Quote:
Originally Posted by roflwaffle
I wouldn't even go as far as to say uphill, unless the driver never goes downhill.  Crr>W. There, I said it.  |
Only of you coast uphill
Otherwise, wasted power :/
Crr does indeed go up with weight... But we're talking of a scalar on the order of hundredths (Crr) that is used in linear a linear equation with respect to a constant (weight) versus a scalar in the tenths (cD) which is used in a quadratic equation with respect to a variable (velocity).
So lets say the car weights 2500lb and cRR is ~.01 (partially arbitrary). That gives us a RR of 25 pounds. This is constant. So, we can linearly reduce that value until we have no wieght in the car whatsoever (not possible - but just for discussion's sake).
So lets say the car originally had a cD of .3 and an area around 25 square feet. That's a cDA of 7.75ft.^2....
That means - given a cD of .3 versus .17.... A difference of 25 pounds is had at 53mph. Of course, we can't remove all of the car from the car... So lets take 500lb off. The even breakpoint would then be at ~24mph (5lbs).
Lets say he goes crazy - and pulls out 900 pounds of car off the car.... He'd then have to go ~32mph to get an equivalent difference in gains (9lbs)
Now, I made some assumptions on cRR, A and original cD. I know they're somewhere - I just used handy numbers from the coast down instructable and an estimated weight of 2500 lbs. I'll eventually come across the more correct numbers and throw it into a spreadsheet
I can also post the converse if you'd like (what reduction in cD is require to break even at some given velocity)...
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To reiterate - removing weight is good (and shouldn't be overlooked)... But reducing aero losses has the potential for greater gains in drag reduction. The biggest (and cheapest), of course, is just driving slower
And finally - an example case... Basjoos' car