I know the Z has got a very good one, in the real world though does it make any difference? I mean of course its not square like a truck so theres alot less drag but say compared to another car in its class, S2000?

 

does it make any difference? well, that depends on what you want to do.

first and foremost, the coefficient of drag (Cd) is just that, a coefficient. it's a constant that goes into some equation. the actual resistance created by drag is also dependant on another very important detail, the cross-sectional area of the object (that is, the size of the object when you're looking at it head-on).

if you're comparing against an s2000, you may have a better Cd, but his car is shorter (height wise) and narrower, so he's got a smaller surface area pointed at the wind. in the end, i'm betting the difference in force of drag will be too small to make any significant difference.

the other point is that drag doesn't really play any significant part until you reach high speeds. drag increases by the square of the velocity. that is, if you double your speed, you quadruple the force of drag. at around 70 mph it starts to be pretty significant. if you're going on a long, high speed run, a low Cd may help you win (think indy cars).

to compute the force generated by drag (Fd), use the following equation.

Fd = (1/2)(Cd)(p)(v^2)(A)

p is the density of air. it's 1.225 kg per m^3 at sea level and 25 degrees C.
v is your velocity, in meters per second.
A is the cross-sectional area, in meters squared.

here's a couple of computations. i used a rough measure for A by taking width of the z times the height. the true value is somewhat less than this.

height: 51.9 in. (1.31826 m)
width: 71.5 in. (1.8161 meters)
A = w*h = 2.394091986 m^2

at 50 kilometers per hour (31 mph)
Fd = .5 (.29) (1.225 kg/m^3) (13.8888889 m/s) (13.8888889 m/s) (2.4 m^2)
= .5 * .29 * 1.225 * 14 * 14 * 2.4 (kg m) / ( s^2 )
= 83.5548 Newtons

a force of 83 Newtons is the same as the amount of force you have to use to hold an 8.5kg (18.6 pound) object in the air without falling. that's about the weight of a heavy bowling ball. this is the force of the air pushing your car back and slowing you down.

at 150 kilometers per hour (93 mph)
Fd = .5 (.29) (1.255) (41.6666667) (41.6666667) (2.4)
759.4428186 N

a force of 760 Newtons is about the same as the amount of force you would need to hold a 78 kg (170 pound) object in the air. quite alot more than a bowling ball (more like 10 of them). At high speed, drag becomes a real issue.

so, the real lesson here is that a lower Cd is one of the things that'll help you reach higher speeds (other things would be a thinner and shorter body, and of course more horsepower). if you're racing someone about your size at top speeds, it may give you the edge.


man, i have got to get me some sleep.
rands

 

Excellent post! I'll be reading that over a few times lol thank alot!

 

actually indy cars have AWEFUL drag. they use drag as a means to produce downforce through all of their spoilers.

sedans have better(less) drag than sports cars because in the perfect world the length should be like 11x that of the height, it just helps the air flow smoothly.

last thing not mentioned, Cd is fine and dandy. but in reality 1PSI low in your tires and your probably creating more drag than the entire car. tires have tons of drag(think of how hard it is to push a car on a flat surface, its not the air making it hard)

 

Nope, A is the surface area in contact with air. Cross-sectional is accounted for in the drag coef.

 

RANDS, great post! That should be a sticky.

"actually indy cars have AWEFUL drag."
True, but they translate that drag into downforce with great efficiency. Essentially, they're using their 600HP engine to help plant the car into the pavement. Aerodynamics are critical in the world of racing. F1 teams spend millions of dollars in computer modeling and simulation to tweak every square inch on the car. (By the way, these simulations take days to compute on even the fastest processors that money can buy!!).

Actually, on street cars most of the aerodynamic drag goes towards lift. (A car roughly resembles an airplane wing). The 350Z has zero-lift due to proper air control under and around the car...this is why this car has a far better Cd than the similarly shaped Audi TT.

Here's a real world application that goes with RANDS post:
Aero engineers typically refer to drag in "counts." A 'count' is a Cd of 0.001. They estimate that 10 counts difference in the Cd is about 1 MPG at highway speed, give or take. So if you compare the following cars:

Audi TT: Cd = 0.34
350Z (track): Cd = 0.29

If everything else were equal between these two cars, the 350Z would get 5 MPG more than the Audi just from the reduction in drag. With a 20 gallon tank, that's an additional 100 miles!

 

Actually, RANDS is correct. The formula he used does not take into account the surface drag. That is a whole different (and very complex) formula. Unless the object is unusually long or wide, surface drag is pretty negligable compared to the effective frontal area.

 

from here

http://www.centennialofflight.gov/es...t/drag/TH4.htm

 

Ahhhh..... I just finished a 65 hour lab report studying drag force on automobiles and how it's all related to required engine output.... I'm finally on trimester break and I don't even want to look at those resisting force equations.

 

Silverstone, are you Mech E? Me too.

Funny how I don't use or see 80% of what I learned in Mech E classes.

 

Yea I'm a mechanical engineer... So you're telling me I wasted 65 hours workin' my @$$ off when I could have really just been driving my Z some more before storing her for winter? ahhhhh

hahah

 

i was using a simplified drag equation like the ones presented here:

http://www.ac.wwu.edu/~vawter/Physic...DragForce.html

it probably wouldn't do for fluid mechanics, but it's definitely good enough for my purposes here (especially considering i don't have anything close to the relevant numbers for surface area). no need to go overboard.

 

Trimester? What school do you go to?

 

yikes! I'm bein' schooled!

 

oh come on, you know what I mean. Mech E has several fields. Some of which are

Aero & Fluid Mechanics
Statics
Dynamics & control
Thermodynamics
Design

You'll never see most of it again.

 

You guys are the biggest NERDS. And my IDOLS...

 

I go to the Milwaukee School of Engineering... it's an insanely hard school that goes really quick. 10 weeks per trimester, and then all of a sudden it's finals.

 

Nope, what your thinking of, such as various conditions of where air hits the surface and shape of the car etc etc, is taken account in Cd(Drag Coefficent). Which is a complicated series of calculation which to do calculate requires a lot of ccomputer time, to find accurately requires doing experiment in wind tunnel.

 

MOZEE is in the hizzouse! I kinda had a feeling that you went there. Me? MSOE Alum, 2000. Well, it's good to hear that they're still brutalizing impressionable young minds over there! That school is no joke...I personally think it is one of the best (and hardest) engineering schools in the country, MIT included. I'm back in town at the moment, so I'm thinking of stopping down at the bookstore to get an Alumni sticker for the new Z...just to show everyone who paid for it!

Well, good luck. Just keep your eyes on the prize...oh, and it IS possible to graduate from MSOE...it's only the "Einsteins" that make it out in 4.

350ZROADSTER, I think you're right...I'll have to look up on this. I know that it makes a big difference on whether the flow is laminar or not (which has to do with the shape of the object and the linear speed of the flow)...hmmm maybe it was the equation for Cd that was the real ******#$&!...

 

hahah that's good to hear... so far it looks like i'm on pace to make it in 4 and a trimester. anyways, how long are you in town? we should take a cruise in our Z's. I'm like 1 of 3 Z's I've seen around this entire milwaukee area.