The advantages of a NA engine?

[Serious]

Originally Posted by bruce.augenstein@comcast.

I'm not an engineer, but in a full on race, more horsepower puts more torque to the wheels, and wins. Perhaps more simply, you need more horsepower to go faster - in a completely linear fashion, ignoring wind and rolling resistance. Torque is just a force. Horsepower includes speed in the equation. More speed equals more work being done, which needs more horsepower.

In short: Two cars of equal weight, the one with more horsepower at a given instant will accelerate faster than the other car, no matter the torque figures or gearing.

Torque is not a force. Torque is force * radius.

More HP doesn't = more torque at the wheels, in fact it's usually quite the opposite. HP is just (torque *engine speed)/constant so, you can get the same number of HP'z by either having high torque and low engine speed or high engine speed and low torque.

You're last statement is completely untrue. Imagine riding a multiple speed bicycle. Now you're legs only strong enough to make a specific force, that force is imparted through the lever length of the crankshaft & pedal assembly to create a torque... that torque is then multiplied a specific amount depending on what gear you're in.

(low gears you pedal crazy fast but don't have to use much strength; whereas in high gears you pedal slowly but use a lot of energy pushing the pedals.)

Now are you really going to tell me you will be just as fast up a steep hill in 18th gear when you can't even push the pedals as you are in 3rd? Or that how strong the biker is irrelevant? Not a chance.This is why gearing is matters, and torque is an absolutely critical measurement in analyzing car acceleration.

Where most of you are hung up is that you only look at peak HP & torque numbers instead of area's under the curves and don't understand the relationship between forces, torque, work, power and how they interrelate to each other in terms of physics.

[Dodge2Dub]

Originally Posted by Serious

Torque is not a force. Torque is force * radius.

More HP doesn't = more torque at the wheels, in fact it's usually quite the opposite. HP is just (torque *engine speed)/constant so, you can get the same number of HP'z by either having high torque and low engine speed or high engine speed and low torque.

You're last statement is completely untrue. Imagine riding a multiple speed bicycle. Now you're legs only strong enough to make a specific force, that force is imparted through the lever length of the crankshaft & pedal assembly to create a torque... that torque is then multiplied a specific amount depending on what gear you're in.

(low gears you pedal crazy fast but don't have to use much strength; whereas in high gears you pedal slowly but use a lot of energy pushing the pedals.)

Now are you really going to tell me you will be just as fast up a steep hill in 18th gear when you can't even push the pedals as you are in 3rd? Or that how strong the biker is irrelevant? Not a chance.This is why gearing is matters, and torque is an absolutely critical measurement in analyzing car acceleration.

Where most of you are hung up is that you only look at peak HP & torque numbers instead of area's under the curves and don't understand the relationship between forces, torque, work, power and how they interrelate to each other in terms of physics.

Another great post.

With all of the information in this thread, along with some wiki searches, we are all going to be the hit at the next cocktail party!

[Sedan_Clan]

Originally Posted by Serious

Torque is not a force. Torque is force * radius.

More HP doesn't = more torque at the wheels, in fact it's usually quite the opposite. HP is just (torque *engine speed)/constant so, you can get the same number of HP'z by either having high torque and low engine speed or high engine speed and low torque.

You're last statement is completely untrue. Imagine riding a multiple speed bicycle. Now you're legs only strong enough to make a specific force, that force is imparted through the lever length of the crankshaft & pedal assembly to create a torque... that torque is then multiplied a specific amount depending on what gear you're in.

(low gears you pedal crazy fast but don't have to use much strength; whereas in high gears you pedal slowly but use a lot of energy pushing the pedals.)

Now are you really going to tell me you will be just as fast up a steep hill in 18th gear when you can't even push the pedals as you are in 3rd? Or that how strong the biker is irrelevant? Not a chance.This is why gearing is matters, and torque is an absolutely critical measurement in analyzing car acceleration.

Where most of you are hung up is that you only look at peak HP & torque numbers instead of area's under the curves and don't understand the relationship between forces, torque, work, power and how they interrelate to each other in terms of physics.

