Race cars - Torque vs Hp - The Undiscovered Country (for many)

[zanick]

We have some unbelievable discussions regarding the acceleration of race cars.

At first glance, many just look at what they remember, F=ma and go on long rants regarding engines that have high levels of torque. However, even when you rearrange the term to a=F/m, its still a Force at the tires that causes acceleration, not at the engine before the gear box.

Many different types of race cars have all sorts of engines. some with high torque and some with low torque, even some of those might have equal HP output. Since HP is a rate of doing work, a rate of change of kinetic energy, if a two cars were to be compared and both had the same HP, yet one had half the torque output, if their HP curve shapes were the same, the rate of acceleration at any vehicle speed would be the same. Correct?
This rate of acceleration is proportional to Power at any vehicle speed. (all other things being equal, such as the car's weight, speed)

Acceleration=Power/(mass x velocity)

Can you guys here enlighten the folks in the racing community regarding equations for acceleration of a race car and the relationship of their engines torque output to the torque and force found a the wheels to accelerate it at any vehicle speed, anywhere on the track.

I posted a set of engine HP/torque curves that showed that even with lower engine torque values, one engine can produce more rear wheel torque and force at any speed, because its HP curve was boader. Not the rule, but the exception that a good percentage of the racing community cannot accept.

Here is the graph that was posted based on the original question, "What would be better on the race track if the two engines in question, both had the same HP but one had more torque than the other. " From the data provided from this person posing the question, I put it in a graph and looked like as follows:

 

[S_Happens]

There is a reason I don't participate in these types of discussions on the automotive forums that I used to, or still do frequent. It's a waste of time. The people who understand have no interest in participating. The ones who are uneducated and believe the answers can be found in a handful of numbers are drawn like moths to a flame, and for the most part will resist anything different.

However, even when you rearrange the term to a=F/m, its still a Force at the tires that causes acceleration, not at the engine before the gear box.

Yes, but to make your comparison (the graph later) you have to ignore any mechanical advantage. The correct comparison for what you want is chassis dyno info (it makes no difference really from your engine graph, but it seems silly to discuss forces at the tire with flywheel data), which is done in a 1:1 gear ratio.

Many different types of race cars have all sorts of engines. some with high torque and some with low torque, even some of those might have equal HP output. Since HP is a rate of doing work, a rate of change of kinetic energy, if a two cars were to be compared and both had the same HP, yet one had half the torque output, if their HP curve shapes were the same, the rate of acceleration at any vehicle speed would be the same. Correct?
This rate of acceleration is proportional to Power at any vehicle speed. (all other things being equal, such as the car's weight, speed

You demonstrate later in your post that you know peak numbers don't mean much, so why do you allow yourself to use peak numbers to describe entire curves here? This mistake is the same one that you are arguing against with your car buddies.

What you would be concerned with as far as HP curves for equal acceleration would be the area under the curves being equivalent. This could be done either by having the exact same curve shifted along the rpm axis (accounting for differences in peak torque), or with a different curve that has the same area below it (allowing for more narrow or broad rpm ranges).

Can you guys here enlighten the folks in the racing community regarding equations for acceleration of a race car and the relationship of their engines torque output to the torque and force found a the wheels to accelerate it at any vehicle speed, anywhere on the track.

I posted a set of engine HP/torque curves that showed that even with lower engine torque values, one engine can produce more rear wheel torque and force at any speed, because its HP curve was boader. Not the rule, but the exception that a good percentage of the racing community cannot accept.

I understand what you mean, but again your wording is poor. The difference in torque at the flywheel and the wheels (in 1:1 ratio) is simply a difference of drivetrain losses and will not change. HP is a function of torque and rpm and therefore you can supplement lower torque with higher rpms. Again, the important thing will be area under the curve.

That's all I have at 4:30 AM :zzz: . I'm trying not to be overly critial of your post because you recognized an important concept about the curves whereas most never will. Try not to concern yourself with what others on the internet don't/can't understand.

 

[russ_watters]

All you really need to do with engine torque is multiply it by the gear ratio to get the torque at the wheels, then divide by wheel diameter to get the force.

 

[zanick]

Thanks for the response. Actually, I could have been more clear. the curve I posted was set from the guidlines of one of the posters on that discussion. IT IS REAR WHEEL HP AND TORQUE values. It would open an entire set of variables if we start talking gear box efficiencies, etc. I used the peak torque and HP, because those were the only requirements, making it easy for me to debunk the idea that engine torque matters at all as a judging value. am I correct in saying that the HP at any vehicle speed will determine the rear wheel torque and thus forces if the tires are the same diameter? I think so and its been the the foundation of the argument. In otherwords, it doesn't really matter what the engine torque peak or value is, if the HP is the same at any vehicle speed. (meaning the shape of the HP curve is the same for both) then, both vehicles will acceleratae at the same rate at any vehicle speed.

I used the flywheel data, as measured at the rear wheels by a dyno. we have to assume that the gear rations of the comparison for the two engines of equal HP, will be adjusted to have the same final vehicle speed at each maximum rpm point on the engine, AND the same % rpm drop for each shift, to keep the argument centered around what the accelerative potential is of the two engines.

sure, we need to look at the optimal use of HP to create the maximum amount of HP available to the rear wheels, using gear box ratios. I don't think this is done by looking at the area under the curve, because wouldn't the time spent at the different HP levels be more accurate? So, I start thinking in terms of "HP-seconds", unit measures of work, because at the higher rpm levels, near maximum engine speeds, the acceleration rates will be less and more time will be spent at those levels. I know the concepts are simple, but the application is relatively complex, correct. you have variable power on the curve, non-constant jerk, (is that a term?) and a host of other variables we are not even talking about regarding the acceleration of a vehicle on a race track.


Getting back to your last paragraph about my wording. maybe i wasnt making my point well enough. my point was that the engine power determines rear wheel torque as it is multiplied through the gear box at any vehicle speed. sure engine torque will be multiplied through the gear box to get rear wheel torque based on the reduction ratios for a given speed. So, as an example, if both cars are at max hp, the rear wheel torque will be the same, even if the engine torque is 1/2 for one vs the other. (because the rpm are traded off for gear reduction). of course it is consistant with the equation, Acceleration = power/(Mass x velocity) where the power determines the acceleration rate at any vehicle speed, not the engine torque numerical value.

Thanks for the replies

Mk

There is a reason I don't participate in these types of discussions on the automotive forums that I used to, or still do frequent. It's a waste of time. The people who understand have no interest in participating. The ones who are uneducated and believe the answers can be found in a handful of numbers are drawn like moths to a flame, and for the most part will resist anything different.



