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Old 12-26-08, 02:05 AM   #46 (permalink)
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Thanks for the nit-picks Helps with Accuracy

Joule isn't a unit of power and i never said it was...I said joule was a unit...that unit with a direction is called a watt.

Joule is equal to...

Quote:
Originally Posted by Wiki
work done to produce power of one watt continuously for one second; or one watt second (compare kilowatt hour), with the symbol Ws. Thus a kilowatt hour is 3,600,000 joules or 3.6 megajoules
Quote:
Originally Posted by B+O
it's pretty much energy capable of being done in an amount of time, just like a Watt (or similarly a Joule with direction)
I should have said Joule with a direction and Time frame of 1 second... My bad on that one.
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Old 12-26-08, 02:15 AM   #47 (permalink)
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Quote:
Originally Posted by sap View Post
Just a bit of friendly nitpicking on a good sticky Some of my points are admittedly pedantic. I won't claim to have perfect knowledge on this, so any discussion and/or disagreements are welcome.



The units are actually Newton meters, kilogram force meters, and pound feet. It's a product of how much force you apply and how far away it's being applied. I'd also suggest specifying kilograms force (kgf) meter instead of kilogram meter, since a kilogram is usually unambiguously a mass.

I explain this below in my comments

Edit: I see this has been discussed before ... but reading through I can't even tell who was arguing what The torque produced is a product of a distance and a force, not a ratio. Because of this there should be no "per" in the unit name and the order of the distance and force units is not significant (though they could be strongly ingrained, I know I would never say "meter Newtons"). This is consistent with 5ft x 100 lbs = 500 ft. lbs.

same as previous

A Joule, besides being a scalar quantity and not having direction, isn't a unit of power.

Answered in previous post

I think this would be more clear if you just stuck to defining power as an amount of work done per unit time, leaving energy out of it. Purely opinion, of course. Also, as noted above for Joule, neither energy nor work have direction.



In this sample, and the one in the previous paragraph, it should be "the ability to lift at constant speed" as opposed to "the ability to move" since the force applied is what's used in calculating work. By lifting 550 lbs without accelerating it you're actually applying 550 lbs. of force, which isn't the case for pushing a 550 lb. block.

this is true as a constant downward force would be the only way to replicate...however, in keeping with simplicity i neglected to make this point...it is, however understood, thank you for making sure it was poignantly clear


No direction. Work done or energy produced/transferred over a set time gives power.

Mentioned in previous post

I think "the work done per second increases" would be more clear (emphasizing that power is the rate of doing work).

makes sense, but i believe that was already established as the time frame and that i made certain that work was a function of time and force



answered below:
I think this is missing a key opportunity to introduce gearing, and being able to put more torque to the axles than you have at the flywheel. It's not only saving shift times that makes high revving engines fast.



Gearing comes into play here as well. If you gear that 19000 RPM down to 9500 RPM you're now getting 525 ft.lbs of torque ...


Maybe a note that torque at the wheels is related to force on the road by the wheel radius?



If i wanted to get into gear multiplication i would have in another thread...let's not confuse the two issues...

we're speaking specifically of Engine HP/Torque...

Complicating the matters with Torque Multiplication and Mechanical Advantage will just confuse most of the people looking for a simple explanation of torque and hp.

And I'm not understanding your "Gear down to 9500rpm to 525lbs/ft"

If you gear an engine higher (that is to reduce, numerically) you've REDUCED Mechanical advantage and thereby REDUCED power to the roade, so you power output would actually be LESS... if we were specifically talking about a 1:1 drive ratio.

This is the sole purpose of a transmission. you're delivering less than the engine's power in an "overdrive" situation. Underdrive you're delivering MORE power to the wheels.


