>
>Whoa.. that's a whacked way to determine CID. The simplest way to
>determine cubic inches is to use 1 cubic inch = 0.016387 liters as
>determined with a Webster's Dictionary Weights and Measures.
>
Sorry, I was in the Mojave Desert, and had to pull McGiver (?sp?)...
Didn't have a slide rule with me, so had to do it in my head (knowing
that a 8.0L V-10 is 488cu, 5.9L V-8 is 360, 5.2L V-8 is 318...) You
could as easily done ratios with the 8.0L to a 3.9L... and come up with
a 237.9... so a 238 cu (rounded). 3.9Lx488cu, divided by 8.0L=
237.9--238 cu.
Let's do another example:
Let's say you have a Ford (I know) 289... How many liters is it??
Take a KNOWN value-- Viper 8.0L, 488 CU... or a Cheby 5.7L, 350 CU.
Take the Viper first-- 8.0L(viper)x 289 cu(ford)=2312 Now divide 2312
by 488 cu(viper)= 4.7377...liters. A 289 is 4.74 liters (4.7)
Now take your formula and 289x0.016387= 4.73584... Hmm looks a lot like
the 4.7 I have... :P
Take now take the Cheby-- 5.7L(Cheby)x 289cu=1647.4 Now divide 1647.4
by 350cu (Cheby)= 4.706... Hey, still 4.7L here. Now both the L and CU
of all manufacturers are rounded (some round up even it's not more than
50% to next bigger number).
The exact CU or exact L is irrelevant because they are (most) always
rounded to the nearest number. Ever seen a 4.78L??? Me neather. How
about a 4.948874L Mustang??
(off of soap box now).
Sam '95 SLT
Mathematician...
Riddle: You leave San Diego 7am on a Friday, in a 5.2L, automatic, 3.92
geared truck, headed to Kansas City. You average 71.428 mph, driving
nonstop (inflight refueling) for the 1500mi trip, averaging 18 mpg.
1) What is your average (constant) fuel flow per minute(any format)?
2) How far did you travel in that minute (Feet rounded)?
3) and what time do you get there?
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