Yes, but to make your comparison (the graph later) you have to ignore any mechanical advantage. The correct comparison for what you want is chassis dyno info (it makes no difference really from your engine graph, but it seems silly to discuss forces at the tire with flywheel data), which is done in a 1:1 gear ratio.



You demonstrate later in your post that you know peak numbers don't mean much, so why do you allow yourself to use peak numbers to describe entire curves here? This mistake is the same one that you are arguing against with your car buddies.

What you would be concerned with as far as HP curves for equal acceleration would be the area under the curves being equivalent. This could be done either by having the exact same curve shifted along the rpm axis (accounting for differences in peak torque), or with a different curve that has the same area below it (allowing for more narrow or broad rpm ranges).



I understand what you mean, but again your wording is poor. The difference in torque at the flywheel and the wheels (in 1:1 ratio) is simply a difference of drivetrain losses and will not change. HP is a function of torque and rpm and therefore you can supplement lower torque with higher rpms. Again, the important thing will be area under the curve.

That's all I have at 4:30 AM :zzz: . I'm trying not to be overly critial of your post because you recognized an important concept about the curves whereas most never will. Try not to concern yourself with what others on the internet don't/can't understand.

 

[zanick]

Yeah, I think they all get this, but some how can't get their arms around the power being the driving factor. The main point here is that torque is a factor of power. power determines what torque you are going to have available at any vehicle speed. Seems most get it backward, after all power is the rate of doing work. heck, force is just a factor of work.

What is the formula for acceleration of a mass over a speed range based on power?
since power is the rate of change of kinetic energy, Is there a formula that ties the two together in one neat equation?

Thanks,

All you really need to do with engine torque is multiply it by the gear ratio to get the torque at the wheels, then divide by wheel diameter to get the force.

 

[Bob S]

Power in watts = torque (in Newton-meters) times RPM x 2 pi/60
power in HP = power in watts/746

So a torque of 300 Newton meters at 6000 RPM is 188,500 watts or 253 HP. Getting the right torque to the wheels is selecting the right gearbox ratios.

If torque is in pound feet, multiply by 1.356 to get Newton meters:

So a torque of 300 pound feet is 407 Newton meters

407 Newton meters x 6000 RPM x 2 pi/60 = 255,700 watts = 343 HP

 

[zanick]

Im not sure what your point is below. You just gave he difference in power output using the same rpm but different amounts of torque (i.e 300ft-lbs is different than 300NM)

The point of all this is that with the same power output, the rear wheel torque at any speed on the road will be the same. true? Power dictates power to the rear wheels at any vehicle speed. Obviously, engine torque is going to change inversely with RPM needed to acheieve the same vehicle speed, if both cars have the same power output at any same vehicle speed.

For example, if we have a total gear ratio of 8:1 for a high torque engine (most first gears of most cars) and have the same size diameter tires, at 300hp, and 300ftlbs of torque at 4000rpm, i would have 2,400ft-lbs of torque at the tires. A same hp 300hp car with 150ft-lbs of engine torque and 8000rpm would have 2,400ftlbs of torque because it would be running a 16:1 gear ratio in that same gear. All other gears would follow suit, because the RPM drop % would be the same (very common) and the HP curve shape is the same.

In some cases, the curve might be more optimal if broader for a lower torque engine, so it would have more accelerative forces at any vehicle speed. this is the crux of the argument

I'm trying to show those with a little less basic physics 101 knowedge than I have :), how acceleration is determined by HP available, not engine torque at any same speed.
With distilling out a couple of Newtonian identities, a=F/M and P=Fv, I use a=p/(mv) to show that acceleration is proportional to power at any same vehicle speed. (and same car ).
This isn't going over well with folks that think and say "Torque gets you out of turns and hp gets you down the straight". :)

Thanks for the comments!

Power in watts = torque (in Newton-meters) times RPM x 2 pi/60
power in HP = power in watts/746

So a torque of 300 Newton meters at 6000 RPM is 188,500 watts or 253 HP. Getting the right torque to the wheels is selecting the right gearbox ratios.

If torque is in pound feet, multiply by 1.356 to get Newton meters:

So a torque of 300 pound feet is 407 Newton meters

407 Newton meters x 6000 RPM x 2 pi/60 = 255,700 watts = 343 HP

 

[S_Happens]

my point was that the engine power determines rear wheel torque as it is multiplied through the gear box at any vehicle speed. sure engine torque will be multiplied through the gear box to get rear wheel torque based on the reduction ratios for a given speed. So, as an example, if both cars are at max hp, the rear wheel torque will be the same, even if the engine torque is 1/2 for one vs the other. (because the rpm are traded off for gear reduction). of course it is consistant with the equation, Acceleration = power/(Mass x velocity) where the power determines the acceleration rate at any vehicle speed, not the engine torque numerical value.

You are confusing and misusing terms, as I tried to be polite about in my first post. HP is a function of torque and rpm, period. There is no way for two equivalent HP values at different rpms to have the same torque. If you are considering your graphs to be at the wheels, then you already know exacty what the torque is at the wheels.

I know what you are trying to say, but you are doing a poor job of it and as is it is technically incorrect. I was trying not to focus on that, as it is not what is really important at this point.

 

[zanick]

Dont worry, I am conferring with you guys because I want it to be technically correct. so, I appreciate the input. If you read what i wrote again, you can see that I said that two cars with equal Hp, one having more "Engine" torque than the other, will still produce the same rear wheel torque or rear wheel Force (as measured after the gear reductions) at the same vehicle speed. (since we use 24" diameter tires, the force would be equal to the torque found at the rear wheel input shaft :) )

Maybe its confusing because there are two types of rear wheel torque. Rear wheel torque that is calcuated by the dyno, relating to flywheel torque, and the rear wheel torque that is responsible for the rear wheel force on the pavment. I'm trying to be clear with the two, and is why I say, "Rear wheel torque, as multiplied through the gear box", when I am speaking about the end force that is developed at any same engine HP level. So, when I say, "Engine HP determines rear wheel torque" I'm saying that Engine HP detrermines Rear wheel Forces at any vehicle speed.

Again, the terms are:

Engine HP= obvious
flywheel HP= obvious

Rear wheel Hp= that which is measured at the dyno
Rear wheel torque= that is which is measured at the dyno,but relates to torque at the flywheel as measured and calculated at the rear wheels using engine RPM

Rear wheel torque/force= that which is found at the rear wheels as acting on the pavement. This value is engine torque multiplied by gear ratios at any given vehicle speed. Maybe this should be called "Wheel Force" and forget about torque as it can be confused with rear wheel torque that the dyno calculates.



