Anyone have any idea what the Cd is of a typical Winnebago type RV? A recent discussion with someone talking about the FE of their RV mentioned how much better FE the RV gets at 55 than at 65. I was going to calculate the actual FE difference based upon the Cd alone mathematically, but can't find good info for the Cd range.
They probably never test them, because its expensive to do and they wouldn't want to advertise the result. I found a paper for Class 7 and 8 Trucks which says Cd = ~.60 is typical for a large truck. This would probably be a reasonable estimate for something like a Winnebago as well. The SAE paper can be found here: Information Bridge: DOE Scientific and Technical Information - Sponsored by OSTI
No. A motorhome is long. Length (in the direction of air flow) lowers Cd. My guess is somewhere between .45 and .6, and dependent on how much extra crap is hanging out in the wind, like big rear view mirrors, awnings, roof A/Cs, etc. Besides increasing the frontal area, these items raise the Cd as they aren't the same shape as the vehicle (drag increases by more than just the percentage increase in frontal area). A couple web sites claim the old GMC motorhome from the '70s had a Cd of .31.
I actually meant the refrigerator lying on its belly with wheels. But you are right. I didn't say that. I'd be more interested to know the FE of those monstrous things. I'd like to hike the Muir Trail, or the Appalachian trail with one of those as my support vehicle: I'd hike, with a day pack, and each night the RV would be at my camp site. Like inn-to-inn hiking but you never have to pack or unpack. You have your own bed every night, and a hot shower and a proper roof. (Yes, I know there's no road access to the camp sites. It's just a silly fantasy, like having a Bengal tiger for a pet.) But how much fuel do they burn per mile? Oh, and they cost more than my house did!
Not sure what you mean. With a few gross assumptions - drag force Fd = .5 (rho) v^2 Cd A power to overcome drag Pd = Fd v = .5 (rho) v^3 Cd A energy used in time, T, Ed = .5 (rho) v^3 Cd A T total energy v T / FE c = Ed + Eo = Ed (1-a) where c is some conversion factor relating gallons of gas to enerby and Eo is other energy (rolling resistance, inefficiencies, etc). The last step is true only in the limit of most of the energy being used to overcome drag and probably not very accurate even at freeway speeds So, FE ~ K/(Cd A v^2), where K is all that other stuff. So a 20% change in speed (65/55) results in a 40% change in fuel efficiency. But a like change in Cd or A, hard to achieve, gives a 20% change in FE. Oh, almost forgot, you could drop the winnebago or a refrigerator off a cliff, measure its terminal velocity, the speed where drag equals the force of gravity and back figure the Cd!
Sounds like fun! We could drop a Winnebago and a refrigerator and then we'd know if (as I suspect) their Cds are about the same.
But, what would the Winnie's fuel economy be on the way down? Then, I imagine the instantaneous MPG will be purt-near zero ( or maybe even minus infinity) when it exploded on impact.:faint:
