Saturday, January 17, 2009

Walkable cities aren't conservative but they are crowded

Fat Knowledge previously posted on a site gauging the "walk score" of several cities. The formula is actually pretty complex, but in short, the closer things you'd usually drive to are to one another, the higher the walkability.

My first thought was that this might be a chichi way of ranking cities based on population density. That's essentially the case (click on the graphic to view it more clearly). The correlation between a city's walkability and its population density is .85 (p=0). Cram everything together, including people and the places they go to, and walking becomes more feasible.

More interestingly, when looking at the city rankings for walkability, one might as easily be looking at a ranking of cities whiterpeople like the most. They're progressive like Seattle, trendy like New York City, or both, like the walkable capital of the country, San Francisco. But they're not conservative. The correlation between Bush's share of the vote in 2004* and walkability is an inverse .79 (p=0).

Densely populated places are expensive places. Expensive places with few open spaces are not conducive to raising families. When family formation is unaffordable, Democrats do well. This is another example of the validity of Steve Sailer's affordable family formation revelation that Republicans, just beginning to really smart from a disastrous eight year fall from grace, would be wise to make a tactical centerpiece of their electoral strategy going forward.

Data are here.

* Because election results are tracked by county, that's how I tracked voting patterns. In cities primarily in one county but with some extension into another county (or counties), like Fort Worth, only the primary county was used. When a substantial contingent of the population was spread across multiple counties, like NYC, each county was included in determining the total.

Using county data proxies well for city electoral preferences, as the city's population constitutes the majority of the county's population in most cases. Los Angeles county, encompassing the cities of both Los Angeles and Long Beach, is an exception. It has 10 million people to the 4 million actually living within LA. King county is another exception. In addition to Seattle, it houses affluent suburbs to the north and east of the city. Consequently, all three of these quite walkable cities are assumed in the analysis to have voted more heavily for Bush (36% for LA and Long Beach, 34% for Seattle) than they actually did. That the correlation remains as vigorous as it does despite these artificial confounding data points (and there are surely other more minor occurences of the same) suggests that the real relationship is even stronger than it appears to be above.


Fat Knowledge said...


I hear what you are saying about being more expensive, but if you can live without a car than it gets much less expensive relatively speaking.

Does voting for Bush correlate more with walkability or population density? Maybe not much of a difference.

I wonder if part of the difference has to do that when you live in a dense population that you get more for your money with gov't services. If you live in a rural area, your library doesn't have many books, so you still need to purchase many yourself. But, if you live in a city than your library has tons of books then you don't need to purchase many yourself. For the same dollar amount of taxes per person, you get more value in a dense population. Same thing with public transportation and museums. I am then not surprised that city dwellers vote for the party of larger government.

Audacious Epigone said...


Population density and Bush's share of the vote in '04 correlate at a relatively modest (and inverse) .69, compared to the .79 correlation between walkability and level of Bush support. That's a meaningful difference, given how walkability and population density basically proxy for one another.