Question about hills

[StorminMatt]

Originally Posted by KARR

I've done a search trying to find the exponential dependence of drag with speed but I didn't find it. The only formula I know is



Can you tell the formula of the exponential dependence please?

Alot of people look at the exponent of two, and assume that the relationship is exponential. But an exponential relationship would REALLY mean that there is a constant raised to the power of the speed.

There is one thing you need to keep in mind, though. MANY people like to use this relationship to beat people over the head with to get them to drive as slow as possible. But you have to keep in mind that drag coefficient is NOT the constant that this relationship makes it out to be. It can literally do ANYTHING as you go faster. Decreasing Cd with speed is not at all uncommon. And sometimes, it drops DRASTICALLY at a given speed. This is why you just can't say that 55MPH is ALWAYS the most efficient speed for EVERY car. Of course, changes in engine efficiency with RPM and load also complicate the issue of speed vs efficiency.

Originally Posted by Greyg

I have a 92 Daytona 5 speed 2.2 turbo DOHC

Daytona IROC? Now THERE'S one RARE car. Chrysler did not do too many good things with the 2.2 (and building it in the first place wasn't one of those things). But by outsourcing to Lotus for a DOHC head, they REALLY turned a turd into a winner!

[KARR]

Originally Posted by lowbridescape

Sorry. Misunderstood. I think of it as an exponent.

Yes, it is, but when the exponent is constant and the base is the variable we have a power, when the exponent is the variable and the base is constant we have an exponential function. I just tried to help distinguishing the two functions because they behave differently as the variable increases. If drag force were a exponential v-dependence it would be very difficult to travel above average maximum speed at highways for example.



Sorry for the off-topic.