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Discussion Starter #1
The goal of EPA mileage tests is to control all test conditions to be able to reliably compare different vehicles. The tests are run indoors, on a dynamometer, by professional drivers, with antiquated 55 mph era driving profiles. The maximum highway speed is 60 mph.

I drive my vehicle outdoors (has an average headwind of 7 mph), on a real road surface, I’m in a hurry to get home, so I gun the gas, there are other vehicles on the road so I have to speed up and down to accommodate them, I travel on a highway at 70 mph, and I have my A/C on. Normal driving can never give a result anywhere close to the EPA rating.

Using my dynamic model cruise range (See attachment) and assumptions for the Volt, I estimate that an average 7 mph headwind, a typical road surface, and A/C (accessory power increased from 100W to 700W) would each reduce Single Charge range @45mph by 15%, 7.7%, and 14% respectively. The combined effect (not the simple sum of individual components) is a 30% reduction. If I increase the profile speed by 5 mph to 50 mph, the total reduction is 36%. A 40 mile EPA rating translated to the real world is estimated to reduce the single charge rating to 25.6 miles. If anybody has a better way of doing this estimate, let me know.

For example, see http://www.wired.com/cars/futuretransport/news/2007/05/hybrid_mpg

Tom
 

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Sounds good to me Tom. Consider this... When I first get my Volt (after the required break-in period of course) I'm going to be burning rubber all over the place. Out pacing any car I can and generally acting like a child. I expect to get around 10 miles of all-electric range per charge or less! Afterwards, when I'm more concerned about not burning petroleum, I will drive like my grandmother. I think people will grasp this concept and be OK with it. Drive like the devil and kill your range or drive like a grandmother and save the world. It might even become cool to drive slow. ;)
 

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Discussion Starter #4
It depends on where you live.

Wind is a two dimensional vector with elements that are normally distributed, but uncorrelated. It is skewed with excess kurtosis. Statistically, it is the average of a Rayleigh distribution. It depends on where you live. Window glass has to be thick in Chicago.

See http://www.awea.org/faq/usresource.html
 

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Tom, I am referring to your, not the wind, direction. Usually for me when I have a headwind going somewhere, I have a tailwind on the way back.
 

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"headwind" happens much more often than a tailwind, because even a sidewind cause an increase in aerodrag, and has the same effect as a headwind. In other words, (example only; I'm not an engineer) a 20mph sidewind may have the effect of a 5, 10, 15mph headwind, depending on the exact direction.

This explains why one encounters some degree of headwind (or the equivalent) much more frequently than a tailwind.
 

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Discussion Starter #7
It doesn't average out like flipping a coin

Resolution of forces: The motor applies a traction force, i.e. this is a force that is tangential to the angle of the road. This is the primary vehicle force, which at constant velocity cruise equals the opposing dissipative forces. All other forces have to be resolved relative to this road tangential force. Ordinarily there is some average wind velocity blowing, say 7 mph. To make things easy, let’assume that the road is at the same angle as the local topology of the land. Then the wind velocity blows at the same angle as the local topology, i.e. the road. The wind force is a hydrodynamic fluid drag. Let’s look at this drag force relative to the vehicle, i.e. translate to a frame of reference in which the vehicle is standing still. Then when the car is traveling at say, 50 mph, in the vehicle frame of reference, the air is traveling at 50 mph toward the vehicle. In this frame the total drag force velocity is 50 mph plus the headwind speed. Tailwind reverses effect. The aerodynamic drag force depends on the square of this total velocity. It’s effect is greatest for highway driving. Headwind effect is proportionally (50 + 7)^2 = 3025 while tailwind effect is (50-7)^2 = 2025. Head wind effect @ 50 mph is 76% greater than tailwind effect. It’s not like flipping a coin. It does not average out.

Also, as pnieder mentioned, there is a cross wind effect. This depends on wind direction and is far more complicated to assess. The drag coefficient is smaller, but the cross sectional area is larger. It also adds to tire rolling resistance (yaw) losses. On average, this loss occurs regardless of wind direction.

There are some proposed changes to EPA test methods to more realistically assess mpg.
See “Fuel economy labeling of motor vehicles revisions to improve calculation”

http://books.google.com/books?id=9CnIlO94DqgC&pg=PA91&lpg=PA91&dq=+tailwind++drag++vehicle&source=web&ots=Uet2RP79BW&sig=USfpDrh3JwbQOKS4ZWTW-jeGGZ4&hl=en
 

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I can empirically vouch for Tom's headwind/tailwind analysis. Our cycling club sometimes does an event called a "time trial", where we ride a fixed distance one way on a road, then turn around and return to the start. This gives us both headwinds and tailwinds of equal distance on the ride. The higher the average wind velocity, the slower the average time, regardless of which direction the wind is blowing.

Rolling hills have a similar effect. We gain potential energy as we climb and convert potential energy into kinetic energy as we descend, but because we descend faster than we climbed, we lose some of that potential energy to increased air and rolling friction. A flat route, with no wind, is always the most energy-efficient.
 

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"headwind" happens much more often than a tailwind, because even a sidewind cause an increase in aerodrag, and has the same effect as a headwind. In other words, (example only; I'm not an engineer) a 20mph sidewind may have the effect of a 5, 10, 15mph headwind, depending on the exact direction.