UNITS:

(and i corrected your labeling because earlier in the thread I was using incomplete explanations of why it's Lbs/ft)

as a matter of fact an engine's power is determined 1 foot away from its rotating axis, that would be the Crankshaft. If you used the units of force ft.lbs instead of the units of torque lbs/ft, an engine would get more powerful the further away from the crankshaft you got.

just the same as brakes us ft.lbs as a unit, working in an opposite direction, you'd LOSE torque...so in a simple Machine an axle/wheel combination would be lbs/ft because as you increase the radius of the wheel (the Feetses), you reduce the force (the "poundage") delivered to the outer most surface...using ft.lbs to denote that would be wrong as you're calculating the mechanical advantage a lever has on a fulcrum versus the fulcrum has on a lever.

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Old 12-26-08, 12:51 PM   #48 (permalink)
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Quote:
Originally Posted by B+O View Post
UNITS:

(and i corrected your labeling because earlier in the thread I was using incomplete explanations of why it's Lbs/ft)

as a matter of fact an engine's power is determined 1 foot away from its rotating axis, that would be the Crankshaft. If you used the units of force ft.lbs instead of the units of torque lbs/ft, an engine would get more powerful the further away from the crankshaft you got.
And it would, just like you produce more torque with a breaker bar you do more work with a longer stroke.

Quote:
just the same as brakes us ft.lbs as a unit, working in an opposite direction, you'd LOSE torque...so in a simple Machine an axle/wheel combination would be lbs/ft because as you increase the radius of the wheel (the Feetses), you reduce the force (the "poundage") delivered to the outer most surface...using ft.lbs to denote that would be wrong as you're calculating the mechanical advantage a lever has on a fulcrum versus the fulcrum has on a lever.
You don't lose torque with a bigger wheel (well, you can make an argument about angular acceleration of a larger wheel taking more torque, but that's not really relevant to this discussion). The toque is the same no matter what size the wheel is and you lose force on the ground because you're trying to apply your torque at a further distance. The relationship for force on the ground is torque/radius, but that still means torque = radius * force.
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Old 12-26-08, 01:05 PM   #49 (permalink)
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Quote:
Originally Posted by B+O View Post
If i wanted to get into gear multiplication i would have in another thread...let's not confuse the two issues...

we're speaking specifically of Engine HP/Torque...

Complicating the matters with Torque Multiplication and Mechanical Advantage will just confuse most of the people looking for a simple explanation of torque and hp.
That's fair enough, but making torque at high RPM does have more advantage than not having to shift as quickly (see below).

Quote:
And I'm not understanding your "Gear down to 9500rpm to 525lbs/ft"

If you gear an engine higher (that is to reduce, numerically) you've REDUCED Mechanical advantage and thereby REDUCED power to the roade, so you power output would actually be LESS... if we were specifically talking about a 1:1 drive ratio.
You don't put power to the road, nor torque for that matter (I know what you mean though, just one of those things I can't stop myself from pointing out).

What I mean is that you get twice the torque out of the output shaft if the speed is reduced 2:1 through a reducer, and the power output is unaffected. So if you've got an engine running at 19000 RPM driving a gearbox with an output shaft at 9500 RPM you get twice the torque on the output shaft that you had on the engine.

Quote:
This is the sole purpose of a transmission. you're delivering less than the engine's power in an "overdrive" situation. Underdrive you're delivering MORE power to the wheels.
A gear reduction, as I understand it, is to have an output of a lower speed than the input, but that might be a notation issue on my behalf. If you have a slower output shaft you've got more torque available. My understanding is that the underdrive gears are the ones reduced.
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Old 12-26-08, 01:11 PM   #50 (permalink)
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Quote:
Originally Posted by B+O View Post
Thanks for the nit-picks Helps with Accuracy

Joule isn't a unit of power and i never said it was...I said joule was a unit...that unit with a direction is called a watt.

Joule is equal to...

I should have said Joule with a direction and Time frame of 1 second... My bad on that one.
Yeah, a Joule in a unit time is exactly what I meant.

Energy and power don't have direction though, they're strictly scalar quantities.
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Old 12-26-08, 04:13 PM   #51 (permalink)
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Quote:
Originally Posted by sap View Post
And it would, just like you produce more torque with a breaker bar you do more work with a longer stroke.



