Let’s start with a simple 23 miles per gallon. You’ll have to excuse me for switching between imperial and metric — the sizes we’ll get to make less intuitive sense with inches. Just bear with me from the standpoint that miles and inches are convertable to meters and mm.


No problem…

In[41]:= UnitDimensions[%%]
Out[41]= {{LengthUnit,3}}

Oh — this is strange. This works out to an inverse area. What’s going on here? The nice thing about units is you can take the reciprocal and it still makes sense — that is gallons per mile.

In[47]:= UnitConvert[1/efficiency,gal/mi]//N
Out[47]= 0.0434783gal/mi

This also makes sense, no? That you use 0.43 gallons to go a mile. This still leaves us with a regular area:

In[48]:= UnitDimensions[%%]
Out[48]= {{LengthUnit,3}}

So what does that mean? How can fuel efficiency be measured by an area of all things? What does this work out to any way?

In[44]:= UnitConvert[1/efficiency,((mm)^2)]//N
Out[44]= 0.102267(mm)^2

Or for the inperial crowd:

In[45]:= UnitConvert[1/efficiency,((in)^2)]//N
Out[45]= 0.000158514(in)^2

Yeah, you can intuit a tenth of a square millimeter better than a thousandth of a square inch, right? ;-) Here’s where it gets kind of cool though.

Imagine a really thin straw full of gasoline that’s 0.1mm^2. Yes, this is a thin straw. But as your car progresses down the road this is the amount of fuel it uses to keep moving!

Don’t believe me? Well, let’s take that area and multiply it by 23 miles which will get us back to a volume…

In[46]:= UnitConvert[ %%*23mi, gal]

If you sweep that tiny little area over road, that’s how much fuel you’re using at any instant!

MilesPerGallon PDF From Mathematica