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      09-08-2007, 08:31 AM   #1
southlight
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M3 Dynotest

Here's the full dynotest result of the new M3:
http://www.rri.se/popup/performanceg...p?ChartsID=768

JustMe mentioned the results some days ago. Not bad...

For comparison:
RS4: http://www.rri.se/popup/performanceg...p?ChartsID=769
M5: http://www.rri.se/popup/performanceg...p?ChartsID=153
335i: http://www.rri.se/popup/performanceg...p?ChartsID=647

Best regards, south
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      09-08-2007, 08:52 AM   #2
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Quote:
Originally Posted by southlight View Post
Here's the full dynotest result of the new M3:
http://www.rri.se/popup/performanceg...p?ChartsID=768

JustMe mentioned the results some days ago. Not bad...

For comparison:
RS4: http://www.rri.se/popup/performanceg...p?ChartsID=769
M5: http://www.rri.se/popup/performanceg...p?ChartsID=153
335i: http://www.rri.se/popup/performanceg...p?ChartsID=647

Best regards, south
Very interesting stats. But for those of us who are not as technically astute, could you translate these results so that numb heads like me will understand? Thanks.
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      09-08-2007, 09:20 AM   #3
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Quote:
Originally Posted by carnuts3 View Post
Very interesting stats. But for those of us who are not as technically astute, could you translate these results so that numb heads like me will understand? Thanks.
The disparity of power -applied to the pavement- between the 335 and new M is less than the disparity of the claimed power. One of two, (or both) things occurred; that BMW has understated the 335's power much more than the M; assuming that the M is even understated. Secondly, the 335 has less drivetrain loss; the former is the more likely.

Regardless, the drivetrain loss of the M is very minimal. In other words, it's putting down some serious power. A typical drivetrain loss might average 15%; the M has lost only 10%, or like I said, BMW underestimated the M's power as well.

As you can see, the RS4 doesn't get a good percentage of power to the ground. One reason is that it's AWD has a parasitic effect on power; robbing another 5%, or so. Not to trash Audi, but the chart is not to impressive.

The 335 and M5 are excellent examples.
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      09-08-2007, 09:25 AM   #4
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Sure.
Assuming that both cars had the stated engine power the conclusion is that the M3 has much less drivetrain loss than the RS4.
The M3 brings 45hp more power to the wheels (both have the same stated engine power, 414hp) and even the same amount of torque to the wheels (RS4 is stated with 317 lb/ft, M3 with 295 lb/ft).

Anyway the drivetrain loss of the M3 is higher than the M5's and 335's, which means either that M3's drivetrain is less efficient or that M5 and 335i are underrated (which is more likely).

Best regards, south
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      09-08-2007, 10:53 AM   #5
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Wasn't this discussed in depth on another thread? Someone referenced that dataset before on this forum.
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      09-08-2007, 11:05 AM   #6
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Quote:
Originally Posted by lucid View Post
Wasn't this discussed in depth on another thread? Someone referenced that dataset before on this forum.
So you're calling repost in a friendly manner?

EDIT: Found the thread: http://www.m3post.com/forums/showthread.php?t=79356

As I said in my first post, Just_Me mentioned already the dynoed torque number of M3, but now we have the full test result aswell as a second RS4 test...
[Maybe Jason or Mark could merge these two threads]

Best regards, south
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      09-08-2007, 11:17 AM   #7
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Quote:
Originally Posted by southlight View Post
So you're calling repost in a friendly manner?

EDIT: Found the thread: http://www.m3post.com/forums/showthread.php?t=79356

As I said in my first post, Just_Me mentioned already the dynoed torque number of M3, but now we have the full test result aswell as a second RS4 test...
[Maybe Jason or Mark could merge these two threads]

Best regards, south
See, I reserve judgement when I am not sure, but that doesn't keep me from asking.

I knew this had been discussed somewhat, but also kind of thought your post had some knew info, so I didn't call repost, but asked instead.
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      09-08-2007, 11:19 AM   #8
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Power peak

Quote:
Originally Posted by southlight View Post
Sure.
Assuming that both cars had the stated engine power the conclusion is that the M3 has much less drivetrain loss than the RS4.
The M3 brings 45hp more power to the wheels (both have the same stated engine power, 414hp) and even the same amount of torque to the wheels (RS4 is stated with 317 lb/ft, M3 with 295 lb/ft).

Anyway the drivetrain loss of the M3 is higher than the M5's and 335's, which means either that M3's drivetrain is less efficient or that M5 and 335i are underrated (which is more likely).

Best regards, south
Why do you suppose peak power on the M3 was measured at 7,808 rpm vs. 8,300 as spec'd?
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      09-08-2007, 11:21 AM   #9
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Quote:
Originally Posted by lucid View Post
See, I reserve judgement when I am not sure, but that doesn't keep me from asking.

I knew this had been discussed somewhat, but also kind of thought your post had some knew info, so I didn't call repost, but asked instead.
Very kind, but you're right. Honestly it wasn't necessary to start a new thread...

