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My departure from college has left me with an itch for physics-y problems to solve, and my boredom and free time gave way to this approximation of the gallons of gas I use for the trip to my girlfriends house.

  1. First approximated miles per gallon (mpg) as a function of velocity [mph]

$\boxed{mpg(v)}$:

MPG(v)

(Function approximated from this image from the Wikipedia page on Fuel Economy)

MPG2(v)

  1. Then approximated the speed (mph) as a function of time (hrs) for the trip. (n.b. this graph assumed no traffic, and is therefor horribly inaccurate as I live in LA, and there's literally always traffic in LA)

$\boxed{v(t)}$:

v(t)

  1. Checked the approximation by integrating $v(t) \ dt$ over the trip and checking that the total distance travelled was right (it was :D)

$\boxed{x(t)}$:

x(t)

  1. Then used $mpg(v)$ and $v(t)$ to get $mpg(v(t))=mpg(t)$.

$\boxed{mpg(t)}$:

mpg(t)

  1. The reciprocal of miles per gallon is gallons per mile, i.e. $\dfrac{1}{mpg(t)}=gpm(t)$, so I assumed that gallons per mile times miles per hour would give gallons per hour as a function of time $\dfrac{v(t)}{mpg(t)}=gph(t)$

$\boxed{gph(t)}$:

gph(t)

  1. Then integrated $gph(t) \ dt$ over the whole trip to get gallons used:

gallons

However, this seems quite low. From personal experience, I'd say I use closer to 2 gallons for this trip. I'd say my assumption of no traffic probably had some effect, but even with that correction, it's still only around 1.5.

Did I make any glaringly false assumptions? Any way I could make this approx. more accurate?

Also, for anyone who knows cars, why does it seem like I use gas more quickly when my gas tank is less full than when its more full? One would think that less gas=less weight=better fuel economy... Something to do with the shape of the tank?

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  • $\begingroup$ Mpg(v) is steady state? What about the cost of acceleration? $\endgroup$ – Jim Garrison Dec 6 '15 at 3:05
  • $\begingroup$ "why does it seem like I use gas more quickly when my gas tank is less full than when its more full? One would think that less gas=less weight=better fuel economy... Something to do with the shape of the tank?" Probably because the gauge is not a linear function of the volume of remaining fuel. $\endgroup$ – DanielSank Dec 6 '15 at 7:15

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