On 2008 Jan 28 , at 2:12 PM, Ziser, Jesse wrote:
There appears to be a subtle logical flaw here. Namely, I'm not
sure the concept of
"fleet average fuel efficiency" is useful, using either form of
averaging.
Jesse is quite right. When averaging things that are ratios,such as
speed in kilometres per hour, or or rates of gasoline consumption in
litres per kilometre or any other ratio, the average of two or more
values can only be averaged by the simple rule* if the quantities in
the denominator are equal. (In averaging speeds, the two speeds being
averaged must be for the same time since speed is a ratio of distance
over time. In averaging two rates of gasoline consumption, the two
rates being averaged must have been measured over the same distance.)
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*The "simple rule" for averaging, referred to above, is:
"Add up the values and divide by the number of values."
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If Johnny has $30 and Mary has $50, then on average they have $40,
because
(30+50)/2 = 40
BUT
if I travel 30 km/h for half of a trip (let's say 150 km) and 50 km/h
for the second half (another 150 km), my average speed for the entire
300 km trip will NOT be 40 km/h.
Try it and you'll see.
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Traveling the first 150 km at 30 km/h will take 5 hours.
Traveling the second 150 km at 50 km/h will take 3 hours.
The average speed is the total distance (300 km) divided by the total
time (8 hours).
Dividing that distance by that time gives 37.5 km/h (NOT 40 km/h).
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There's a well known elementary physics problem that gives even more
surprising results by simply reorganizing the above problem. Here it is:
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A driver starts out on a 300 km trip. If she drives at 30 km/h for the
first half of the trip, how fast must she drive for the second half of
the trip to average 60 km/h.
The answer is:
She can't! (It's impossible!)
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"The proof is left up to the student", as they say, but I will supply
the solution to anyone asks.
Regards,
Bill Hooper
Fernandina Beach, Florida, USA
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Make It Simple; Make It Metric!
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