So, with your first objection, I'm talking about the same HP of two engines making the same rear wheel torque. (the 2,400ft-lbs value) made up from either 300ft-lbs at the engine at 4000rpm or the 150ft-lbs made by the other engine at 8,000rpm. Both engines have the same HP output at that vehicles same speed, and thus will have the same rear wheel forces acting on the pavement.

If there is a better way to explain this, I surely would like to hear it. Its all in an effort to show that power of the engine determines acceleration at any same vehicle speed.
Is ok, I have a thick skin. I want to be techically correct and explain it in a way that is not confusing.


Thanks

You are confusing and misusing terms, as I tried to be polite about in my first post. HP is a function of torque and rpm, period. There is no way for two equivalent HP values at different rpms to have the same torque. If you are considering your graphs to be at the wheels, then you already know exacty what the torque is at the wheels.

I know what you are trying to say, but you are doing a poor job of it and as is it is technically incorrect. I was trying not to focus on that, as it is not what is really important at this point.

 

[russ_watters]

Going back to your first post:

Many different types of race cars have all sorts of engines. some with high torque and some with low torque, even some of those might have equal HP output. Since HP is a rate of doing work, a rate of change of kinetic energy, if a two cars were to be compared and both had the same HP, yet one had half the torque output, if their HP curve shapes were the same, the rate of acceleration at any vehicle speed would be the same. Correct?

From the equation that relates torque and horsepower, getting half the torque at the same horsepower requires double the rpm. You tell me - does a car that requires double the rpm to generate the same horsepower accelerate slower or faster?

Or perhaps more telling: if these two cars get the same horsepower, torque, and acceleration at a given speed (their hp to the wheels is the same) and to do that, one has to operate at twice the rpm, what does that tell us about the horsepower at the engine?

And unless one car redlines at twice the rpm as the other, what does that tell us about their max hp and max speed?

 

[xxChrisxx]

WE need Mike in here.

With regard to one of your questions in the OP.

"What would be better on the race track if the two engines in question, both had the same HP but one had more torque than the other. "

The one with more torque, torque adds drivability to an engine. Thats an easy question to answer.

But what if the question was "What would be better on the race track a turbo 4pot 500HP with 200Nm (avg) torque, or somewithing like a v8 with 450HP and 350Nm (avg) torque (these are figures I just plucked out of the air)"

The answer would depend on, what's the track, what's the gearing of both cars, how good is the driver etc etcetc. In this case its not as simple as the first, as the car with more torque would accelerate harder in any given gear lower down, but has less ultimate grunt, the engine would feel 'torquey'. Its also importent to remember that the higher torque gives an 80% max power reading across a wider rev range, increasing drivability.

With the other engine, all the power and acceleration comes at he top end. However this isn't as bad as it seems for acceleration as some of this low down loss can be eiminated with clever selection of gears.

So what would you want here. If the track had long open straights, and relatively few slow corner to fast straight, you'd be better with the HP of the turbo 4pot. If the track had lots of slow corners mid length strights you'd want the extra torque of the v8 for inital acceleration.

I'll try to find a nice link that explains this better.

 

[rcgldr]

In the way you're asking about this, it doesn't really matter. Assume that the horsepower versus RPM curves look the same, with the only difference being the rpm scale at the bottom of the chart. Say the chart for the higher torque engine shows rpm from 0 to 5000 rpm, spaced 1000 rpm per inch across, while the one for the lower torque engine shows rpm from 0 to 10000 rpm, spaced 2000 rpm per inch across. If the curves appear to be the same, then the performance will be the same. If you then overlay these charts with torque versus rpm, then the first torque curve will be twice as high as the second torque curve, but as you stated, the actual torque curve doesn't matter, it's the power curve.

The shape of the torque curve matters, but this would show up in the power versus rpm curve.

 

[xxChrisxx]

Eh... Jeff that's not really comprehensive.

Are you saying if the power curves look the same it'll go the same?

Edit: also people don't take into account the fact that its the gearbox that's the critial factor of how the engine power gets put down. All people seem to talk about is single value readings, this is a horrible way of determining performance.

 

[russ_watters]

In the way you're asking about this, it doesn't really matter.

I don't agree, but the difference might be in the testing...

Assume that the horsepower versus RPM curves look the same, with the only difference being the rpm scale at the bottom of the chart.

Yes, assuming one can top out at a much higher rpm than the other.

But what is being tested to get that curve? Are we talking about a car sitting on a dyno in first gear? Do they usually go through all the gears? Or is this just the engine output curve?

-If it is the engine output curve, then the one that operates at lower rpm has lower drive losses and performs better when installed in a car.
-If this is a dyno run in 1st gear and the curves produced are identical except for the engine rpm scale at the bottom, drive losses decrease as you go through the gears, so the car that operates at higher engine rpm accelerates faster...and generates more horsepower at the wheels than the other engine, as both accelerate.

As with a lot of questions, most of the problem here is with the question being not specific enough. If you make the question specific, the answers are pretty straightforward - these are relatively simple issues.

 

[xxChrisxx]

Normally you'd do a dyno test (all the ones I've done) with as close to 1:1 drive through the gearbox as possible (usually 4th gear). Yes I know about the final drive at the diff, I don't really know why, i'd guess its so it doesn't rev up to the limiter too quickly.

 

[rcgldr]

Normally you'd do a dyno test (all the ones I've done) with as close to 1:1 drive through the gearbox as possible (usually 4th gear). Yes I know about the final drive at the diff, I don't really know why, i'd guess its so it doesn't rev up to the limiter too quickly.

The reason the gear closest to 1:1 ratio (usually 4th gear) is chosen, is that normally the drivetrain losses are less in that gear than other gears.

Going back to my point about power versus rpm curves, assume that the dyno power curves (force versus speed) are the same. The car with the lower torque, double the rpm engine, would have a final drive ratio with twice the reduction factor.

I think the real issue here is the shape of the power or torque versus rpm curve. Generally more power can be acheived with a peakier curve, at the sacrifice of a smaller rpm range where that high power is acheived. The spacing between gear ratios and shift speed are factors in how narrow the powerband can be. Formula 1 cars have 7 closely spaced gears, with computerized shifts that take 30ms to 50ms, so a fairly narrow power band can be used. A race car with fewer gears and a larger spacing between gear ratios will need a wider power band. Nascar race cars are only allowed 4 gears for example, and they run at a few road courses, like Infineon, each season. In race classes where computerized shifting isn't allowed, shift times become significant, and translate into fewer and wider spaced gearing.

