BMW M3 M4 RPM Redline Visual, Finally

[solstice]

Originally Posted by CanAutM3

Agreed, compared to today's engines they are not that extraordinary. Just shows how much engine technology has evolved in the last decades. However, I still think they were pretty exciting engines at the time they came out in the mid/late nineties .

Important point, it's how special the car is when it's launched that counts. I don't think an IBM AT is very impressive now but when it was launched it was the tits. Is the S55 as special in today's market? Maybe not taken it's two predecessors.

[CanAutM3]

Originally Posted by swamp2

I think there is a combination of a slight lack of clarity on my behalf and you assuming the worst about my comments. Without identifying each individual component of the entire drivetrain some must be re-accelerated after each shift and some do not. However, during acceleration the entire drivetrain must be accelerated (angularly obviously). Anyone who knows anything about how a car works can tell the difference between which component fits which description.

Thanks for clarifying, we agree on that point

Originally Posted by swamp2

The devil is in the details. A force (torque) balance through the entire drivetrain also involves I's for the driveshaft, axles and wheels. The I's for those components affect the actual acceleration force delivered to the wheels and thus the shift points. Perhaps think of it this way. We all know lighter wheels are great because they essentially act like more power. No they are not actually more power but it is VERY similar as the car has more acclerative force. Different forces affect when the curves cross and hence shift points. Place massively heavy wheels on your vehicle and ideal shift points would change.

On this one, I am not sure I agree. IMO, heavier wheels will have a negative impact on overall acceleration, but not on optimal shift points.

My understanding:

The angular velocity of the wheels is directly proportional to the speed of the vehicle regardless of the selected gear. In other words, their equivalent mass is independent of the gearbox ratio. The same concept applies for all the rotating components downstream of the gearbox (and some inside the gearbox). Only the engine's (and part of the transmission's) angular acceleration relative to the car's linear acceleration varies as a function of the selected gear.

The mass of the car does not need to be considered for optimal shift points because it does not vary with different gears, same applies to rotating components downstream of the gearbox. For a given torque generated at the gearbox output, the same proportion of that torque will be used to accelerate the downstream rotating components regardless of which gear is selected.

Originally Posted by swamp2

I do not recall any simulations I provided that showed a vehicle shifting always at or around 7200 rpm. We talked about and older 5er here in this thread as well as the M4. Both share the trend of a at or near redline shift in the first gear (or two) followed by an ever decreasing ideal shift point.

The one in post #195. The shift points for every gear are around 7200. Since the inertial effect of the engine decreases with higher gears, I expect to see higher shift points in the higher gears...

[CanAutM3]

Originally Posted by solstice

Important point, it's how special the car is when it's launched that counts. I don't think an IBM AT is very impressive now but when it was launched it was the tits. Is the S55 as special in today's market? Maybe not taken it's two predecessors.

I still need to wait and see...

The fact that it promises such a broad power plateau and if the engineers are able to eliminate lag through some smart technical trick could well make the S55 a special engine for its time.

[solstice]

Originally Posted by CanAutM3

I still need to wait and see...

The fact that it promises such a broad power plateau and if the engineers are able to eliminate lag through some smart technical trick could well make the S55 a special engine for its time.

I agree, it all comes down if it's revolutionary free from lag. If it is, then yes it will be very special indeed. I clearly remember being blown away by every M cars launched to date first by specifications and then in real performance. This time the specifications isn't really mind blowing with a typical FI power curve and for today's M cars moderate power. The real life performance remain to be felt. No lag, some lag or a lot of lag? We'll see.

[CanAutM3]

Originally Posted by solstice

I agree, it all comes down if it's revolutionary free from lag. If it is, then yes it will be very special indeed. I clearly remember being blown away by every M cars launched to date first by specifications and then in real performance. This time the specifications isn't really mind blowing with a typical FI power curve and for today's M cars moderate power. The real life performance remain to be felt. No lag, some lag or a lot of lag? We'll see.

Agreed, we're on the same page. I was super excited about every previous M3 with only what I saw on paper; I bought my M3s without even test driving them and was never disappointed. This one is leaving me lukewarm on paper. I will reserve my final judgment when I can drive one myself.

[pkimM3r]

Originally Posted by swamp2

Results are not at all as you predicted. Shift points are all well below redline. This is with the exact power curve, gear ratios and redline you have used.

Torque is simply falling too fast above ~5500 rpm to "have room" for the force curves not to cross!