Great explanation. The bicycle example REALLY puts it into perspective (..and a format for someone to get the gist of HP/torque/force/etc.). Well done!

[2m3]

Originally Posted by bruce.augenstein@comcast.

That's a quote from me, and it exists in this forum as well.

That is also a quote from Ray Korman that I read in 1996.

[Dodge2Dub]

What is the actual difference between the wheel torque calculated on a Dyno (in the hundreds) and the actual wheel torque (force) calculated by looking at the final gear ratio, axle ratio, tire circumference, speed, etc. (in the thousands)? Is the dyno wheel torque figure the force that is exerted on the dyno wheel and the actual wheel torque the force exhibited by the car to the wheel itself?

Basically, is the dyno number the true wheel torque and the calculated number (in the thousands) the force. Shouldn't we differentiate between the two instead of calling both of them the wheel torque (ie. Dyno = Wheel Torque and Calculated Tq = Wheel Force)?

Torque is the tendency of a force to rotate the body to which it is applied.Torque is always specified with regard to the axis of rotation. It is equal to the magnitude of the component of the force lying in the plane perpendicular to the axis of rotation, multiplied by the shortest distance between the axis and the direction of the force component. Torque is the force that affects rotational motion; the greater the torque, the greater the change in this motion.

Whereas,force is the agency that alters the direction, speed, or shape that a body would exhibit in the absence of any external influence.It is a vector quantity, having both magnitude and direction. Force is commonly explained in terms of Newton's laws of motion. All known natural forces can be traced to the fundamental interactions. Force is measured in newtons (N).

[bruce.augenstein@comcast.]

Originally Posted by mkoesel

This is hard for me to accept, Bruce. Can you explain why this is so?

My immediate thought is that a corrolary to this would be that if we took your example, and then gave the two cars the same horsepower at a given instant then they will accelerate at the same rate, correct?

If so, then that would mean that even if these two cars happen to be identical (say two stock M3's of the same specs), and one car were in 1st gear and one were in 4th gear, then they'd accelerate at the same rate for some given RPM. But obviously that makes no sense. So what I missing?

You're missing what I didn't originally state, but have now added to my post. Sorry.

The missing item was that the two cars are side by side, i.e., the same speed.

In that case, the car making better power to weight will accelerate harder than the other car, and, again, torque and gearing simply don't matter. Better power to weight is the only thing that matters at any given instant, ignoring and differences in rotational inertia, wind resistance, etc.

In fact, you can look at horsepower as a great simplifier in this context. You can mess about with torque at the drive wheels (as I did for years, calling it "the Dunderbex factor", because it sound vaguely technical ), but you don't need to bother. Horsepower and weight are the factors that actually matter at any given instant in time.

[Serious]

Originally Posted by Dodge2Dub

What is the actual difference between the wheel torque calculated on a Dyno (in the hundreds) and the actual wheel torque (force) calculated by looking at the final gear ratio, axle ratio, tire circumference, speed, etc. (in the thousands)? Is the dyno wheel torque figure the force that is exerted on the dyno wheel and the actual wheel torque the force exhibited by the car to the wheel itself?

Basically, is the dyno number the true wheel torque and the calculated number (in the thousands) the force. Shouldn't we differentiate between the two instead of calling both of them the wheel torque (ie. Dyno = Wheel Torque and Calculated Tq = Wheel Force)?

Torque is the tendency of a force to rotate the body to which it is applied.Torque is always specified with regard to the axis of rotation. It is equal to the magnitude of the component of the force lying in the plane perpendicular to the axis of rotation, multiplied by the shortest distance between the axis and the direction of the force component. Torque is the force that affects rotational motion; the greater the torque, the greater the change in this motion.

Whereas,force is the agency that alters the direction, speed, or shape that a body would exhibit in the absence of any external influence.It is a vector quantity, having both magnitude and direction. Force is commonly explained in terms of Newton's laws of motion. All known natural forces can be traced to the fundamental interactions. Force is measured in newtons (N).

