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E46 Performance (Here we go again) (archive)

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Posted by Richard Peterson on December 23, 1999 at 17:40:23:

Making generalizations based on the M5, the M3 should have (approximately) the following performance.

The M5 produces 395ft*lbs of torque with an engine displacement of 303ci. Which equals (395/303=) 1.303ft*lbs/ci. (A high figure for a naturally aspirated engine!)

Assuming that the new M3 gets the 3.4L engine, it will have a displacement of about 207ci of displacement, equating to (207*1.303) 268ft*lbs of torque.

Now, let’s make the assumption that the gear ratios are the same (this is a HUGE assumption, because a change in gear ratios essentially changes the amount of torque applied to the rear wheels, and ultimately the road!!!!!) Also, let’s make the assumption that the rear wheel contact patch is identical, i.e. provide the same traction.

Now the weight of the M5 is 3792lbs. That makes the torque to weight ratio (3792/395) 9.6lbs/ft*lbs.

The weight of the M3 is estimated to be similar to the 328 coupe, approx. 3230lbs. That makes the torque to weight ratio (3230/268) 12.05lbs/ft*lbs.

With above two assumptions, the M3 will be considerably slower than the M5. Definitely no faster than 5 seconds 0-60. Probably closer to 5.5.

But let’s change our assumptions.

The M5 has a final drive ratio of 2.60:1. If the transmission is in a gear that has a direct drive (1:1 gear ratio) the torque at the rear wheels will be approx. (395*2.60) 1027ft*lbs. Equating to a (3792/1027) 3.69lbs/ft*lbs torque to weight ratio.
For the M3 to have the same torque to weight ratio at the rear wheels, the final drive ratio could be raised to (3230/3.69 = 875.338 required torque) (875.338/268) 3.266.

This change in drive ratio would provide the exact 0-60 acceleration!!! (Assuming the same traction).

Even if the final drive ratio is identical, they could shorten 1st and 2nd gear (and third if needed) and the result would be the same.

So how does HP role into this equation? Understanding that HP is a function of torque and the RPM the engine is turning (hp = (torque * RPM) / 5252), it has almost no bearing on 0-60 acceleration, but it will have a large impact on ¼ mile times. Why?

The HP rating is an indication of how fast the engine can rotate and still provide maximum torque. (All of this is a function of airflow, but that is another story) Once the engine reaches its airflow capacity, it no is longer able to fill the cylinder with maximum volume, decreasing torque. The point where the torque decreases at a rate greater than the ratio of RPM/5252 increases, the HP curve will begin to decrease.

So, to compare the e36 M3 US to Euro spec, their torque outputs are similar, while their HP is significantly different. (Assume for math, their torque is equal.) If the peak torque is 240ft*lbs on both engines we can assume because of the HP ratings (240US to 320 Euro) that the HP peak will occur at about 5252rpm for the US engine ((5252*240)/5252) and at ((7000*240)/5252) 7000rpm for the euro model. So, above 5252rpm the euro engine will have an advantage. Now ask yourself how long the engine remains in above 5252prm at full acceleration in first gear? (Maybe .5 seconds) Then in second? (Maybe 1 second) By the time you are out of second gear, you are going about 60 MPH! So, in first, the euro engine has an advantage for about .5 seconds, and in second it has an advantage for about 1 second. This advantage may result in a measurable gain in acceleration, but it will be very small! (Maybe 2 tenths!) But when you hit 3rd and 4th on your way to a 1/4 mile time, the amount of time spent above that RPM is significant! Knocking off a good portion of time.

Anyway, let’s hope the new M3 will be under 5sec. (I can’t imagine it being slower than the M roadster). And if we’re all lucky, we will see the V8!!

Richard Peterson



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