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Re:Proposition C by HuShifang on Friday October 19, 2018 @01:52PM (#18258559)
Prima facie this strikes me as an apples-to-oranges comparison. New York and Los Angeles are large cities with millions of residents and dominate their geographical areas (LA perhaps less than New York, but still). San Francisco has something like 884,000 residents, which makes it the second largest city in the Bay Area (after San Jose, with 1,035,000 or so), which has something like 8.7 million residents if you go by Combined Statistical Area (CSA) definition, which makes intuitive sense to me having lived here. By comparison New York has 8.6 million out of 24 million for its CSA; LA has 4 million out of 18.7 million for its CSA.

So really, a more meaningful comparison (if one is resigned to handwaving about tax bases, without delving into details of city finances) for SF would be Boston, with 865,000 people out of 8.2 million for its CSA.

Re:Proposition C by pj_mukh on Friday October 19, 2018 @01:55PM (#18258591)
AFAIK, homeless problem scales with population size. So if you divide by population size (which those values do), the value should stay roughly the same.

What are the factors that make this untrue?

Re:Proposition C by aeternus on Friday October 19, 2018 @02:08PM (#18258694)
Cost of housing is a big one. Low housing costs and plenty of space available make the homeless problem both less likely and easier to deal with.
Re:Proposition C by mturmon on Friday October 19, 2018 @06:32PM (#18260575)
Yep, there are plenty of outlying areas in LA that are still not expensive in the same way SF is - LA is just much, much bigger. So you can hang on with significantly less money. Still, it's an interesting question.
Re:Proposition C by cpr on Friday October 19, 2018 @05:14PM (#18260121)
Much of it has to be geographical--SF has mostly bearable weather year-round, Boston doesn't.