A dyno (drum style dyno like a dynojet) measures lbft at the wheels and then uses input from the operator to divide to measured torque at the wheels to calculate lbft at the flywheel.

If the gearing is input incorrectly dyno numbers will be skewed. This is one reason most dyno operators use 1:1 gears.

[bruce.augenstein@comcast.]

Originally Posted by MVF4Rrider

It's hard to accept because it's incorrect. You can't explain acceleration without gearing (among other things). Horsepower is just the idea of engine power in terms of a correlation of torque and rpms. You still need to deliver that power to the ground which is gearing, and it's measured in lb-ft in the thousands (not hundreds). It's the compilation of multiplied torque across the rpm range [in use per gear] over time and distance which explains acceleration. You can leave horsepower out of the argument.

As I just mentioned, I omitted the key phrase indicating that the two cars have to be at the same speed. In that case, my statement that power to weight wins is 100% true, and as I said, torque and gearing are immaterial.

It's not that you can't duplicate an acceleration curve using torque and gearing. As I just mentioned, I did that for years, before I learned that power is the great shorthand in that context.

[Dodge2Dub]

Originally Posted by Serious

A dyno (drum style dyno like a dynojet) measures lbft at the wheels and then uses input from the operator to divide to measured torque at the wheels to calculate lbft at the flywheel.

If the gearing is input incorrectly dyno numbers will be skewed. This is one reason most dyno operators use 1:1 gears.

If we know the wheel torque through the dyno, why do we care about the wheel force? Separately, if wheel force is what actually matters, why do we even bother with reporting out and comparing wheel torque ratings? Is it just an easy way of saying my car makes X power vs. having to say my car makes X wheel force at Y speed in Z gear?....just trying to understand.

[Serious]

Originally Posted by bruce.augenstein@comcast.

You're missing what I didn't originally state, but have now added to my post. Sorry.

The missing item was that the two cars are side by side, i.e., the same speed.

In that case, the car making better power to weight will accelerate harder than the other car, and, again, torque and gearing simply don't matter. Better power to weight is the only thing that matters at any given instant, ignoring and differences in rotational inertia, wind resistance, etc.

In fact, you can look at horsepower as a great simplifier in this context. You can mess about with torque at the drive wheels (as I did for years, calling it "the Dunderbex factor", because it sound vaguely technical ), but you don't need to bother. Horsepower and weight are the factors that actually matter at any given instant in time.

sorry but you're still wrong.

Take an electric motor powered which makes constant torque throughout the rpm range, pair it to a normal 6spd gearbox (w/ theoretical instant shifts)& differential and do a 30mph rolling race between two equal cars.

Car #1 starts in 5th gear and the 2nd car starts in 2nd gear... the torque transmitted to the wheels of the 2nd car will be significantly higher and thus the acceleration will be superior to the first car.

[Serious]

Originally Posted by Dodge2Dub

If we know the wheel torque through the dyno, why do we care about the wheel force? Separately, if wheel force is what actually matters, why do we even bother with reporting out and comparing wheel torque ratings? Is it just an easy way of saying my car makes X power vs. having to say my car makes X wheel force at Y speed in Z gear?....just trying to understand.

Sorry I should've been more clear... the inertia dyno works by measuring the acceleration of the drum and then using newton's second law F=m*a to calculate force (mass of drum is known).

Then it actually calculates work by force * distance (circumference of drum * rotations) and then converts work into HP through power=work/time (all factoring in correct unit conversions). Torque is then calculated through HP=(torque * RPM/5252).

So technically an inertia style dyno actually calculates power and then derives torque.

[Dodge2Dub]

Originally Posted by Serious

Sorry I should've been more clear... the inertia dyno works by measuring the acceleration of the drum and then using newton's second law F=m*a to calculate force (mass of drum is known).

Then it actually calculates work by force * distance (circumference of drum * rotations) and then converts work into HP through power=work/time (all factoring in correct unit conversions). Torque is then calculated through HP=(torque * RPM/5252).

So technically an inertia style dyno actually calculates power and then derives torque.

