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Old 04-04-2008, 07:55 PM   #11 (permalink)
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Great find, fabrio. Very interesting.

Quote:
Originally Posted by Otto View Post
I wish I knew how to post scanned images here. If so, I could show you illustrations out of Dr. S. Hoerner's "Fluid-Dynamic Drag" page 9-16, with cites and sources.
The easiest way is to post your image to a file hosting service, such as http://xs.to/ and then obtain the file link and use the little "link" button above the text editor to add it to your post.

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Old 04-05-2008, 05:38 PM   #12 (permalink)
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Quote:
Originally Posted by Otto View Post
I wish I knew how to post scanned images here. If so, I could show you illustrations out of Dr. S. Hoerner's "Fluid-Dynamic Drag" page 9-16, with cites and sources.

The optimum outlet is not as shown in the BMW photo, or anything close. It is more like the bottom drawing D in the above illustration, but with the outlet channel as close to parallel to outside flow as possible, and with the downstream intersection radiused.

Fast flow of the outside air means low pressure, which in turn sucks the outlet discharge into the free stream. If the outlet is as close to parallel with the free stream as possible, and the downstream intersection radiused, the outlet air can merge with the free stream with the least possible turbulence and greatest efficiency. Think of it as an on-ramp to a freeway, where the cars represent air molecules.

BTW, some folks use NACA submerged inlets and turn them around backward as outlets. Ungood. These NACA arrowhead-shaped inlets look sexy and work pretty good as inlets, but are not efficient as outlets.

Bottom line: If outlets and internal ducting are good, the stagnation point inlet at the nose can be quite small.
fantastic!

for posting your photos, yuo can also to use http://imageshack.us/ which to permit to add to your browser an pratical and profit toolbar for instantaneous upload of images and videos.

I attend innovation
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Old 04-05-2008, 06:24 PM   #13 (permalink)
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about the image posted by me, I have a question.
In the explanation text, it can be read that the Cd is in relaction on exposed area!
So, in the last case (C), the area it is equal to zero.
Now, if the area is zero, also the Cd is zero.
But, how can to konw the effective drag impact for last solution?
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Old 04-05-2008, 08:45 PM   #14 (permalink)
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We know that air flowing in/out of holes in a surface can directly affect Cd without changing projected area- eg. vacuum holes/slots in airplane wings; "active" pneumatic aerodynamic devices at the rear of some experimental vehicles.
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Old 04-05-2008, 09:19 PM   #15 (permalink)
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Quote:
Originally Posted by fabrio
about the image posted by me, I have a question.
In the explanation text, it can be read that the Cd is in relaction on exposed area!
So, in the last case (C), the area it is equal to zero.
Now, if the area is zero, also the Cd is zero.
But, how can to konw the effective drag impact for last solution?
This has kind of thrown me for a loop...

Form drag = Cd*A

Total Drag
Cd = cd + Cd,i

Profile Drag
cd = cd,f + cd,p + cd,w

Since cars aren't experiencing wave drag, they just combat profile drag (cd,f + cd,p)...that is skin friction drag and pressure drag.

Since that flat plate is operating in viscous flow, it will be subject to shear stress...therefore it will have skin friction drag.

cd,f = Df/Q∞*S

where S is surface area, not frontal area.

That plate is dealing with pressure drag due to the exit duct.

Cp = p-p∞/q∞

I believe cd,p requires the integration of all pressure coefficients across the surface area of the body and then the horizontal component solved. I'm about 2% sure on this one, though...

The Cl is determined from the pressure coefficients the same way, but lift is determined normal (perpendicular) to the body. I imagine the duct creates some form of lift, so I don't know how else this could be determined outside of experimentation??? The Cd,i is then determined from the Cl via

Cd,i = Cl^2/pi e AR

Sum it all up to form Cd and then we come to an area of contention...A. I believe in flat plates that A is determined from surface area. I take that from here:

Quote:
where A is a suitable body area, either the surface area of the body or the area of the body when projected onto a plane normal to the flow
So form drag = Cd * A

That's all I could come up with...a long and convoluted answer... In conclusion, I'm not 100% sure...but those are all the tools I have to work with. I look forward to anyone knowing the true answer posting a response...I hope this isn't a common pitfall of aerodynamics: applying equations where they don't belong.

Quote:
Originally Posted by MetroMPG View Post
We know that air flowing in/out of holes in a surface can directly affect Cd without changing projected area- eg. vacuum holes/slots in airplane wings; "active" pneumatic aerodynamic devices at the rear of some experimental vehicles.
What about form drag, though?

Form drag = 0.002 * 0
Form drag = 0?

That doesn't seem right, for some reason. Maybe it is?

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Old 04-06-2008, 04:31 PM   #16 (permalink)
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LostCause, if you spoke Italian, it would be all easier!!!!!!
sorry, but this is over my knoledge at the moment, again I do not have understand some symbols that you ave used.
Do you tasks that cd of 0.02 it is possible?
If yes, another interesting question is an surveying of quantity of airflow through solutions displayed in the scheme
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Old 04-06-2008, 05:39 PM   #17 (permalink)
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Don't worry, I didn't really answer anything

I do think that a Cd of 0.02 is possible. I think that if the duct is designed properly, the exit flow should come close to the freestream airflow. Any difference in speed will show up as drag.

