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02202013, 03:39 AM  #45  
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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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02202013, 06:41 AM  #46  
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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; 02202013 at 09:06 AM. 

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02202013, 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 boltons, loved it, but there was something anemic about it in the upper rev limits, and this explains it well.
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02202013, 10:10 AM  #48  
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02202013, 10:57 AM  #49  
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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; 02202013 at 11:07 AM. 

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02202013, 11:21 AM  #50  
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But, Power = F*V, so how do you figure out the Force? Last edited by HighandDry; 02202013 at 11:41 AM. 

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02202013, 11:33 AM  #51 
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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; 02202013 at 11:39 AM. 
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02202013, 11:45 AM  #52 
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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; 02202013 at 02:50 PM. 
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