The latest sunset I can find for today is in the vicinity of Morro Bay at 4:53 pm; two minutes later than here in San Luis Obispo.
Yah, but either way it’s too close to call. The OP may as well pick any hilltop from Point Reyes to Vandenberg for what they are looking for, as the differences will be minimal, at best.
I guess the angle of the CA coast is such that nearly every point sees sunset at about the same time.
Yup, there’s a long stretch of CA coast that’s almost a straight line and runs at just the right angle SSE that the S exactly offsets the E in winter.
Ahh - but check the Farallon Islands! 4:54pm today - the latest I can find! However, it may be problematic getting there to see the sunset on any given day, plus the risk of fog.
Okay, using the horizon formula, if I’ve done it right, from the top of Mt Olympus at 8000’, the horizon is 110 miles away. That is double the distance to the most westerly point at the coast, about 2 deg longitude or 8 minutes additional time to sunset. So in summer… maybe Mount Olympus is indeed the winner (aside from the road access stipulation).
Unless you want to make it complex and factor in atmospheric refraction. The view from the mountain passes through more atmosphere than the sea-level point at the horizon. Would that make the apparent sunset slightly later from the mountain?
ETA: Or earlier, because after the horizon, it’s passing from higher density into lower density?
Yeah, they certainly don’t publicize that it’s only during a certain time of the year. I just happened to stumble across that particular detail at this website.
Sorry for the delay in getting back to this one folks, but life intervened last night. Anyway, I’m not particularly hung up on the specific dates of October 7 through March 6 but, in general, the closer you get to summer the farther north that point is going to move. Once you get past the Arctic Circle, there is no sunrise or sunset at all for several months and the point become moot. Of course, I ruled out Alaska in the OP for just that reason.
That said, if there’s a point somewhere along the California/Oregon/Washington coast that can lay claim to the last rays of sunlight for a significant number of consecutive days, that would probably be acceptable. That will probably push us up into the Olympic Peninsula for a significant portion of the summer, but I suppose that can be OK too. How about if we designate a summer champ (for the lower 48) as well as a winter champ?
Regarding the higher mountains inland, I did think about that but assumed they would fall into the shadows of lower peaks that were closer to the coast. Unfortunately I failed at coming up with an easy way to calculate the necessary geometry and turned to the combined intelligence and skills of the Teeming Millions. My initial gut reaction was to look at somewhere like Mount Tamalpais north of San Francisco, or maybe one of the ridges in Humboldt County in the Eureka/Arcata/Fortuna area. That seems to be the westernmost point of California and being that far north certainly wouldn’t hurt.
For what it is worth, from this article:
Not All Sunrises are Created Equal - Farmers’ Almanac (farmersalmanac.com)
Of course, Fortuna is not the last place in the U.S. to see the Sun go down. That happens in Cape Flattery, Washington , the northwesternmost point in the contiguous U.S. Because Cape Flattery isn’t as far from the eastern edge of the Pacific Time Zone as Fortuna is from the eastern edge of its time zone, sunrise and sunset times come a little earlier there. Its latest sunrise occurs at 8:09 a.m. late December, and the latest sunset happens at 9:23 p.m. late June.
OK, based on the suggestions above I did a little data gathering. Starting from the north: there appears to be a Forest Service road (or something similar) approximately 2.5 miles inland from Cape Alava, WA that climbs to a peak elevation of 1013’ at 48.201321 Lat, -124.642637 Lon. Is that enough elevation change to offset the 2.5 miles further east?
Moving down to Oregon we have the Cape Blanco Lighthouse at 42.836999 Lat, -124.563639 Lon. It sits on a point 59’ above sea level and is 245’ tall, giving a total height above sea level of 304’.
Heading south into California, we first come to Cape Mendocino. Again, there appears to be a (possibly dirt) road leading up to an elevation of 1067’ approximately .6 miles (3200’) inland from the coast at 40.441495 Lat, -124.39770 Lon.
Next, near the Point Reyes Lighthouse visitor center, there is a spot at 37.996237 Lat, -123.023103 Lon. and 570’ elevation. Would that be appreciably different than the lighthouse itself at 37.995660 Lat, -123.023110 Lon, and 319’ total elevation?
And the last spot I checked (for now), Mount Tamalpais sits at 37.927379 Lat, -122.579976 Lon, with 2356’ of elevation and approximately 4 miles inland from the coast.
With that information, is it possible to calculate the differential in sunset times for those points as sun’s path moves southward from the fall to the spring and determine which spot holds the distinction of getting the last lingering rays of sunshine for the longest period of time? And if someone can share the appropriate methodology and mathematical formulas necessary, I will certainly be happy to do the grunt work of plugging in the requisite numbers.
Thanks to all who have provided input so far!
Thinking about the whole mountain-top thing got me thinking about some tall mountains close to the coast. Perhaps Cone Peak, which is nearly a mile above sea level and only about 3 miles from the shore, would have the last rays of sun - it has an unobstructed view of the ocean. Or nearby Junipero Serra Peak, which is 700 feet taller (but farther inland). Both are in the Big Sur region of California, and both are accessible.
