darkoshi: (Default)
Follow-up to this post.

In the prior post, I pondered about angles. In particular, about the ~13 degrees per day that the moon orbits around the earth, and whether that angle would still look like 13 degrees to me, when measured from the surface of the earth.

The angle would *not* be exactly the same. However, because of how far away the moon is compared to the size of the earth, the difference in angle is very small. That difference can be calculated using trigonometry.

Here's a new diagram. All figures mentioned below are approximations or averages.

The angle measured from the center of the earth is 13 degrees.
"X" is the corresponding angle measured from the surface, which will be calculated.
"D" is the distance from the center of the earth to the moon: 384400 km
"R" is the radius of the earth : 6371 km

I've drawn 2 right triangles such that both have the same "opposite" side, with length "O".
The length of the adjacent side for the X-angled triangle is "A".
The length of the hypotenuse for the X-angled triangle is "B".
The length of the adjacent side for the 13-degree-angled triangle is A + R.
The length of the hypotenuse for the 13-degree angled triangle is D.

sin 13 degrees = O / D
O = D * sin 13 = 384400km * 0.2249511 = 86471 km

cos 13 degrees = (A + R) / D
A = (cos 13) * D - R = (0.9743701 * 384400) - 6371 = 368176.85

tan X = O / A
X = arctan(O / A) = arctan (86471 / 368176.85) = 13.217

So, the corresponding angle from the surface of the earth is ~13.2 degrees.
darkoshi: (Default)
For the last 3 days, I've seen the crescent moon in the sky during the late morning.

2017/08/16, 10:16am EDT:

2017/08/17, 8:43am EDT:

(On 2017/08/18, I saw the moon around 10:30am, but didn't think of taking a photo.)

But I have been unable to find the moon in the sky around 2:30pm (during my lunch breaks). I've been wondering why I can't find it in the afternoon.

(No wonder I've never paid much attention to the path of the moon in the sky. At night, I'm usually inside or asleep. In the daytime, even when the moon is in the sky, it's hard to see.)

On all 3 days, it's been partly cloudy, with today being the least cloudy. So it's possible the moon was behind a cloud. But as much as I've searched the sky, it seems unlikely it's *always* been behind a cloud.

As of today (2017/08/18) at my location, per the NOAA solar calculator (Find Sunrise, Sunset, Solar Noon and Solar Position for Any Place on Earth), solar noon is around 1:30pm. So at 2:30pm, the sun is still fairly high overhead.

On 8/16, it was 5 days before new moon and the eclipse, so the moon would have been about 5 * 13 = 65 degrees away from the sun. So that was most likely too near the horizon for me to see, as there are some trees and buildings around.

On 8/17, the moon would have been 4 * 13 = 52 degrees away from the sun. I think I should have been able to see it at that angle.

Today on 8/18, the moon would have been 3 * 13 = 39 degrees away from the sun. Surely I should have been able to see it at that angle.

The closer we get to the new moon, the thinner the crescent is. So the harder it is to see. It is hard to find a tiny arc of white in a light blue sky, and even more so when there are distracting white clouds around. But is that the only reason I haven't found it?

Per this page: Finding the Moon, crescent moons are "not observable" except right before sunset or after dawn. But I've seen it at 10:30am which isn't right after dawn. So I think it would be more accurate to say "not easily observable".

If I can see it at 10:30am when the sun is already bright in the sky, why shouldn't I be able to see it at 2:30pm?

I got to wondering whether how I think of the angles in the sky is wrong. I am thinking of 45 degrees as being the distance from straight overhead to a point halfway to the horizon. But the 13 degrees that the moon moves per day is in relation to the center of the earth, not to my spot on the surface of the earth. Therefore, is how I'm visualizing the angles in the sky wrong?

When the moon orbits 45 degrees around the earth, is that a much greater distance than the distance I see from overhead to halfway to the horizon?

But... as can be seen in the diagram, the larger you draw the earth, the closer the 45 degrees gets to one's visible horizon, and it would eventually even pass below the horizon. Yet I've been able to see the moon in the mornings, and the distance between it and the sun hasn't seemed such a large angle. So surely the above diagram can't be right.

