I'm looking for a formula which will help me define how the the fuel
efficiency of a car varies with speed |
Request for Question Clarification by
omnivorous-ga
on
10 Nov 2006 05:40 PST
Daniel --
A precise formula is unlikely, as there are engine factors and the
aerodynamics are also a significant factor. Since cars differ greatly
in aerodynamics (wind resistance increases with the squaring of the
speed) you're more likely to get an example of fuel consumption vs.
speed, rather than a formula.
Is that acceptable?
Best regards,
Omnivoous-GA
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Request for Question Clarification by
sublime1-ga
on
10 Nov 2006 13:13 PST
In addition to variations in engine design and aerodynamics,
there are numerous other variables which preclude the possibility
of a generic formula for all vehicles, such as vehicle weight,
differences in tire traction, road conditions, idle speed, etc.
However the US government has a website about fuel economy, and
they have a graph there that provides a visual estimate of fuel
economy vs speed for the average vehicle:
"While each vehicle reaches its optimal fuel economy at a
different speed (or range of speeds), gas mileage usually
decreases rapidly at speeds above 60 mph.
As a rule of thumb, you can assume that each 5 mph you drive over
60 mph is like paying an additional $0.20 per gallon for gas."
http://www.fueleconomy.gov/feg/driveHabits.shtml
You can also see that between 35 mph and 60 mph the fuel economy
is relatively stable.
Let me know if this satisfies your interests...
sublime1-ga
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Request for Question Clarification by
sublime1-ga
on
10 Nov 2006 13:30 PST
daniel...
On this page from Wikipedia, they provide the formula for the
power required to overcome aerodynamic drag, which is only one
of the factors involved in fuel economy, but has a major impact:
"Note that the power needed to push an object through a fluid
increases as the cube of the velocity. A car cruising on a
highway at 50 mph (80 km/h) may require only 10 horsepower
(7.5 kW) to overcome air drag, but that same car at 100 mph
(160 km/h) requires 80 hp (60 kW). With a doubling of speed
the drag (force) quadruples per the formula. Exerting four
times the force over a fixed distance produces four times
as much work. At twice the speed the work (resulting in
displacement over a fixed distance) is done twice faster.
Since power is the rate of doing work, four times a work
in half the time requires eight times the power."
http://en.wikipedia.org/wiki/Air_resistance
Let me know if this satisfies your question...
sublime1-ga
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Clarification of Question by
danielgrimes-ga
on
11 Nov 2006 08:19 PST
Thanks - I've seen the graph at fuel economy.gov - this is what I'm
trying to replicate using real data.
It seems to me that there are three phasesto the fuel economy - less
than 20mph the engine is not running at it's optimum, and above 55mph
wind resistance kicks in.
Much of the variables are available - weight, a published figure for
fuel consumption (indicative of engine efficiency) shape of car (SUV,
compact etc - indicative of aerodynamics), weather, tyre type.
So I'm happy for the figure to be a medium level of accuracy, but I
want to try and plot with real data - so need a formula probably of
the format:
f(x)=speed( f(wind resistance), f(engine efficiency), f(weight),
f(tyre resistance))
thanks for so far !
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