# How aerodynamic is a TT?



## Dr_Parmar (May 10, 2002)

as i was waalking back to my car today in canary wharf car park, 8) i was looking at the BMWs the TVRs the Porsches and the zillion TTs which seem to have sprung up around here, and i thought.. how aerodynamic is a TT when compared to the Boxster or to a Z3? anybody know what the drag coefficient is? and more importantly, anyone understand what a drag coefficient is? ???


----------



## Guest (May 14, 2002)

Not sure how accurate this is - check out http://www.thecollection.com/new/audi/tt_spec.htm


----------



## Stu-Oxfordshire (May 7, 2002)

I seem to remember a great picture being posted on the old old forum over a year ago of a TT in a wind tunnel with smoke being passed over the car in a simulated 80mph wind.

Associated with the pics were all the drag coefficients etc.....don;t know whether we can still search on the old old forum? Jae?


----------



## Ruffles (May 6, 2002)

IIRC, drag coefficients are a number only and an expression of the efficiency of the shape travelling through the air. I am not 100% sure why they give the frontal area as I don't see any particular relevance unless they are using this as the "definitive characteristic" for comparative purposes.

In terms of comparisons, I managed to sniff out the following:

BMW Z3 - 0.37
BMW X5 - 0.38
Porche Boxster - 0.33
Lotus Elise - 0.28
Mercedes CLK320 - 0.31

For those interested, I also managed to find this:

http://www.teknett.com/pwp/drmayf/dragcd~1.htm


----------



## justtin (May 6, 2002)

From "The TT Story":

Drag coefficient Coupe cD=0.34 Roadster cD=0.37

Frontal area of both is 1.99 m2

Justin


----------



## Ajs (May 7, 2002)

my understanding of drag coefficient and frontal area is that its a comparrison to the drag on a flat plate of equal area
if a flat plate of equal area (1.99m2) is placed in a wind tunnel, measure the force against it. Put a tt in there and measure the force against it.
ie a tt has 66% less force exserted on it than a flat plate, giving a cd of 0.34

Can someone add anymore


----------



## scottm (May 7, 2002)

AJS, that's interesting! So does that mean that a car with a drag coefficient of (say) 0.3 but a frontal area of 2.5m2 actually has more overall drag than a car with 0.35cd but an area of 2m2, even though the latter has a higher drag coefficient?

If so, then what's the point of magazines, manufacturers, etc, publishing drag coefficients, when they're totally meaningless without knowing the frontal area, which is hardly ever published?


----------



## Ajs (May 7, 2002)

ScottM
haven't got a clue mate, I was dredging the depths
of my memory just to get that, 
could be total bo***cks
???


----------



## mrfish (May 8, 2002)

Stolen from http://www.orst.edu/instruct/exss323/week9.htm as I have blotted the experience of my engineering fluid dynamics course from my mind.

"Profile drag force can be determined using the formula:

Cd* A * p *V^2
F = ---------------------
2

where Cd is "drag coefficient" a unitless quantity which depends on the shape of an object.

A is the frontal area (silhouette area) in square meters. 
p is greek letter rho, and is the density of the medium in kg per cubic meter. 
V is the flow velocity or relative velocity of the medium with respect to the object."

Example: What is the drag force acting on a an Audi TT Coupe moving at 150 mph? What engine power needed to go this fast?

Cd = 0.34
A = 2 m2 (approximately)
p = 1.2 kg/m3 (at sea level)
V = 66.75m/s (approximately)

(.34)(2)(1.2)(66.75^2)
drag force = --------------------------------- = 1818 N
2

Power needed = Force * Velocity
= 1818 *66.75
= 121 KW

Your TT 225 produces 165kW, yet has a max speed of 150mph => the gearbox, quaTTro system, tyres vortex drag & other friction account for the rest

Hope that helps.

Cheers,

Andrew


----------



## vagman (Sep 6, 2002)

Thanks Andrew, it's as clear as day now.


----------



## jgoodman00 (May 6, 2002)

Regarding what Andrew said, does anybody know how much horsepower is lost by the 4-wheel drive system??


----------



## mrfish (May 8, 2002)

The losses in the quattro system can be estimated from a dyno plot. During a run the dyno measures power at the wheel. Once maximum speed is reached, the car is put into neutral to coast down. This then gives the power needed to turn over the drive system. Adding this power back onto the power at the wheel gives the quoted power at the engine.

If you have access to a dyno plot, you'll be able to read this off. I'd expect losses will be about 10-15%, as typical efficiency of 2 meshing cogs is something like 95%. Just to put this into perspective, a low-tech bike chain is about 98% efficient. All this adds up to the FWD 180 car performing similarly to the 4WD 225 (the weight affects things too).

Andrew


----------



## Dr_Parmar (May 10, 2002)

ummmm original question..

how aerodynamic is a tt?

a)very
b)slightly
c) its awright
d) like a brick

;D


----------



## mrfish (May 8, 2002)

the answer is b.

It looks like a nice bar of soap so it must be b ;D


----------