(Update #2, 2017/08/20: I've figured it out. The diagram is basically correct, but my assumption about the 45 degree line eventually passing below the horizon was wrong (just because I don't draw the horizon line to infinity, doesn't mean it doesn't go to infinity). If the angle to the moon as measured from the center of the earth is 45 degrees (from directly overhead), then the angle as measured from the surface of the earth would be more than 45 degrees. But because the distance to the moon is so large in comparison to the size of the earth, the angle is only slightly more. See follow-up post.)

On the same topic, I got to wondering how much of the sky / celestial sphere am I actually capable of observing from a point on the earth, at any moment in time. Ie. if I turn all the way around, looking towards the horizon, and up above me, how much of the sphere of the sky which surrounds the earth, am I seeing?

Based on the diagram, the amount of sky seen would not be half the sphere, as I've previously assumed. Yet again, the larger one draws the earth, the less of the sky one would seem to see. Surely that can't be right?

Based on these answers, it sounds like you should be able to see half of the sky at any time. But I don't understand the formulas and calculations listed.

Update (afternoon of 2017/08/19):

Today, the morning of 8/19, around 7:40am and again at 10:20am, I wasn't able to find the moon in the sky, even though it was clear with no clouds. So as of 2 days before new moon, the crescent must be too small and faint to see in the daytime. Perhaps a clear sky being so much brighter than a partially cloudy sky, also makes it harder to see.

MoonCalc.org - shows you the current position of the moon in the sky, and moonrise/moonset directions, for any position you select on the map.

Sun Locator Lite - a free app which lets you find the sun and moon by pointing the phone at the sky (as long as the phone has an internal compass/magnetometer - mine doesn't, but Qiao's does). The Pro version lets you get information for any day and time of the year.

Today, 2 days before the eclipse, the moon should be about 2 * 13 = 26 degrees from the sun. I used the above Sun Locator app to find the position of the moon and compare it to the sun's position, and estimated the angle between them. If anything, it seemed less than 26 degrees, not more. So that indicates that there's something wrong with my thinking in terms of the above diagram. But where have I gone wrong? I still haven't figured that out.
(And even with the app to show me its exact location, I still can't see the crescent moon in the afternoon sky.)

But I did have an epiphany on how much of the sky is visible from a point on earth at a single moment in time. It depends on what I'm calling the "sky". I think of the sky as a sphere centered around the earth, upon which I see moon, sun, stars, clouds, etc. But there are many such possible spheres around the earth, different distances from the center of the earth.

How much of the sky is seen depends on which of those spheres one considers. If one considers a sphere which is say, 10 kilometers above sea level, you can calculate the surface area of that sphere. The earth's diameter is 12,742 km. So the sphere's diameter would be 12,752 km, its radius (r) would be 6376 km, and it's surface area would be 4*pi*r^2.

[ another interesting thought... For an infinitely thin sphere, the size of the inside and outside surface areas should be the same, right? But how can that be? I can't visualize them being the same size. ]

Imagine that we cut a small slice, 10 km deep, from the top of that sphere. We can then calculate the surface area of that slice (with some formula, which I would have to look up.) That would tell us how much of the whole sphere we can see at a single moment, and it would be a fairly small portion.

But now, consider a sky-sphere with a much larger radius of 5 light-years - reaching the nearest stars - or even larger. At such distances, the diameter of the earth is minute in comparison - it can be considered negligible. A plane which touches the surface of the earth at one point is practically the same as another parallel plane which intersects both the center of the earth and the sphere. Either way, half of the sphere is above the plane, and half below. So the person can see half of that sky-sphere.

Now, what about a sphere with radius of 150 million km (about the distance from the earth to the sun)? In comparison to that distance, the earth's diameter is roughly 0.01%.* So again, it's basically negligible, and we can see practically half of the sphere at any moment in time.


Other interesting tidbits:

How far away is the horizon? Short answer: About 4 to 5 kilometers away, at standing eye-level for an average-height adult.

I see the moon: introducing our nearest neighbour - has several good diagrams/images.
Per this page, the moon's orbital plane is tilted 5 degrees from the ecliptic. That's not as much as I imagined. But when you add in the 23.5 angle of the earth's axis, the moon can orbit up to 29 degrees above or below the earth's equator.

