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## MIDDLE EAST UPDATES (if there are any to speak of) August 9 - 15, 2016

### The Ultimate Solar-Distance Proof

The figures in the last chapter were revamped for over a week after letting it go public, for the purpose of needing more precision. If you loaded the page at any time before seven days after it was out, you didn't get the final copy, and you may have gotten the page only partially revamped, which would have had some inconsistent numbers. I found some typing mistakes amongst the numbers too, sorry (hoping all has been made well).

NASA is away on leave when we Google-search "length of the earth's shadow." Is this a problem for the astronomers? Yes, it can be, because they don't give the correct length? And I can tell you why. The eclipse data told me that they have sun lines spreading wider from the earth's shadow than is the reality, and I've come to realize why this is so. It's just a matter of looking at a drawing with lines from the sun's edges to the earth's edges; let the lines come to a tip behind the earth and call it the E-M (evening-morning) triangle. Call the small triangle behind the earth the umbra. The total eclipse of July, 2000, told me that these lines are about .218 degrees, or, depending on where one places the zero-degree line, .218 less than 90 degrees. The tip is to be viewed as twice .218 when the zero-degree line is from the tip through the earth's core.

I had these solar lines found at about .218 (not fully refined yet) in summer, when the sun is always at its greatest distance. In both the spring and fall, the sun achieves its average distance. The last chapter claimed that the average-sun angle is about .2221 degree. This is where I'm at thus far in this study, where I have the astronomers under my magnifying glass. If you imagine the lines, at .2221 degree, going out as far as 93 million miles, the sun's edge's should be placed exactly between the lines, if astronomers wish to have the incorrect distance to the sun in order to create and support their long, cosmic-evolutionary picture. In the least, they could leave the sun between the two lines if they are going to make the sun out that far. Why don't they? What would be the problem if they did?

The answer is easy. They committed to putting the sun 93 million miles away in a different triangle, the one with a tip at the eye, which is the one at .5331 degree on average. They called this the angular size of the sun. In this way, the sun has edges that are beyond the lunar-eclipse lines one the latter have reached 93 million miles. Anyone who knew how to work with triangles would realize that the sun could not be 93 million miles away if the angular-size lines spread out at .5331 while the E.M lines spread out at .2221. It's easy to figure where these lines meet as they extend outward from the earth, and it's in the neighborhood of 5 million, not 93. Therefore, their solar distance-and-diameter ratio is not to scale with a sun that is placed between the E-M lines exactly. I had an average-sized sun to scale in the last chapter, at 5.1 million away, and about 47,508 miles wide. The sun was both between the E-M lines and the one going out at .5331 degree. That's how it should be.

The umbra turned out to be 1.037 million miles long, yet they have it at a maximum of about 885,000 miles...because they have the sun wider than the E-M lines allow. When the sun is made wider than these lines, the shadow is going to come out shorter.

Here is my claim: they are lying to us when they make us assume that they have measured the E-M lines from opposite sides of the earth. They claim that a line from both edges, both to the center of the sun, is at .0024 degrees (they call this the solar parallax). This is the number they need to have a sun at 93 million miles. In other words, rather than reporting the actual angle, they report the angle from their math with 93 million as part of their computation. But they themselves claim that it's impossible to measure this angle because they claim that the sun refracts. That is, they say that the sun is not seen where it really is. They claim that the sun is seen further out from the lines than it really is. It's perfectly convenient for their 93-million-mile hoax. This situation probably goes back to when the hoax was established, before they had the ability to do the hands-on measurement. They merely did the math to get the solar-parallax angle, and they had astronomers coming in with different figures until they settled on an angle that had the sun 93 million away.

Later, astronomy developed to the point of making a pickle out of this hoax. For example, sunrise and sunset times could expose it, wherefore they had to control sunrise/sunset information that goes out to the masses, and they claimed that the evening-morning lines were at an angle wider than the reality, and conducive to their 93-million scheme. I've not ruled out that some refraction occurs, but not as much as they claim. They are expected to claim as much as needed to protect their scheme.

http://www.cleavebooks.co.uk/scol/calrtri.htm

has to be at the same angle as the E-M triangle.

Let's try the same at the average solar distance. The E-M triangle is now 92,955,000 + 864,000 = 93,819,000 miles long; it works out to be .528 degree. Let's put it this way, that at that angle, the shadow works out to be 864,598 miles long (put .528 in the angle-A box with the solar diameter in the edge-a box to find the shadow length is the edge-b box).

To find what they ought to claim for the shadow-length on their average solar distance, the angle of their E-M triangle needs to match the angle of the shadow-alone angle. In this case, the shadow length works out to about 858,818.5 miles long, for when we add that number to 92,955807 to get 93,814,625.5 the latter finds an angle of 52823206. (using their solar diameter in the edge-a box). Meanwhile, with 858,822 in the edge-b box along with the earth diameter in edge-a, the angle comes out as .52823167, a virtual match. There will be a match at some fraction of 858,818.

I was unable to find any claimed shadow size for any particular sun size amongst their major, three sun sizes (maximum, minimum and average). If we had just one of those, we could do some further exposing, which can explain why not much comes up on the topic in a Google search. The less the astronomers say about this, the better for them. Now that we know how long their shadow is for an average sun, we can take any eclipse with a sun at its average distance, and check to see whether they have the correct umbra diameter for the lunar path. This is a great tool, but I did not realize it until working on this chapter.