Best regads, south
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      09-08-2007, 11:23 AM   #10
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Quote:
Originally Posted by GregW / Oregon View Post
Why do you suppose peak power on the M3 was measured at 7,808 rpm vs. 8,300 as spec'd?
Tranmission losses are non-linear. Meaning, the %loss might be higher @8300 than @7800. They actually did not go above 8100 in the test. I downloaded the whole sheet, and here are the details:

@7808: 377.9 hp; 340.0 Nm
@8108: 377.8 hp; 327.6 Nm

It seems like they were taking measurements at 300 rpm intervals and stopped once Tq started dropping seriously and Hp kind of leveled off. I still would have like to see a datapoint at 8300 though.
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      09-08-2007, 11:24 AM   #11
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Quote:
Originally Posted by GregW / Oregon View Post
Why do you suppose peak power on the M3 was measured at 7,808 rpm vs. 8,300 as spec'd?
Don't know what that means. My first thought was: An M5 engine stays an M5 engine, even with 2 cylinder less... (BTW nobody says cylinder, what's the right term?)

Best regards, south
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      09-08-2007, 11:28 AM   #12
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Quote:
Originally Posted by southlight View Post
(BTW nobody says cylinder, what's the right term?)
Cylinder is correct in US usage (6-cylinder, 8-cylinder, etc.)
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      09-08-2007, 11:28 AM   #13
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Quote:
Originally Posted by southlight View Post
Don't know what that means. My first thought was: An M5 engine stays an M5 engine, even with 2 cylinder less... (BTW nobody says cylinder, what's the right term?)

Best regards, south
Are you thinking of pistons?
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      09-08-2007, 11:36 AM   #14
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Quote:
Originally Posted by GregW / Oregon View Post
Cylinder is correct in US usage (6-cylinder, 8-cylinder, etc.)
Thanks!

Quote:
Originally Posted by Epacy View Post
Are you thinking of pistons?
When you would say the new M3 has 8 pistons instead of it's a 8-cylinder, then I'm thinking of pistons. That's one of these differences I'm unsure about. In German you say: It's a 8-cylinder aswell as it's V8, but nobody says that in English, does he? (Now I'm thinking of a Royal with Cheese)

Best regards, south
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      09-08-2007, 11:41 AM   #15
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Yeah, we say it all.
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      09-08-2007, 01:38 PM   #16
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Quote:
Originally Posted by southlight View Post
Thanks!


When you would say the new M3 has 8 pistons instead of it's a 8-cylinder, then I'm thinking of pistons. That's one of these differences I'm unsure about. In German you say: It's a 8-cylinder aswell as it's V8, but nobody says that in English, does he? (Now I'm thinking of a Royal with Cheese)

Best regards, south
V8 is used as much as 8-cylinder. Most tech heads say V8 though.
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      09-08-2007, 01:46 PM   #17
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Quote:
Originally Posted by Garrett View Post
V8 is used as much as 8-cylinder. Most tech heads say V8 though.
Thanks!

Best regards, south
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      09-08-2007, 02:55 PM   #18
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Hey Southern thanx so much for the info. I've been waiting to see a full dyno on the M3.
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      09-08-2007, 05:44 PM   #19
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Quote:
Originally Posted by devo View Post
The disparity of power -applied to the pavement- between the 335 and new M is less than the disparity of the claimed power. One of two, (or both) things occurred; that BMW has understated the 335's power much more than the M; assuming that the M is even understated. Secondly, the 335 has less drivetrain loss; the former is the more likely.

Regardless, the drivetrain loss of the M is very minimal. In other words, it's putting down some serious power. A typical drivetrain loss might average 15%; the M has lost only 10%, or like I said, BMW underestimated the M's power as well.

As you can see, the RS4 doesn't get a good percentage of power to the ground. One reason is that it's AWD has a parasitic effect on power; robbing another 5%, or so. Not to trash Audi, but the chart is not to impressive.

The 335 and M5 are excellent examples.
Drive train losses are not a fixed percentage, but rather fixed number.
Parasitic Torque losses= Friction coefficient X Rotating Mass X Radius.
Frictional coefficient is higher at lower RPM, and does change with RPM.
By using lighter materials in the transmission, you can reduce the drive-train losses.

Audi has higher losses because the rotating mass (Quattro drive train) is more than the that o RWD system.
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      09-08-2007, 08:10 PM   #20
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Quote:
Originally Posted by chonko View Post
Drive train losses are not a fixed percentage, but rather fixed number.
Parasitic Torque losses= Friction coefficient X Rotating Mass X Radius.
Frictional coefficient is higher at lower RPM, and does change with RPM.
By using lighter materials in the transmission, you can reduce the drive-train losses.

