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)
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 alma mater's mailing list sometimes includes a mathematical brain teaser in their emails. Often, when I first look at the brain teaser, it seems like it should be easy to solve. Sometimes, it is. Other times, it takes me a long time to figure out (or I may even give up, deciding I have other pressing things to do with my time).

I wonder if I could have solved these problems quicker/easier, when I was younger. Or if they would have been the same difficult for me, back then.

Latest brain teaser:
There is a number less than 3,000 that when divided by 2 leaves a remainder of 1, when divided by 3 leaves a remainder of 2, when divided by 4 leaves a remainder of 3, when divided by 5 leaves a remainder of 4, when divided by 6 leaves a remainder of 5, and so on up to nine.
What is that number?

How I finally solved it )

(no subject)

Monday, February 16th, 2009 09:59 pm
darkoshi: (Default)
This mathematical thing looks awfully interesting and is full of stuff I've never heard of before. I came across the term E8 on this page... still not totally sure if that is something real or a spoof or something.

Octopus survives 5 days on the run... seems that octopi are smarter than I realized.

Both links from [livejournal.com profile] fayanora.

(no subject)

Friday, August 1st, 2008 08:51 pm
darkoshi: (Default)
I've been thinking about it more, and the conclusion I've come to is that the formula given on that webpage is not an accurate calculation of the probably of someone being male or female, even taking for granted that the input variables r_n are accurate.

It is simply a formula which appears to give a reasonable result for various ratios. But it is not necessarily accurate given n > 1.

If the male-to-female ratio of visitors to a website is 2-to-1, then the probability of a website visitor being female will be 33.3%. However, given another website with the same ratio, and given that a person has visited both websites, this does not conclusively indicate that the probability of that person being female is then 20%, as the formula suggests. In order to determine the actual probability, we would need to know the actual M/F ratio of people who have visited both websites; however this piece of data is not given nor can it be determined from the separate ratios for the individual websites.

For example, suppose that all the people who visited the first website are the same group of people who visited the 2nd website. In that situation, the overall probability of one of those people being female would remain 33.3% even though they had been to both websites.

As a another example, suppose that a certain topic is mainly of interest to females, but that a small percentage of males are also interested in that topic. Websites devoted to that topic will tend to have a high F/M ratio. However, the males that are likely to visit one of those websites will also be likely to visit other of those websites. Just because a person visits multiple of those websites, does not make them less and less likely to be male.

The formula does seem to make sense in a general way, as a guess as to whether a person is male or female, based on whether the result is more or less than 50%. However, the actual percentage numbers given are not logically accurate.


Update on 08/02/2008 (still haven't been able to drop this line of thought):

Ok, so if you base your calculation on the assumption that each website is completely distinct topic-wise, from every other website, and so that the M/F ratio at any website is completely unrelated to the M/F ratio at another website... That there is just some inherent quality of the website that attracts Ms and Fs in different ratios, and that quality is completely unrelated to all the other websites' inherent qualities of M/F attraction...

Ok, so then maybe that formula does make sense.

Given website #1, if X females visit it, then X * R1 males visit it (as R1 = the M/F ratio for that website).

Given website #2, suppose that all the visitors from website #1 "bump" into website #2 (where bumping into the site isn't visiting it, it is just coming across it and being given the decision to visit it or not).

Given that R2 is completely unrelated to R1, then we can expect that of the people from website #1 who have bumped into website #2, that the ratio of males to females who decide to visit #2 will be R2.

Suppose that of the X females who've visited #1, Y percent of them decide to visit #2. So X * Y = the number of females who visit both websites.

The percentage of males who've visited #1 and also decide to visit #2 will be Y * R2, since R2 is the generic M/F ratio for website #2.

Therefore the number of males who visit both websites will be the number who've visited #1 (X * R1) multiplied by the percentage of those that visit #2 (Y * R2). That equals X * Y * R1 * R2.