Cape Flattery, while westernmost, is not that much west of the coast, and the northern latitude plus sea-level elevation could cancel-out any western advantage? Mt Olympus is tall, but pretty far inland so may be blocked by other mountains, and is very inaccessible (not to mention blanketed by cloud a lot).
It was pointed out above that it takes several thousand feet to improve the “last rays of sunset” time by a minute. It should be possible to identify the height gain versus distance to the east loss of “last rays” time relationship and filter the candidate peaks.
Also, the article I found reminded me that there is a limit to “westernmost” set by the time zone lines. Too far west and suddenly it is an hour earlier at any point you are observing the sunset.
Finally, on a tangential note, when I saw the thread title “West Coast Equivalent to Cadillac Mountain in Acadia NP”, my immediate reaction was “What is a hill, Alex?” 
This was me, based on a quick Google at the time, and I think it was incorrect. @Chronos suggested using the horizon distance formula which sounds correct. Based on that, I come up with the following.
Horizon formula Horizon - Wikipedia
d^2 = 2Rh + h^2
which rearranges to h = SQRT( R^2 + d^2) - R
where
R is radius of Earth for which I’m using 3959 miles
h is height
d is distance to horizon
I’m then solving for h with d in 12 mile increments that correspond to 1 minute increments at latitude 45N where 1 degree of longitude is 48 miles
So I get the following, where the first column is minutes (time), the second is horizontal miles to the horizon at latitude 45N, the third is feet elevation. Divide the first column by 4 if you want to work in degrees longitude.
| 1 | 12 | 96 |
|---|---|---|
| 2 | 24 | 384 |
| 3 | 36 | 864 |
| 4 | 48 | 1536 |
| 5 | 60 | 2400 |
| 6 | 72 | 3457 |
| 7 | 84 | 4705 |
| 8 | 96 | 6145 |
| 9 | 108 | 7777 |
| 10 | 120 | 9600 |
So if you are comparing to a point on the west coast at sea level, if you could find a point that is (for example) 48 miles east of there with a clear view of the horizon, it would need to be at an elevation of more than 1536’ to see the sun set later.
Thanks @Riemann, that helps a lot. Gives me a place to start, at least. Now I need to figure a way to factor in the change in the angle of the sun as the seasons change to pinpoint when a particular point is the last one staying sunlit. Anyone have any ideas on that part?
If one googles sunrise times for Miami and Cadillac Mountain for today, I find it to be 7:01 am and 7:04 am respectively even though Miami is 12° further west than Acadia National Park. At the time of the winter solstice, Miami has about 100 more minutes of daylight than Acadia. This anomaly only occurs within two weeks of the winter solstice.
You mean the exact bearing of sunset? At these latitudes, this will go from due W at the equinoxes to roughly SW in winter and NW in summer. You can find the exact bearing for any location and date in the “sun” section here:
The altitude of the sun at sunset is obviously always zero, but if you’re concerned about how high it will be in the sky (say) an hour before sunset, that’s also on that site - go to the table of dates lower down on the page, click on a date to expand it, and you get a graph with the altitude and bearing of the sun throughout that day.
never mind
Actually, a little Googling turned up a link to the NOAA Solar Calculator which allows one to " Find Sunrise, Sunset, Solar Noon and Solar Position for Any Place on Earth" for any specified date using latitude & longitude coordinates. It also has a handy-dandy “Show Sunset” check box which draws a line to the point where the sun reaches the horizon on the specified date.
I believe I can use that tool to create a table using the coordinates for various points along the coast and the calculated sunset times for a targeted range of dates to see what the optimal points would be at various times of the year. I’m assuming the calculator doesn’t take elevation into consideration but I can check that by comparing sunset times listed for nearby sites at different elevations and see if it varies. If so, good. If not, I can adjust as necessary for elevation differences and use the drawn line to look for interferences. I’ll report back with what I find, if it does indeed work like I expect it to.
Go for it, and thanks! I look forward to your findings. I love these sorts of topics, and if I can’t go down the rabbit hole headfirst myself, watching/following you and other heroic participants in this thread who do so is great fun too.
Boy, I hate it when work gets in the way of getting important things like this done, but I was finally able to find a good tool for including elevation in the calculation and put together a table for comparison. The tool I found was Photo Ephemeris which is targeted at photographers for just this purpose. How handy!
Anyway, after playing around with it for a while I was able to determine what spots got the last rays of sunlight on the west coast at different times of the year. Note: while Mount Olympus actually wins for pretty much the whole summer, I dropped it from the calculation because of the difficulty reaching the summit for these purposes. That said, here is what I came up with (drum roll please):
From November 12 through February 6, the winner is Cone Peak, CA. Junipero Serra Peak is actually a couple of hundred feet taller, but reaching the summit is quite a bit more challenging and the time difference wasn’t that much, so I went with Cone Peak.
Next we have Cape Mendocino, CA wearing the crown from February 7 through March 17, and again from September 28 through November 11.
And finally, as expected, Cape Alava, WA as the most northwesterly point on the lower 48 meets my specified criteria for the summertime, starting March 18 through September 27.
If anyone would like a copy of the spreadsheet I put together to make these calculations, PM me your email address and I will gladly send you a copy.
Thanks to all who provided input. It certainly helped pinpoint the most likely spots and saved quite a bit of time and effort. Happy Holidays to one and all!