Quote:
Originally Posted by JQmile
This look right to you guys?
Power (watts) = 1/2 (1.2(air density))(velocity cubed(m/s))(frontal area(m/sq))(drag coefficient).
I know the car goes 80mph (35.2 m/s) and it has made 36hp at the wheels (1hp=745.7 watts) so ........
36(745.7)=26845
26845=(.5)(1.2)(35.2(cubed))(1.9)(X=Cd)
26845=49720X
26845/49720=.53!
So by this math, that means my Cd is .53!! Looks like my fuel economy is all light weight and small engine, and no aero.
Thoughts?
I have an G-Tech, so I might try a coastdown measurement later provided I can do the math LOL
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I think the formula needs a value which brings force per unit time into the equation.I'll have to compare to my US standards stuff at home.Sorry,I'm not metric!-------------------------- Also,you need the entire road load of the car at top speed which will include power absorbed by the tires.-------------------------- Finally,does the 36-hp value include powertrain efficiency loss?--------------------------------- Hucho,in his book,provides a rule of thumb for drag reduction vs top speed: 30% drag reduction = 10% increase in top speed.-------------- If you have the original top speed,then modify the shape of the car(ONLY),and you realize a 10% improvement in speed.then you've cut your drag coefficient by 30%.