One can do the same hit-and-miss calculation (as got the 858,818.5) with the sun at aphelion, where it was, roughly, at the 2000 lunar eclipse. NASA tells us that the sun's radius on that day was 15'44.2", which is 15.73667 arc-minutes = .262277 degree (just divide the arc-minutes by 60). Multiply by two to get .52455 degree as the tip of the triangle (this is not the E-M triangle). To compare that angle with their aphelion figure of 94,500,000 miles, we put the latter in edge-b along with the solar diameter in edge-a to find an angle of .5244012 degree. The latter angle is supposed to be the smallest possible, and it is smaller than .52455, wherefore we are good. The numbers are reliable enough, and because they are virtually the same angle, we can go forward seeking the shadow length on the day of the eclipse. After we find that, we can test things against the eclipse data.

It should be noted that there is ZERO REFRACTION of light in these calculations, yet they get the figures that astronomy uses...meaning that the astronomers really do not include refraction in their scheme, even though they claim to, when needed. They have refraction set up to bend around the earth so as to make the umbra shorter than it would otherwise be, yet the calculations here are of straight, unrefracted lines...from THEIR solar dimensions across the edge of the earth.

Here is the result with an umbra length of 873,080:

94,500,000 + 873,080 = 95,373,085 miles, at an angle of .51960087 degree;
The shadow alone at 873,085 miles long works out to .51960066 degree, a match.

Knowing now that their scheme has a shadow of very close to 873,085 miles long for the eclipse, we would like to know how far along that shadow the eclipse took place. Their eclipse page (below) has a round-about way to discover their shadow path as 5,738 miles wide. I need to show how to arrive to this, which was done a few chapters ago using the data on the eclipse page below:

Gamma is given as .0301, just 119 miles between the lunar and umbral centers...The umbral magnitude, 1.7731, means that there are 1,670 miles from the outer edge of the umbra to the outer edge of the moon. The reported umbral diameter is therefore twice as much as 1,670 + 1080 + 119 = 5,738 miles, ridiculous.
http://eclipse.gsfc.nasa.gov/LEplot1/LEplot1951/LE2000Jul16T.GIF

The times given for the eclipse proved that 5,738 was too low a figure for the reality. I proved from their own figures (not one of them my own) that it was at least 5,998 miles wide. Just so you can verify the gamma figure above, it refers to .0301 of the earth radius, or .0301 x 3,959 = 119 miles. To find 5,738, one adds their umbral magnitude, from edge of shadow to edge of moon, then goes to the center of the moon by adding half a lunar diameter of 1,080 miles, and then goes from the moon's center to the shadow center for a total of 1,670 + 1080 + 119 = 5,738 miles. This is the eclipse where, if one goes by their data, the moon's velocity works out to much lower than the stated, lowest lunar velocity (I've read 2,163 mph).

We can now prove that their eclipse figure is wrong by a method I have now known until now. This is the ultimate proof. We know (or assume) the shadow to be 7,918 miles wide at the earth's edges, and we know it to be zero miles at the tip. Can we find how far from the earth the shadow is 5,738 miles wide when it's 873,085 miles long? I think we can, easily. For example, we know that the shadow is 3,959 miles wide at a distance from the tip that is 873,085 / 2 = miles, and 1,979.5 miles wide at a distance from the tip that is 873,085 / 4 miles. What does the 4 mean? It means 1/4. The shadow is 1,979.5 miles wide when it's 1/4 the way to the earth from the tip.

The math thus becomes obvious using any number. If want to know how far from the tip the 1,979.5-mile width is, we would use: 1979.5 / 7,918 = .25. Therefore, we do, 5,738 / 7,918 = .7247 from the tip, which is 873,085 x .724638 = 632,671 miles (from the tip). This is fantastic. To find the distance from the center of the earth: 873,085 - 632,671 = 240,414 miles! Busted!!!

Something is wrong: their scheme gets the moon at only 240,414 miles for this eclipse, yet the perigee and apogee converter below tells that the moon, on July 15 (2000), was 406,199 kilometers = 252,400 miles away, while the eclipse was less than one day later. One can estimate the moon, one day later, to be at least 250,000 miles away. The reason that the math didn't work for their scheme is that they artificially increased the sun's diameter to more than what the solar lines allowed, thus reducing the length of the earth's shadow to less than the reality. It needs to be longer if the math above is to provide 252,000.
https://www.fourmilab.ch/earthview/pacalc.html

Let's try the same math with the 1.039 million miles that my scheme got (in the last chapter) for the length of this same umbra. I had the moon's path where the umbra was 5,998 miles wide. This was the bare minimum (i.e. it could have been 6,000 or more). The shadow-path math now looks like this:

5,998 / 7,918 = .757514 from the tip. We multiply the latter by 1.037 million to find 785,542 miles from the tip, or 1.037 million - 785,542 = 252,457 miles from the earth.

We are now a little too high, but right in the ballpark, very revealing for showing the error of the astronomers. Don't kid yourself; they crossed this very problem, and, apparently, denied what it told them.

