Starship Asterisk*

I wonder where this part of the thread is going to end up
From what i can find, kicky-wicky only has to do with Shakespeare, and seems to mean a wife, and there's even less about the unseen box.

... by the way, a very original idea and great picture to celebrate solstice!

Hi All, sorry to have gotten off on this curved path idea. I checked a bit more and you are all correct; the path only changes less than 1/6 the diameter of the sun here, so is quite small. I was just remembering a time lapse of star trails setting and how straight the ones on the equatorial plane looked and how curved others on either side looked. Must have been a much wider field of view. Also, I wasn't implying that Danilo had manipulated the image...just that there are so many new photographic creations these days it is sometimes hard to figure out what is what while enjoying the beauty.

Fritz Stumpges wrote:
Hi All, sorry to have gotten off on this curved path idea. I checked a bit more and you are all correct; the path only changes less than 1/6 the diameter of the sun here, so is quite small. I was just remembering a time lapse of star trails setting and how straight the ones on the equatorial plane looked and how curved others on either side looked. Must have been a much wider field of view.

A single hour's worth of a single star trail is generally pretty straight:

http://apod.nasa.gov/apod/ap110805.html
http://apod.nasa.gov/apod/ap111014.html

But it is fun to go off on tangents.

I am confused as to when the image sequence was acquired. In the explanation it states it was taken "June solstice day of 2012". There is no date information with the image (in Danilo Pivato's website) that was brought up through the "This composite image" in the explanation. However, in the 'Sequences & Sunset' in the 'Miscellany' section of the website there is a small version of the image that has "Seq. & Selfportrit - S.Severa 1986". I'm . It is though a good interesting image.

neufer wrote:

admirer wrote:
Can we deduce the latitude of the observer from the angle between the path of the sun and the horizon, knowing the date?

Yes. And the angle = the latitude at equinox.

Just a minor correction: The acute angle (0 ≤ Θ ≤ 90°) between the sun path and the horizon = 90 - Latitude at the equinox, i.e. at the equator, the sunset path is vertical.

Chris Peterson wrote:... Furthermore, the angle with respect to the horizon is exactly as it should be for the date (note that the angle of the ecliptic at sunset is equal to the latitude of the observer, 42°, but appears somewhat more vertical in this image because the ecliptic doesn't intersect the images of the Sun; it changes position quite a bit over an hour and a half).

Your comment about ecliptic angle = observer latitude caught my interest because, although very close to true in this picture, I didn't believe it was generally so. At the Solstice, I modeled the spherical trig details for refraction-free cases at arbitrary latitudes. I looked at the derivative of altitude wrt azimuth for an azimuth range of 15°before the sunset az (inclusive). Here's what I found:
1. Due to a changing ecliptic orientation wrt the horizon (your comment), the refraction-free sunset path has a "sag" (concave appearance) that amounts to about 2/3 the sag for the sunset path which includes refraction. So although small, the bulk of the path curvature appears to originate from the changing ecliptic orientation due to Earth's rotation.
2. At the solstice, I looked at the ecliptic angle wrt the horizon at the moment of sunset, and at latitudes ranging from 0° to 60°. I found that the ecliptic angle = latitude fortuitously at ~42°, and nowhere else. It turns out the "fundamental" angle from which to derive the ecliptic angle variation with latitude is still "90-Latitude", not Latitude. Within a few degrees error in predicting the ecliptic angle at sunset (over a 60° latitude range), I found an interesting (though arbitrary) equation: EclAng ≈ (90-Lat)·½·[cos(√Lat)+cos(Lat)]. So at sunset on the Solstice, the ecliptic angle ≈ 90-Lat for low latitudes. At Lat ~42°, the ecliptic angle and latitude are equal. At Lat = 60°, the ecliptic angle ≈ 20°.

cos(√Lat)

alter-ego wrote:Your comment about ecliptic angle = observer latitude caught my interest because, although very close to true in this picture, I didn't believe it was generally so...

I didn't try to follow your analysis, so I won't comment beyond saying that whether we use the latitude or (90° - latitude) in the calculations depends on the arbitrary choice of whether we measure with respect to the horizon or the perpendicular to the horizon. As a meteor scientist, I normally use the later in referring to angles in cases like this.

That said, my comment about the angle was mainly directed to the observation that the line of suns in the image do not lie along the ecliptic, because the ecliptic is moving with respect to the horizon. So the angle made by the line of suns isn't the angle of the ecliptic at sunset. That can be seen in these simulations, which show both the Sun at each time as well as the ecliptic at each time. The first image shows each frame, the second just the first and last. An image like this gives the impression that the ecliptic is steeper with respect to the horizon than it actually is.

neufer wrote:cos(√Lat)

Good point, Art. A square root of an angle, that's pretty funky. There is a simpler function that better describes the ecliptic angle wrt the horizon at solstice sunset: (90-Lat)·√cos(Lat) i.e. square root of the cosine function. This is straight forward and doesn't require any fancy footwork

Chris Peterson wrote:

alter-ego wrote:Your comment about ecliptic angle = observer latitude caught my interest because, although very close to true in this picture, I didn't believe it was generally so...

I didn't try to follow your analysis, so I won't comment beyond saying that whether we use the latitude or (90° - latitude) in the calculations depends on the arbitrary choice of whether we measure with respect to the horizon or the perpendicular to the horizon. As a meteor scientist, I normally use the later in referring to angles in cases like this.

That said, my comment about the angle was mainly directed to the observation that the line of suns in the image do not lie along the ecliptic, because the ecliptic is moving with respect to the horizon. So the angle made by the line of suns isn't the angle of the ecliptic at sunset. That can be seen in these simulations, which show both the Sun at each time as well as the ecliptic at each time. The first image shows each frame, the second just the first and last. An image like this gives the impression that the ecliptic is steeper with respect to the horizon than it actually is.

Yes, I understood, and agree with, your comments about the ecliptic and sunset path angle appearing steeper. Your pictures showing the sag (concave-up curvature) in the sunset path visually express your statement nicely (the value of a picture!). Prior to my analysis, I made the exact pictures both with and w/o refraction to see just how much impact the atmosphere has (then of course I became interested in more detail ). The changing ecliptic slope is why I chose to calculate the ecliptic angle at the moment of sunset (altitude = 0°) Your certainly right, too, about choice of axis for zero degrees. Recognizing this, I tried to be clear with my angle definition which I thought logically agreed admirer's view. As a meteor scientist, it makes sense that you would naturally reference angles normal to the horizon.