On a side note, with improvments in metal alloys and engine design, piston speeds in sport vehicle engines have increased to result in more power, especially motorcycles. The Corvette Z06 has a 7 liter (427.6 in^3) engine that redlines a bit over 7000 rpm, to provide a minimum (SAE) of 505 hp at the flywheel, translating into 440 hp a the rear wheels (the average is more like 450 rwhp). Note the Z06 engine is liter than the turbocharged Porsche 911 3.6 liter engine (480 hp for turbo 911, 530hp for the GT2), and gets better gas mileage.

In the case of motorcycles, a 1 liter road racer replica will have about the same 1/4 mile performance as a 1.3 liter superbike (Hayabusa), but the 1.3 liter bike gets 80% of peak torque for the upper 2/3rds of it's rpm range, while the 1.0 liter bike gets 80% of peak torque only in the upper 1/3rd of it's rpm range, half the powerband width. When crusing on a freeway at around 65mph, in top gear (6th), the liter bike won't be able to accelerate as fast as the superbike (1.3 liter engine) without downshifting to get the rpms up. In the case of Suzuki, the 1 liter engine redlines at 14,000 rpm, while the 1.3 liter engine at 11,000 rpm.

 

[xxChrisxx]

Hmm thinking about it, that's obvious really as the 4th gear locks the input and output shafts. Bit of a brain fade moment :P

 

[rcgldr]

Hmm thinking about it, that's obvious really as the 4th gear locks the input and output shafts.

Not in the case of a manual transmission. It's just another pair of gears, in this case equally sized gears that generally have a bit less power loss than the other gear combinations. Motorcycles transmissions never bother with this, losses with a tranny and chain drive are about 8%.

Strength of the gears is a factor in determining gear ratios. The bigger the difference in gear size, the less contact (fewer gear teeth engaged) between gears. To compensate the gear widths are varied. Back in the 70's and 80's rear end ratios were less than 3.00, and first gear ratio over 3.00, but since the 1990's, this was swapped to reduce gear width in the trannies of the higher powered cars (Z06, Viper) to allow more gears (6 instead of 4) and to be able to handle more engine torque. Using the C6 Z06 as an example, 1st gear ratio is 2.66, while the rear end ratio is 3.42.

 

[xxChrisxx]

In MT it locks the shaft. The layshaft isn't used so it has to.

EDIT: most the gearboxes I've seen have run direct drive as its better then running a ratio of 1:1

 

[rcgldr]

In MT it locks the shaft. The layshaft isn't used so it has to.
EDIT: most the gearboxes I've seen have run direct drive as its better then running a ratio of 1:1

I forgot that some MT's still lock the shaft. I doubt racing cars transmissions do this though. I don't think the Tremec T56 or T6060 do this either.

 

[xxChrisxx]

Yeah, i'll admit I don't know a super amount about racing boxes. Its been a while since I've done anything with transmissions.

Getting slightly back on topic. I'm glad this thread has swung around to what really matters for acceleration. The gear ratios! Yes peak torque and peak horsepower figures are great for willy waving in the paddock, but careful selection of gear ratios and usefulness of the powerband is king.

 

[zanick]

Sorry about that, I left out that detail. Yes, the rpm would obviosly have to be a variable. we would be trading rpm at any vehicle speed, for lower torque, and the gear ratio would be proportionally changed as well.

mk

Going back to your first post: From the equation that relates torque and horsepower, getting half the torque at the same horsepower requires double the rpm. You tell me - does a car that requires double the rpm to generate the same horsepower accelerate slower or faster?

Or perhaps more telling: if these two cars get the same horsepower, torque, and acceleration at a given speed (their hp to the wheels is the same) and to do that, one has to operate at twice the rpm, what does that tell us about the horsepower at the engine?

And unless one car redlines at twice the rpm as the other, what does that tell us about their max hp and max speed?

 

[zanick]

Actually, that's why I came to you guys.

First of all, the hp curves would be the same shape. even though one had a 8000rpm redline vs the other at 6000rpm, the HP curves being the same shape, would create the same exact rear wheel forces at any speed on the track. coming off a turn, down a straight , anywhere. This is because the HP curve is the same shape. Think of the gear ratios as the same speeds in each gear. (e.g. 50mph 1st, 80mph 2nd, 120mph 3rd, etc)
this way, the gear spacing is the same and the only difference is the numerical torque values at the engine. at the rear wheels, since HP is determining the rear wheel force at any vehicle speed, the torque or force would be the same.

what I am asking is how do we explain this so it is easy to understand and fight the perception that the greater engine torque will be "stronger" off a turn, or the higher rpm engine is better down the straight. :)

Thanks

Mark

WE need Mike in here.

With regard to one of your questions in the OP.

"What would be better on the race track if the two engines in question, both had the same HP but one had more torque than the other. "

The one with more torque, torque adds drivability to an engine. Thats an easy question to answer.

But what if the question was "What would be better on the race track a turbo 4pot 500HP with 200Nm (avg) torque, or somewithing like a v8 with 450HP and 350Nm (avg) torque (these are figures I just plucked out of the air)"

The answer would depend on, what's the track, what's the gearing of both cars, how good is the driver etc etcetc. In this case its not as simple as the first, as the car with more torque would accelerate harder in any given gear lower down, but has less ultimate grunt, the engine would feel 'torquey'. Its also importent to remember that the higher torque gives an 80% max power reading across a wider rev range, increasing drivability.

With the other engine, all the power and acceleration comes at he top end. However this isn't as bad as it seems for acceleration as some of this low down loss can be eiminated with clever selection of gears.

So what would you want here. If the track had long open straights, and relatively few slow corner to fast straight, you'd be better with the HP of the turbo 4pot. If the track had lots of slow corners mid length strights you'd want the extra torque of the v8 for inital acceleration.

I'll try to find a nice link that explains this better.

 

[zanick]

I think you got it, exactly right. Now if I can get the terminology down, it might be better to explain. One of the things I struggle with is the comparing of different HP curves. if the HP curves have the same area under the curve (integration), it doent mean it will move the mass at the same acceleration rate at a given speed. Its a matter of what I call, Hp-seconds, or watt-seconds. obviously, its better to have the higher hp at the top of the HP curve at the higher speeds, because more time is spent there in accelerating over the operational speed range in each gear.

Mark

In the way you're asking about this, it doesn't really matter. Assume that the horsepower versus RPM curves look the same, with the only difference being the rpm scale at the bottom of the chart. Say the chart for the higher torque engine shows rpm from 0 to 5000 rpm, spaced 1000 rpm per inch across, while the one for the lower torque engine shows rpm from 0 to 10000 rpm, spaced 2000 rpm per inch across. If the curves appear to be the same, then the performance will be the same. If you then overlay these charts with torque versus rpm, then the first torque curve will be twice as high as the second torque curve, but as you stated, the actual torque curve doesn't matter, it's the power curve.