Thanks, that makes sense to me. So, we never have to default to the wheel force if we have the dyno run handy when performing bench test comparisons....which leads to the question of why we are talking about wheel force when it is directly correlated to the measured torque on an inertia style dyno?

[Serious]

Originally Posted by Dodge2Dub

Thanks, that makes sense to me. So, we never have to default to the wheel force if we have the dyno run handy when performing bench test comparisons....which leads to the question of why we are talking about wheel force when it is directly correlated to the measured torque on an inertia style dyno?

I believe it's because force is easier to understand in terms of linear acceleration (ie we know the mass of the car and it's rate of acceleration thus we know the total force necessary to propel it) whereas torque is in terms of a rotating object (either a wheel & tire or flywheel depending on how you approach it).

In reality all this thread is a big unit conversion circle jerk. If you know some 100 level physics equations you can get force, power, torque from any of the other known quantities... it's just what makes the most sense and is the easiest concept to understand to each individual person, but that doesn't mean you can break the laws of physics (as some have tried in this thread) and say that power is torque or is a force; they can be found and converted into each other given a bit of additional information but they aren't all the same thing.

[bruce.augenstein@comcast.]

Originally Posted by Serious

Torque is not a force. Torque is force * radius.

More HP doesn't = more torque at the wheels, in fact it's usually quite the opposite. HP is just (torque *engine speed)/constant so, you can get the same number of HP'z by either having high torque and low engine speed or high engine speed and low torque.

You're last statement is completely untrue. Imagine riding a multiple speed bicycle. Now you're legs only strong enough to make a specific force, that force is imparted through the lever length of the crankshaft & pedal assembly to create a torque... that torque is then multiplied a specific amount depending on what gear you're in.

(low gears you pedal crazy fast but don't have to use much strength; whereas in high gears you pedal slowly but use a lot of energy pushing the pedals.)

Now are you really going to tell me you will be just as fast up a steep hill in 18th gear when you can't even push the pedals as you are in 3rd? Or that how strong the biker is irrelevant? Not a chance.This is why gearing is matters, and torque is an absolutely critical measurement in analyzing car acceleration.

Where most of you are hung up is that you only look at peak HP & torque numbers instead of area's under the curves and don't understand the relationship between forces, torque, work, power and how they interrelate to each other in terms of physics.

With the caveat that I forgot to mention same speed for the two cars in my note (now changed), my statement stands.

Your example is lovely (really), but the fact is that you're going to have to make a bunch more torque to the pedals in 18th gear in order to make the same power, because rpm is down. Capiche?

Think of a giant truck engine making 2950 pound feet of torque at 390 rpm, matched up against an M3 at 3900 alongside. Both vehicles are making 219 HP at that speed, and, given similar weight (think eight sumo wrestlers aboard the bimmer), they'll accelerate the same.

[bruce.augenstein@comcast.]

Originally Posted by 2m3

That is also a quote from Ray Korman that I read in 1996.

I claim dibs, since I wrote mine somewhere in '93, or early '94, and the website proliferation began almost immediately.

[bruce.augenstein@comcast.]

Originally Posted by Serious

A dyno (drum style dyno like a dynojet) measures lbft at the wheels and then uses input from the operator to divide to measured torque at the wheels to calculate lbft at the flywheel.

If the gearing is input incorrectly dyno numbers will be skewed. This is one reason most dyno operators use 1:1 gears.

This is completely incorrect. A dynojet cares only about the acceleration of the drum and the speed. That's all you need to calculate power since the rotational inertia of the drum is a known quantity. Typically, you'll get very close to identical numbers, regardless of the gear. Any variance will be due to tire slippage, the change in rolling resistance as speeds climb, and transmission efficiency which varies slightly by gear.

As a proof point, if the operators can't find a spark lead pickup to get rpm, they'll still generate a power graph. Torque won't be included because the unit will not be able to calculate it from rpm.

[bruce.augenstein@comcast.]

Originally Posted by Serious

sorry but you're still wrong.

Take an electric motor powered which makes constant torque throughout the rpm range, pair it to a normal 6spd gearbox (w/ theoretical instant shifts)& differential and do a 30mph rolling race between two equal cars.