This is starting to shape up excellently, providing what looks like a sure-fire way to adjust the unknown numbers used in calculation of the last chapter. Don't think that the 252,457 figure above is correct, because it's still an approximation. That figure depends on the 1.037 million figure together with the 5,998, but there are other options. After re-vamping the last chapter a few times over a week due to need for precision, I ended up with this:

The 3.05 million figure was obtained with 3960 - 6,208/2 = 856, but with a shadow 5,998 miles instead of 6,208, the 856 becomes 960. The latter number in the edge-a box along with 251,500 in the edge-b box gets .21868 degree, and with the solar radius on that day at .524555 / 2 = .2622778 degree, it's a difference of .2622778 / .21868 = 1.19937 times. The solar distance using the 960 figure is like this: 3,959 / (.004578 - (960 / 251,500)) = 5.203 million miles. We multiply the latter by 1.19937 above to find that the E-M triangle is now 6.24 million miles long, giving an earth shadow of 6.24 - 5.203 = 1.037312 million, and the right-angle calculator has it at 1.037282 million when at .21868 degree.

We have the option of changing numbers in the first two sentences because NASA won't give us the correct data; we are left to fend for ourselves in arriving to the correct numbers. Here is a good option for changing the numbers, using 252,390:

With a shadow 5,991.4 miles wide, the rise from shadow edge to earth edge is 963.3 miles. The latter number in the edge-a box along with 252,392 in the edge-b box gets .21868 degree, and with the solar radius on that day at .524555 / 2 = .2622778 degree, it's a difference of .2622778 / .21868 = 1.199368 times. The solar distance using the 963.3 figure is like this: 3,959 / (.004578 - (963.3 / 252,390)) = 5,200,404 miles. We multiply the latter by 1.199368 above to find that the E-M triangle is now 6,237,198 miles long, giving an earth shadow of 6,237,198 - 5,200,404 = 1,037,942, and the right-angle calculator has it at 1,037,282 when at .21867853 degree (put 3959 in the edge-a box).

5,991.4 / 7,918 = .756,681. We multiply the latter by 1,037,282 to find 784,892 miles from the tip, or 1,037,282 million - 784,892 = 252,390 miles from the earth.

We now have 252,390 versus 252,392. Close enough. Expect this scenario to be 252,391, therefore. We can see that the shadow is either 1,037,942 or 1,037,282. Not using all decimal points should explain the difference, and for these calculations I think the calculator version should be used. But below is another scenario that uses a different number system yet retains the 5.2 million. Let's see how this one squares with the shadow-path math:

With a shadow 6,004 miles wide, the rise from shadow edge to earth edge is 957 miles. The latter number in the edge-a box along with 250,732 in the edge-b box gets .218685 degree, and with the solar radius on that day at .524555 / 2 = .2622778 degree, it's a difference of .2622778 / .218685 = 1.19934 times. The solar distance using the 957 figure is like this: 3,959 / (.004578 - (957 / 250,732)) = 5,201,169 miles. We multiply the latter by 1.19934 above to find that the E-M triangle is now 6,237,973 miles long, giving an earth shadow of 6,238,490 - 5,201,169 = 1,037,321, and the right-angle calculator has it at 1,037,259 when at .218685 degree (put 3959 in the edge-a box).

As we lower the diameter of the shadow path, the lunar distance must increase in order to keep the same 5.2 million distance, and the reverse is also true. We have yet to check whether this second scenario lines up with the shadow-path math:

6004 / 7,918 = .758,272. We multiply the latter by 1,037,259 to find 786,524 miles from the tip, or 1,037,259 - 786,524 = 250,735 miles from the earth.

It's close, 250,735 versus 250,732. Expect this scenario to be 250,733, therefore.

Although both scenarios retained the 5.2 million, one comes in at 252,391 for the lunar distance, and the other with 250,733. How many other scenarios can there be that retain 5.2 million? I suppose one can alter the shadow-path distance slightly and just adjust the lunar distance accordingly to get many, but one way to know the factual one is to get NASA to cough up the real shadow-path diameter and/or the lunar distance of this eclipse. In all the scenarios that I went through, the umbra remained at about 1.0373 million miles. This absolutely destroys the 93-million sun, for by the time one gets the 1.0373 down to where they need it, at about .88 million, the lunar distance is far shy of the 250,000 miles needed for the eclipse.

Ultimately, one should not tailor the math to coordinate with 5.2 million. That's the figure I like for providing the best magic-number scenario. The best way to do this is to find how long it took for the moon to pass through one diameter of the shadow. NASA changed the times for parts of this eclipse so that we cannot know. NASA changed the times (from the reality) and made the moon move much slower than its slowest speed. That's how we can claim that they changed the times. If they did not, the lunar velocity would have been correct. Sick and demented NASA. With the planets all at the wrong distances from one another, all their stated masses of the planets are erroneous. There is virtually nothing that they have correct, and the signs of this must be all over their math. Yet, they continue to move forward with the 93-million-mile sun. Think of the respect NASA would lose to one day say, we were wrong about 93 million, it's only 5.2.