You don't lose torque with a bigger wheel (well, you can make an argument about angular acceleration of a larger wheel taking more torque, but that's not really relevant to this discussion). The toque is the same no matter what size the wheel is and you lose force on the ground because you're trying to apply your torque at a further distance. The relationship for force on the ground is torque/radius, but that still means torque = radius * force.
Here's the thing about "angular acceleration"

It's all but irrelevant when you're speaking about Vehicle Acceleration. The main cause of reduced acceleration when someone upgrades to larger wheels is the rolling diameter...this is because it's a reverse lever action...

You can play with theoretical formula, but I normally deal with real world physics on a daily basis...The torque application in true form emanates from the fulcrum. Regardless of what each piston is doing, the final full output of the engine before it runs through the driveline is measured, as a unit, from the fulcrum of its rotating assembly.


You brought out the breaker bar, and that's what i was hoping you'd do...In true form, this is an example of reverse actuation the fulcrum is the rotating point, not the Breakerbar which is applying a twisting force. the Torque originates from a lever action of the connecting rods. now if you were to calculate the power from EACH CYLINDER based on the explosive content of the mixture then yea, you'd use Ft.Lbs because the downward pressure is what rotates the crankshaft in relation to the distance from the crank centerline (offset of the rod bearing)...

as you measure an engines total explosive output, that output is denoted as a unit of torque from the fulcrum to a lever. Not a Lever to a fulcrum... because, in essence, you have multiple explosions happening at one time and there is always a variation in the output. if you were going to use the measurement Ft.Lbs, you'd be better off denoting it as the Average of Ft.lbs per number of cylinders at a certain RPM. because you'd be calculating the downward force of those individual explosions...

the engine itself, simplified to a simple machine in a dynamic process, doesn't posses a lever. This machine produces a twisting force the sum of its multiple levers, and as such, this machine will then be rated in lbs/ft because its force eminates from the fulcrum, versus this machine being a Lever and fulcrum.

The farther away you go from the crank shaft the easier it is to stall the engine if you were to..."toss a stick in the spokes" if you will....


EDIT:

Hope that made sense...I was half trying to relax after a really long day and keep my brain on boil for the last couple of hours

Last edited by B+O; 12-26-08 at 04:25 PM.
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Old 12-27-08, 01:37 AM   #52 (permalink)
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Quote:
Originally Posted by B+O View Post
Here's the thing about "angular acceleration"

It's all but irrelevant when you're speaking about Vehicle Acceleration. The main cause of reduced acceleration when someone upgrades to larger wheels is the rolling diameter...this is because it's a reverse lever action...
Agreed, sometimes I just can't make a statement without making a disclaimer

Quote:
You can play with theoretical formula, but I normally deal with real world physics on a daily basis...The torque application in true form emanates from the fulcrum. Regardless of what each piston is doing, the final full output of the engine before it runs through the driveline is measured, as a unit, from the fulcrum of its rotating assembly.

You brought out the breaker bar, and that's what i was hoping you'd do...In true form, this is an example of reverse actuation the fulcrum is the rotating point, not the Breakerbar which is applying a twisting force. the Torque originates from a lever action of the connecting rods. now if you were to calculate the power from EACH CYLINDER based on the explosive content of the mixture then yea, you'd use Ft.Lbs because the downward pressure is what rotates the crankshaft in relation to the distance from the crank centerline (offset of the rod bearing)...
Right, each piston exerts a downward force through the connecting rod to the crankshaft.

You're not the only one who plays w/ real world physics on a daily basis, though I'll admit this isn't specifically the area my day-to-day deals with

Quote:
as you measure an engines total explosive output, that output is denoted as a unit of torque from the fulcrum to a lever. Not a Lever to a fulcrum...
I disagree with this. The output of the engine produces a force on the crankshaft which acts as a lever around the fulcrum, with the net result being a twisting moment about the fulcrum. The force on the crankshaft from a cylinder is just like a force on a breaker bar (or six pulsing forces on connected breaker bars if we want to look at every cylinder).

