It sounds like a joke, but it's not: a new study proclaims Los Angeles to be the least sprawly large metro in the US. The study, conducted by a sociology doctoral student at NYU, uses aerial images of Census tracts in the 150 largest US metros (according to 2010 data) to estimate the portion of the metro population living below three "thresholds": 3,500, 8,500, and 20,000 people per square mile. Then, explains CityLab, those thresholds are averaged to get a metro's index number—the higher the index number, the more sprawl in that metro. Of the 150 largest metros, Los Angeles has the lowest index number, meaning that despite its spread-out rep, it's actually not so bad at all.
Los Angeles's "sheer lack of very low-density development" helped the region get its highly non-sprawly status, which is great because the analyst goes on to quantify all the negative effects of sprawl: "For every 10 percent increase in sprawl, there is an approximately 5.7 percent increase in per capita carbon emissions, a 9.6 percent increase in per capita hazardous pollution, and a 4.1 percent and 2.9 percent reduction in the owner and renter housing affordability index, respectively." Ooof, bit of a double-edged sword there, sprawl. It makes sense that less polluted, less sprawly places are more expensive to live in—who wouldn't want to live in them? And that certainly does ring true in LA, which is both one of the most unaffordable places to live and, by this study's count, the least sprawling large metro in the nation. (You can also see how its many suburbs cancel out some of the benefits of its urban density.)
If this sounds bogus, keep in mind that this is not the first time LA's been found surprisingly non-sprawls; an analysis last year by Smart Growth America called LA the "biggest success story" in the scaling-back of sprawl and Census data from 2012 showed it was the most densely populated urban area in the US.