View Single Post
Master EcoModder

Join Date: Jan 2008
Location: Sanger,Texas,U.S.A.
Posts: 14,509
Thanks: 22,513
Thanked 6,710 Times in 4,250 Posts
Cd 0.13

Quote:
 Originally Posted by JulianEdgar Looking at the table I posted, and remembering that required power goes up with the cube of the speed (ie your nominated speed of 50 km/h will require 2.7 times the 36 km/h aero power shown in the table), I wouldn't think you'd be able to get the Cd down sufficiently to achieve that outcome. Using the table, at 36 km/h, a bike with a frontal area of 0.35 square metres and a Cd of 0.13 requires a total (aero + rolling resistance) power of 50W (of that 24W is aero power, so a total of 90W at 50 km/h). So that would imply that to meet your criteria, you'd need a Cd of something like 0.13 - pretty hard. Happy to have someone check my maths - never my best area.
Cd 0.13 seems very plausible.
I was unsure of the all-up weight, and used data for the 0.40 m-squared rig, thinking Grant-53 might be able to tuck his head inside the enclosure.
* If the R-R was 38-Watts @ 36 km/h, and tires were below standing wave, and increasing resistance arithmetically, then @ 50 km/h, R-R power would rise to 52.7-Watts. Leaving 47.3- Watts budget at the higher speed for aero.
If it's 47.3 @ 50- km/h, then it'd be 17.65-Watts @ 36 according to the cube law.
One of the bikes had a 24.2-W aero power consumption with CdA = 0.044 m-sq.
Adjusting for 17.65-W available power, the CdA would need to fall to 0.0325 m-sq.
Dividing by 0.40-m-sq, Cd 0.0813 falls out of the equation.
Really low!
At Battle Mountain, Nevada, the 'laminar' bikes doing 85-mph ( 137-km/h ) are around Cd 0.11 according to the university teams.
It's okay checking my math as well. I stayed with metric, where typically I switch to US standards, then convert back.
__________________