Lunar Orbital Libration
Libration definition: "a real or apparent oscillatory motion, especially of the moon."

Altitude and Azimuth

* A lot of these numbers are rough calculations I've done, and they may have errors. Please don't rely on any numbers I've posted, without verifying them. If you find an error, please let me know so that I can correct it.
darkoshi: (Default)
I thought of an easier way to explain why the eclipse shadow travels west to east, even though the moon travels east to west through the sky.

First, here's the general picture from the perspective of the sun, when looking down at the solar plane from above:
The earth revolves around the sun in a counter-clockwise direction, completing a full circuit about every 365 days.
The moon revolves around the earth in a counter-clockwise direction, completing a full circuit about every 28 days.
The earth rotates around its own axis in a counter-clockwise direction, completing a full turn every 24 hours.

Here's the general picture from the perspective of a spot on the earth at the equator, when looking up at the sky:
The sun revolves around the earth in an east to west direction, completing a full circuit every 24 hours.
The moon revolves around the earth in an east to west direction, completing a full circuit about every 24.5 hours (I hope I calcuated that right)

Now to explain why the solar eclipse shadow goes west to east:

Imagine you are standing on the north side of an east-west street, facing south.
The moon is a person walking on that street from east to west.
The sun is another person walking on that street from east to west, except that they are walking slightly faster than the moon, and emitting a bright light.

When the sun is still a fair bit behind the moon, the shadow that is cast from the moon due to the sun's light will point towards the west.
As the sun starts overtaking the moon, walking behind the moon compared to the observer, the shadow that is cast points towards west-northwest.
As the sun continues passing behind the moon, that shadow changes direction, towards to the northwest, then north, then northeast, then east-northeast.
So even though both the sun and moon are going east-to-west, the shadow goes west-to-east.

Maybe that is totally obvious to other people? I mean, it seems pretty obvious to myself now that I've explained it.


It's actually more complicated than that, of course.

The sun's path does go from east to west rather consistently, even though during the summer, the path is higher in the sky (northeast -> northwest) than during winter (southeast -> southwest).

But the moon's path is more dynamic, as it doesn't revolve in the plane of the equator. It may rise in the southeast and set in the northwest. Or it may rise in the northeast and set in the southwest. (right? I haven't ever really paid much attention to the moon's path, but I must have learned that somewhere.)

Because of that, based on the images I've seen, instead of the moon crossing the sun from right to left, during this eclipse, it will cross it from lower right towards the upper left.

So in the above example, the moon would be on a different street, at an angle to the other street, and the streets would happen to cross each other right at the point where the sun was walking behind the moon.
(although what angles the streets need to be at, and which direction the moon is going on its own street is a bit difficult for me to visualize right now.)

I guess it's time to create a new eclipse tag for all these entries, and to rename the eclipse tag I used on a single other post in reference to the software called "Eclipse".
darkoshi: (Default)
This morning, I started wondering why the eclipse will be seen on the west coast before it is seen on the east coast.

I know that the earth spins counter-clockwise (towards the east) when looking at it from above the north pole. And that the moon travels around the earth in the same counter-clockwise direction. And that the earth spins relatively faster. It does a complete rotation (360 degrees) in 24 hours, whereas the moon only travels 1/28th of the way around the earth (360 / 28 ~= 13 degrees) during that time.

So how can the moon's shadow travel from west to east? Isn't the earth spinning into the shadow and out of it in a clockwise west to east direction (the same as it always spins), and so the shadow should appear to move from east to west, just like the moon appears to do in the sky?

According to the answer on PhysLink.com, it has something to do with the moon's orbital velocity being greater than the earth's. But according to this orbital velocity formula, it seems that anything closer to the earth (ie. the earth's surface) would have a greater orbital velocity than something further away (ie. the moon). So that answer seems to be wrong or badly worded, maybe. Of course, if we simply consider velocity, the moon does travel a further distance through space than the earth's surface does, in the same amount of time. But what does that have to do with the eclipse? If it were a race, the earth would still win, rotating faster than the moon revolves.

Here's another page (cached, as the original eclipse2017.org page isn't responding - the website must be swamped) that tries to answer the question. Some of the commenters on that page seem to have the same confusion as I do.