The shape of the torque curve matters, but this would show up in the power versus rpm curve.

 

[xxChrisxx]

I don't get your last statement, why would we want to fight the perception of toruqe being better off the turn and power down the straight when that is correct?

You are saying the curves look the same but one goes to a higher RPM and produces the same power then by definition they don't look the same.

You don't integrate the HP curve, that would be pointless. For the work acutally done you'd integrate the torque curve.

 

[zanick]

I think he is and he is right. But, I will say the gear box is important, but we are assuming same gear spacing and same MPH speed in each gear. even if they were not the same speed in each gear, the differences are subtle and are trade offs. IN fact, any change might favor one track, while being not optimal on an other. Gear boxes don't create hp, they just optimize the HP available for a given speed range.

mk

Eh... Jeff that's not really comprehensive.

Are you saying if the power curves look the same it'll go the same?

Edit: also people don't take into account the fact that its the gearbox that's the critial factor of how the engine power gets put down. All people seem to talk about is single value readings, this is a horrible way of determining performance.

 

[xxChrisxx]

I think he is and he is right. But, I will say the gear box is important, but we are assuming same gear spacing and same MPH speed in each gear. even if they were not the same speed in each gear, the differences are subtle and are trade offs. IN fact, any change might favor one track, while being not optimal on an other. Gear boxes don't create hp, they just optimize the HP available for a given speed range.

mk

All of your statemnts so far have been riddled with things that are contradictive and incredibly confusing.

If you set the gear ratios the same, and you have more torque with one engine you will get a higher mph for that given gear.

 

[zanick]

As Jeff said, dynos are done in a few of the top gears to avoid wheel spin. usually, 3rd or 4th gear. This just gives the rear wheel HP and calculated engine torque values. Now, if ths same car had a different engine and transmission installed and it did the same test, in its 4th gear, but the ratio was higher due to the proportionally higher rpm, due to the lower engine torque, the power would be the same, and the acceleration of the vehicle would be the same. I wouldn't expect the effeciency to be less in lower gears. if there was a difference, it would be canceled out by the fact that the initial dyno runs used proportionally different total ratios, even if it was 1:1 in the gear box, the final drive is likely 2-4:1 additioally.

What we are looking for is the correct terminology to explain to folks that still confuse engine torque with rear wheel forces and torque though the gear box.
We know that Acceleration = Power/(mass x velocity), so acceleration is proportional to power at any vehicle speed. This also indicates that power will also determine rear wheel torque , or longitudinal force at any speed as well. How do we say this?


mk

I don't agree, but the difference might be in the testing... Yes, assuming one can top out at a much higher rpm than the other.

But what is being tested to get that curve? Are we talking about a car sitting on a dyno in first gear? Do they usually go through all the gears? Or is this just the engine output curve?

-If it is the engine output curve, then the one that operates at lower rpm has lower drive losses and performs better when installed in a car.
-If this is a dyno run in 1st gear and the curves produced are identical except for the engine rpm scale at the bottom, drive losses decrease as you go through the gears, so the car that operates at higher engine rpm accelerates faster...and generates more horsepower at the wheels than the other engine, as both accelerate.

As with a lot of questions, most of the problem here is with the question being not specific enough. If you make the question specific, the answers are pretty straightforward - these are relatively simple issues.

 

[xxChrisxx]

This is getting slightly out of hand with multiple different replys going on at the same time, and I think we need to go back to square one. Can you restate the question you want answered. Preferably in 2 scentences or less.

And what do you think a dyno acutally measures? The variables.

 

[zanick]

well, certainly it matters what the HP curve looks like, but the torque curve, (if we are talking about different engines, like 4 bangers vs z06 vet 7 liters and 6 bangers.) is not usefull for a couple of reasons. First, its not even an indication of performance, as most of the time, the operational range of an engine will be between 4500rpm and 6500rpm for a V8. at 5250rpm, the torque is heading down anyway, and discounts all the "flat torque curve talk". in fact, the porsche GT3 has the peakiest HP curve you would ever see, but the torque is a flat 275 from down low almost until the top rpm :).

I have two engines that have the same hp, one with 65ft-lbs more than the other. their HP curves are close, but the lower torque , higher rpm engine has a more broad HP curve. It actually would put down more rear wheel forces at ANY vehicle speed, anywhere on any track. How do we convey this to the "torqueez" out there? they still think in that case, the higher torque engine has more "grunt" off turns, yet on the dyno runs, the curves show that even at the lower rpm, the lower torque engine has more HP or at least the same.

How do we say, "HP determines rear wheel torque at any vehicle speed"?

also when looking at different shapped HP curves, how do we talk about the differences in time spent at the different segmements of speed in any individual gear? watt-seconds? HP-seconds? what makes the most sense?

Thanks,

Mark

Yeah, i'll admit I don't know a super amount about racing boxes. Its been a while since I've done anything with transmissions.

Getting slightly back on topic. I'm glad this thread has swung around to what really matters for acceleration. The gear ratios! Yes peak torque and peak horsepower figures are great for willy waving in the paddock, but careful selection of gear ratios and usefulness of the powerband is king.

 

[zanick]

I said same MPH range in each comparable gear. (each gear might be a 27% rpm drop for each shift ) This means the actual ratio will be in proportion to the increased or lower rpm levels to reach that same MPH range in each gear.

mk

All of your statemnts so far have been riddled with things that are contradictive and incredibly confusing.

If you set the gear ratios the same, and you have more torque with one engine you will get a higher mph for that given gear.

 

[zanick]

Well, I think we are getting somewhere now. yes, the curves "shape" look the same (same shape over the rpm range used)

why do folks always talk about area under the HP curve (which is integration of the curve, correct?)

It seems that intgrating the torque curve would be pointless, as at the engine, without knowing final drive ratios, you would end up with nothing useful, correct? However, if you could integrate the torque curve at the rear wheels , AFTER the gear ratios, so you could get value of work done, correct? I would think that the HP curve integration would give an average power, and if you have time tied to it, (something like watt-seconds) you could get an idea of work done that way as well. This is the part that is causing the most confustion.

Thanks,

Mark

I don't get your last statement, why would we want to fight the perception of toruqe being better off the turn and power down the straight when that is correct?

You are saying the curves look the same but one goes to a higher RPM and produces the same power then by definition they don't look the same.