Shifting twice to get to 60 mph doesnt seem that great. Im going to feel like im in an evo. Oh wait ill be getting dct the next time around because of the dumb auto rev match crap and have my e90 for some mechanical noise fun!

[pkimM3r]

Originally Posted by CanAutM3

Thanks for clarifying, we agree on that point

On this one, I am not sure I agree. IMO, heavier wheels will have a negative impact on overall acceleration, but not on optimal shift points.

My understanding:

The angular velocity of the wheels is directly proportional to the speed of the vehicle regardless of the selected gear. In other words, their equivalent mass is independent of the gearbox ratio. The same concept applies for all the rotating components downstream of the gearbox (and some inside the gearbox). Only the engine's (and part of the transmission's) angular acceleration relative to the car's linear acceleration varies as a function of the selected gear.

The mass of the car does not need to be considered for optimal shift points because it does not vary with different gears, same applies to rotating components downstream of the gearbox. For a given torque generated at the gearbox output, the same proportion of that torque will be used to accelerate the downstream rotating components regardless of which gear is selected.

The one in post #195. The shift points for every gear are around 7200. Since the inertial effect of the engine decreases with higher gears, I expect to see higher shift points in the higher gears...

So second gear nets you 14mph?!

Something wrong here.

[CanAutM3]

Originally Posted by pkim1079

So second gear nets you 14mph?!

Something wrong here.

Those were theoretical gear ratios to optimize the power plateau. With the nearly instantaneous shifts provided by the DCT, more frequent shifts are not as inconvenient to optimize performance. When I calculated those ratios, I did not factor in the impact of the engine inertia on shift points. When I have some time to spare, I will recalculate them. It should not make a huge difference, especially in the higher gears. These ratios are unlikely to make it to the actual car.

FD 3.154 RPM km/h
1 4.310 7500 70
2 3.218 7500 94
3 2.403 7500 126
4 1.794 7500 169
5 1.340 7500 226
6 1.000 7500 303
7 0.747 1850 100

[Rookie84]

Originally Posted by pkim1079

Shifting twice to get to 60 mph doesnt seem that great. Im going to feel like im in an evo. Oh wait ill be getting dct the next time around because of the dumb auto rev match crap and have my e90 for some mechanical noise fun!

Wouldn't the auto rev-match have an on-off switch? 370Z has the auto rev-match and has a switch to turn it off.

[pkimM3r]

Originally Posted by CanAutM3

Those were theoretical gear ratios to optimize the power plateau. With the nearly instantaneous shifts provided by the DCT, more frequent shifts are not as inconvenient to optimize performance. When I calculated those ratios, I did not factor in the impact of the engine inertia on shift points. When I have some time to spare, I will recalculate them. It should not make a huge difference, especially in the higher gears. These ratios are unlikely to make it to the actual car.

FD 3.154 RPM km/h
1 4.310 7500 70
2 3.218 7500 94
3 2.403 7500 126
4 1.794 7500 169
5 1.340 7500 226
6 1.000 7500 303
7 0.747 1850 100

But is it optimizing overall speed and acceleration as in the triple digits or was it to optimize quarter mile times or just based on the powerband. Because no matter how much you are in the powerband, if you have a long 7th gear, wouldnt it be pretty slow? Just curious and asking.

[pkimM3r]

Originally Posted by Rookie84

Wouldn't the auto rev-match have an on-off switch? 370Z has the auto rev-match and has a switch to turn it off.

From reading other threads it turns on when in sport+ but hopefully its programmable. In any case ive always wanted both an MT and a DCT m3 so i can keep the e90 and get the new m3/4 in dct. Also want a coupe and a sedan so will opt for m4 this time.

[CanAutM3]

Originally Posted by pkim1079

But is it optimizing overall speed and acceleration as in the triple digits or was it to optimize quarter mile times or just based on the powerband. Because no matter how much you are in the powerband, if you have a long 7th gear, wouldnt it be pretty slow? Just curious and asking.

Basic theory is the longer time you spend in the power plateau, the better acceleration you will have. The intent for this gearing was to optimize acceleration over the entire speed range while still requiring redline shifts (as mentioned, it did not take into consideration engine interia). To have even better acceleration, even tighter ratios should be used to keep the RPM in the 5700-7300 power plateau (instead of the 5500-7500 band) and shifting short of redline; but this would hurt MPG due to a shorter 7th gear.