As they changed the times, we can't do any proper math. But by changing the times, they shoot themselves because the times stated, along with their own lowest lunar velocity, gets a shadow diameter of 6,208 miles at minimum. I had found a way to guess at how they changed the times, and by using that logic, brought the miles down to about 6,000. To find the umbra diameter, find the slowest lunar speed claimed by astronomy (can we trust it?), and use it for the one lunar diameter between U1 (11:57:17am) and U2 (13:02:05pm) on the eclipse page. Then, use that same velocity between U1 and U3 (14:49:06pm), that being virtually the full umbra diameter (actually, it's a wee-bit less than a full diameter because the moon was not passing through the perfect middle of the umbra). The times given between U1 and U3 (2.87 hours) multiplied by the slowest lunar velocity (2,163) gets a shadow about 6,208 miles wide. You can burn NASA in this way. And it deserves to be burnt. It does not deserve to be an educator over us. Their time for U1 - U2 has the moon working out to at 2,000 mph.
http://eclipse.gsfc.nasa.gov/LEplot1/LEplot1951/LE2000Jul16T.GIF

You can do the same math with the central-path eclipse of July, 2018, where they have the lunar velocity even slower than 2,000 mph. As this page may disappear, let me record its times for U1, U2 and U3: 18:24:27, 19:30:15, 21:13:12. The lunar path is not as central in this one, but still very central.

Their smallest moon size, so far as I can find, is said to be 29.3 or 29.43 arc-minutes. There are a few webpages repeating 29.3 versus 34.1 as the smallest and largest sizes. Having 250732 in the edge-b box along with 2160 in edge-a, the angle is found as .49335 degree x 60 = 29.6 arc-minutes. That's over 29.43 and therefore allowable. If we use the other figure found for the lunar average, 252,392, in the edge-b box, the angle comes out as .490105 degree x 60 = 29.41 arc-minutes, which is smaller than 29.43. With their angle (29.43 / 60 = .4905) in the angle-A box, their furthest lunar distance is found as 252,189 miles. If anyone has disagreement with that latter distance, they need to change the angular size of 29.43 arc-minutes. If 29.3 / 60 = .488333 degree is fed to the angle-A box, the lunar distance comes out as 253,308, which makes my figure, 252,392, allowable. Thus, if 29.3 = .488333 is most-likely to be the correct figure, the lunar average should tentatively be made: (29.3 + 33.5) / 2 = 31.4 arc-minutes = .52333 degree. At that angle, the average lunar distance comes out as 236,367. This is a mess for precision work.

The NASA data for the eclipse has the moon at 29.44 arc-minutes, and this requires the distance to be 252,103 miles. I've not seen anyone reporting a figure that low for the furthest apogee. It appears that their 29.43 figure is not low enough for the smallest size, and this would make the lunar average a little smaller too, which makes things worse for NASA's average figure, doesn't it?

It's notable that 5.2 million divided by the shadow length is 5.01 times, a virtual round number. In other words, the distance between the earth and sun is 5 times the shadow length. I've only just discovered that 1.0373 million divided by the earth diameter (7918) = 131.005. Or, if we take the number above, 1,037,250, and divide it by the earth diameter, it's 130.9989896.

Now that we have the ultimate proof that the shadow was 1.0373 million miles long, on July 16, 2000, when the moon was about 252,300 miles away, we find it's angle with half the earth diameter in the edge-a box and 1037300 in the edge-b box. The angle is .218676 degree. Just draw this line, and when it meet one line for the angular size of the sun on the day of the eclipse, that's where the sun sits on July 16. The sun at the eclipse was at .52455 degree, wherefore each of its angular-size lines are half of that, or .262275 degree.

With .218676 in the angle-A box, put 1 in the edge-b box to find that the E-M line spreads by .00381664 mile per mile away from the earth. With .262275 in the angle-A box, put 1 in the edge-b box to find that the angular-size line spreads by .0045776 mile per mile away from the earth. Start the race at the earth. The E-M line starts at the top or bottom edge of the earth, and the angular-size line starts at the center of the earth. The angular-size line needs to spread by only half the earth diameter before it meets the E-M line, wherefore the math is this: 3,959 / (.004577 - .00381664) = 5,206,744 miles away.

### The Evasive Lunar Distance

One should be able to find a lunar-distance calculator online, where anyone can have the distance of the moon at any minute of any day. But if you search for "lunar distance calculator," you get all sorts of junk, for example, one that calculates the distance between the moon and the planets, another that calculates the distance of two points on the moon's surface, and others that need to be downloaded not knowing what you'll get...all as if Google pretends that it doesn't know what you want. Of course the Google people know what you want; except that they programmed the system not to get it for you when you ask. Google clearly doesn't want people to be able to know lunar distances any more than the garbage you'll find when asking for it.

Searching "moon distance calculator" doesn't give what one obviously wants with that query. At the bottom of the page, Google does not suggest any search method to get anyone such a calculator. Googling "find the distance to the moon," the bottom of the page has suggestions, but, guess what? No calculator suggestions. We on the street are NOT to know the lunar distances at any one time. Be very suspicious of your astronomical superiors.

They are not reporting to us the correct lunar distances. Repeat from the end of the last chapter:

They have a real problem to explain. If their average lunar distance is really 238,700, as they claim, their furthest distance would be found by first dividing .524 by .49 to get a 1.069 difference in size, wherefore we get the furthest distance of 238,700 x 1.069 = 255,263 miles. Yet, the eclipse of July of 2000 has the smallest-possible moon at only 252,412 miles. Clearly, their own facts support an average lunar distance considerably less than 238,700...There is a difference of 2,851 miles between 255,263 and 252,412, suggesting 238,700 - 2,851 = 235,849 as a more-correct lunar average...

