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Old 04-19-2007, 03:35 PM   #1
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Calling All Aerodynamics Scholars.

Question: In the wake my car leaves at 60 MPH how much negative pressure is sucking my car backwards? In total pounds, PSI, PSF, or all.

I have figured this myself and the numbers are so huge I can't be very confident of my answer. So I need and would appreciate any help and/or suggestions.

I am not an aerodynamics scholar. I am an aerodynamics fumbler. I do what seems right, looks right, feels right and what I and others have researched and pretty well proven. Occasionally you will have to forgive my methods.

I have a 1999 Ford Escort ZX2 Coupe. Overall length 175.2 in. Overal width 67.4 in. Overall height 52.3 in. Wheelbase 99.4 in. Track frt. and rear 56.5 in. Factory CD .33. Wt. 2560.

To the frontal area I have added a 5 inch air dam the width of the car. I am figuring I have in the area of 20 sq. ft. frontal area. For purposes of calculation I have been figuring this entire frontal area as being involved in producing the wake.

I have been wearing out a stretch of highway near me lately with plastic tubes taped and dangling all over my car. Connected to direct read WC (water column) gauges. I have made progress that I will tell about later.

I do have other aero items. Full body-flush front and rear fender skirts down to the body sills, sealed all around, front grille and openings are totally blocked and sealed, tire air deflectors front and rear, both front and rear of tires. Have a partial belly pan from the front bumper to the front wheels, completely sealed front and sides.

Before you grind me up too bad, please note my latest tank gaslog entry was 73.70 MPG. I have averaged near 70 MPG tanks for near 6 months. New EPA rating for my car is 22/30/25. With few engine and driveline mods, something is working. But looking for more.

Anxious to hear from you. Thanks CO ZX2.
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Old 04-19-2007, 03:43 PM   #2
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Somewhere somehow I ran across some utility/program/something that indicated that for a car like mine, very roughly, there was some 60-70 lbs of force being applied due to air pressure at 60 miles per hour or so. But big disclaimer.... my memory could be failing me, the estimate could be BS.... who knows. I remember thinking that that wasn't all that much... or maybe it was/did seem about right after thinking about it.
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Old 04-19-2007, 04:05 PM   #3
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You could calculate the force by the fuel burn rate but it would vary upon the efficiency of the engine. Air pressure in PSI x Frontal area in square inches = force in pounds. It will be a LOT of force - take a bathroom scale on a 2x4 with a square foot of board on the top of it and hold it out in the wind at 60 mph but have someone else hold onto it with both hands. Then put a rounded shape on it and see how that works.

1 psi x 20sqft x 12x12 sqin/sqft = 2880 pounds

I don't think it will be 1 psi of pressure.
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Old 04-19-2007, 06:16 PM   #4
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It is the net pressure differential that counts. At rest, the pressure is 14 lbs sq in, all around the vehicle on all surfaces. At speed there is an increase in pressure in the front (but shape based) and a more important decrease in pressure in back (again, dependent on shape). The theoretically perfect shape has zero drag, yielding zero pressure differential all around.

Hold your hand out the window at 60 and you can easily keep your palm open, suggesting the combined force on your hand is not that great..... that's maybe 1/4 sq foot and maybe what.... 5 lbs or less? Just guessing, but it isn't very high or it would slam your arm back. Extend your arm over the edge of a table and try to hold 100 lbs....
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Old 04-19-2007, 06:16 PM   #5
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Quote:
Originally Posted by CO ZX2 View Post
[FONT=Verdana]

I have been wearing out a stretch of highway near me lately with plastic tubes taped and dangling all over my car. Connected to direct read WC (water column) gauges. I have made progress that I will tell about later.
That's the way to do it... Hehe - instead of CFD (computational fluid dynamics) you're doing EFD (experimental fluid dynamics) <-- I'm fairly sure I just made that up...

But you're on the right track.... If you can measure the pressure equidistantly, you can make a pressure map of the rear end of your car... You know the pressure at discrete points (a rather large matrix), you know the surface area of those discrete points. Then, it's a matter of solving that matrix - which is just matrix algebra. At least, given the 30 seconds of thought I have put into this, I think that's how it could be done.

The concept of that calculation isn't so much aero as it is finite analysis (taking a big problem - breaking it up - solving the little pieces - putting the pieces back together). I can explain it more in detail if you want I'm actually curious of what my car's pressure field looks like :P I wish I had a vacuum gauge sensitive enough to measure

Now, if you knew the force pulling you back -- we could solve this backwards A slightly more involved process, but I think it's a lot of fun (oh, crap... I can't believe I just said that!)
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Old 04-19-2007, 08:19 PM   #6
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C'mon guys. Gimme some kind of a guess. There has to be a general comparison for a fairly standard, commonly designed car.
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Old 04-20-2007, 07:36 AM   #7
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I did. I remember some web site or utility saying about 70 lbs total for a VX. After thinking about it, it made some sense. It can't be too high, else the work over 100's of miles would be ridiculous. Some very simple physics should get you a sense.... many of these problems are best solved by guessing the answer 1st, then working backward. Guess 100 lbs and caclulate power output over an hour at a certain speed. Then compare to the energy potential of a gallon of gas. If you get within 10x you are on the right track. 5X close. 2X.... maybe right on the money.
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Old 04-20-2007, 09:34 AM   #8
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9000 wh/gal * 5280 ft/mile / 40 mile/gal * 60 mile/hr /550ft-lbs/sec /745 w/hp = 173.9 lbs

Assuming a 25&#37; efficient engine resulting in 9kwh of energy per gallon approximate (9kwh /liter)

Also could calculate the downward slope of a hill and the car's terminal speed to determine the force involved much easier.
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Old 04-20-2007, 12:18 PM   #9
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With about 1/2 that going to overcome rolling resistance, etc and the other half the drag.... ie. 80-90 lbs of air resistance for the drag.

Or did I miss something?
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Old 04-20-2007, 12:21 PM   #10
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Let's see. Something like 1" of water over a 20 sqft area, uuhhh, works out to, uhh, doing this in my head here in the airport, about 100 lbs. It's in that range. Probably more like 0.5 inches of water.

Use your manometer to measure some pressures at the rear, come up with an average, multiply that by the area, and then convert to pounds.
for example:
0.5 inches of water x 24 square feet = 1 cu ft of water = 62 lbs of force.
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