[pkimM3r]

Originally Posted by CanAutM3

Basic theory is the longer time you spend in the power plateau, the better acceleration you will have. The intent for this gearing was to optimize acceleration over the entire speed range while still requiring redline shifts (as mentioned, it did not take into consideration engine interia). To have even better acceleration, even tighter ratios should be used to keep the RPM in the 5700-7300 power plateau (instead of the 5500-7500 band) and shifting short of redline; but this would hurt MPG due to a shorter 7th gear.

dont you have to take in top speed to account as well?

[CanAutM3]

Originally Posted by pkim1079

dont you have to take in top speed to account as well?

Yes, you usually do. But the power plateau is so broad on the S55 that it is a non-issue. With the theoretical ratios I provided, you can reach 300km/h in 6th with 7th to spare...

[pkimM3r]

Originally Posted by CanAutM3

Yes, you usually do. But the power plateau is so broad on the S55 that it is a non-issue. With the theoretical ratios I provided, you can reach 300km/h in 6th with 7th to spare...

should be good....one thing i guess that having torque down low is not having to downshift to 2nd on tracks. but i guess its even more throttle modulation or just power slides

[swamp2]

Originally Posted by CanAutM3

On this one, I am not sure I agree. IMO, heavier wheels will have a negative impact on overall acceleration, but not on optimal shift points.

My understanding:

The angular velocity of the wheels is directly proportional to the speed of the vehicle regardless of the selected gear. In other words, their equivalent mass is independent of the gearbox ratio. The same concept applies for all the rotating components downstream of the gearbox (and some inside the gearbox). Only the engine's (and part of the transmission's) angular acceleration relative to the car's linear acceleration varies as a function of the selected gear.

The mass of the car does not need to be considered for optimal shift points because it does not vary with different gears, same applies to rotating components downstream of the gearbox. For a given torque generated at the gearbox output, the same proportion of that torque will be used to accelerate the downstream rotating components regardless of which gear is selected.

Let's cut right to the equations here...



This is the total propulsive force accelerating a vehicle. The first term is the real basic one obvious to most, torque x combined gear ratio x combined transmission loss. The entire second term is the "inertial term", I (e,t,d,w) are moments of inertia of the engine, transmission, driveshafts and wheels. Nf is the final drive ratio, ax is the vehicle acceleration and r is the drive wheel radius. This of course is not F = ma for the entire vehicle this is just the propulsive force. This is equation 2-9b from Gillespie as referenced prior.

This entire 2nd term is most often completely ignored, even I have been guilty of this in the past.

Now let me try to show in a somewhat qualitative way how wheel inertia can affect shift point.

The first term just gives us basically an upside down "lump" shape mirroring the shape of the engines torque curve (assume plotting vs. speed). Now since we agree that ideal shifting is based on when these propulsive force curves from different gears intersect it should start to be obvious already. Imagine the effect of the second term. How can it affect shift points (as you already agree it will) but then Iw can not? This term with an Iw changed from one value to another will drop the "lump" in an rpm dependent (i.e. speed dependent) fashion because it is explicitly acceleration dependent (ax term)! It will drop the curves in each gear by a different amount (clearly from the Ntf term). Thus how could it possibly be that the curves with two different Iw's will intersect at the exact same rpm? They won't, period. Yes it is a small correction but it is a real effect. I see why you feel it is a bit counter-intuitive but that's the reason for the precision of the equations! Thus to correct your reasoning above try to forget about which terms are gear dependent and focus on how they are acceleration dependent and thus implicitly speed dependent and imagine how the intersection of the curves will be affected.

Unfortunately, CarTest does not seem to capture this subtlety and in fact neither does my huge complex spreadsheet that performs the same basic calculation. CarTest does capture the effect of wheel inertia on acceleration but does not capture its effect on shift points. It is also clearly captures the I (e, t and d) terms and of course the losses as well. My own spreadsheet uses a slightly different approach (really a "cheat" of sorts) of effective mass as outlined in Gillespie to include all inertial effects both on acceleration and on shift points. But due to the way that is implemented Iw can not be explicitly changed and see the effect directly.

Fortunately these Iw corrections are small, both on the shift points, on the overall acceleration and from any change from this interaction (changing acceleration by changed shift point(s)).

But again the total corrections are anything but small. Recall that the overall acceleration in 1st gear for a car like an M3 is about 20% lower (in 1st gear) when including the inertial term or not including it. These effects are quite substantial overall. It's like 750 lb, clearly a heck of a lot more than the component weigh! In higher gears the percent "correction" of the second term is closer to 5-10%.