Then the ratio of M/F for people who've visited both websites will be the number of males divided by the number of females:
X * Y * R1 * R2 / (X * Y) = R1 * R2.

For each additional website that is added to the calculation, the total M/F ratio will be R1 * R2 * ... * Rn. And that is what the formula on that other page uses.

But it is based on the assumption that the Rn numbers are accurate, and that all websites have unrelated M/F qualities of attraction.

(no subject)

Friday, August 1st, 2008 07:54 am
darkoshi: (Default)
I started thinking about it in the shower again, gah!

I'm not grasping the concept of multiplying ratios together. I understand multiplying percentages, but not ratios. Although I can see that multiplying the percentages in this case doesn't give a reasonable result, and that the given calculation does give reasonable results.

The calculation makes sense where n=1. For one website, the probability that you are female would be equal to the number of females out of the total number of website visitors. When the ratio of male to female visitors is 2 to 1, then there would be one female out of 3 visitors (1 / 1 + r_n). But I'm not grasping the rest of it. If you visit 2 websites, and both have a m/f ratio of 2/1... the calculation is saying that there is one female out of 5 visitors... as if the 2 males at both websites are added together, while the one female remains the same person... but how do we know that some of the males aren't the same person, too? What does it logically result in, when you multiply ratios together? .... It seems like the answer is so close, and yet so far away.

(no subject)

Thursday, July 31st, 2008 08:50 pm
darkoshi: (Default)
A mathematical equation is bothering me. I can't figure out the logic behind it.

"if you had been to two sites that had a 2-1 ratio of men to women, the probability of you being female would be: 1 / (1 + 2 * 2) = 1/5 = 20%"
Why 20%? Why 1 / (1 + 2 * 2) ?

I used to be good at math. Supposedly.

.

It was thundering a bit before, the sky was darkening, and it was getting windy. But then it seemed to pass, without climaxing... without rain. I was going to write that. But a while later, it did start to rain.

.

I just can't stop thinking about it. It's been bugging me since yesterday.
So the probably of you being male would be (1 - the prob of you being female) which would be
((1 + 2*2) - 1) / (1 + 2*2)
which is 2*2 / 1 + 2*2

But Why?
Why multiply those numbers together, and why add one?

punny

Tuesday, January 6th, 2004 09:58 pm
darkoshi: (Default)
(found on the web somewhere...)

At New York's Kennedy airport today, an individual later discovered to be a public school teacher was arrested trying to board a flight while in possession of a ruler, a protractor, a set square, a slide rule, and a calculator. At a morning press conference, Attorney General John Ashcroft said he believes the man is a member of the notorious al-gebra movement. He is being charged by the FBI with carrying weapons of math instruction.


"Al-gebra is a fearsome cult", Ashcroft said, "They desire average solutions by means and extremes, and sometimes go off on tangents in a search of absolute value. They use secret code names like "x" and "y" and refer to themselves as "unknowns", but we have determined they belong to a common denominator of the axis of medieval with coordinates in every country.

"As the Greek philanderer Isosceles used to say, there are 3 sides to every triangle," Ashcroft declared. When asked to comment on the arrest, President Bush said, "If God had wanted us to have better weapons of math instruction, He would have given us more fingers and toes.

"I am gratified that our government has given us a sine that it is intent on protracting us from these math-dogs who are willing to disintegrate us with calculus disregard. Murky statisticians love to inflict plane on every sphere of influence," the President said, adding: "Under the circumferences, we must differentiate their root, make our point, and draw the line."

President Bush warned, "These weapons of math instruction have the potential to decimal everything in their math on a scalene never before seen unless we become exponents of a Higher Power and begin to factor-in random facts of vertex."

Attorney General Ashcroft said, "As our Great Leader would say, read my ellipse. Here is one principle he is uncertainty of: though they continue to multiply, their days are numbered as the hypotenuse tightens around their necks.