Jim bailey - 928 International

On the other hand the early XKE roadster and fast back coupe the roadster with top up was better ! As I said it is NOT intuitive.

docmirror

It almost certainly will. That does not mean that there is less total drag on a 747, but the Cd will probably be quite low. It is a very efficient design, flush riveted over the entire surface, fillets, and foils in turbulent areas, exemplary design all around.

Doc

mark kibort

i wouldnt go as far to say a box will outperform sleek lines, but i think i know what you are saying. actually, a blunt facing object, with a teardroped rear end will be more aerodynamic than a bullet with a flat rear end! The point i think you trying to make here, is that C/D is a dimensionless factor. this means, the CD for a 747, is pretty close to a model of one many times smaller.

the reason a BMW has a lower CD , is because its much narrower as proportioned to the 928 and with its little wing and shape, gives it lower drag numbers. A great example of this is the 24hours of daytona last year. there were two GT3 RS porsches. from the side view, you couldnt really tell them apart. same wings, spitters , etc. however, the big flares, which made the GT3RSR about 6" wider, made for a much less aerodyanamic shape. for the narrow body GT3, which was fitted with all the GT3RS equipment, it was able to get 10mph faster down the straights due to the lower CD. certainly would hurt a little being narrow on handling at tracks like laguna and sears, but at daytona, it was the right trade off. My rear end on my 928 is near 76" wide. the BMW is in the 66" range in width. thats 10" wider!!!!! now, if you made a 928 as narrow as a BMW, it would blow it away as far as C/D. because the shape is sleek, but not perfect by any means. Ive heard the 928 is better with drag if it was pointed backwards! (measured from the rear!)

mk

heinrich

Mark, a box will ONLY outperform a sleek design IF it has the right GEARING ....

Nicole

Back to the original question: How it is measured...

When the cars are put into the wind tunnel, they sit on very sensitive scales that can measure forces in many directions. This way, it can be measured how hard the car is pushed back, and how much each wheel is pushing down or not.

Remember that aerodynamics for street driven cars include a lot of other issues that might not be as relevant to race cars. For example, they test for cross wind performance (car is turned sidevways at various degrees), wind noise, how water/dirt moved (to keep the side windows clean), and a host of other things.

mark kibort

you mean if a box has more HP???



there is a comparison from my aero days that talked aobut the drag of a round strut (like on a biplane) vs an aerodyamic, air foil shapped strut. the round strut had the same drag as a aerodynamic shape 1/5 its size!!. so yes, a box can have the same drag as a air foil, (bmw vs a porsche) but it takes a much larger air foil to match the drag of a box! someone once said, " give me enough power and ill fly a rock" (thrust to weight ratio)

MK

heinrich

Louie Ott likes to call (and I agree with him) the Subaru's Phone Booths. I kinda think maybe we're in search of a more slippery phone booth here with the appropriate gearing of course

Jim bailey - 928 International

A bit of borrowed input .."Different shapes have very different drag. The size of the drag for a given shape is usually measured by the drag coefficient, CD, which is defined as the drag force per square metre frontal area of the object. *)
The picture shows a hollow hemisphere open towards the wind, just like the cup of a cup anemometer or like a parachute. Such a shape has a very high CD of 1.42, whereas its CD is only 0.38 if you turn it 180° around a vertical axis. A modern automobile typically has a CD in the range 0.27-0.35. A runner has a CD about 0.5, a racing bicyclist about 0.4. An airfoil shape used on aircraft wings or rotor blades, typically has an extremely small CD about 0.04. **) " I seem to recall that a square meter of a flat material (no real thickness) perpendicular to the flow gives the base 1.0 CD figure. One you add thickness and shape to the back of that flat square meter everything starts to change. But always compare frontal areas.

IcemanG17

Jims gotta pretty good.....essentially a flat surface perpendicular to the wind would be 1.0.....some jet fighters are in the low .2 range.....

The other aspect of drag is frontal area...you can have a horrible drag # of say .40...but with 5 sq ft of area the total frontal area is 2 sq feet...vs a very aerodynamic .2 drag with 10 sq ft frontal area also ends up with the same effective drag area of 2 sq feet? So you really need both #'s to determine a cars really aerodynamic drag.....

