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      02-20-2013, 03:39 AM   #45
urBan_dK
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Quote:
Originally Posted by bruce.augenstein@comcast. View Post

A nit: The power necessary to overcome aero drag actually increases with the square of speed.

Bruce
This is false. CanAutM3 is correct.

The power required to overcome drag is given by 1/2ρ * v^3 * A * Cd where:

ρ is the fluid density (air in this case)
v is velocity
Cd is coeffieient of drag
A is frontal area

Bruce, you were probably thinking of force required to overcome drag, not power.
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      02-20-2013, 06:41 AM   #46
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Quote:
Originally Posted by bruce.augenstein@comcast. View Post
Great note. Those of us who misspent our youth way back when often learned this the hard way, when we were able to easily dust a Caddy or Lincoln off the line, but then had to watch helplessly as he steadily began reeling us in, and then just kept motoring away.

A nit: The power necessary to overcome aero drag actually increases with the square of speed. You get to keep the cube, however, because the need for power increases in a linear way as speeds climb. That is, ignoring aero and rolling resistance, you'll need five times the power to accelerate at the same rate at 100 mph as you needed at 20.

Bruce
I think we are saying the same thing but just to nitpick:

It is actually the force of drag that increases with the square of speed.

Power = work / time = force x distance / time = force x speed

So power needed to overcome aero drag does increases with the cube of speed.

So it takes 125 times the power to overcome aero drag at 100mph compared to 20mph .

Last edited by CanAutM3; 02-20-2013 at 09:06 AM.
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      02-20-2013, 06:59 AM   #47
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Great post! Looking at the data this way makes me appreciate the lower torque M3 engine philosophy. That's why the car's power is so linear/smooth. I had a 335i myself with some bolt-ons, loved it, but there was something anemic about it in the upper rev limits, and this explains it well.

Quote:
Originally Posted by e1000 View Post
i disagree. Cars with high torque values compared to their power output typically have early peak torque, which then trails off. Perfect example of this is the N54/N55. (Blue line)



Of course, with the N54/N55, this is due to the turbo's being optimized for a certain rev range, and once it gets past this range, the fun tapers off quickly. This reflects the driving characteristics of the car, a strong initial shove and then that acceleration tapering off in the high rpm range.

While that's all happening on the 335i side, The M3's torque curve looks like this.



The almost completely flat torque curve means that the S65 pulls right up until rev-line.

I know what you're thinking, "but that's a FI vs NA car, surely this is a symptom of being turbocharged!"

Not so, take for example, the M156 6.3L AMG engine. Again, relatively high torque vs the horsepower it produces.



Again, we see the drop off occur starting at 5.5k rpm.

Why does this happen? This happens because regardless of FI or NA, horsepower and torque will always have a relationship with one another, and if a high amount of torque is created in the early to mid range, the engine will have to slump at the end. The answer to this of course, is what M did with the S65, and what Ferrari likes to do. Cut the peak torque to achieve an almost completely flat torque curve.

From there, you and I both know why the torque number itself means little to nothing. Throw in the proper cogs and the equivalent hp/low torque engine will almost surely run away with the race.
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      02-20-2013, 10:10 AM   #48
bruce.augenstein@comcast.
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Quote:
Originally Posted by urBan_dK View Post
This is false. CanAutM3 is correct.

The power required to overcome drag is given by 1/2ρ * v^3 * A * Cd where:

ρ is the fluid density (air in this case)
v is velocity
Cd is coeffieient of drag
A is frontal area

Bruce, you were probably thinking of force required to overcome drag, not power.
Quote:
Originally Posted by CanAutM3 View Post
I think we are saying the same thing but just to nitpick:

It is actually the force of drag that increases with the square of speed.

Power = work / time = force x distance / time = force x speed

So power needed to overcome aero drag does increases with the cube of speed.

So it takes 125 times the power to overcome aero drag at 100mph compared to 20mph .
Deal. I stand corrected.

Bruce
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      02-20-2013, 10:57 AM   #49
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Quote:
Originally Posted by HighandDry View Post
LOL

True, HP is always king, but IF you want to figure out the instantaneous acceleration than torque will give you that.

Force determines your acceleration and torque is directly proportional to force for that RPM.

F=M*A

Work=F*DIstance or F=Work/Distance

Therefore Acceleration=Work/(M*Distance) or something like that
Actually you don't. You can figure instantenuous acceleration directly from power with the following formula:

acceleration = available power / (mass x speed)

Available power is the power generated at that given moment after you have deducted the power required to overcome drag and resistance losses.

Last edited by CanAutM3; 02-20-2013 at 11:07 AM.
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      02-20-2013, 11:21 AM   #50
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Quote:
Originally Posted by CanAutM3 View Post
Actually you don't. You can figure instantenuous acceleration directly from power with the following formula:

acceleration = available power / (mass x speed)

Available power is the power generated at that given moment after you have deducted the power required to overcome drag and resistance losses.

But, Power = F*V, so how do you figure out the Force?

Last edited by HighandDry; 02-20-2013 at 11:41 AM.
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      02-20-2013, 11:33 AM   #51
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Quote:
Originally Posted by HighandDry View Post
But how do you calculate available power...
How would you calculate available torque? Same principle applies to power:

At a given moment in time, the car is traveling at a given road speed. With the gear ratio used, that road speed can be correlated to an engine speed. Finally, from the engine power curve you can figure the power that the engine is generating at that moment for that engine speed. What is more complex is the calculatation of the power used to overcome drag and resistance losses.

However the exact same complexity lies in establishing the forces required to overcome drag and resistance losses when using torque to calculate acceleration.

Last edited by CanAutM3; 02-20-2013 at 11:39 AM.
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      02-20-2013, 11:45 AM   #52
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Quote:
Originally Posted by HighandDry View Post
But, Power = F*V, so how do you figure out the Force?
You don't need to:

P = F * v and F = m * a

substituting F in the first equation:

P = m * a * v

then isolating a:

a = P / (m * v)

Last edited by CanAutM3; 02-20-2013 at 02:50 PM.
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      02-20-2013, 11:53 AM   #53
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LOL I'm no engineer, but I did stay at a Holiday Inn Express
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