I realize this is wild speculation, but what are your thoughts on timing? When will the virus be widespread in most communities?
An estimate of that is important for travel planning.
It depends on the natural infectiousness of the disease and the efficacy of measures taken to control it. This study of the early period in Wuhan
https://www.nejm.org/doi/full/10.1056/NEJMoa2001316 claimed that the number of cases doubled every 7.4 days. In the early days in Wuhan, they really didn't know what they were dealing with and control measures were not in-place, but for the sake of argument, let's use that rate
So, for the entire US, starting with 1 non-isolated patient, you can easily do the math Y=2 ^N, where Y is the cumulative number of cases and N is the number of 7.4 day periods. So log
2Y=N. Assume that Y =1% of the US population (~3.3 million), then log
23,300,000 = 21.654 periods or 160 days. So if you consider a 1% infection percentage to be widespread, you would get there by about mid August. And you could get to 100% in another 40-50 days after that.
HOWEVER, this assumes free spread of the disease with no social isolation/disinfection measures. To the extent these measures are at all effective, they will extend the doubling time. This also assumes, of course, that a vaccine will not be developed. Additionally, the doubling rate will naturally decrease as there become fewer uninfected people to become newly infected.
That's as close as I can estimate right now.
Edit to add: As far as I can see, the newest case data from China is not showing a doubling every 7.4 days. In fact, this Johns Hopkins chart suggests that is has taken 20 days for the latest doubling in China and the rate is slowing further every day.
https://gisanddata.maps.arcgis.com/apps/opsdashboard/index.html#/bda7594740fd40299423467b48e9ecf6
As newer, more accurate data arrives, you can just use a longer period to multiply by N in the equation above. The math works the same.