M4 dyno test by Sportauto 465hp/577 nm

[Boss330]

Originally Posted by CanAutM3

Not quite. Power represents how fast work can be done . P = W/t

Torque is a rotating force applied. A given amount of torque applied on a given amount of revolutions equates to work. How fast the the revolutions on which the torque is applied are done equates to power.

I am just nitpicking over semantics here, I know you understand the concept better than the majority here .

Originally Posted by CanAutM3

Again here, just nitpicking, but you need to convert the minutes into seconds for the formula to work:

P = Torque x 2pi x Rpm / 60

P = Power in watt
Torque is in Nm

[bradleyland]

Originally Posted by swamp2

Although I agree with you in essence, I do also see how folks can get confused.

Thanks. I should be clear in my intent though. My goal isn't to dispatch with torque altogether. As you pointed out, many common formulae require a value for force, so working from power requires you to derive force. Obviously using only power results in needless calculus exercises.

Rather, my goal is to point out to the members of the Church of Torque that you can't simply compare two engines on the basis of torque alone (ignoring RPM), and once you start comparing torque and RPM, you might as well be comparing power.

Put more directly, torque and horsepower aren't mutually exclusive values, and shouldn't be treated as such.

[swamp2]

Originally Posted by CanAutM3

Not quite. Power represents how fast work can be done . P = W/t

Not exactly either. It is not how fast work can be done but simply the rate of change of work, which can be any value. Of course the formula you gave is only true when power is constant which isn't at all a terrible approximation under a WOT type of acceleration run in a car (an even better approximation for engines like the M3 S55). That being said, when power is dynamic i.e. a function of time the more general definition is simply the time derivative:

P = dW/dt

Then when power is constant it is technically the work done in any time interval divided by the duration of the time interval

P = W/Δt

[Boss330]

Originally Posted by swamp2

Not exactly either. It is not how fast work can be done but simply the rate of change of work, which can be any value. Of course the formula you gave is only true when power is constant which isn't at all a terrible approximation under a WOT type of acceleration run in a car (an even better approximation for engines like the M3 S55). That being said, when power is dynamic i.e. a function of time the more general definition is simply the time derivative:

P = dW/dt

Then when power is constant it is technically the work done in any time interval divided by the duration of the time interval

P = W/Δt

[grimlock]

Originally Posted by swamp2

Not exactly either. It is not how fast work can be done but simply the rate of change of work, which can be any value. Of course the formula you gave is only true when power is constant which isn't at all a terrible approximation under a WOT type of acceleration run in a car (an even better approximation for engines like the M3 S55). That being said, when power is dynamic i.e. a function of time the more general definition is simply the time derivative:

P = dW/dt

Then when power is constant it is technically the work done in any time interval divided by the duration of the time interval

P = W/Δt

W=P*t
P=w/t
dP=dw/dt

either I unknowingly have misunderstood physics+math all my life, I have no idea how you got that first equation, Swamp

why do you need to know dP (rate of change of power) anyways?
the total work done, or the area under a power/time chart, is simply P*T, or the integral of power wrt time.

[ASAP]

There already is a dynojet graph out... let's just say there will be a lot of e92 m3's on ebay soon lol.

[JoeFromPA]

I'm thinking between the reviews that are coming in and the initial dyno results, I'm very glad I placed an order beforehand. It won't be limited production like a 1M, but I wouldn't be surprised if early M3 values hold extremely high.

[regular guy]

First F80 M3 dyno's in new S55 dyno database:

http://www.s55dynos.com

[CanAutM3]

Originally Posted by regular guy

First F80 M3 dyno's in new S55 dyno database:

http://www.s55dynos.com

Thanks for this.

The SportAuto Maha dyno result should also be added to the database.

[swamp2]

Originally Posted by grimlock

either I unknowingly have misunderstood physics+math all my life, I have no idea how you got that first equation, Swamp

why do you need to know dP (rate of change of power) anyways?
the total work done, or the area under a power/time chart, is simply P*T, or the integral of power wrt time.

Probably not the best place for physics or math 101... However,

P = dW/dt

from which most of the other formula follow is not derived but a definition!

[regular guy]

Originally Posted by CanAutM3

Thanks for this.

The SportAuto Maha dyno result should also be added to the database.

I will get to that later today.

