According to this site:
Current Local Time in Montréal, Quebec, Canada the length of the day today in Montreal is 12 hours and 8 minutes. Now I can calculate that approximately 2.8 minutes of that is due to the width of the sun and another 1.6 minutes comes from the fact that the day is getting about 3.2 minutes longer every day at this time of year. That still leaves at least 3 minutes unaccounted for. Is there really .75 degrees of refraction from the atmosphere?
Would having the equinox not fall exactly at noon have something to do with it? I would expect that if it falls in the early morning, the day will be slightly less than 12 hours because the majority of the day is in winter, and the opposite if it falls in late evening.
At the horizon, yes, there is almost that much–0.57 degrees, to be exact. I don’t know enough to check for other errors in your math–maybe you had the sine of the angle of sunrise wrong for Montreal’s latitude, or something.
How long is the day during the fall equinox? If it’s 12:03, the difference would be attributed to latitude.
Could it not be that the Earth moves about one degree along its orbit per day? So the sun appears to moves slightly to the east (or is it west?) every day.
Playing with this some more . . .
The longer-day factor tends to shorten today’s day-length, relative to a 12-hour norm, because the vernal equinox isn’t until 19:07 Eastern Standard Time. That’s about a third of a day after noon, so that factor alone moves you to an 11:59 norm. You’re correct in allowing 2.8 minutes due to the width of the sun. That gets you to about 12:01:48.
But you over-estimated the amount of refraction necessary. As I say, the true figure is 34 minutes of arc or 0.57 degrees.
Montreal is at latitude 45.5 degrees, so the sun rises and sets (at the equinox) at a 44.5-degree angle. The sine of 44.5 degrees is 0.7. (0.57/0.7) = 0.81 degrees of arc that the sun needs to travel to compensate. At 4 minutes per degree, 4 * 2 (once each for sunrise and sunset) * 0.81 = 6.48 additional minutes of daylight.
Voila, that brings your total to about 12:08:15.
Just to clarify …
The Vernal Equinox for 2007 begins on Wednesday, 21 March 2007, at 0007 UTC.
Use this conversion table (and add seven minutes) to arrive at when it occurs for your location.
You cannot count both the rising and setting, only one of them. I used 45 deg, which is close enough, to get the 2.8 minutes. (The sun subtends .5 degrees and each degree takes 4 minutes, the number of minutes in 24 hours divided by 360 and the cosecant of 45 deg is 1.4.) Then i halved the 3.2 minutes by which the day lengthens to get 1.6. The .57 deg of refraction adds about 1.1 minutes (actually 1.14) and the total is 5.5. This still leaves 2.5 minutes unaccounted for.
Why not? Refraction makes sunrise earlier and sunset later.
I was talking about the time it takes the sun to rise and set. Only one of them can be counted. But you are right about the refraction. So the computation comes to 2.8 minutes for rising (or setting), 2.3 minutes for refraction, 1.6 minutes for the fact that the day was lenghtening during the day, which gives a total of 6.7. This still falls a bit short of the eight minutes of my original post, but another site:
http://weatheroffice.ec.gc.ca/city/pages/qc-147_metric_e.html
gave the day length as 12:06 minutes, so perhaps the remaining discrepancy is round-off error.
No. You subtract, not add, for day lengthening. Daytime Tuesday was before the equinox. Before=shorter day. And your number for refraction is wrong. See my Post#6 for the correct calculation.