By elemental definition ---> POWER = FORCE x SPEED ( x conversion factor ).

I then humbly suggest that ---> POWER required to overcome air drag is proportion to SPEED cubed (not squared), since air drag FORCE is proportion to speed squared.

If POWER required at 70 mph is 31.71 hp, it does not point to a SPEED squared proportionality. With an automobile, POWER required at any speed is needed to overcome other forms of retarding FORCE ( rolling friction, transmission friction, heating and even acoustical losses, etc ) that increase at rates less than that of air drag. Air drag is only a component in the overall retarding FORCE.

Good Day,

Philip Lam

****************************************************************

Message: 5

Date: Thu, 28 Nov 2002 08:08:12 -0000

From: "Jack Hohner" <axama@...>

Subject: Re: oil coolers

--- In AirVW@y..., "Clare Snyder" <clare@s...> wrote:

> From my E V days, the calculated HP for 40MPH on a Fiat 128 coupe

was 10.14

> HP

> Wheel torque required at 40 was 77.7 ft lb. At 70 it goes up to

150.1 Tire

> RPM at 40 is 634 RPM.. With 91% efficiency in the drivetrain it

works out

> right.

> Take 70mph, and 1109 rpm, and it is 31.71hp at the wheels, with

gearbox

> efficiency down at about 89%, for about 35.6 motor hp required

Thanks for the data Clare. We know the air drag goes up with the

square of the velocity. It seems this rule also applies to the

overall drag and friction. Taking the square of the 70 mph and

dividing by the square of the 40mph and mulitplying by your 10.14 hp

at 40 we get 31.05 hp required at 70mph. Very close to the 31.71 you

reported.

I will be lucky if my Heath Parasol goes 70 mph so I am figuring my

VW is about right. But then again, I am also asking it to lift my

210 lbs into the air. Hopefully less by the time my project is

finished....but then there is that turkey feast tomorrow.

Happy Thanksgiving

Jack Hohner