You don't integrate the HP curve, that would be pointless. For the work acutally done you'd integrate the torque curve.

 

[xxChrisxx]

It acutally seems we're getting nowhere. You seem obsessed about the shape of the HP curve. What governs the shape of the hp curve.

What does a dyno measure?

 

[zanick]

Sorry, didnt see this post until I had answered the others.

the dyno measures the rate of change of kinetic energy. (by knowing the change of speed of the drum and the weight and size of the drum). without a spark (rpm) signal of the engine, the output is only power vs MPH. I suppose it could produce rear wheel forces, but it doesn't output that. the engine rpm could be 3000 to 6000rpm or 6000rpm to 12,000rpm. once you have a spark signal, the dyno then outputs an engine torque value.

The question is how to we describe engine power determining rear wheel torque, as found at the rear tires at any vehicle speed.?

When comparing two equal HP engines, but one has more torque than the other, it can depend on the shape of the HP curve in determining which one will develope more accelerative forces at the driving tires. I am asking what is the proper way to describe this reality?

why wouldn't the area under the HP curve or getting a HP-seconds, or watt-seconds value to determine which Hp curve is more optimal for acceleration based on different shapes.

Thanks for being patient and helping out here.

Mark

This is getting slightly out of hand with multiple different replys going on at the same time, and I think we need to go back to square one. Can you restate the question you want answered. Preferably in 2 scentences or less.

And what do you think a dyno acutally measures? The variables.

 

[xxChrisxx]

We're going to have to deal with this one question at a time I'm having trouble following many mish mashed questions at one. Sorry to be a bit pedantic about this, but following your reasoning is very difficult.

You haven't clearly stated what the dyno measures. Does it measure the torque or horsepower that the engine puts out?

 

[zanick]

Im obsessed with the "shape of the HP curve" because it seems its the only thing that determines rear wheel torque (after the gear ratio reductions). I know you could get the same info through the torque curve and multiplying it through the gears manually for any speed, but it seems that HP incorporates more information and is easier to work with. Plus, when you compare identical HP engines but with different shaped hp curve, the lower torque, higher rpm, could have more torque at the rear wheels through the gear box at any vehicle speed.

What governs the shape of the HP curve is the engines ability to burn air and fuel at any rpm.

I think I answered the dyno question above, but it measures rate of change of kinetic energy, by looking at drum speed changes and knowing mass and size of the drum. (we are talking about wheel dynos here.)

mk

It acutally seems we're getting nowhere. You seem obsessed about the shape of the HP curve. What governs the shape of the hp curve.

What does a dyno measure?

 

[xxChrisxx]

ok i'll give you the answer. a dyno measures torque

 

[zanick]

techncally, it measures speed change of the drums. by knowign the size and weight of the drum, you end up with torque (thats what you were getting at, i know) But, its torque at the rear wheel, and of course, rate of change of kinetic energy. Beause you have to know speed, speed change, and drum inertia, by definition, you have Hp. You can't even attempt to figure out engine torque unless you someone tells you the rpm running. A spark signal, allows for the computer to then work backward and provide engine torque.

mk

We're going to have to deal with this one question at a time I'm having trouble following many mish mashed questions at one. Sorry to be a bit pedantic about this, but following your reasoning is very difficult.

You haven't clearly stated what the dyno measures. Does it measure the torque or horsepower that the engine puts out?

 

[xxChrisxx]

oh my god.

no no no.

You run the dyno up through the rev range. It MEASURES the torque output at each rpm, it then CALCULATES the horsepower reading.

Edit I apologise if that seemed a little bit brash, I am just getting a wee bit sleepy.

 

[zanick]

How is this possible if the dyno is not getting a rpm signal. without an rpm signal, it has no way of knowing what the engine rpm or torque is, correct? the output is HP and MPH.

No problem, I just want to get the right information.

Im sure the computer can caluate the HP just as easily as it can calculate torque. It has to calculate torque, because all it is doing is measuring the rate of speed change of the rollers. by knowign the speed, and the speed change, it has rear wheel torque for any speed change period. But, isn't it just as easy for it to out put HP first? rate of change of kinetic energy is power too, right??

mk

oh my god.

no no no.

You run the dyno up through the rev range. It MEASURES the torque output at each rpm, it then CALCULATES the horsepower reading.

Edit I apologise if that seemed a little bit brash, I am just getting a wee bit sleepy.

 

[xxChrisxx]

oh dear.

it can't output horsepower first. as that is the calcualted value.
the variable measured by the acutal apparatus is torque.

You are totally 100% incorrect in what you have said in the previous post.

http://auto.howstuffworks.com/horsepower1.htm
http://en.wikipedia.org/wiki/Dynamometer

I understand you are proabably more used to the 'rolling road' but this is giving you a skewed understanding of what is going on.

 

[zanick]

Isn't it all calculated? By the way, I am talking about a dynojet 248e, which is a chassi dyno that has giant rollers, known diameter and weight (like 2000lbs or something). Not water brake, electromagnetic brake, or engine dyno.

Help me understand. You have drums with a known size and mass. they are being accelerated. Its not lifting anything, and there is no drag working against the drivng force.
Worst case, you have acceleration by the difference and time between two known rotational speeds, gives you a rate of change of kinetic energy. Isnt that by definition, Power?
Sure you can find torque too, but knowing the rate of speed change at some rotatioal speed, (dv/dt) it gives torque on the drum, correct? Still need to converte to engine torque, but ONLY if you have the engine rpm. again, that has to be calculated as well. Hp in that case could be found as easy as drum torque, right?

Im 100% wrong? wow, I really don't understand. Help me understand.

See if you can be as detailed in your response as possible.

Of course we are getting a little off the track here. But that's ok.


Thanks,

Mk

oh dear.

it can't output horsepower first. as that is the calcualted value.
the variable measured by the acutal apparatus is torque.

You are totally 100% incorrect in what you have said in the previous post.

http://auto.howstuffworks.com/horsepower1.htm
http://en.wikipedia.org/wiki/Dynamometer

 

[xxChrisxx]

But this thinking in horsepower is giving you a false sense of what is going on. The force on the rollers (that contributes to their acceleration) can be defined as a wheel torque.

As Torque = Force * distance.

So in this case the inertia and mass of the drum is known. The change in the speed of the drum in the amount of time is the acceleration. from this the force is known. From the size of the wheel the torque at the wheel is known.

So you see the torque is really still the measured 'thing'.

After the run up, the clutch is dipped and the run down let's you know the losses through the transmission. This will give you the torque at the flywheel. The revs are found from a tachometer iirc.

EDIT: I'll apologive now, as I know I am making a pigs ear out of trying to explain this.

Well you weren't 100% wrong, sorry for saying that. What i meant was, your reading of what was going on was wrong. Yours knowledge of how the rolling road works is spot on.

 

[zanick]

Well, at least I have the effects on the road correct. :) thanks for that. that's most important, but just want to get the terminology right.

I think its coming into a philosophical debate now, which came first, the force or the power that created it. I think I understand, that even if the power creates the force, its the force that is causing the acceleration, even though the power of the act is known at the same time. (how is that for a compromise?).

So, how do we address the hp curve and the progressive time spent at different rpm as the engine accelerates over a given vehicle speed range when comparing different shaped HP curves of two engines, in the same car, with the same max HP?

Mark

But this thinking in horsepower is giving you a false sense of what is going on. The force on the rollers (that contributes to their acceleration) can be defined as a wheel torque.

As Torque = Force * distance.

So in this case the inertia and mass of the drum is known. The change in the speed of the drum in the amount of time is the acceleration. from this the force is known. From the size of the wheel the torque at the wheel is known.

So you see the torque is really still the measured 'thing'.

After the run up, the clutch is dipped and the run down let's you know the losses through the transmission. This will give you the torque at the flywheel. The revs are found from a tachometer iirc.

Well you weren't 100% wrong, sorry for saying that. What i meant was, your reading of what was going on was wrong. Yours knowledge of how the rolling road works is spot on.

 

[xxChrisxx]

Its the force (engine) that creates the power (engine) :P hehe.

Torque - Ability of engine to do work
Power - Rate at which said work can be done.

Well the horsepower curve is a function of the torque. I believe the angle you are coming from is that: two cars traveling a constant speed along the road (the one with less torque requires it to be in a higher rpm band) both floor it. The answer would very corrctly be that they would both probably accelerate at the same rate.

However, the other car would require different gearing to do this. This was why I was harping on about it earlier.

This didnt come across in your original question. As two identical engines but one with more torque to me means identical gearing and ataring form the same rpm. In which case the engine with the larger torque would win out. (This was why your arguements were quite confusing)

The practical upshot is that, high torque engines accelerate very well in the bottom end of the rev range. but higher revvers accelarate better in the upper rev range. In very slow corners (such as a hairpin) it's quite likely that a car with a narrow power band is likely to fall out of the optimal range. This is very true with sometihng like a formula 1 car where the power band is about 1500RPM.

This led me to my point about the slow corners thing. A huge V8 would out accelerate a smaller high revving engine initally, but then the high revver would have a relative increase in acceleration as it gets down the straight.

This then leads into useful power band and that Jeff said about lots of narrow gears are needed for the high revver as its peaky but a big lazy V8 would need less to get the same job done.

EDIT: I think we've just argued ourselves to a similar conclusion! Damn! You thinking was pretty much ok, but like you suspected its your terminoligy that was incorrect (and what got me all :S). I'd like to say I am sorry for being a bit curt with you earlier.

One thing you always have to keep in mind though and this is the most important thing. Its NOT the power that's accelerating you, its still the torque produced at the engine that is determining the rear wheel force and therefore acceleration. So don't think of it as 'RWHP determinign torque, its torque determining RWHP!

 

[Gnosis]

Well, at least I have the effects on the road correct. :) thanks for that. that's most important, but just want to get the terminology right.

I think its coming into a philosophical debate now, which came first, the force or the power that created it. I think I understand, that even if the power creates the force, its the force that is causing the acceleration, even though the power of the act is known at the same time. (how is that for a compromise?).

So, how do we address the hp curve and the progressive time spent at different rpm as the engine accelerates over a given vehicle speed range when comparing different shaped HP curves of two engines, in the same car, with the same max HP?

Mark

Zanick, the matter hasn’t been reduced to a philosophical debate in the least. Your supposition concerning horsepower and torque is quite simply backwards. Torque isn’t a factor of horsepower rather; horsepower is a derivative of torque. You must first produce torque if you are to “calculate” a “horsepower rating” for the engine.

Horsepower is an intangible quantity that cannot be measured directly, but more importantly, horsepower isn’t a “force”. Horsepower is simply a rating of how much work can be performed over time and distance. It is “calculated” per the torque measured since torque CAN be measured directly in ft lbs. For instance, when an engine is stated as having 200 HP, it is merely a rating of how much work can be performed at a given RPM per the engine’s available torque.

The dyno measures torque by applying a known braking force to the dyno’s rotating drum. When the vehicle’s drive wheels accelerate the dyno’s drum under this braking force up to a given RPM in a given time, it reveals the torque being applied by the vehicle’s drive wheels, as it would require a given torque to have accomplished this rate of drum acceleration in the time recorded.

Force produced in a Combustion Engine

Consider where the combustion engine actually produces its force. It is per the optimized A/F ratio ignited per each power-stroke, as the piston and its connecting rod are forced downward by the expanding hot gases in the cylinder with the other end of the connecting rod applying its force to the crankshaft in a manner that provides leverage to rotate the crankshaft and produce crankshaft torque. So, torque is the accelerative component, not horsepower.

You need to get horsepower out of your head and acknowledge the actual source of the vehicle’s acceleration; that being torque. Horsepower is only a rating of how much work can potentially be performed at a given RPM per the available torque.

Horsepower is a “rating” of potential work that can be accomplished, but it is NOT an actual force

For instance, my workouts had me moving approx 30,000 pounds of free weights a distance of 3 feet in an hour’s time. This can be given a horsepower rating HOWEVER, that horsepower rating does not imply an actual force used. It merely specifies a total of 30,000 pounds was moved 3 feet over a span of one hour (a given weight moved a given distance over time). In fact, some weight movement was several hundred pounds per rep while things like curls were more around the 95 to 115 pounds. Again, horsepower doesn’t specify a given force rather; it implies a given amount of work that can potentially be accomplished.

Hopefully this will provide the insight you need to resolve your confusion associated with horsepower. Unfortunately, those who tend to adore horsepower and have long confused its meaning have a most difficult time relinquishing their love for its mighty-sounding namesake, when it’s actually “torque” that should be held in highest accord.

I demonstrated the example below back in my mini-bike building days. It demonstrates the significance of an engine’s torque.

Firsthand Example:

I used a 1.5 HP lawnmower engine to power a mini-bike and I geared it to yield a top-speed of 35 MPH while on flat ground with my body weight of 140 pounds. However, while riding at 35 MPH on flat ground, I came to one of the steepest streets in my neighborhood. My mini-bike steadily lost speed and would not make it all the way up to the top of the steep hill. When I hit the hill, the engine was at its upper rev range therefore producing its 1.5 HP, so why could it achieve 35 MPH on flat ground, but not make it to the top of the steep hill?

