Do you win? Heck, I don't know. I'll take a shot at it. Watch me get
it spectacularly wrong.

The 1.9L engine has 475cc displacement per cylinder. I think two
cylinders fire per revolution, so at 900 rpm we got 900 x 2 x 475 =
855cc of air per minute being sucked into the engine. If the mixture
is a good stoichiometric one, we'd have about a 1:15 fuel:air mass
ratio. Air has a mass of 1.3g per liter, so that's .855 x 1.3 = 1.1kg
of air per minute, with being one-fifteenth of that, or 74g per
minute. At 60 minutes that's 4.4kg of fuel. Gasoline masses roughly
740 grams per /liter, so that's 6 liters, roughly 1.6 gallons U.S. per
hour.

Those are the numbers I come up with and I bet they're within an order
of magnitude of being right.

Mike "Rocket J Squirrel" Elliott

On Nov 24, 2007 3:52 PM, Matthew Snook <matt@snooksband.com> wrote:
> 0.9375?  Is that right?  Do I win?
>
> :)
>
> At 60mph, mine's turning @ 3200rpm.  It will burn 3 gallons doing that.  3
> gallons per hour at 3200rpm.  But it idles at 1000rpm.  So that comes to
> 0.9375 gallons per hour at 1000rpm.  Of course there's no wind resistance at
> that speed, so maybe less...
>
> Matthew Snook (@ ~3000 ft)
>
>
>
> -----Original Message-----
> From: Michael Elliott
> Subject: Calculate fuel consumption when idling?
>
> A properly-tuned 1.9L WBX engine would consume how many gallons of
> gasoline per hour when idling at sea level? Would this be significantly
> different at 6,000 feet?
>
>