In looking at the paths of solar eclipses, I’ve noticed that the shape of the moon’s shadow on the Earth changes from an oval/ellipse to a circle and back to an oval/ellipse.
See here:
My question is: If one is within the zone of totality, is there any difference when observing the eclipse if the moon’s shadow is circular (like in Mexico for the upcoming eclipse) versus being oval/elliptical (like in New England for the upcoming eclipse)?
I believe that the only difference occurs if you’re near the inside edge of the zone versus the center line. The reason the totality time is shorter near the edge is that the elliptical shadow takes longer to reach you than the shadow nearer the center, and then goes away sooner, all because you’re just on the fringe of the shadow. But if you’re within the zone of totality, when it actually occurs there should be no difference at all in what you observe.
I don’t think so.
The difference between oval and circle has to do with the angle of the earth’s surface WRT the sun, and therefore the local time. The shadow will be circular near local real solar noon (not DST) and oval-shaped before and after that. That obviously correlates with the local viewing angle, i.e. how high the sun is in the sky.
Hold a ball between a light source and a piece of cardboard, then change the angle of the board. When it’s perpendicular to the line from the light to the ball, the shadow is a circle. When you angle it, either way, it’s an oval.
It’d never actually be a circle except when the Sun (and Moon) are directly overhead, which never happens except in the tropics. What you’re seeing as circles are actually just less-extreme ellipses.
But yeah, where it’s closest to a circle is where it’s closest to directly overhead.
There are two different factors here. I’m describing (perhaps somewhat incidentally to the OP’s question) why the totality time is shorter at locations near the edge of the umbra than closer to the center line. What I believe you’re describing is why the total time of some total eclipses is different from others.
I’m here in Mazatlán. Landfall centerline will occur 20mi/32km southeast of me. At my location it’s 4m16s of totality and right on centerline it’s 4m26s. I’ll spot them 10 seconds to save a drive out into the wilds of the Mexican countryside with lots of other people.
The altitude of the sun at max eclipse will be a hair under 70 degrees. We’re at 23.25 degrees north latitude.106.45 West longitude. Solar noon on eclipse day happens at 1207 local, and the eclipse max happens at 1109 local.
So at eclipse max the sun will be ~20 degrees south and ~15 degrees east of local zenith. So not quite straight overhead, but pretty close. Leading to a very nearly circular umbra. And one hell of a lot of neck craning.
Here’s hoping on the weather! I’m stoked.
I’m sorry. My “I don’t think so” was a reply to the OP, not you, and the rest of my post was answering his questions about the shape of the shadow.
Sorry for the confusion.
Yeah, ideally, you want to be right on the centerline, but as long as you’re reasonably close, it’s pretty good. Halfway from the edge to the centerline gives you 71% of the centerline duration.
EDIT: I shouldn’t try to do trig in my head. Halfway from the edge gives you 85% duration. 71% of the way from the centerline gives you 71% duration.
The totality duration on centerline at landfall is 266 seconds. Where I am at landfall I’ll have 256 seconds = 96.24 of the local max.
It appears from playing with the OP’s cited map that the duration of totality is about 3m46s or 226 seconds on centerline where it enters the Atlantic in New Brunswick. A couple of spot-checks along the path in the central US suggests the longest duration over land is right here.
The path width here is 199km. So the half-width is 100km. I just realized I mismeasured my lateral offset from the centerline in my previous post. It’s more like 25km ~= 15mi. So about 25% of the way from totality centerline towards totality edge.
I’m not sure where you’re getting this.
Per the linked Wikipedia page above, the maximum eclipse is at Nazas, Durango, Mexico where the eclipse duration is 268 sec (4m 28s).
Using this calculator, in Kerrville, TX it will last 267 sec (4m 27s). In St. Albans, VT it will last 214 sec (3m 34s), while in Blackville, NB it will last 199 sec (3m 19s). Note that all of these locations are at or very near the centerline of the eclipse.
