Drivetrain loss
06-30-2010, 10:01 PM
#1
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Location: sc
Posts: 457
Drivetrain loss
We all know about the Mustang losing 15 - 20% parasitic power loss from the flywheel to the rear wheels, right?? That means my Mustang is putting out 490hp at the flywheel in order to produce the 417hp I show on the Dyno.
( 490X.85 = 416.5hp ).
Now, theoretically, does that mean that a SuperCar putting out 900hp at the flywheel is losing 135hp to 180hp due to parasitic loss??? That's 15 - 20 %.
That does not make any sense.
Does this formula apply only to cars in the mid HP range like our Mustangs?
Or is the whole 15- 20 % thing a pile of rubbish? What about the Corvette Z06???
Someone who definitely knows the answer, will you please enlighten me?
No so called " theories ", OK? I've heard enough of those.
Is this whole formula flawed, an "Old Wives Tale"?
A loss of 20% of 900hp, 180hp, makes no sense.
The truth must be out there.
Thanks guys and gals.
( 490X.85 = 416.5hp ).
Now, theoretically, does that mean that a SuperCar putting out 900hp at the flywheel is losing 135hp to 180hp due to parasitic loss??? That's 15 - 20 %.
That does not make any sense.
Does this formula apply only to cars in the mid HP range like our Mustangs?
Or is the whole 15- 20 % thing a pile of rubbish? What about the Corvette Z06???
Someone who definitely knows the answer, will you please enlighten me?
No so called " theories ", OK? I've heard enough of those.
Is this whole formula flawed, an "Old Wives Tale"?
A loss of 20% of 900hp, 180hp, makes no sense.
The truth must be out there.
Thanks guys and gals.
07-01-2010, 01:30 AM
#2
Banned
Join Date: Aug 2007
Location: AZ
Posts: 2,463
We all know about the Mustang losing 15 - 20% parasitic power loss from the flywheel to the rear wheels, right?? That means my Mustang is putting out 490hp at the flywheel in order to produce the 417hp I show on the Dyno.
( 490X.85 = 416.5hp ).
Now, theoretically, does that mean that a SuperCar putting out 900hp at the flywheel is losing 135hp to 180hp due to parasitic loss??? That's 15 - 20 %.
That does not make any sense.
Does this formula apply only to cars in the mid HP range like our Mustangs?
Or is the whole 15- 20 % thing a pile of rubbish? What about the Corvette Z06???
Someone who definitely knows the answer, will you please enlighten me?
No so called " theories ", OK? I've heard enough of those.
Is this whole formula flawed, an "Old Wives Tale"?
A loss of 20% of 900hp, 180hp, makes no sense.
The truth must be out there.
Thanks guys and gals.
( 490X.85 = 416.5hp ).
Now, theoretically, does that mean that a SuperCar putting out 900hp at the flywheel is losing 135hp to 180hp due to parasitic loss??? That's 15 - 20 %.
That does not make any sense.
Does this formula apply only to cars in the mid HP range like our Mustangs?
Or is the whole 15- 20 % thing a pile of rubbish? What about the Corvette Z06???
Someone who definitely knows the answer, will you please enlighten me?
No so called " theories ", OK? I've heard enough of those.
Is this whole formula flawed, an "Old Wives Tale"?
A loss of 20% of 900hp, 180hp, makes no sense.
The truth must be out there.
Thanks guys and gals.
It is about accelerating that drivetrain. The faster you are accelerating that drive train, the more hp it robs, hence why it is on in percentages. If you knew ANYTHING about the physics of motion, you would understand.
So, simply put, the more acceleration you apply to the moving parts of the drivetrain, the more hp they rob. Layman's terms at least. THAT, my friend, is why it is a PERCENTAGE, NOT a constant. It's not about the amount of hp it takes to move the drivetrain a certain speed. Take a calculus class, as well as a university physics class, and you will understand.
Last edited by mfj; 07-01-2010 at 01:53 AM.
07-01-2010, 02:07 AM
#3
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Join Date: Aug 2007
Location: AZ
Posts: 2,463
The greater the acceleration, the more force (power) it requires to accomplish that amount of acceleration. Simple law of physics. Force = mass times acceleration (F=M x A). That being said, the quicker you are rotationally accelerating each of the drivetrain components, the more force that is required to spin them AT THAT RATE OF ACCELERATION.
