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09082007, 08:31 AM  #1 
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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 
09082007, 08:52 AM  #2  
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09082007, 09:20 AM  #3  
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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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09082007, 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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09082007, 11:05 AM  #6  
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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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09082007, 11:17 AM  #7  
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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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09082007, 11:19 AM  #8  
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09082007, 11:21 AM  #9  
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Best regads, south 

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09082007, 11:23 AM  #10  
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@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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09082007, 11:24 AM  #11  
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Best regards, south 

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09082007, 11:28 AM  #12 
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Cylinder is correct in US usage (6cylinder, 8cylinder, etc.)
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09082007, 11:28 AM  #13 
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Are you thinking of pistons?
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09082007, 11:36 AM  #14  
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When you would say the new M3 has 8 pistons instead of it's a 8cylinder, then I'm thinking of pistons. That's one of these differences I'm unsure about. In German you say: It's a 8cylinder 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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09082007, 11:41 AM  #15 
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Yeah, we say it all.
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09082007, 01:38 PM  #16  
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09082007, 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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09082007, 05:44 PM  #19  
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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 drivetrain losses. Audi has higher losses because the rotating mass (Quattro drive train) is more than the that o RWD system. 

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09082007, 08:10 PM  #20  
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09082007, 08:52 PM  #21  
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I was trying to make the distinction between using a fixed percentage of drivetrain 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:
So like you rightly pointed out, the cumulative frictional coefficient across the drivetrain 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; 09082007 at 09:16 PM. 

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09082007, 09:50 PM  #22  
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..... but this one goes to eleven! 

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