ETA: I think you might be assuming that the speed of the eclipse shadow is constant, but it is not. Per this animation, it speeds up considerably once you get past the maximum eclipse, so that even though the shadow elongates into an more eccentric ellipse, this is more than counteracted by the increasing speed of the shadow (which shortens the eclipse duration). See video here:
Being on the centre line yields the longest totality. But being offset can yield better Bailey’s Beads. Part of the edge of the moon will graze along the Sun rather than suddenly slam into it. Given how long totality is for this one, being exactly on the centre line is a lot less important. Given a free choice I would be offsetting myself a bit.
I’m making no assumptions or calculations about any speeds.
I’m using Javier’s map in the OP. I’ve clicked on spots along the totality line and read the start & stop times off his pop-up. Perhaps I’ve flunked subtracting minutes and seconds, perhaps there’s some differences in various calculators for various websites.
I’m definitely not claiming expertise much less infallibility here.
You are apparently referring to Xavier Jubier’s excellent interactive map, which was linked to in the OP this other thread, not my OP above in this thread.
Anyway, your math is off, and is not necessary regardless. The interactive map shows the eclipse duration for any location you pick along with the start and stop times. It is not necessary to do any subtractions.
Per the interactive map, the duration of totality where it enters the Atlantic in New Brunswick is 195 sec (3m 15s). This is certainly not the longest duration over land, nor is it in the central U.S. Instead, the longest duration is in Mexico (highlighted as “Eclipse City” on Jubier’s map.
ETA: Thank you for reminding me of the existence of this map. That other thread has been going for so long I completely forgot about it. It’s a lot easier to use than that cumbersome eclipse calculator I linked to in my last post.
The shape of the umbra (totality zone) will depend on the angle of the sun and moon, relative to the earth. “Azimuth” is a word that’s used a lot, and I’ll leave its definition to people who actually know something about it; all I know is that it has something to do with angles.
You’re totally right about which map I’d been using, and thanks for pointing out the totality duration is already calculated & displayed. I’d missed that number.
I was apparently writing a bit confusingly. Because you really misunderstood what I was trying to say. Although my math errors sure didn’t help. Gaah!
At Mazatlan Mexico where the eclipse centerline makes landfall from the Pacific ocean the duration is 4m26s as I’d said. At the opposite end, in New Brunswick Canada where the eclipse centerline exits land and enters the Atlantic ocean the duration is, as you say, 3m15s. My defective subtraction came up with 3m46 and that’s what I wrote. Oops on me.
Noticing that Mexico’s duration was longer than New Brunswick’s, I took a couple of samples of totality duration along the track centerline in the central USA near Dallas and in Ohio and found they were intermediate durations with Ohio’s shorter than Dallas’s. Suggesting to me that totality duration got shorter all along the land portion of the track. Without offering a theory on why that was so.
As you rightly say, that’s not quite true. Jubier (thanks for the name correction) points out the actual max duration is partway across Mexico at the little red dot on the map that I had not noticed until you’d pointed it out. Thank you.
We’ll get all the nits picked along about next Tuesday I figure. 
Thanks again for the help.
If you are in the zone of totality, how exactly would things be affected if it were a cloudier day? I presume things would darken but the sun may not be so visible?
Yes, exactly. So you miss the corona and the hole in the sky where the sun should be.
@robby nailed it. I’ll add that seeing the sun is a big part of the experience, so if it is blocked by clouds, top me that would be a big disappointment. Try to find clear skies if you can.
Azimuth is one of the two angles that describes the position of something in the sky, but it’s not actually the relevant one here. The shape of the umbra is due to the other angle, altitude.
Basically, azimuth is the angle measured on a compass, from 0º at North, to 90º at East, 180º at South, and 270º at West. Altitude is then how high above the horizon something is.
With the caveat that I don’t think any of the weather forecasts take the eclipse itself into account (it totally is relevant), and so none of the forecasts are going to be all that accurate.
My understanding is that the eclipse shadow does produce a local cooling effect which can dissipate clouds. However, if a given area is completely overcast, I don’t think there will be much of an effect.