The .524 figure was mine from finding the midway figure between Wikipedia's smallest and lowest angular sizes of the moon. However, when we re-do the math using .5183 degree (their figure) instead of .524, we get .5183 / .49 = 1.057 times, which, when multiplied by 238,700 gets 252,340 miles, bang-on in the neighborhood of their furthest distance. The reason that .5183 is their figure is because it's the angle in degrees for 31.1 arc-minutes, that being the average angular size as reported by astronomy. I was getting my figure from the midway of 29.43 and 33.5, which is 31.465 = .524417 degree. It begs the question on whether my figure, or theirs, is the correct average figure. Mine is midway; theirs is not. Why not?

When we put .524417 degree in the angle-A box and their lunar diameter of 2,159 in the edge-a box, the lunar distance comes out at 235,877 miles, close to the 235,896 that I used in the last chapter. But if we put 2,159.2 in the box, the lunar distance works out to 235,899. Why don't the people reporting the average lunar distance use a figure like the ones here? The angle calculator doesn't lie; they do.

Solarsystemnasa.gov: "The average distance to the moon is 382,500 kilometers." What? That's 382,500 x .62137 = 237,674 miles, not even close to 238,700. The Google page for "average distance to the moon" has 384,400 kilometers, not even close to NASA's number above. Where are all of these figures originating if not from astronomers changing their numbers over the decades? Why have they been changing the lunar distance if the Apollo space program put reflectors on the moon capable of obtaining the lunar distance down to the inch???

University.com gives a refined number as per Google's 384,400: "The short answer is, the average distance to the Moon is 384,403 km (238,857 miles)...At its closest point (known as perigee) the Moon is only 363,104 km (225,622 miles) away. And at its most distant point (called apogee) the Moon gets to a distance of 406,696 km (252,088 miles). It appears that the 406,696 should be 405,696 = 252,088 kilometers. If one adds the smallest and largest figures, then divides by 2 to find the middle number, it get the stated averages.

Go ahead and try to find NASA's figure for the moon's angular sizes by googling " nasa 'moon's angular size' ". If you're thinking it will be right up-front on the first page, not so. On the page above, Google suggests that we look at "moon's angular size in arcminutes", but when we click to it, there's nothing on the topic as per the exact figures. I smell a cover-up because there is no consistency. NASA is responsible for providing consistency, and the correct figures, but here Google comes up blank. Surely, those lunar reflectors that NASA boasts about should provide consistency and accuracy???

Wikipedia: "the Moon's angular diameter can vary from 29.43 arc minutes at apogee to 33.5 arc minutes at perigee." Someone else writes: "According to Wikipedia the Moon's angular size varies from 29.3 arc minutes (1,758 arc seconds) to 34.1 arc minutes..." That's not the same as 33.5. What's going on? "Acording to the nautical almanac, the Moon's semidiameter was predicted to be about 16.6' today." That gets 33.2, which we do not see in the figures above. An educational page says: "Using Eq. 2, it follows the Moon's mean angular size is 31.09 arc minutes." What a mess. How can anyone depend on these figures to find accuracy? Why doesn't NASA together with Google come save us from this embarrassing "education"? From the last chapter: "Believe it or not, by googling "aphelion angular size," Google brings up the Flat Earth Society first of all." Where's NASA hiding??? Are we now going to be educated by the Flat Earth Society?

So, which is it? Is the average lunar distance midway between the nearest and furthest distances? Or not? Here's from the last chapter: "Earth's closest approach to the sun, called perihelion, comes in early January and is about 91 million miles (146 million km). The farthest from the sun Earth gets is called aphelion. It comes in early July and is about 94.5 million miles (152 million km)." One assumes correctly that the smallest and largest angular sizes of the sun correspond exactly to these figures, but we would like to do better than approximated distances. It turns out that the perihelion figure is more-exactly 91.4, or so someone says: "Earth is about 147.1 million kilometers (91.4 million miles) from the Sun at perihelion in early January, in contrast to about 152.1 million kilometers (94.5 million miles) at aphelion in early July." From the 152.1 figure, we see that they have gone a little further than a round-number approximation, and it works out to 94.51 million miles. The middle number between it and 91.4 is 92.955, their stated average solar distance bang-on. It tends to show that the average lunar distance should indeed be the midway figure. And that's why we can now be sure that the lunar average distance, corresponding to the average angular size, is in the ballpark of 235,896 (my new figure starting in the last chapter).

All I need to do is find the average angular size of the moon that NASA uses. It may be like finding a needle in a cosmic haystack, if Google can help NASA. The educational page below gives all three angular sizes along with corresponding lunar distances. Note that angular size can be measured from the surface versus the center of the planet: "...this angular diameter as measured from the center of the Earth works out to be about 0.518 degrees (31.1 arc minutes; a small correction would be needed to account for the fact that the observer lies somewhere on the Earth's surface). However, the Moon's center may be as far away as 406,600 km and as near as 356,600 km from the Earth's center. These angular diameters are 0.490 degrees (29.4 arc minutes) and 0.558 degrees (33.5 arc minutes), as measured there. This difference in angular diameter of roughly 4.1 arc minutes (about 13% of the Moon's average angular size) is noticeable to the human eye." (From the earth's surface, the angles would be greater.) Lunar distances are typically from the earth's core to lunar core. [The 13 percent means the difference between the smallest and largest numbers.]
http://homepages.wmich.edu/~korista/moon-illus.html

Okay, so we now have an average moon at 31.1 arc-minutes, if I'm reading it correctly, and someone refined it, as we saw earlier, at 31.09 arc-minutes.