Torque doesn't act from fulcrum to lever, or vice versa, it only acts as a twisting about the fulcrum (the vector representation is perpendicular to the plane passing through the force vector and the moment arm vector). 500 ft.lbs of torque produces a 100 lb force if you're 5 ft away, 100 lbs of force on a 5 ft breaker bar produces 500 ft. lbs.

Quote:
because, in essence, you have multiple explosions happening at one time and there is always a variation in the output. if you were going to use the measurement Ft.Lbs, you'd be better off denoting it as the Average of Ft.lbs per number of cylinders at a certain RPM. because you'd be calculating the downward force of those individual explosions...
The output of the engine is varying, but it's also pretty heavily damped by the rotating mass. Measured torque is essentially an average of the torque produced by those 6 varying downward forces.

Quote:
the engine itself, simplified to a simple machine in a dynamic process, doesn't posses a lever. This machine produces a twisting force the sum of its multiple levers, and as such, this machine will then be rated in lbs/ft because its force eminates from the fulcrum, versus this machine being a Lever and fulcrum.

The farther away you go from the crank shaft the easier it is to stall the engine if you were to..."toss a stick in the spokes" if you will....
Once you've got a value for torque output for the engine you can completely ignore the piston-crankshaft activities inside the engine. All you've got is a shaft turning at a certain speed and producing a certain amount of torque. When you attach this shaft to a tire the tire becomes the lever to put a force on the ground. Make the tire taller and you've got a longer lever and the same amount of torque, which means less force on the ground.

Let me bring back your earlier example. If I write out the solution it clearly works out with units on the torque of ft.lbs. on the torque. If you're up for it I'll reverse your challenge and ask you to do the same



This is also perfectly consistent with your "stick in the spokes" example. The stick is some given force at a distance from the fulcrum. To stop the wheel turning you have to either hold the stick with enough force, or hold it far enough away, to produce an equal countering torque. Increasing either the distance or the force increases the torque, because the torque is proportional to both the force and the distance. In order to have units of lbs./ft. it would have to be inversely proportional to the distance.

Quote:
EDIT:

Hope that made sense...I was half trying to relax after a really long day and keep my brain on boil for the last couple of hours
Heh. You're making sense, and I'm enjoying the conversation, but you should know that I'm as close to absolutely certain of this as I think I'm capable of

I'm glad you don't seem to be bothered by having this discussion over again. I'm always concerned about coming across as argumentative or disagreeable when I'm talking with people who are consistently reasonable (like yourself).
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Old 12-27-08, 02:01 AM   #53 (permalink)
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As an aside, does anyone have a source I can check that would agree with torque units being force per unit distance?

Despite noting that I'm fairly certain, the fact is that I'm *never* so certain that I can't be swayed by reliable sources. Everything I've checked so far (including one textbook I happened to have sitting at home) agrees with force times distance.
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Old 01-02-09, 02:29 PM   #54 (permalink)
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He he he he, I knew this would make it start all over again.
My point was that with cars like the legend you won't get anywhere with only one of them.... I mean you need both torque and HP, but the high Torque rating is typical for really heavy vehicles because it needs power to pull the many kiloes/pounds but they don't have as high an amount of horsepowers, so they don't go that fast.
I have a big diesel company car and it has a very big amount of torque(380 NM) but only (140 din HP), so it will start off with a hard pull and then it will quickly reach it's max power and therefore it won't be that fast.
My 6-spd Legend Coupe, only has a torque of 279 nm and 233 din HP and it won't have the same quick hard pull but it will have power all the way to the top in a continuesly stronger and stronger amount of power. So it makes my Legend way faster than the high torque rated car.

That was my point.
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Old 01-02-09, 02:55 PM   #55 (permalink)
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Originally Posted by sap View Post
I haven't read through this thread, but Gothche seems either terribly confused or really bad at explaining things.

