Engineering

[joe strummer]

Interesting, the LS430 has a lower coefficient of drag than does the FR-S, 0.26 vs 0.27.

Wikipedia

[jawn]

The LS family of cars have always been really well engineered. Toyota does not mess around when it comes to the LS cars.

Another note, drag coefficients have to be multiplied by the reference area in order to really evaluate how much drag force is being imparted. Ultimately, you can't cheat big.

[humdizzle]

you have to factor in the frontal area too. scroll down to drag area on that page. thats a better representation of how 'aerodynamic' a car is. Cd only tells part of the story. '

[joe strummer]

Originally Posted by humdizzle

you have to factor in the frontal area too. scroll down to drag area on that page. thats a better representation of how 'aerodynamic' a car is. Cd only tells part of the story. '

So assuming the Lexus has a larger frontal area than the Scion (sure looks like it), it has more drag at a particular speed than the FR-S does at the same speed?

[jawn]

Originally Posted by joe strummer

So assuming the Lexus has a larger frontal area than the Scion (sure looks like it), it has more drag at a particular speed than the FR-S does at the same speed?

Yes.



The formula is fairly straight forward. That A is for area.

[joe strummer]

Very interesting. Thanks.

However, Wiki has this formula for drag coefficient-

is defined as:

where:
is the drag force, which is by definition the force component in the direction of the flow velocity, is the mass density of the fluid is the flow speed of the object relative to the fluid, is the reference area.

It appears that (frontal) area is already calculated into the result. So, although the FR-S may have a smaller frontal area than the LS, its drag coefficient is higher, and therefore its total drag at a given speed is higher than the LS.

What am I missing?

[jawn]

The reference areas of the two cars are different.

EDIT: You're dividing by the full area value of each car to arrive at the Cd. The total drag force is different for each car, too. It's a little circuitous, because in practice you'd likely measure the force in order to calculate Cd. Cd's just a handy way of saying how slippery something is per unit area.

[humdizzle]

i think he's saying the 'area' is already incorporated into the Cd number by the equation. So it doesn't make sense to multiply the Cd by Area, because you are essentially canceling them out.

.. unless the terms 'reference area' and 'frontal area' are different?

[jawn]

Originally Posted by humdizzle

i think he's saying the 'area' is already incorporated into the Cd number by the equation. So it doesn't make sense to multiply the Cd by Area, because you are essentially canceling them out.

.. unless the terms 'reference area' and 'frontal area' are different?

That's the goal. You want force, so you can't have A in the denominator.

[joe strummer]

Originally Posted by jawn

That's the goal. You want force, so you can't have A in the denominator.

But in the equation it is in the denominator.

[ZDan]

It's in the denominator of the equation for Cd, which is of course the *exact same equation* rearranged.

Of COURSE area is a big player in drag. Same Cd, twice the frontal area => twice the drag.

Cd*A is what you want to compare to say which car has more drag. LS has slightly lower Cd, but bigger frontal area will more than offset that and it will experience more drag at a given speed.

[jawn]

Originally Posted by joe strummer

But in the equation it is in the denominator.

Oh jeez. As mentioned, it is the same formula. The question is whether you want to figure out drag force or coefficient of drag. Cd is dimensionless, but it is useful for figuring out how much Force is caused by drag.

By the way, when something is in the denominator, it doesn't mean it tells you anything about the totality of the object. Density has volume in the denominator, but density tells you nothing about the total volume, or the total mass of an object. Likewise, if we want to know about the total drag force, we need to multiply by total area. Cd is a dimensionless number.

[joe strummer]

Originally Posted by jawn

Oh jeez. As mentioned, it is the same formula. The question is whether you want to figure out drag force or coefficient of drag. Cd is dimensionless, but it is useful for figuring out how much Force is caused by drag.

By the way, when something is in the denominator, it doesn't mean it tells you anything about the totality of the object. Density has volume in the denominator, but density tells you nothing about the total volume, or the total mass of an object. Likewise, if we want to know about the total drag force, we need to multiply by total area. Cd is a dimensionless number.

Despite your condescension I appreciate that you took the time to explain. Thanks.

[PMPB]

There is also a good chance that the FR-s has some designed in downforce, as evidenced by the rear diffuser, and the front lower lip, under body panels, and low ground clearance.

All of those things increases drag coefficient.