Hope this helps.

[Kadema]

Where has CanAutM3's last post gone??

[CanAutM3]

Originally Posted by swamp2

Let's cut right to the equations here...



This is the total propulsive force accelerating a vehicle. The first term is the real basic one obvious to most, torque x combined gear ratio x combined transmission loss. The entire second term is the "inertial term", I (e,t,d,w) are moments of inertia of the engine, transmission, driveshafts and wheels. Nf is the final drive ratio, ax is the vehicle acceleration and r is the drive wheel radius. This of course is not F = ma for the entire vehicle this is just the propulsive force. This is equation 2-9b from Gillespie as referenced prior.

This entire 2nd term is most often completely ignored, even I have been guilty of this in the past.

Now let me try to show in a somewhat qualitative way how wheel inertia can affect shift point.

The first term just gives us basically an upside down "lump" shape mirroring the shape of the engines torque curve (assume plotting vs. speed). Now since we agree that ideal shifting is based on when these propulsive force curves from different gears intersect it should start to be obvious already. Imagine the effect of the second term. How can it affect shift points (as you already agree it will) but then Iw can not? This term with an Iw changed from one value to another will drop the "lump" in an rpm dependent (i.e. speed dependent) fashion because it is explicitly acceleration dependent (ax term)! It will drop the curves in each gear by a different amount (clearly from the Ntf term). Thus how could it possibly be that the curves with two different Iw's will intersect at the exact same rpm? They won't, period. Yes it is a small correction but it is a real effect. I see why you feel it is a bit counter-intuitive but that's the reason for the precision of the equations! Thus to correct your reasoning above try to forget about which terms are gear dependent and focus on how they are acceleration dependent and thus implicitly speed dependent and imagine how the intersection of the curves will be affected.

Unfortunately, CarTest does not seem to capture this subtlety and in fact neither does my huge complex spreadsheet that performs the same basic calculation. CarTest does capture the effect of wheel inertia on acceleration but does not capture its effect on shift points. It is also clearly captures the I (e, t and d) terms and of course the losses as well. My own spreadsheet uses a slightly different approach (really a "cheat" of sorts) of effective mass as outlined in Gillespie to include all inertial effects both on acceleration and on shift points. But due to the way that is implemented Iw can not be explicitly changed and see the effect directly.

Fortunately these Iw corrections are small, both on the shift points, on the overall acceleration and from any change from this interaction (changing acceleration by changed shift point(s)).

But again the total corrections are anything but small. Recall that the overall acceleration in 1st gear for a car like an M3 is about 20% lower (in 1st gear) when including the inertial term or not including it. These effects are quite substantial overall. It's like 750 lb, clearly a heck of a lot more than the component weigh! In higher gears the percent "correction" of the second term is closer to 5-10%.

Hope this helps.

This is getting interesting. I doodled with the math and thought I had it nailed by being able to simplify the Iw term out of the shift point equations (see attached PDF).

However, acceleration is still part of the resulting simplified equation; and acceleration does vary with Iw. If it is the case, it also implies that optimal shift points will vary depending on the total mass of the car. So if you have 3 passengers, you might end up with different optimal shift points compared to when you are driving by yourself.

I still find this a bit counter intuitive and will play with the math some more. I'll post my findings ...if any

While I underestimated the impact of drivetrain inertia on optimal shift points, I still think that Cartest overestimates the impact by using the mass factor rather than the actual inertia.

[Ezio]

Originally Posted by swamp2

Let's cut right to the equations here...



This is the total propulsive force accelerating a vehicle. The first term is the real basic one obvious to most, torque x combined gear ratio x combined transmission loss. The entire second term is the "inertial term", I (e,t,d,w) are moments of inertia of the engine, transmission, driveshafts and wheels. Nf is the final drive ratio, ax is the vehicle acceleration and r is the drive wheel radius. This of course is not F = ma for the entire vehicle this is just the propulsive force. This is equation 2-9b from Gillespie as referenced prior.

This entire 2nd term is most often completely ignored, even I have been guilty of this in the past.

Now let me try to show in a somewhat qualitative way how wheel inertia can affect shift point.

