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Old 04-06-2008, 02:47 PM   #1 (permalink)
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Is aero. vs. MPG change linear ?

If you have two identical cars, but one gets 10 MPG and the other 100 MPG, how will the change in fuel economy be affected by a 50 % reduction in drag ?
Is there a set figure, or does the amount change based on the initial MPG figure ?

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Old 04-06-2008, 03:37 PM   #2 (permalink)
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two identical cars, but affectet by various FE?
Identical road, same speed?
Perhaps different gear ratio or perhaps, I do not understand!
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Old 04-06-2008, 04:26 PM   #3 (permalink)
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( If all variables are identical except fuel economy, how will the fuel economy be affected by aerodynamic changes if one car has an engine that gets lower fuel economy than the other in ... other words .)
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Old 04-06-2008, 04:28 PM   #4 (permalink)
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( It is a hypothetical question )
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Old 04-06-2008, 04:50 PM   #5 (permalink)
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ok, if I undertand correctly, you consider two equals veicles without the same engine.

Te answer can be find from one simulation.
Download excel attached.
In the section "input e calcoli" in orange the BSFC, in green fuel consuption in KmLiter, in "C" colum, the Cd coefficient.
You can change the Cd whit the same BSFC and read the changement in the Km/L
note, impact of Cd on Km/L it is consequence of speed.
Answer at your question, change in function of veicle speed.
I think that now, you have all answers.

bye
Attached Files
File Type: xls aero2.XLS (48.5 KB, 43 views)
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Old 04-06-2008, 05:22 PM   #6 (permalink)
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It all depends on the balance of drag forces on your car. Generally, there are three drag forces that use up the energy being applied to the wheels: rolling resistance, air drag, and changes in inertia (starting, stopping, etc.).

When you are plodding along at a steady 5mph, almost all of the drag on your car is rolling resistance. When you accelerate quickly from 0 to 15mph, the drag is split between inertia and rolling resistance. When driving at a steady 65mph, the majority of the drag is air resistance, followed by rolling resistance. Do you see how different scenarios have different balances of drag?

To determine the affect an aerodynamic chage will have, you have to determine what percentage of total drag is attributed to air resistance. Lets say it is 60% of total drag at the moment, the other 40% being rolling resistance (i.e. highway cruising). A 50% reduction in air drag will halve the amount of air resistance.

The vehicle now experiences 70% (30% + 40%) of the original total drag, 43% (30% / 70%) of which is now from air resistance. Theoretically, you should see a 30% improvement in mileage in those conditions. I hope the math is correct...

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Old 04-06-2008, 06:04 PM   #7 (permalink)
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hahah LostCause Iwanted answer like you, but do not have the ability with Engliss
many suimple with the xls file
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Old 04-09-2008, 07:09 PM   #8 (permalink)
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Cd vs MPG

According to the S.A.E.Journals and Transactions,a 10% drag reduction will yield: 5% increased mpg at 55-miles per hour,and 6% increased mpg at 70-miles per hour.This weekend's testing has given me some data to suggest that a 10% drag reduction is good for 11% better mpg at 100-miles per hour.The results are sketchy,so lets not take that to the bank yet.The data for 55 and 70 is good and General Motors Aero Lab is comfortable using the relationships.( I believe they did the original research ).
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Old 04-09-2008, 10:13 PM   #9 (permalink)
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Considering only drag induced by the fluid P=F*v=0.5*(density of air)*(velocity)^3*(frontal area)*(coefficient of drag)

The truth is always in the equations. Power has a linear relationship with the coefficient of drag, all other variables being constant. The equation doesn't always really work out because of frontal area approximations for irregular shapes, Cd's are generally never exact, and density is truly variable depending on the fluid conditions and the boundary layer, both thermal and hydraulic.

(I know it's been said, but I and others may place our faith in the math)
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Old 04-09-2008, 11:07 PM   #10 (permalink)
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Quote:
Originally Posted by GenKreton View Post
Considering only drag induced by the fluid P=F*v=0.5*(density of air)*(velocity)^3*(frontal area)*(coefficient of drag)

The truth is always in the equations. Power has a linear relationship with the coefficient of drag, all other variables being constant. The equation doesn't always really work out because of frontal area approximations for irregular shapes, Cd's are generally never exact, and density is truly variable depending on the fluid conditions and the boundary layer, both thermal and hydraulic.

(I know it's been said, but I and others may place our faith in the math)
I don't agree... The truth is always in the empirical evidence (of quality methodology). The solution (and understanding) is always in the equation It's semantics with significance

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