This explains why one encounters some degree of headwind (or the equivalent) much more frequently than a tailwind.

Would that not be parasitic drag as opposed to headwind?
 

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In sailing, they call it "apparent wind", the resultant of vector addition of the air velocity vector parallel to the direction of travel and the air velocity vector perpendicular to travel. For drag force calculation purposes, I think they only use the parallel velocity vector component.
 

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Well, the new EPA tests are a bit tougher, thereby lowering their MPG ratings for a certain vehicle. Here is where my problem with the EPA lies..no matter what they say a vehicle is supposed to attain in their mileage tests, every single car and truck I have ever owned has gotten better than their rating. My old '05 Scion tC was rated at, iirc, 29MPG hwy. Driving on the highway (I95) doing the speed limit using cruise control, I managed to get 37MPG. No joke, I thought I did the math wrong and had my wife recalculate and I was right. Our '03 1.8T Jetta is averaging 34MPG in mixed driving, my '05 Nissan Frontier 4.0 is averaging 20-22MPG in mixed driving. So basically, what I'm trying to say here is that the EPAs ratings are virtually worthless. It all depends on how you drive, and how well you maintain your vehicle, that will dictate how good mileage you are going to get.
 

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As JJSKI78 mentioned the EPA did just redefine their testing parameters for cars starting with 2008 model year. I believe the new testing procedures account for some circumstances Tom mentioned, so the same speeds as the testing should get you much closer to the ratings than 30%. Although the ratings are more realistic now, they are still just 2 discrete results. In this day and age, they should include a graph like Tom's with their city/highway rating test points highlighted. This way, individuals could reasonably determine what mileages they should expect.

Tom, thanks for the explanation of the "effective" headwind. I now understand what you mean by this, but 7mph doesn't seem right. Let's use an average actual wind speed of 10mph for an example. If traveling at 50mph directly into and directly with this wind for the same amount of time, we would experience an "effective" headwind of 0.99mph (sq root of [((50+10)^2 + (50-10)^2)/2] - 50). This doesn't even account for the fact that the winds direction can be at any angle to the direction of motion. I believe we would accurately use the actual average wind speed times cos45. Unless I'm missing something, 0.9mph seems like a realistic average "effective" headwind.
 

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Looks good to me. Are you going to incorporate this into your range model since that had Vw=7mph in drag force equation, correct?

Did your revised models reduce the auxiliary loads for low power accessories since GM has indicated the plan to have a separate battery for 12V loads?
 

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Discussion Starter #16
Supplemental 12V Storage Battery: Good for 1.6 Li Ion charge cycles.

Koz,

Yes I will have to make modifications. Assume we recharge a 200 Ahr 12V battery every day. We don't want to discharge it more than 50% because that will impact lifetime. So we have 100 A-hr per charge. From the list below, we see that Average Accessory Power may be 740W = 62A = 1.6 hour. With a 40 mile range per single charge of a Li Nanotch battery at an average of 40 mph gives 1 hour per single Li battery charge. The 12V is good for 1.6 Li Ion charge cycles.

Accessory, Power, DutyCycle, Avg_Power
A/C – Heating 700 0.8 560
Idle Control Electronics 50 1 50
HVAC Air Blower 10 0.8 8
Display Lights 3 1 3
Headlights – Low Beam 55*2 0.2 22
Headlights - High 120*2 0.1 24
Tailgate LED Lights 5*4 1 20
Interior Lights 1*4 0.2 1.6
Power Steering 10 0.5 5
Brake Hydraulic Pump? 10 0.5 5
Battery Case Blower 10 1 10
Cooling Fan 20 1 20
Wipers 10 0.1 1
Stereo 20 0.5 10
DVD 10 0.2 2

Average Total Power= 740W
 

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Are you attempting to come up a worst case average load or average load for average American driver? What ambient conditions are you considering for this average load?

-Headlights will most likely be HID, so their full load s/b 76W.
-24W seems way too high for current generation LED running tailights.
-Last I've seen is that there will liquid thermal regulation for the battery, so I believe there would be a separate compressor & pump for that.
-I see 700W load for cabin heating/cooling. I think GM had said they were not planning on using resistive heating, so for a reversible compressive using the same load for heat/cool would closer to actual use. 700W seems reasonable but I don't have a good frame of reference for automotive loads. Is this what you've found for a high efficiency system?
 

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Figure 3/4 ton nominal of AC cooling at peak conditions. At 1 HP/ton, that's 3/4 HP of compressor shaft power. 700 watts of electrical would be about right or maybe a bit on the conservative side if GM does some fancy compressor/motor efficiency tricks.
 

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I average 45mpg, I do not drive much in excess of 65mph or whatever the speed limit may be and I use my cruise control alot.
This also give me around 48-50 miles of electrical power (as long as I don't try to climb any mountains).
I personally find the EPA rating to be too low, I know of others who get more electrical mileage and better gas mileage than I.
I almost always drive with my A/C in ECO at fan level 1 which makes the car comfortable yet not too chilly.
 

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Lets not forget the EPA figures are derived from non ethanol gasoline which has a higher energy content than gasoline with 10% Ethanol which is nearly 99% of automotive gasoline sold in the U.S. Here in Oregon it is mandatory to use 10% Ethanol in gasoline autos. Many people claim they get as much as 3-5 mpg more with non ethanol gasoline.
 
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