Answer: The engine lacked sufficient torque to accomplish the climbing of the steep incline. The incline imposed additional loading on the crankshaft, so the engine’s crankshaft steadily lost RPM. As its crankshaft slowed, less A/F mixtures were ignited per second as a result, so less over all energy was provided per second, which further slowed my ascent. It must be realized that “IF” the engine’s torque were sufficient to climb the steep hill in the first place, the engine would have never slowed down. A 2.5 HP combustion engine remedied the issue, as it produced greater torque and even had enough extra torque to start out from a dead stop up the steep hill. From a dead stop (engine just above idle), it couldn’t be the extra HP yielding the more than adequate acceleration up the hill; it was a direct result of the extra torque provided by the 2.5 HP engine.

Point: A given task requires a given minimal torque to accomplish the task. If less torque is produced, then it makes no difference what the “horsepower rating” is, the task will not be achievable.

 

[xxChrisxx]

Thanks Gnosis, that's acutally a pretty damn good explination. You did far better in one post than I manages in lots!

People may say that 'all that matters is power'. This is to an extent true, but you've got to go back to how that 'more power' is produced. You either spin in more quickly, or generate more torque. So to say that torque is irrelevant is completely erroneous (I know you didnt say this, but I've seen it as an argument on quite a few forums). An engine that generates no torque generates no power.

The acceleration can be calculated both ways, by using the power method or torque method. I'm a massive proponent of the torque method because torque is the acutal measurable driving force.

 

[rcgldr]

integrating hp versus rpm curves

This make more sense if the calculated area is divided by the range of rpm, which calculates the average hp across a range of rpm. Even this information may not provide a good estimate of acceleration if one curve is very peaky compared to another. Torque versus rpm curves are easier to read because it eliminates the rpm multiplier effect which adds a linear slope to the power versus rpm curve.

torque ... power

A torque can be applied to a non moving object, such a wrench applying a torque to a bolt; in this case zero work is being done. Power defines a rate of work. In the case of an idealized CVT (continously variable transmission), then as previously mentioned, acceleration = power / (mass x speed). (power = force x speed = mass x acceleration x speed).

Note, torque = force times radial distance, not distance moved by a tire.

chassis dynos

Not all chassis dynos are inertial. Some use a variable load via a mechanical or electrical brake. The torque on the chassis drum equals the force from the driven tire contact patch times the radius of the drum. The angular speed of the chassis drum equals the driven tire contact patch speed divided by the radius of the drum. The dyno converts this information into contact patch force times contact patch speed, which equals contact patch power. The driven wheel torque can't be calculated from this unless the effective radius of the driven tires, or the engine rpm and overall gearing is included with the sampled data. The effective engine torque can be calculated given engine rpm data without requiring overall gearing data.

 

[zanick]

Ill start with you Chris, because your post is shorter. Maybe I didnt make myself clear in the beginning. when i mentioned same engine HP but different torque values. (one High rpm, and the other lower rpm) the gearing had to be appropriate for the same speeds in each of the cars respective gears. So, yes, not "same gear ratios" but proportionaly the same gear ratios. Does that make it more clear. If you come from the torque side of things, only looking at engine torque, you could be way off, as a lower torque engine, could make more power in certain sections of its HP curve and thus, yeild better acceleration to to MORE avaialbe torque at the rear wheels as it is seen through the gear box. correct?

You are also right, I've never said torque doesn't matter, I only said that the value is not an indication of better or worse acceleration based on a comparison with two equal peak HP engines.

So, going back to the original question, can you look at hp and say in the right venacular, that HP dictates the torque at the rear wheels at any vehicle speed? (remember, not engine torque, multiplied rear wheel torque and ultimately force to the ground).

Also, looking at area under the torque curves is a popular concept, but that can't be right, as you might have the most area under the curve shown by integrating the curve, from 1000 to 2000rpm. certainly, the mulplied force at the wheels wouldn't reflect optimum use of power at that engine rpm range. You would have to look at gear ratio factors as well. Also, area under the HP curve is better, because it would point to max accelerative potential at any vehicle speed, but it doesn't address the progressive time element that is seen as a car accelerates through the higher rpm. In otherwords, two HP curves could share the same HP peak, but one might have more area in the lower hp range, and the other in the higher hp range. the one with area in the higher HP range will have more accelerative potential. Thats why I was asking about HP-seconds or watt-seconds as a way to explain it or quantify accelerative potential.

Then, going back to Acceleration=power/momentium, doesn't this say that acceleration, and thus rear wheel Force will be proportional to power at any given vehicle speed?

Thoughts on this line of thinking or terminology.

thanks

Mark

Thanks Gnosis, that's acutally a pretty damn good explination. You did far better in one post than I manages in lots!

People may say that 'all that matters is power'. This is to an extent true, but you've got to go back to how that 'more power' is produced. You either spin in more quickly, or generate more torque. So to say that torque is irrelevant is completely erroneous (I know you didnt say this, but I've seen it as an argument on quite a few forums). An engine that generates no torque generates no power.

The acceleration can be calculated both ways, by using the power method or torque method. I'm a massive proponent of the torque method because torque is the acutal measurable driving force.

 

[zanick]

Chris,

But, in those two examples of two equal HP engines, one high rpm, lower torque, and the other high torque, lower rpm. If the lower torque engine had a broader HP curve, (and this is definitely possible as I have seen), the lower torque engine can produce more rear wheel torque at the same vehicle speed, because it is making more HP at that same speed. (again, gear ratios kept proportional to keep things even for the comparison. in otherwords, both cars have 1st gear to 50mph, 2nd to 80mph, 3rd to 120mph for example)

So, we know how engine torque creates rear wheel HP and rear wheel torqe as muliplied through the gear box, BUT, if there is more available hp at any vehicle speed with another engine, it will produce more rear wheel forces, correct?

I agree that the force is what accelerates the cars through the torque at the rear tires.
But, the engine torque is not an indication of what the total force will be when comparing two engines, unless you have the gear ratios or rpm handy. Hp values from an engine can determine what the rear wheel torque numbers will be at any vehicle speed, right?

mk

Its the force (engine) that creates the power (engine) :P hehe.

Torque - Ability of engine to do work
Power - Rate at which said work can be done.

One thing you always have to keep in mind though and this is the most important thing. Its NOT the power that's accelerating you, its still the torque produced at the engine that is determining the rear wheel force and therefore acceleration. So don't think of it as 'RWHP determinign torque, its torque determining RWHP!