[bradleyland]

Originally Posted by grimlock

W=P*t
P=w/t
dP=dw/dt

either I unknowingly have misunderstood physics+math all my life, I have no idea how you got that first equation, Swamp

why do you need to know dP (rate of change of power) anyways?
the total work done, or the area under a power/time chart, is simply P*T, or the integral of power wrt time.

I'm not sure where the rate of change is getting pulled in to this, but P = ∆W/∆t is the formula for average power. The only caveat I'd add is that the P in this form of the formula is usually followed by a subscript "avg" to denote that P is not instantaneous, but an average.

It's not worth going down this rabbit hole though, or the next thing you know we'll be adding limits. When all is said and done, we'll spend all evening doing calculus. And as much fun as that sound like, I'm sure everyone else will be satisfied with accepting that we're all pretty much on the same page when it comes to the measure we're talking about

[swamp2]

Originally Posted by bradleyland

I'm sure everyone else will be satisfied with accepting that we're all pretty much on the same page when it comes to the measure we're talking about

Of course we are, as every such discussion eventually ends. That is except for the "torque rulz" diehard fans whom can never be convinced of reality and facts...

[CanAutM3]

Originally Posted by swamp2

Not exactly either. It is not how fast work can be done but simply the rate of change of work, which can be any value. Of course the formula you gave is only true when power is constant which isn't at all a terrible approximation under a WOT type of acceleration run in a car (an even better approximation for engines like the M3 S55). That being said, when power is dynamic i.e. a function of time the more general definition is simply the time derivative:

P = dW/dt

Then when power is constant it is technically the work done in any time interval divided by the duration of the time interval

P = W/Δt

I was trying to keep it simple since I am pretty sure that the majority of forum readers are not fully versed in differential equations.

I agree, P=dW/dt or Pavg=W/t would have been more mathematically accurate. But was it necessary in the context?

Thinking about it some more, I am not sure I agree with your definition. We are going deep into semantics here, but power is the "time rate of doing work" (Engineering Mechanics Vol2 Dynamics). The "rate of change" of the energy level of an object equates to power, not the rate of change of work itself. Work can be done at a constant rate (no change of rate at which work is being done) and power is not zero. Hence my definition of "how fast (or slowly) work is being done" is still appropriate (albeit a bit crude).

[bradleyland]

Originally Posted by CanAutM3

Thinking about it some more, I am not sure I agree with your definition. We are going deep into semantics here...

Yeah, I know this is really confusing, but "rate of change of work" is a pretty standard definition of power. I know it does't "sound right", but that's the way someone in the field of mathematics would describe the most basic definition of power. The reason is because it is the best method of describing the most basic formula for power:

P = W/t

"Work over time" is a "rate of change" from a calculus perspective. Don't think of it in practical terms, think of it in math terms lol.

[Audio Fan]

I don't think anyone is really saying torque in a vacuum is all that matters. When talking engine torque, most assume that the torque is applied accordingly. Most of us can feel the need for low end torque, high end power, and a good mix throughout for a car that shines throughout the rev range.

[CanAutM3]

Originally Posted by bradleyland

Yeah, I know this is really confusing, but "rate of change of work" is a pretty standard definition of power. I know it does't "sound right", but that's the way someone in the field of mathematics would describe the most basic definition of power. The reason is because it is the best method of describing the most basic formula for power:

P = W/t

"Work over time" is a "rate of change" from a calculus perspective. Don't think of it in practical terms, think of it in math terms lol.

Going through all my old engineering textbooks and even googling a little, those are the definitions of power that come up the most:

"the time rate of doing work"
"the rate at which work is done"
"the rate of energy transfer by work"
"the time rate at which work is done or energy is transferred"

But I could not find a single reference to "rate of change of work"...

But in the end, I think we all violently agree on the basic principle

[m981addicts]

Originally Posted by M4TW

I suspect that is more of a problem with w3rkn. That, or he is struggling with typing it.



Still. I don't see 400m racing taking off anytime soon.

Cuz USA got that one firmly under control...imagine that--imperial bodies dominating metric distances!

http://en.wikipedia.org/wiki/List_of...ts_(men)#400_m

[swamp2]

Those not interested in the gory details of work, energy and power please skip...