Astronomers can draw two lines spreading at any degree, and know how wide the lines are at any distance across. You can find the same at the calculator below by entering the distance across in the edge-b box, and the distance of line spread in the edge-a box. The angle appears in the angle-A box.
http://www.cleavebooks.co.uk/scol/calrtri.htm

Here is an example: for two lines to the moon at 252,708 miles across, the angle calculator gives .489492 degree (with 2159 in the edge-a box). One can use what I call the 57-method for this: 57.295 / 252,708 x 2,159 = .489497 degree. It's not precise with larger numbers, but there is a good reason to know the 57 method: we can use it to find unknowns by shifting the math around (the right-hand calculator won't tell you how to shift the numbers around).

Let's settle on the figure, .4895, which works out to be 29.37 arc-minutes, which would have been rounded off to 29.4 if someone were rounding it off, meaning that it cannot be correct if one Wikipedia writer is correct with 29.3. When the distance is unknown, the 57 method, shifted around, comes in handy; it's now: 57.295 / .489497 x 2,159 = 252,708. To get precision, use the .489492 from the right-angle calculator.

Using the same math, here's the lunar distance when the angular size is 29.3: 57.295 / (29.3 / 60) x 2,159 = 253,328. In other words, using their 29.3 gets a maximum lunar distance of almost 900 miles further than the apogee of the 2018 eclipse. I don't think this can be correct. NASA is away without leave. Where are you, NASA, our savior? Come and save us from this error. Use your lunar reflectors to wash away this mess.

Another Wikipedia writer used 29.43 instead of 29.3, which makes a big difference in the distance: 57.295 / (29.43 / 60) x 2,159 = 252,219. This is now less than the 2018 apogee, and is inconsistent with NASA yet again. How did things ever get to this?

Let's use 31.09 arc-minutes to find the average lunar distance by that figure: 57.295 / (31.09 / 60) x 2,159 = 238,726. This must be it. This 31.09 angle must be the one that gets them their 238,700 figure. Where did the 31.09 figure come from? We can't know until NASA verifies, with its lunar reflectors, whether the figure is right or wrong. Someone at NASA just needs to point its lasers at one reflector when the moon is 31.09 arc-minutes in the sky. It shouldn't take longer than about 2 seconds, and the world could have the answer real-straightaway. But NASA is out to lunch, unconcerned. Google isn't helping matters. Solarsystemnasa.gov, which we assume is using NASA figures, claims: "The average distance to the moon is 382,500 kilometers." And here it is from NASA's own distance-to-the-moon (for students) page: "The average distance to the Moon is 382,500 km."
https://www.nasa.gov/pdf/180561main_ETM.Distance.Moon.pdf

As that is 237,674 miles, we find the angle like so: 57.295 / 237,674 x 2159 = .520 degree = 31.22 arc-minutes. That's not 31.09. Who's taking the angles? Trainees? Children? Why can't we have this straight? The page says that the perigee is "approximately 360,000 km," and apogee is "approximately 405,000 km," not exactly precise figures. The page is well short on facts, just as though NASA doesn't want to talk about this. But this is the page that Google brings up as exemplary when searching " NASA 'distance to the moon' ". The page above says: "Distance from Earth to the Moon for a given date can be obtained by asking a local planetarium staff." Why not give the world a calculator so that we can do it ourselves? What if we would like a dozen cases? Call the planetarium 12 times? For serious study, we need a lunar-distance calculator.

NASA's fact sheet on the moon says the lunar diameter is 1,896 arc-seconds, which, when divided by 60, is 31.6 arc-minutes. Whatever this number is, there is no other, nothing in the ballpark of 31.1.
http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html

Wikipedia's "Lunar Distance" article: "The mean semi-major axis has a value of 384,402 km (238,856 mi). The time-averaged distance between Earth and Moon centers is 385,000.6 km (239,228.3 mi). The actual distance varies over the course of the orbit of the moon, from 356,500 km (221,500 mi) at the perigee to 406,700 km (252,700 mi) at apogee..." By "time-averaged," I assume it's the middle ground on where the moon is spending its time. Elsewhere on the page: "A lunar distance, 384,402 km, is the Moon's average distance to Earth." Why isn't it NASA's 382,500???

Let's find the angle for Wikipedia's 384,402 figure, = 238,856 miles: 57.295 / 238,856 x 2,159 = .5179 degree = 31.064 arc-minutes. It's a whole new figure once again. If they can't even agree on the three main angles of the moon, the stars are laughing at their triangulation angles. This average of 238,856 is not the middle number between 221,500 and 252,700 on the same page.

The kilometer figures at the Lunar Distance article are improved upon by this: "The mean distance to the Moon is d=384399km. This gives a (mean) angular diameter of a=1.90065°. At Perigee the distance is d=362600km. This gives an angular diameter of a=2.01482°. At Apogee the distance is d=405400km. This gives an angular diameter of a=1.80226°" I don't know what these angular-diameter figures are; perhaps a blunder. The middle number now is 384,000 kilometers = 238,606 miles. Can't NASA do something about this mess? Or has it created it?
http://astronomy.stackexchange.com/questions/7736/what-is-the-angular-diameter-of-earth-as-seen-from-the-moon

What can we trust in all this? NOTHING. No one above mentions that the figures they use are not agreed upon by others. They all release their "facts" as facts. Then we have this unacceptable / laughable statement: "Modern laser guided measurements have shown that the average distance to the Moon is 382,000km." Why round the figure off if lasers to the reflectors achieves great accuracy? I've been hunting the real average for hours, but have yet to read once what the laser-begotten figures for lunar distance are. Why doesn't Google have the laser-tested figures on its moon-distance pages??? We can only assume that NASA would use them, but even NASA's page (as offered by the first Google page) has a big, fat round number, certainly not down to the inch.