The only really great explanation of HP and torque I've ever found is this page



Generally speaking, kilograms are a measure of mass and pounds are a measure of force (weight is a force). It just so happens that over the surface of the earth there's a fairly constant proportion between weight and mass (it's called gravity) so we tend to interchange them. Technically if I step on a bathroom scale I'm measuring my weight (force) directly, and if I step on a balance scale I'm measuring my mass directly.

It's common to see imperial mass measured in "pounds mass", with 1 lbm. equal to the mass that weighs 1 lbf (pound force, at sea level IIRC). Conversely there is a kilogram of force, which is the weight of 1 kg.. The only time a pound and a kilogram are measures of the same property is if you're comparing lbf to kgf, or kg to lbm. It'd be very helpful if lbf. and lbm. were specified when talking about "pounds".

The pounds portion of torque is lbf.



Horsepower is simply a combination measure of how fast the engine is turning and how much torque it's making. 200 ft.lbs at 6000 RPM is more power than 200 ft.lbs at 5000RPM. That's all there is to it. In speed applications torque at higher RPM is more valuable than torque at lower RPM because you can gear down the higher RPMs to get more force to the ground, and this is why people compare HP.



What are your definition of "faster" and "quickest". If "faster" is in terms of max speed, see below.

If we're talking about acceleration it's much more complex. The entire torque vs. RPM curve comes into play, but generally speaking you'll be putting more force on the ground if you can keep producing torque up to a higher RPM (which means higher HP).



No, no, no, no, no.

The maximum speed is determined by how fast you can get the engine to rev and the gearing. If you have two cars that both rev to 6000 with identical gearing (and can get to redline in the highest gear) they'll have exactly the same top speed, even if one has more horsepower. Higher horsepower engines are usually higher revving, so picking it to have a better top speed is reasonable, but HP is not the max speed available.





Torque has nothing to do with how efficiently you're using power. Nothing.



This is true, but poorly put. They need to produce lots of torque at low RPM. The tradeoff to this is their ability to rev up, which simultaneously limits their top speed and power. Being able to run an engine at high RPM is good for both torque and top speed. They don't miss out on top speed because of low HP, they miss out on both HP and top speed because they don't rev well.

The water wheel example on the page I linked at the start of this post is a fantastic example of what you're trying to show.



It's hard to explain because there aren't as many absolutes as people would like to believe. Making a pile of torque is useless if you can't rev your engine up to speed, making a pile of horsepower is useless if your low rpm torque isn't enough to get the car moving, but in reality every car is somewhere in between.



Just out of curiosity, what "another way" of fast is the difference between F1 and drag cars? I honestly don't know how much torque an F1 car produces, but putting a lot of force on the road is the goal in every type of racing.



How much horsepower you've got is a direct function of how much torque you can make and how high you can make it in the RPM band. If you increase torque across the band you're improving your horsepower as well.

Talk of HP vs. Torque isn't really silly, because there are very practical differences in driving depending on if you make your torque at high RPM or low RPM, and this is what people are really talking about in HP vs. Torque.
Had I just put more energy into explaining it well....
Yes you are right.
Your comment about the F1 cars and puttinh power to the ground is like this.
Because the car(F1) is very light it doesn't need the same amount of torque, that is why the F1's have less torque and much higher HP.
But thanks for explaining it so well..... yea, I'm not from an english speaking country so the meaning can get blurred out.

And it's not that simple about the reving.... Some turbo cars like the Rover 620ti produces 197HP(max) at 6000rpm and 174(max) lbs-ft at 2100rpm and it will accelerate much faster than our legend until 40-50mph because of that early max torque rating. That is because of the early torque peak not other way around. Our Legends have a powerband of 1200rpm(the amount of rpm's between the max torque and max power) compared to the Rover 620ti which has a powerband of 3900rpm, but the car won't keep the max output of torque all the way from 2100rpm till the max HP rating at 6000rpm, so the way the engine works is less stabile and less consistant. So even though the Rover is way faster till about 40-50mph our Legend type2 manual is still faster/quicker in the end(0-60mph).