Car #1 starts in 5th gear and the 2nd car starts in 2nd gear... the torque transmitted to the wheels of the 2nd car will be significantly higher and thus the acceleration will be superior to the first car.

Every so often I have to get into explaning the basics, yet again.

Your analogy breaks down completely, since the the 2nd gear vehicle is turning much greater rpm than the 5th gear car, thereby making much higher power at the same torque level, at the same speed.

I'm dead serious, Serious. You need to think long and hard about this.

Much more of this and I'm threatening to call Swamp or lucid in.

[bruce.augenstein@comcast.]

Originally Posted by Serious

Sorry I should've been more clear... the inertia dyno works by measuring the acceleration of the drum and then using newton's second law F=m*a to calculate force (mass of drum is known).

Then it actually calculates work by force * distance (circumference of drum * rotations) and then converts work into HP through power=work/time (all factoring in correct unit conversions). Torque is then calculated through HP=(torque * RPM/5252).

So technically an inertia style dyno actually calculates power and then derives torque.

Agree 100%

[scalbert]

Originally Posted by Serious

(low gears you pedal crazy fast but don't have to use much strength; whereas in high gears you pedal slowly but use a lot of energy pushing the pedals.)

IMO, and is often shared, you want to maintain pedal speed in the 70 - 90 RPM range regardless of road speed. That is why the gears are used; to maintain cadence but vary speed. So, no, you don't pedal crazy fast in the low gears and slower in the higher gears. You pedal the same and let gearing do what it is supposed to do.

I understand what you were tryiong to point out though but wanted to clarify.

[swamp2]

Originally Posted by Edward

BTW, someone needs to throw in a real engineer's anwer to torque multiplication. I'm no engineer or physicist, but I do realize that the gearing on the M3 allows it to deliver much more torque to the rear tires than the same amount of torque in a 335 (if it had the same engine flywheel torque) would deliver through the powertrain.

Hmmm if I understand your point you want the short simple formula based way to see/understand gearing torque multiplication (under approximations suitable to the discussion)? We could write the equation for rear wheel torque but ultimately it is vehicle acceleration anyone cares about. If two cars have the same wheel size and weight then you can just compare rear wheel torque to rear wheel torque.

a = (loss x g x FD x Te)/(rw x m)

This simply comes from F = m x a and T = r x F (torque = radius x force) - basically Newton's law and the definition of torque. In plain english:

accelerative force at rear wheels =( drivetrain loss (torque loss) x gear ratio x final drive ratio x engine torque )/ ( rear wheel radius x vehicle mass )

(Note: you must use ALL SI units for these formulae if you want to put some numbers in - plug in common US units and you will get nonsensically sized results)

Of course some terms are rpm dependent such at Te and even the loss, but the key factor here for understanding torque multiplication is simply the g and FD terms. g as well is gear dependent (obviously).

Just for example:

1st gear M3: g x FD = 4.78 x 3.15 = 15.1
1st gear 335i: g x FD = 4.06 x 3.08 = 12.5

That is a whopping 21% advantage for the M3. A 335i needs about 360 ft lb just to make up the difference in 1st gear. Many other posts have plotted this out across gear and across rpm to see the entire picture. This has been done for a variety of cars not just the 335i and M3 actually right here in the forum. Without more graphs, which seem to scare some folks off this should suffice to demonstrate the principal.

Lastly as Bruce always points out you don't really need to do this calculation anyway. As hp more or less does the calculation for you. It it not perfect nor precise but engine torque combined with gearing is basically hp. Here we don't mean mathematically, just from understanding the performance of a car.

Hope that is useful to some folks.