Originally Posted by CanAutM3

Thinking about it some more, I am not sure I agree with your definition. We are going deep into semantics here, but power is the "time rate of doing work" (Engineering Mechanics Vol2 Dynamics). The "rate of change" of the energy level of an object equates to power, not the rate of change of work itself. Work can be done at a constant rate (no change of rate at which work is being done) and power is not zero. Hence my definition of "how fast (or slowly) work is being done" is still appropriate (albeit a bit crude).

You don't agree or just want to BS and aren't sure? You can't have your cake and eat it too. Anyway, here we go...

Rate of change of (fill in your favorite quantity) is indeed as bradleyland (sorry for the prior misspellings of your handle...) pointed out very standard terminology for d/dt of (that same quantity). Although this may be an ever so slightly more "math-y" than "physics-y" way to say it, it is rigorously and precisely correct, language and the math. Again it is a fundamental definition of power.

That being said, your original statement, which I corrected was

Originally Posted by CanAutM3

Power represents how fast work can be done ;-). P= W/t

And it is still imprecise unless a rigorous definition of how fast something can be done is equivalent to its time derivate. I don't think so. In addition you provided the formula P = W/t which also is neither precise nor general. It was this latter part of the definition that is actually more problematic.

In a system with no changing potential energy (no big hills for a car) the work-energy theorem states that the change in kinetic energy is equal to the work done thus we can replace W with E in the formula. Thus,

P = dW/dt = dE/dt

You will likely further confuse yourself with language like "work being performed at a constant rate". This is similar to the common sloppy language used to describe speed, "a high rate of speed", which is common language to mean just a high speed but could also be reasonably interpreted as a high acceleration...

So does "work being performed at a constant rate" mean W = constant or slope of W vs. t is constant? The first is like a force pushing against an immovable object, F≠0 but v=0 therefore W=F*v=0 and P=0 (no if it is you pushing the power and work done by you are obviously not zero but the work done on the wall is indeed zero). However if W vs. t has a constant slope then power is constant (again a typical decent approximation for WOT with cars) and acceleration continually tapers as t^-(1/2).

[grimlock]

Ok, forgive me for not letting up on the topic.

I think the basic issue is Swamp's statement:

P=dW/dt
"Power is the rate of change of work" (wrt time)

My common sense tells me its a language issue, which both CanAutM3 and I understand the statement to mean (impossibly) that Power is the change in the rate of work.
'Normal' people, as CanAutM3 and I, would interpret that to mean acceleration, or change in the rate of work.. in other words it's a very autistic (for lack of a better word) way of stating it, which surely you cannot expect everyone else to conform to this use of language!
Why can't you say Power is the rate at which work is done?
"rate of change or work" = acceleration (or the integral wrt time) to most people, does it not??
You engineers/mathemeticians/physicists/whatever have a funny way with words.

add:
btw
'd' terminology is understood to mean differential
and the triangle is for an amount of the quantity
so the statement
P=dw/dt
should be
P=∆w/∆t

[bradleyland]

Originally Posted by grimlock

'Normal' people, as CanAutM3 and I, would interpret that to mean acceleration, or change in the rate of work.. in other words it's a very autistic (for lack of a better word) way of stating it, which surely you cannot expect everyone else to conform to this use of language!

We can if we're talking mathematics

Originally Posted by grimlock

You engineers/mathemeticians/physicists/whatever have a funny way with words.

Indeed. English is a funny language to begin with. English relies heavily on context. In the context of mathematics, the meaning of language changes to describe the maths.

I think what you're really asking is that he use layman terms, which isn't an unreasonable request. I'm not a physicist or an engineer, so lay terms come easily to me, but as someone in a technical field, I can understand the compulsion to use the precise language of my field

[grimlock]

Originally Posted by bradleyland

We can if we're talking mathematics



Indeed. English is a funny language to begin with. English relies heavily on context. In the context of mathematics, the meaning of language changes to describe the maths.

I think what you're really asking is that he use layman terms, which isn't an unreasonable request. I'm not a physicist or an engineer, so lay terms come easily to me, but as someone in a technical field, I can understand the compulsion to use the precise language of my field

No, but the original statement "power is the rate of change work" cannot be true in any sense of language?

rate of change of something is acceleration

edit:
Oh I get it! "rate of change" of work, (wrt) time
Geez, why can't you just say... now I forget how to say it.. damn it.
Seriously, don't use language like this.