. Of course, the higher the power, the higher the torque; but what counts is the power.

For instance, a torque of 200 lb.ft@2500 rpm (=95.2 hp) will create a greater force at the tire's contact patch than a torque of 205 lb.ft@2400rpm (=93.7 hp), AS LONG AS THE VEHICLE SPEED IS THE SAME FOR BOTH EXAMPLES. (And for that, the gear ratio or wheel radius will have to be different for the two examples)

So to sum up:

To get the maximum traction force at the wheel (hence the maximum acceleration) at a given speed:

The whole torque curve: USELESS

Peak power: USELESS (unless you can run a very small RPM range and you rarely need to start from a standstill)

AVERAGE power throughout the USEFUL RPM range: DING! DING! DING! We have a winner!

 

[Fahlin Racing]

Yeah, I know its old an thread, so is the topic itself. I just enjoy in-depth conversation on what I love working with, vehicles I've learned some as well.

It's not difficult to comprehend Jack. It is more or less how someone sees the definition of either. I do agree the average throughout the rpm range is what is desired, but, you will need cylinder pressures to generate torsional rotation to begin & proceed with getting work done the quickest, equating to HP being the most concerned when accelerating and keeping vehicle momentum going.

If the torque curve rises from a manifold change & horsepower (or vice versa) does not on the engine dyno or chassis dyno, what is your view on how the car or truck might be different during operation?

 

[xxChrisxx]

When you say the torque curve rises, but the power curve does not. Can you please clarify what you mean?
Do you mean peak power doesn't change or the acutaly power curve doesn't change?

I've seen these threads for ages on many different forums and it always sparks furious debate where non exists. I find them funny but also dread them. It's like some sort of unstoppable monster of baffling comments and sweeping statements. Given long enough, as sure as the sun will rise, you always get the following two quotes:

“Horsepower sells cars; torque wins races” from the torque crowd.
"Without rotation torque means nothing" from the power crowd.