NASA's average distance of the moon is said by more than one page to be the semi-major axis. "In geometry, the major axis of an ellipse is its longest diameter...The semi-major axis is one half of the major axis..." That is, the largest distance across in an oval is the major axis, and the shortest distance across is the minor axis; the semi-major is the radius between the largest distance. Wikipedia's article on "Orbit of the Moon" has "Semi-major axis 384748 km" followed by "Perigee 362600 km (avg.) (356400 - 370400 km) and "Apogee 405400 km (avg.) (404000 - 406700 km)". How are we to understand this semi-major-axis figure, which works out to 239,071 miles? I'm at a loss to explain.

The average perigee and average apogee is not the middle number between the numbers in the brackets, which assumes that the average figure is taken as the middle of where the moon spends its time. One can therefore expect the average lunar distance to be midway to where the moon spends the accumulation of its time.

To confuse the matter, the semi-axis figure, 384,748, is followed by: "Mean distance 385000 km", followed by "Inverse sine parallax 384400 km". Why are they giving round numbers? Don't they have precision??? NASA's average is 382,500, neither of the above, meaning the Wikipedia writers hashing out of this article didn't trust NASA. Instead, the Wikipedia figures are taken from "M. Chapront-Touzé; J. Chapront (1983)" and "M. Chapront-Touzé; J. Chapront (1988)" respectively. Why are they using figures from decades ago? Shouldn't the latest findings be more accurate? Is this voodoo, the sticking of needles into our brains? In the footnotes for inverse sin parallax: "This often quoted value for the mean distance is actually the inverse of the mean of the inverse of the distance, which is not the same as the mean distance itself." Maybe there has been a conspiracy not to give us the correct average.

What do they mean by having the semi-axis figure 252 kilometers smaller than the mean distance? Both numbers are from the same authors, though I suspect that someone rounded the mean figure. For anyone wishing to do precision work, this is a disaster. No one on the street can do their own work to verify what astronomers tell us. For one reason, we don't know what astronomers believe. They obviously don't believe the same things on lunar distances. We need another way to find the lunar average distance besides asking Google or NASA.

Here is a fact: astronomy can correctly draw on paper the two lines from the core of the earth to the edges of the moon, and the two lines from the core of the earth to edges of the sun. They won't know, by this method alone, how far the sun and moon are, but they will know that the edges of these bodies will be exactly between their respective pair of lines. Any body twice as far as another will have twice the distance between the lines, and the latter is the diameter of the body. For example, if the sun were 5.106 million miles away versus 92.955 million, both between the lines at .5331 degrees, the difference in diameters will be 5.106 / 92.955 = .05493. Therefore, where the astronomers have the sun's diameter pegged at 864,938 miles, its diameter at 5.106 million miles away should work out to 864,938 x .05493 = 47,510.5 miles. The right-angle calculator has 47,510.5 miles with 5106000 in the edge-b box and .533112 in the angle-A box. The number, 47,510.5 is just 2.5 miles off of what I'm calling the magic number. (to get the full .533112, click the page's "significant figures" box).

In the calculator below, feeding the angle-A box with .5331 (astronomer's average solar diameter in degrees), and their 864938 number in the edge-a box, gets 92.955 million miles. They can do the same with the lunar-eclipse lines, and can be expected to get absolute accuracy. They don't need to use lasers bounced off of reflectors on the moon for finding the lunar distance. Along with the lunar diameter, they need only measure the angle, as seen from earth, between both edges of the moon at its smallest and largest. The average, as an ordinary person would see it, or as God would see it, would be the middle size between the two sizes. And they do have their average sky-sun size in the very middle, between the furthest and the closest sun; why are they not doing the same for the moon? What exactly is their problem? Has this to do with their solar lines not fitting between the E-M lines, thus screwing up their math?
http://www.cleavebooks.co.uk/scol/calrtri.htm

According to the rule that a body twice as far is twice as small, they can figure correctly how many times greater the real (i.e. not angular size) solar diameter should be than the real lunar diameter. Astronomers using the solar diameter above will say that the sun is 864,938 / 2159 = 400.62 times larger than the moon. The middle-size moon is now working out to .524417 degree, and the sun at its furthest worked out above at virtually the same angle. It was said: "NASA tells us that the sun's radius on that day was 15'44.2", which is 15.73667 arc-minutes = .262277 degree (just divide the arc-minutes by 60). Multiply by two to get .52455 degree as the tip of the triangle (this is not the E-M triangle). To compare that angle with their aphelion figure of 94,500,000 miles, we put the latter in edge-b along with the solar diameter in edge-a to find an angle of .5244012 degree." Therefore, when the sun is at its furthest, it's exactly the same size as the moon at its average distance of around 235,877 miles. The latter is the figure provided by the right-hand calculator if the lunar diameter is 2,159 miles and the angle is .524417.