07-01-2010, 02:09 AM
#4
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Join Date: Aug 2006
Location: sc
Posts: 457
Wow, you fail at physics. Typical drivetrain loss is 12-20%, so yes, a 1000hp car is losing about 150 hp through it's drivetrain. It's not about how much hp it takes to simply move the drivetrain, as that is a constant.
It is about accelerating that drivetrain. The faster you are accelerating that drive train, the more hp it robs, hence why it is on in percentages. If you knew ANYTHING about the physics of motion, you would understand.
So, simply put, the more acceleration you apply to the moving parts of the drivetrain, the more hp they rob. Layman's terms at least. THAT, my friend, is why it is a PERCENTAGE, NOT a constant. It's not about the amount of hp it takes to move the drivetrain a certain speed. Take a calculus class, as well as a university physics class, and you will understand.
It is about accelerating that drivetrain. The faster you are accelerating that drive train, the more hp it robs, hence why it is on in percentages. If you knew ANYTHING about the physics of motion, you would understand.
So, simply put, the more acceleration you apply to the moving parts of the drivetrain, the more hp they rob. Layman's terms at least. THAT, my friend, is why it is a PERCENTAGE, NOT a constant. It's not about the amount of hp it takes to move the drivetrain a certain speed. Take a calculus class, as well as a university physics class, and you will understand.
It has been 39 years since I graduated High School, so please excuse me for not remembering something I was probably taught nearly 4 decades ago.
Case in point.... my car 'accelerates' it drive train to 100 mph in about 9.7 seconds, a stock Mustang takes about 14 seconds to do the same thing, so what you are saying is that my supercharged Mustang uses a higher PERCENTAGE of power to move it's drive train up to a certain speed than the stock Mustang... for arguments sake let's say 18% vs. 12% to 100mph.
Is that what you are saying?
Thank you
07-01-2010, 02:17 AM
#5
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Join Date: Aug 2007
Location: AZ
Posts: 2,463
I expected to bo semi-flamed for this one, so thank you for not being a jerk and insulting me.
It has been 39 years since I graduated High School, so please excuse me for not remembering something I was probably taught nearly 4 decades ago.
Case in point.... my car 'accelerates' it drive train to 100 mph in about 9.7 seconds, a stock Mustang takes about 14 seconds to do the same thing, so what you are saying is that my supercharged Mustang uses a higher PERCENTAGE of power to move it's drive train up to a certain speed than the stock Mustang... for arguments sake let's say 18% vs. 12% to 100mph.
Is that what you are saying?
Thank you
It has been 39 years since I graduated High School, so please excuse me for not remembering something I was probably taught nearly 4 decades ago.
Case in point.... my car 'accelerates' it drive train to 100 mph in about 9.7 seconds, a stock Mustang takes about 14 seconds to do the same thing, so what you are saying is that my supercharged Mustang uses a higher PERCENTAGE of power to move it's drive train up to a certain speed than the stock Mustang... for arguments sake let's say 18% vs. 12% to 100mph.
Is that what you are saying?
Thank you
07-01-2010, 02:34 AM
#7
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Join Date: Aug 2007
Location: AZ
Posts: 2,463
Drivetrain loss percentages are relatively even across the hp board (for an individual car). Focus on this equation, which is a proven law in physics, Force = Mass x Acceleration. Now let's single out ONLY the drive train parts, and throw in some hypothetical numbers. Say the drive train has a mass of 20 lbs. And it takes "Y"hp (unknown) to move that "c" mass (constant mass never changes) at the rate of acceleration of "X" (X represents the rate of acceleration for example 5mph/sec). Y = C x X. Well if you add hp to the car, then the rate of acceleration of the drivetrain parts MUST increase. Guess what that means? That means it will take more hp (or rob more hp) to move that constant 20lbs at the higher rate of acceleration. Y+5= C x X + 5. That means that since the rate of acceleration increased, and mass is a constant, that means that the Force (Y or HP) also had to increase.
07-01-2010, 08:25 AM
#8
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Nope, you completely missed my point. It's not about speed. It's about acceleration. Speed = distance divided by time (V=d/t)... Acceleration, on the other hand = change in distance divided by the change in time (a = Δv/Δt). You don't NEED to have 500hp to reach x speed, but you DO NEED X AMOUNT OF HP TO REACH Y (the max amount of acceleration)
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