Audi has higher losses because the rotating mass (Quattro drive train) is more than the that o RWD system.
For one you are contradicting yourself here, you say the loss is a fixed number, which is false, then you say in the formula is it a function of rpm, which is correct. Your formula may have some correct features but is wrong as well, a simple check of the units will tell you that (watts is not equal to mass x distance). An AWD has more losses becuase of the number of lossy interfaces, gear interfaces and bearings, etc, as well as the larger rotating mass involved.
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      09-08-2007, 08:52 PM   #21
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Quote:
Originally Posted by swamp2 View Post
For one you are contradicting yourself here, you say the loss is a fixed number, which is false, then you say in the formula is it a function of rpm, which is correct. Your formula may have some correct features but is wrong as well, a simple check of the units will tell you that (watts is not equal to mass x distance).
An AWD has more losses becuase of the number of lossy interfaces, gear interfaces and bearings, etc, as well as the larger rotating mass involved.[/quote]

I was trying to make the distinction between using a fixed percentage of drive-train torque losses as opposed to a given torque loss number.

For the same coefficient of friction, the torque loss at 3000 rpm is the same as the torque loss at 6000 rpm, however your engine torque at both rpm is different. Since the enumerator is the same, with different denominator, it will erroneous to assume that your percentage losses are the same.
You can conversely use the fixed percentage loss and apply it to the various torques at the different rpm, which will give you different Parasitic torque loss. Unfortunately, your torque losses for a given coefficient of friction is not a variable loss, but fixed hence you will be liable to erroneous answers adopting this approach.

With regards to what I wrote as a the formula for frictional losses- it is what we use in the industry, and it forms the basis of managing torque losses.

Torque losses= side force * coefficient of friction*radius of the rotating mass

Side force barring external forces due to bending and curvature on the rotating mass= Normal reaction.
Normal reaction= Weight perpendicular to the plane
Weight perpendicular to the plane(horizontal plane)= Mass * Acceleration due to gravity. (Forgive me for using mass and force loosely didn't realise I was in a physics class: but a mass under the influence of gravity does exert force on a plane)
Side Force*radius = torque.

Torque generally speaking is not a function of rpm, but a function of Force and the radius arm.

Engine Torque does however rely on rpm only because the force generated by the engine is a function of rpm
The engine force= mean effective pressure*cross sectional area of the piston*effective strokes per second*number of cylinders
Torque= Force * Radius.

I can't remember stating Parasitic Power loss, I believe I said "Parasitic Torque losses.
Power represents the ability to do work per unit time.
Torque which I am sure you know very well represents the rotating action of force.

Quote:
Originally Posted by swamp2 View Post
An AWD has more losses becuase of the number of lossy interfaces, gear interfaces and bearings, etc, as well as the larger rotating mass involved.
If you have two surfaces in contact, I believe you will have friction.
So like you rightly pointed out, the cumulative frictional coefficient across the drive-train is a lot more than a RWD system. And the rotating mass is more than what is applicable to a RWD system.

Last edited by chonko; 09-08-2007 at 09:16 PM.
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      09-08-2007, 09:50 PM   #22
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Quote:
Originally Posted by chonko View Post
An AWD has more losses becuase of the number of lossy interfaces, gear interfaces and bearings, etc, as well as the larger rotating mass involved
I was trying to make the distinction between using a fixed percentage of drive-train torque losses as opposed to a given torque loss number.

For the same coefficient of friction, the torque loss at 3000 rpm is the same as the torque loss at 6000 rpm, however your engine torque at both rpm is different. Since the enumerator is the same, with different denominator, it will erroneous to assume that your percentage losses are the same.
You can conversely use the fixed percentage loss and apply it to the various torques at the different rpm, which will give you different Parasitic torque loss. Unfortunately, your torque losses for a given coefficient of friction is not a variable loss, but fixed hence you will be liable to erroneous answers adopting this approach.

With regards to what I wrote as a the formula for frictional losses- it is what we use in the industry, and it forms the basis of managing torque losses.

Torque losses= side force * coefficient of friction*radius of the rotating mass

Side force barring external forces due to bending and curvature on the rotating mass= Normal reaction.
Normal reaction= Weight perpendicular to the plane
Weight perpendicular to the plane(horizontal plane)= Mass * Acceleration due to gravity. (Forgive me for using mass and force loosely didn't realise I was in a physics class: but a mass under the influence of gravity does exert force on a plane)
Side Force*radius = torque.

Torque generally speaking is not a function of rpm, but a function of Force and the radius arm.

Engine Torque does however rely on rpm only because the force generated by the engine is a function of rpm
The engine force= mean effective pressure*cross sectional area of the piston*effective strokes per second*number of cylinders
Torque= Force * Radius.

I can't remember stating Parasitic Power loss, I believe I said "Parasitic Torque losses.
Power represents the ability to do work per unit time.
Torque which I am sure you know very well represents the rotating action of force.



If you have two surfaces in contact, I believe you will have friction.
So like you rightly pointed out, the cumulative frictional coefficient across the drive-train is a lot more than a RWD system. And the rotating mass is more than what is applicable to a RWD system.




..... but this one goes to eleven!




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