[Sedan_Clan]

Originally Posted by swamp2

Hmmm if I understand your point you want the short simple formula based way to see/understand gearing torque multiplication (under approximations suitable to the discussion)? We could write the equation for rear wheel torque but ultimately it is vehicle acceleration anyone cares about. If two cars have the same wheel size and weight then you can just compare rear wheel torque to rear wheel torque.

a = (loss x g x FD x Te)/(rw x m)

This simply comes from F = m x a and T = r x F (torque = radius x force) - basically Newton's law and the definition of torque. In plain english:

accelerative force at rear wheels =( drivetrain loss (torque loss) x gear ratio x final drive ratio x engine torque )/ ( rear wheel radius x vehicle mass )

(Note: you must use ALL SI units for these formulae if you want to put some numbers in - plug in common US units and you will get nonsensically sized results)

Of course some terms are rpm dependent such at Te and even the loss, but the key factor here for understanding torque multiplication is simply the g and FD terms. g as well is gear dependent (obviously).

Just for example:

1st gear M3: g x FD = 4.78 x 3.15 = 15.1
1st gear 335i: g x FD = 4.06 x 3.08 = 12.5

That is a whopping 21% advantage for the M3. A 335i needs about 360 ft lb just to make up the difference in 1st gear. Many other posts have plotted this out across gear and across rpm to see the entire picture. This has been done for a variety of cars not just the 335i and M3 actually right here in the forum. Without more graphs, which seem to scare some folks off this should suffice to demonstrate the principal.

Lastly as Bruce always points out you don't really need to do this calculation anyway. As hp more or less does the calculation for you. It it not perfect nor precise but engine torque combined with gearing is basically hp. Here we don't mean mathematically, just from understanding the performance of a car.

Hope that is useful to some folks.

Thanks swamp. Wow, so in reality, the 335i doesn't really have the advantage in any gear.

[swamp2]

Guys, Bruce is correct here. It indeed takes some serious thought for "car guys" to see it is true. Most of you will not like the simplest proof I will offer but it is simply that:

P = F x v

Power is force x velocity. So two systems at the same speed (their speed, not their wheels, their engines, or anything else) will have a accelerative force that is directly related to the power the system can exert at that speed (power is absolutely a function of speed by the way in a car - probably obvious but worth stating).

This works for a bicycle, a diesel semi, a M3, an electric car, whatever. It is physics and physics is simply correct and predictive.

Here is another way to think about this that complements the above, since few will "buy" the proof above. Contrary to some rumblings in some prior post formulae and physics are VERY meaningful and precise and as long as folks understand and agree on equations they greatly enhance understanding. They can greatly increase confusion as well...

This alternate way to understand is back to gearing. If you floor your car at a low speed in a high gear you get very poor acceleration - it "bogs" as we say. If at the exact same speed you simply put the car into the lowest gear possible (with just a bit of rpm left before redline of course) the car accelerates much more, probably drastically more. Why does the car accelerate so much better? It is because you force the engine rpm higher which makes the engine produce much more power and there you have it. Same speed more power. Do recall that power = torque x rpm (the extra 5252 factor you typically see is just for our terrible english units...).

You can also think in terms of torque multiplication as I talked about just above. The lower gear makes much more torque multiplication. Thus more torque and more force at the drive wheels. Even if your torque curve was dead flat the lower gear and accompanying higher rpm gives you more power.

What tends to complicate this is the "at the same velocity" requirement. We all know as any race or comparison proceeds the two vehicles will not always be at the same speed at the same time. It would be awfully boring if they were. Thus you can only really use this argument for a whole series of imagined rolling drag races each time with the contenders starting at the exact same speed (gear selection and hence power level up to the driver). This can be repeated over and over at slightly increasing starting speeds. But as soon as one contender is going much faster than the other you can not use the basic relationship.

To get around this minor limitation (and the complications of different torque and power curves, and gearing and wheel sizes and redlines and drivetrain loss, etc., etc. etc.) one must actually do a full "transient analysis" and compute the actual force at the wheels, the losses and resistive drag force and the resulting instantaneous acceleration, the resultant increase in velocity from that instantaneous acceleration and then repeat this in very small time steps. This is what CarTest and similar software tools accomplish and that is where my results came from many posts back.

Think hard, ponder it, you will have so much more insight into how and why cars perform and which things are key for more performance (well straight line performance at least...).

The next challenge is final drive modifications. If you also understand why final drive modifications typically increase in gear results but not results across multiple gear you really get it.