Now that we have the moon and sun at the same size in the sky, we multiply 235,877 by 400.62 to arrive to 94,497,043 miles, wherefore the latter is expected to be the full number of their rounded-off 94,500,000. This clearly shows that they should be reporting 235,877 as their average lunar distance. Why haven't they been? What were they trying to hide? They don't even come close to 235,877. Or, if we feed the calculator 2.159.2 for the moon at .524417 degree, their lunar distance is 235,899. This is so easy, yet they don't report this figure.

I've seen someone quoting the largest sky-moon size at 33.50 with the smallest is 29.43, suggesting that the 33.5 figure was not rounded off. The 29.3 figure that is/was used by one Wikipedia article may have been a typing mistake, leaving the '4' out from 29.43. The best I can do at this time (i.e. trusting these numbers) is to find the average moon size with: (29.43 + 33.5) / 2 = 31.465 = .524417 degree. While I don't expect this size to correspond to astronomy's various lunar-average distances, who cares? I can have my own definition of average, and I'm using the middle-number average that everyone can understand and verify.

We can now compare .524417 to the average solar size of .533112 degree. Dividing one from the other should be the same as 92,955,807 / 94,497,043 = .98369, making the average lunar size .98369 smaller than the average sun ("The AU [distance to sun] has been defined as 149,597,870,700 meters (92,955,807 miles).". We can thus argue that the lunar distance should be .98369 of 400.62 (or 394.0859) closer to the earth than the sun. The math: 92,955,807 / 394.0859 = 235,877.02. The math checks out, and we do find that they have the angular sizes of the sun proportional to their solar distances...not at all meaning that their distances are correct. It only proves that their distances are to scale. They have the sun between the lines of their angular sizes, true, but further away than is the reality.

Here is another way to put it, this time using the magic number in the math. As there was reason to size the sun 22 times the lunar diameter, one can suggest that the lunar distance ought to be .524417 / .533112 = .98366 (all three their numbers) of 22 times closer than the sun. But instead of dividing the solar distance by 22, we need to divide it by .98366 of 22.00463 (i.e. 47,508 / 2,159 = 22.00463), which is by 21.6457. The magic solar diameter, 47,508, when half its value is put into the edge-a box, gets a solar distance of 5,105,730, wherefore we divide the latter by 21.6457 to get 235,877. It's that same number again, though it applies only if the moon really is exactly 2,159 miles wide. Repeat: "When we put .524417 degree in the angle-A box and their lunar diameter of 2,159 in the edge-a box, the [average] lunar distance comes out at 235,877 miles."

This all goes to check my math, showing it to be correct. It shows that a sun 5.1057 million away has a diameter at the magic number. And, 5.1057 happens to be just about at the average solar distance when the 2000 eclipse has the sun at 5.2 million miles. Again, 92,955,807 / 94,497,043 = .98369, and when we multiply that latter by 5.2 it gets 5.115 million, only a little larger than 5.1057. This can indicate that the 2000 eclipse scenario should have a distance of slightly more than 5.2 million, and that its numbers can be adjusted accordingly. The 5.1057 figure was used in the last chapter like so:

With a shadow path of 6,089.05 miles [i.e. the diameter of the shadow where the moon passes], the M-E shadow-line spread is 914.475 miles over 235,896 miles. To find that angle in degrees: 57.295 / 235,896 x 914.475 = .22211 degree. With the solar radius at .266556 degree [half of .533112] on its average, it's a difference (between the two triangles) of .266556 / .22211 = 1.20011. The solar distance using the 914.475 figure is: 3,959 / (.004652 - (914.475 / 235,896)) = 5.10575 million miles. We multiply the latter by 1.20011 above to find that the E-M triangle is now 6.1275 million miles long, giving an earth shadow of 6.1275 - 5.10575 = 1.0217 million miles, and the right-angle calculator has it at 1.0213 million with .2221087 degree in the angle-A box (use 3959 in the edge-a box).

It shows that the umbra shrinks from 1.037 million (seen above) to only 1.02 million between the eclipse at the sun's near-furthest and the sun's average distance, wherefore that umbra will never get as small as astronomers claim for it.

The reason that .098369 can be used both for their number of 400.62 and mine of 22.0046 is because both numbers are related to suns between the angular solar lines. No matter that we use their figures, the moon is found 235,877 miles away when it corresponds to a sky-size of midway between the reported sizes of 49.43 and 33.5 degrees. If these sizes change, the average lunar distance (as I define it) changes. And the lunar-distance figure changes when the lunar diameter changes.

On the solar-eclipse page below, which is an event on January 15, and therefore very near the sun's closest distance, the angular size of the sun's radius is said to be 16'15.5" = 16.2583 arc-minutes = .27097 degree. The diameter is therefore .54194 degree. As was said, the lunar-eclipse page under discussion has the sun (July 16), at nearly its furthest distance, as .52455 degree. The middle number here is .53325. As these eclipses are 1.5 days shy of six months apart (= 1/2 the orbit), they represent a good way to get the solar average.
http://eclipse.gsfc.nasa.gov/SEplot/SEplot2001/SE2010Jan15A.GIF

Aphelion falls in the middle of the first week of July, and perihelion in the first week of January. Aphelion was on July 3 in 2000, and perihelion will be on January 3 in 2018.
http://aa.usno.navy.mil/data/docs/EarthSeasons.php

If they can't even agree on the average distance to the moon, what will the stars have to say. Snicker. Are they really millions of light years away, or are the astronomers deluded by the goons who lead them?

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