On that note...S4's are between .34-35-36 or so depending on flaps....but what is the frontal area?

docmirror

I jsut thought he wanted to know if the number quoted was headlight up or down. That question was answered. Cd is a ratio of the flat plate(frontal area in laymans terms)to the drag induced on an object under test. So, a flat plate with sharp edges will have a Cd very close to 1.0 if it is perfectly perpendicular to the relative wind. To determine the Cd of an object of shape, it's flat plate (frontal) area is determined in square meters, and the function plate drag of that shape is tested for the base number. Then the object is subjected to the same relative wind, and the force is measured using strain gauges. The results are the divsor in the Cd value.

Ex: Frontal plate area of 928 = 12.55 SqM and presents xx NM of drag as a plate. Car is tested and results are yy.y NM, then the formula for Cd = yy/xx. It is usually a number less than 1, as Jim pointed out in his research, there are shapes that have higher drag than that of a flat plate.

Very strange things happed when a plane is parallel to the induced wind. Surface conditions can create a boundary layer of air, or there can be quite a bit of surface turbulence, making for interesting drag numbers when the surface is quite large.

Doc

Normy

N <--- 4 years as a flight instructor at Flight Safety Academy, Vero Beach, Florida. I was assistant chief instructor when I left, and taught ground school often...

including aerodynamics.

NOW....

Bear with me while Dr. Normy explains the whole f*cking drag thing to y'all!

Aerodynamic drag is effected by several things: first of all, frontal area. Second, how smooth that "frontal area" actually is. Third, how thick the air is.

Now bear with me: if you double your speed, it has been proven in a wind tunnel that drag quadruples!

So...all that means that the following is true:

D = 1/2 r v2 x Cd

What this says is that "Drag equals 1/2 the air density times the velocity squared times the coefficient of drag"

Now: there are a couple of things I want to point out about the whole drag formula [Here's where Normy "screeches" the chalk on the board; sorry~] to y'all. First off...you see here where it says "V2"? That means V squared. The implication of this is that when you double your velocity you automatically cause the final outcome...the "D" or the total drag to QUADRUPLE! Go from 30 to 90 miles per hour? You are going to see NINE times the aerodynamic drag, registered in pounds or kilograms!

Second...you see that component of the formula "Cd"? That's my way of saying Coefficient of drag. If the car is pretty slick, then it will have a low Cd, of like .29 or something like that. I think the first 928's were pretty draggy, in fact I think that they had far less drag in reverse than they did forward!

[It occurs to me that though that is probably true...the car wasn't designed this way to prevent lift]

If you look at the implications of the Cd on the formula, you recognize that it essentially makes the whole formula greater. If you derive the formula, and are good with algebra, you might even be able to take it all apart and reassemble it in a way that essentially assigns a square, vertical block of stone or metal to a perfect shape of the same frontal area. That is, if you think of it in one way, you can almost call the Cd the "penalty frontal area" for imperfect aerodynamic design.

I don't know what the Cd of the 747 that I fly, but it is certainly pretty low; a draggy DC-3 was reportedly around 0.17, and I suspect that a 747 with its high-speed aerodynamics will be somewhat less. Nevertheless, the 747 will never be able to slide its' 231 foot-length through the air as easily as that DC-3... or for that matter, my 928; simply because the 747 is the size of the average elementary school. My S2 can supposedly go 270 km/h, which is about the liftoff speed of the 747 at max weights. For some reason, I suspect an 833,000lb airplane 195 feet wide and 231 feet long has more drag than a 3300lb car 73 inches wide and 50 inches tall at the same speed....

N!

tomcat

You know Normy, you didn't say it but maybe you were implying it. The profile of a 928 does look like an airfoil backwards (relative to the relative wind). And a very low aspect ratio wing. It definitely would create lift backwards. I don't know if 150mph would be enough to overcome the weight, but I would assume enough to make the car feel a little lighter.

nee.dragger

Yeah, it looks like a fish. A damn ugly fish at that. No way I'd be caught dead driving one of those things. And fish are just food for the hungry. I'd rather have a predator of the sea, a shark is a good start.

Normy

It's not just an implication...the first 928's really weren't any more aerodynamic than the first VW Golf/Rabbit. They indeed tested out to be lower drag going backward than forward! Tony Lapine's design was incredible and timeless, and low drag overall, so we are talking relative numbers here.

N!

SwayBar

Normy, nice write-up!