Here are the symbols I used above:

cd = coefficient of drag
cd,i = coefficient of induced drag
cd,w = coefficient of wave drag
cd,p = coefficient of form drag
cd,f = coefficient of skin friction
Cl = coefficient of lift
Cp = coefficient of pressure
p = pressure at body surface
p∞ = freestream pressure
q∞ = dynamic pressure (1/2 rho V^2)
rho = air density
S = planform area
A = front plate area
pi = 3.14159
e = oswalds efficiency factor (based on lift distribution)
AR = aspect ratio (b^2/S)
b = wingspan
Df = drag force

Any questions? I have plenty...where's trebuchet03 when you need him...

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Old 04-06-2008, 05:57 PM   #18 (permalink)
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fabulous! thanks!

ok form many symbols.
Nok for cd,w, freestream pressure (what is the freestreem?), and planform area? it is the surface of fuselage of airplane (for exhample)?, oswalds efficiency factor and wingspan?

when your write:

cd,f = Df/Q∞*S
Df is the drag force?
and Q∞
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Old 09-27-2008, 12:06 AM   #19 (permalink)
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could you get part numbers for those underbody pieces? I wonder if they could fit any E46.
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Old 09-27-2008, 04:48 PM   #20 (permalink)
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drag

Quote:
Originally Posted by LostCause View Post
This has kind of thrown me for a loop...

Form drag = Cd*A

Total Drag
Cd = cd + Cd,i

Profile Drag
cd = cd,f + cd,p + cd,w

Since cars aren't experiencing wave drag, they just combat profile drag (cd,f + cd,p)...that is skin friction drag and pressure drag.

Since that flat plate is operating in viscous flow, it will be subject to shear stress...therefore it will have skin friction drag.

cd,f = Df/Q∞*S

where S is surface area, not frontal area.

That plate is dealing with pressure drag due to the exit duct.

Cp = p-p∞/q∞

I believe cd,p requires the integration of all pressure coefficients across the surface area of the body and then the horizontal component solved. I'm about 2% sure on this one, though...

The Cl is determined from the pressure coefficients the same way, but lift is determined normal (perpendicular) to the body. I imagine the duct creates some form of lift, so I don't know how else this could be determined outside of experimentation??? The Cd,i is then determined from the Cl via

Cd,i = Cl^2/pi e AR

Sum it all up to form Cd and then we come to an area of contention...A. I believe in flat plates that A is determined from surface area. I take that from here:



So form drag = Cd * A

That's all I could come up with...a long and convoluted answer... In conclusion, I'm not 100% sure...but those are all the tools I have to work with. I look forward to anyone knowing the true answer posting a response...I hope this isn't a common pitfall of aerodynamics: applying equations where they don't belong.



What about form drag, though?

Form drag = 0.002 * 0
Form drag = 0?

That doesn't seem right, for some reason. Maybe it is?

- LostCause
LostCause,I'll try and muddy the waters.The Coefficient of Aerodynamic Drag,encompasses all elements of drag for the body.---------------------------On submarines and aircraft they will use the fully wetted area of the structure,whereas in automobiles,the 1/2,in the drag force equation,converts the drag formula from wetted area to frontal area.This goes back to the "mirroring-effect",where the air actually sees two of your cars,joined at the wheels,one right side up,the other upside down, "below" the ground plane.------------------------ With the Cd,profile drag,skin friction induced drag, internal drag,and interference drag,etc.,are all contained within the Cd.----------------------- And Cd is only a function of shape.So like in my public displays,I have a very large New Beetle model,and a very small New Beetle model.They have identical Cds(given proper Reynold's Numbers),however vary greatly in frontal areas.And remember that a vehicles "drag-factor" is the product of the Cd times it's frontal area,or,CdA.------------------------- With respect to profile drag,it is singularly the largest component of drag for a car,and its something we can do something about,as we can alter it simply by changing the car's shape.------------------------- Pressure drag is a function of profile drag.The pressure behind a 2-to-1 teardrop is essentially the same as the pressure ahead of it.It's"stagnation" pressure is identical to it's "base" pressure acting on it's aftbody.----------------------------- For shapes other than teardrops,separated flow will create a "wake" of turbulence behind the structure which,from the law of conservation of energy dictates,must have a lower pressure than the forward stagnation pressure.The net difference in pressure is the" Delta-P" across the structure which represents it's "pressure drag".------------------------------------- The teardrop has NO form drag,or profile drag,or pressure drag,only skin friction drag,because it has no separation.If it yaws or pitches it may experience induced drag until it straightens back out,otherwise it has the lowest drag of any form yet investigated,excepting perhaps the "Morrelli" form.Thats a different thread.------------------ Hope this helps!

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