Here's another page (Washington Post) that tries to explain it. Again talking about the speed of the moon compared to the earth.

Watching the various NASA visualizations didn't explain it well to me, because of how the videos keep shifting perspectives. Even in this animation, it looks like the sun must be moving from behind the viewer, to make the shadow move like that.

I think I may finally sort of understand it, but my explanation doesn't match any of the answers I read. So it's probably wrong. But... as the moon moves across the sun from right to left (as seen from the earth while facing south), it's shadow as seen from the earth changes direction. First it points towards the west, then straight, then towards the east.
And while the moon itself, from the earth's perspective, doesn't move far in the sky (and due to the earth's rotation, even appears to be going to the west*), it's shadow moves much faster... that must be why the answers keep mentioning the speed of the moon.. they must be trying to say that the speed of the moon's shadow across the face of the earth matches the moon's speed in space. I suppose that is logical, even though it isn't very intuitive to me**.

* But the sun appears to move to the west faster than the moon, so the moon does still cross the sun from right to left, even though they are both moving to the west.

So the moon's shadow moves quickly from the west to the east.

Right? Maybe? Sort of?

** Because the moon doesn't move in a straight line, but rather circles the earth. And those x-thousand miles per hour it moves up in space only correspond to y-hundred miles down on the earth... Oh jeez, now I'll start doubting my above explanation again...

Ok, thinking about it more. The shadow moves west to east like I explained above, because the moon crosses the sun from west to east. The speed of the moon through space around the earth affects the speed of the shadow, but it's not a direct x = y equation. The faster the moon moves across the face of the sun, the faster the shadow sweeps across the land from west to east. Since the shadow is sweeping through an arc (sort of), the far end of the shadow will pass a different distance during that time, depending from how far away you measure it... which for us is based on the distance between the moon and earth. So the speed of the shadow depends on that distance, and on the speed of the moon's revolution, and on the speed of the earth's rotation, and the size of the earth, etc. And it is complicated more because the moon moves in an elliptical orbit, not just straight past the sun, etc.

Now it makes sense to me. If I'm wrong, feel free to tell me which of my logic is wrong.


On a related topic, how long will totality last, across the U.S.? It will start on the west coast around 10:17am (1:17pm eastern time). It will end on the east coast around 2:48pm (eastern time). So for one and a half hours, the shadow will sweep across the country, from coast to coast.

Within that time period, based on the 3 to 4 hour time difference between the coasts, the earth only rotates about half the same distance.

But actually, the earth and the shadow are moving in the same direction... so if the earth weren't turning, the shadow would traverse the distance even faster.

2017/08/16 Corrected some words above. I was mixing up the words "right" and "left", even though I was visualizing it correctly. I'm used to thinking of the west coast as on the left side and the east coast on the right side. For the above, my perspective is from the center of the county looking south. So the west coast is on the right, not the left.
Although since the sun will be pretty high overhead during the eclipse, "left" and "right" aren't good words to use to begin with.
darkoshi: (Default)
My mom called to tell me that she heard that Jupiter is bright tonight - at opposition - and that it might even be possible to see its moons using binoculars. I thought, fat chance of that for me, but I went and looked. First I had to figure out how to adjust the focus on the binoculars; it's been a long time since I used them, and I was turning the diopter adjustment by mistake. Then, when I was looking at Jupiter, I did see another fainter object to the west-south-west (ie, left-bottom-left) of it, about 8 Jupiter-widths away. Could that be a moon? Based on this diagram, it's probably the star Theta Virginis.

Tonight, I kept having a problem with the binoculars, seeing double. I'm not sure if it is an issue with them or my eyes. To begin with, it wasn't double. Then, when it was, every once in a while, the double vision went away.

Since the moon is almost full, I looked at it through the binoculars too. Wow! Maybe I've never looked at it through binoculars before? That is the clearest I've ever seen the moon, that I can recall. The craters and lines/rays and texture/contours (towards the bottom) are visible, and even a crater sticking out on the left edge between light and dark.

Curiosity rover shows new signs of wheel wear - it's still sending back photos from Mars!? Wow, I didn't realize we still had contact with it.