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Old 01-02-09, 02:56 PM   #56 (permalink)
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Quote:
Originally Posted by Gothche View Post
He he he he, I knew this would make it start all over again.
My point was that with cars like the legend you won't get anywhere with only one of them.... I mean you need both torque and HP, but the high Torque rating is typical for really heavy vehicles because it needs power to pull the many kiloes/pounds but they don't have as high an amount of horsepowers, so they don't go that fast.
I have a big diesel company car and it has a very big amount of torque(380 NM) but only (140 din HP), so it will start off with a hard pull and then it will quickly reach it's max power and therefore it won't be that fast.
My 6-spd Legend Coupe, only has a torque of 279 nm and 233 din HP and it won't have the same quick hard pull but it will have power all the way to the top in a continuesly stronger and stronger amount of power. So it makes my Legend way faster than the high torque rated car.

That was my point.
Fair enough.

I guess I'm just of the opinion that we need to be clear and unambiguous in how things are described to avoid misleading people who are just trying to get an understanding of the subject matter. To that end I tend to start picking details apart even if I generally agree with the overall message. Reading back, I also tend to be a bit of a jerk

Not that I'm so certain that I'm actually helping more than I'm hindering the effort
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Old 01-02-09, 03:05 PM   #57 (permalink)
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Excellent info!! I'm soaking it in!
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Old 01-02-09, 03:20 PM   #58 (permalink)
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you guys are going in circles. debating theory. and others coming in the debate practice, and/or specific application.

b+o good original post.
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Old 01-02-09, 03:30 PM   #59 (permalink)
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Quote:
Originally Posted by Gothche View Post
But thanks for explaining it so well..... yea, I'm not from an english speaking country so the meaning can get blurred out.
Heh, your english is good enough that I'm being every bit as hard on you as I would be on a native speaker, so I'd say you're doing fairly well. If your location wasn't listed as Denmark I wouldn't have suspected that it wasn't your native tongue.

Back on the F1 car topic (I'm really sorry if everyone's sick of the talk, I can't help myself ), I think F1 cars are geared down a lot which helps them put more force to the ground in lieu of having more torque at the flywheel. I found some back of the envelope calculations that estimate a speed of 295 km/h at 18500 RPM in the highest gear (7th in this case). Accounting for engine speed through the transmission, final gearing, and to the ground from the tires, I'd have to shift my 6-speed down to 2nd gear to get gearing as good as they have in 7th, and they have 6-gears below that
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Old 01-02-09, 03:59 PM   #60 (permalink)
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Quote:
Originally Posted by dasficktauf View Post
you guys are going in circles. debating theory. and others coming in the debate practice, and/or specific application.

b+o good original post.
This is not theory it's the fact.
I drive three different types of cars pretty often.
My own 1994 Acura Legend LS Coupe 6-spd, my company car and a 1995 Rover 620ti 5-spd(Turbo) and they all work in completely different ways.
I have told you about the company car with the huge amount of torque but only 140HP(slow car with a really hard pull at the first meters), the Rover 620ti with an early torque peak and late HP peak(very fast all the way, but especially in the early rpm's), The Legend slow in the low rpms but very fast at the high rpms). The Legend is a late peaker.
Of all these cars the fastest from 0-60mph is the Legend, but the Rover is way quicker anywhere else in the lower speeds and from stops. That is very simple, it's the relatively high amount of torque at low rpm's and a high amount of HP at the late rpm's. The only problem with the Rover compared to the Legend type2 is that the Rover will loose a bit of momentum in between 2100rpm's and 6000rpm's because the torque will drop a bit(not as much as the HP drops after reaching it's max) before max HP is reached and it will have 33 HP less compared to the Legend. The Legend will start slow and then pull harder and harder all the way til past 6000rpm's, then when you shift gear(go up one gear) on a manual at about 6000rpm, it will only drop 600-800rpm's and still be in the high end of the powerband, so the power will be more consistantly all the way except when you start from a stop.

Last edited by Gothche; 01-02-09 at 04:13 PM.
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