The first term just gives us basically an upside down "lump" shape mirroring the shape of the engines torque curve (assume plotting vs. speed). Now since we agree that ideal shifting is based on when these propulsive force curves from different gears intersect it should start to be obvious already. Imagine the effect of the second term. How can it affect shift points (as you already agree it will) but then Iw can not? This term with an Iw changed from one value to another will drop the "lump" in an rpm dependent (i.e. speed dependent) fashion because it is explicitly acceleration dependent (ax term)! It will drop the curves in each gear by a different amount (clearly from the Ntf term). Thus how could it possibly be that the curves with two different Iw's will intersect at the exact same rpm? They won't, period. Yes it is a small correction but it is a real effect. I see why you feel it is a bit counter-intuitive but that's the reason for the precision of the equations! Thus to correct your reasoning above try to forget about which terms are gear dependent and focus on how they are acceleration dependent and thus implicitly speed dependent and imagine how the intersection of the curves will be affected.

Unfortunately, CarTest does not seem to capture this subtlety and in fact neither does my huge complex spreadsheet that performs the same basic calculation. CarTest does capture the effect of wheel inertia on acceleration but does not capture its effect on shift points. It is also clearly captures the I (e, t and d) terms and of course the losses as well. My own spreadsheet uses a slightly different approach (really a "cheat" of sorts) of effective mass as outlined in Gillespie to include all inertial effects both on acceleration and on shift points. But due to the way that is implemented Iw can not be explicitly changed and see the effect directly.

Fortunately these Iw corrections are small, both on the shift points, on the overall acceleration and from any change from this interaction (changing acceleration by changed shift point(s)).

But again the total corrections are anything but small. Recall that the overall acceleration in 1st gear for a car like an M3 is about 20% lower (in 1st gear) when including the inertial term or not including it. These effects are quite substantial overall. It's like 750 lb, clearly a heck of a lot more than the component weigh! In higher gears the percent "correction" of the second term is closer to 5-10%.

Hope this helps.

.

[catpat8000]

Originally Posted by swamp2

Let's cut right to the equations here...



This is the total propulsive force accelerating a vehicle. The first term is the real basic one obvious to most, torque x combined gear ratio x combined transmission loss. The entire second term is the "inertial term", I (e,t,d,w) are moments of inertia of the engine, transmission, driveshafts and wheels. Nf is the final drive ratio, ax is the vehicle acceleration and r is the drive wheel radius. This of course is not F = ma for the entire vehicle this is just the propulsive force. This is equation 2-9b from Gillespie as referenced prior.

The first term just gives us basically an upside down "lump" shape mirroring the shape of the engines torque curve (assume plotting vs. speed). Now since we agree that ideal shifting is based on when these propulsive force curves from different gears intersect it should start to be obvious already. Imagine the effect of the second term. How can it affect shift points (as you already agree it will) but then Iw can not? This term with an Iw changed from one value to another will drop the "lump" in an rpm dependent (i.e. speed dependent) fashion because it is explicitly acceleration dependent (ax term)! It will drop the curves in each gear by a different amount (clearly from the Ntf term). Thus how could it possibly be that the curves with two different Iw's will intersect at the exact same rpm? They won't, period. Yes it is a small correction but it is a real effect. I see why you feel it is a bit counter-intuitive but that's the reason for the precision of the equations! Thus to correct your reasoning above try to forget about which terms are gear dependent and focus on how they are acceleration dependent and thus implicitly speed dependent and imagine how the intersection of the curves will be affected.

Hope this helps.

Changing wheel radius changes inertial loss in every gear. So I don't see how wheel radius affects shift point from this formula.

If you calculated F in gear 2 at an rpm in which F was higher than in gear 3 at the same velocity, it would all still be true with larger wheels. i.e. the propulsive force in a given gear would drop when increasing Iw but it would drop in all gears and by an amount which kept the optimal shift point constant.

[Kadema]

CanAutM3, I read your PDF-file and found no flaws - which doesn't mean very much of course. So, if you correctly eliminated Iw, you don't have to include it again afterwards only because acceleration is still part of the equation, do you?

[CanAutM3]

Originally Posted by Kadema

CanAutM3, I read your PDF-file and found no flaws - which doesn't mean very much of course. So, if you correctly eliminated Iw, you don't have to include it again afterwards only because acceleration is still part of the equation, do you?

I also think the math is sound. And yes, Iw per se is simplified out of the equation.

But since vehicle acceleration is still part of the equation to be solved, it means Iw still has an impact. Albeit a very very small impact because the wheel inertia becomes burried in the overall equivalent mass of the car (equivalent mass = total mass + inertial effects).

What I did not expect, is for the mass of the car to have an impact on optimal shift points. I am not yet fully convinced of this and will try to doodle some more with the math...