By the way, it's weird that the moon always looks black & white / grayscale, when the rest of the universe is in color.

astronomy! hail!

Friday, May 23rd, 2014 09:49 pm
darkoshi: (Default)
Why do these articles seem so amazing* to me tonight?!

NASA discovers 715 new planets (including an earth-sized one in a habitable zone)

Astronomers find first asteroid with rings

Dwarf planet discovered at solar system's edge

And regarding tonight's meteor shower:
Meteor shower approaches in time for holiday weekend - not just any meteor shower, but a *new* one!

*I know! It's because I got to see quarter-sized hail thunking down out of the sky, just as I was ready to leave work! It was making loud ka-plunking noises on the overhead skylight/ceiling! And then when that ended, there was awesome mist swirling on the pond! And then the sun was so beautifully orange amidst the dark cloudy sky! And then after getting home, there was a little bit more hail, and one of those stationary cloud lightning light shows in the sky!
darkoshi: (Default)
If the Moon was only one Pixel - a tediously accurate scale model of the solar system. Wow. Just wow. I recall being similarly impressed by another scale model, which I thought I had also linked to, but I can't find that one any more...

Ah, here it is - an image showing the earth and moon, and distance between them, to scale.

(no subject)

Thursday, January 3rd, 2008 11:52 am
darkoshi: (Default)
Quadrantid meteor shower tonight - best viewable from western Europe and the Eastern U.S./Canada.

This page has a sky-chart of the radiant location.

It's going to be vewwy vewwy cold here tonight though, not sure if I will venture outdoors long to peek at the sky.

hair milestone

Monday, November 19th, 2007 12:18 am
darkoshi: (Default)
Today was the first day since I started growing my bangs out that my hair felt comfortable enough not to have to pin it back out of my face. In other words, it wasn't scratching my eyeballs or tickling my nose.

It sure is nice to be able to pull shirts over my head again, without it messing up the pins in my hair. And it looks much better without the pins.


I can see the sun as it rises from the yellow room window.

(The sun below the horizon is a window reflection.)
It is surprising how fast the sun rises - on the day I took the above photo, the sun was only halfway over the horizon when I first looked out the window, and that is what I wanted to capture. But in the minute or so that it took me to get my camera from the other room, it had already risen all the way over the horizon, as can be seen in the photo.

I will take a photo every week or so until the solstice, so I can see how the sun moves along the horizon, and what the southernmost sunrise point is... Maybe I'll try to figure out where it should rise on the solstice, based on this data, and then see if I'm right or not.

sunshine stuff

Sunday, May 21st, 2006 02:49 pm
darkoshi: (Default)
Starting in 2007, Daylight Savings Time in the U.S. will start earlier and end later. We will have about 3 more weeks of DST. Our start and end dates still won't match up with Europe's.

From Wiki... "Clocks will be set ahead one hour on the second Sunday of March instead of the current first Sunday of April. Clocks will be set back one hour on the first Sunday in November, rather than the last Sunday of October. ... This will affect accuracy of electronic clocks that had pre-programmed dates for adjusting to daylight saving time. The date for the end of daylight saving time has the effect of increasing evening light on Halloween (October 31)."


My compass doesn't seem to consistently point in the same direction for North. Sigh. How can I use my compass, when I can't even be sure it's pointing in the right direction?

Columbia, SC data
(latitude 34 degrees north, longitude ~81 degrees west)

- Magnetic North is 6 degrees 40' west of geographic North

- On the summer solstice, sunrise azimuth is at ~ 61 degrees, and sunset is at ~298 degrees.
- On the winter solstice, sunrise azimuth is at ~118 degrees, and sunset is at ~241 degrees.

Ie., the sunrise angle varies between 0 and ~29 degrees north and south of Due East
and the sunset angle varies between 0 and ~29 degrees north and south of Due West.

- The sun's highest altitude in the sky is...
on the summer solstice at 80 degrees,
on the equinox at 56 degrees,
and on the winter solstice at 32 degrees.

- In the summer, earlist sunrise is at 6:12am and latest sunset is at 8:40pm (Daylight Savings Time).
- In the winter, latest sunrise is at 7:30 am and earliest sunset is at 5:15pm (Standard Time).