Previous Update

(if there are any to speak of)

October 4 - 10, 2016

Hybrid Solar Eclipse of November, 2013In the eclipse below, three days before the moon was at perigee on November 6, Fred has the lunar shadow touching the earth for only 4.2 seconds. A hybrid eclipse is when the shadow doesn't reach the earth surface in the beginning because the side of the earth is further from the moon than the head-on center of the earth. The shadow then hits the earth at the center, but once again misses as the shadow moves toward the opposite side of the planet. So, for 4.2 seconds, the time that Fred has between U1 and U2, the umbra skimmed along the earth, yet he has the umbra a whopping 57.5 kilometers wide, which doesn't seem right for a mere 4-second kiss of the surface. The eclipse of 2016, the one always under discussion above, was 155 kilometers wide, but was on the earth for over 92 seconds. Something seems very wrong here.

http://eclipse.gsfc.nasa.gov/OH/OHfigures/OH2013-Fig05.pdfWe can calculate how far the tip was, inside the earth, for the 2013 eclipse, if only we had the lunar distance at the time. The perigee calculator (below) gives 365,361 kilometers (227,025 miles) for 9:29 am on November 6 (perigee), and 404,560 kilometers at 14.26 pm on October 25 (= apogee). From the latter date to the eclipse, there were 8 days, 20 hours, 20 minutes = 764,400 seconds. We put the latter into the free-fall calculator below, with .000075505 in the g-box, to find a fall distance of 22,059 kilometers. Therefore, it fell from zero fall speed at 404,560 kilometers up from earth, down to 382,501 kilometers. The acceleration rate of .000075505 is obtained because it's needed with 1,018,980 seconds (the time between apogee and perigee, i.e. October 25 to November 6) in the top box, to find 39,199 kilometers of fall, the total fall between apogee and perigee.

http://keisan.casio.com/exec/system/1224835316You can use the angular-size calculator below as an alternative, but I'll be quoting from the one above:

http://www.1728.org/angsize.htmWe therefore put 382,501 - 1 earth radius = 376130 in the edge-b box, with the lunar radius (1,738.1) minus half of 57.5 in edge-a, to find the angle of the umbra line at .260258 degree. To find how far into the center of the earth this line would go if it could arrive to a tip there, we reset the calculator, put that angle into the angle-A box along with half of 57.5 in edge-a, to find a distance of 6,329 kilometers, almost the entire earth radius so that the tip is expected to strike the earth for far longer than 4.2 seconds. Fred just keeps on making problems for himself. He has an umbra width so wide that the tip would reach nearly the center of the earth, wherefore the umbra should have tracked almost the entire width of the earth's surface. Yet he claims that the tip touched down for only 4.2 seconds.

https://www.fourmilab.ch/earthview/pacalc.htmlNASA's fact sheet has the largest sun size at 1,952 arc-seconds (though the triangle calculator tells me 1,951.8). This eclipse was 62 days before the sun was at the largest size (January 4), but to find the size of the sun at the eclipse, I've got to find the time span between which the sun goes from its average to largest sizes. The aphelion / perihelion page below shows the sun at its largest at midday (UT), January 4, 2014, 62.03 days after the eclipse at 12:47 pm, November 3. Then, the page below has aphelion past midday (14:44) on July 5, 2013. One might use these two dates to arrive to the "more-correct" sun size. There are almost 182.885 days between them. The average sun is therefore predicted at half of that, or 91.44 days, landing at about 2 am on October 5.

http://www.astropixels.com/ephemeris/perap2001.htmlHowever, a day after this update was put online, I found one of Fred's ephemeris pages below that has the average sun somewhere between midnight of October 5 and midnight of October 6. The average sun is when the distance column reads 1.00 astronomical units, which is somewhere between October 5 (he has this day at 1,919.0 arc-seconds) and 6 (1,919.6 arc-seconds). He has the distance at 1.000109 AU at nearly 1 am on the 5th, and .999821 AU at nearly 1 am on the 6th; judging from these numbers, I would peg 1.0 AU at 10:30 am, October 5.

http://www.astropixels.com/ephemeris/sun/sun2013.htmlI've therefore got to count from 10:30 on October 5 to noon on January 4 (2014). a total of 91.063 days. I had a slightly-different number originally, but have modified all numbers in the rest of this chapter as per the math in this paragraph. The eclipse was 62.0 days before the sun was at its largest on January 4, wherefore we divide 62 by 91.063 to find that the eclipse was .68085 of the way (backward in time) between a sun at it largest (January 5) and the average sun (October 5). One finds the total size (or distance) difference by subtracting 1,919.2 from 1.951.8 = 32.6. Then, to find how much of that difference one is to use for the eclipse, it's 32.6 x .68085 = 22.196. Finally, subtract the latter from 1,951.8 to find 1,929.6 arc-seconds (= the estimated angular size of the sun on the day of the eclipse).

So, we divide 1,929.6 arc-seconds by 3,600 to find .536 degree as the angular diameter of the sun, and we can divide by 2 to find the sun radius as a clean .268. The more-correct line to the sun should therefore have been at .268 degree. When we use that in the triangle calculator along with 376,130 (above), we find 1,759.35 in edge-a. To explain what this means, we are not drawing one line, though I use one line alone in the discussion. The full picture draws two lines at .268 degree, one to the top, and one to the bottom, of the moon. In other words, at .268 degree, the two sun lines work out to be 1,759.35 kilometers apart at the lunar distance, which more than 21 kilometers larger than the lunar diameter. That's not a hybrid eclipse. A hybrid needs a sun visually smaller than the moon.

http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.htmlThe angle, .268 degree, is smaller than Fred's solar size for the eclipse.

When an object falls, the drop distance is not proportional to the velocity increase. The latter is proportional to the time of fall. We divide the number of seconds (764,400) between the start of fall to the eclipse by the 1,018,980 seconds to the end of fall, to find .7502. We can therefore try .75 between NASA's slowest and fastest lunar speeds of 2,156 and 2,407 to get a general idea of the lunar velocity at the eclipse. We subtract the smaller from the larger to get a 251 difference, and then multiply it by .75 to find 188.25, which we add to 2,156 for an estimated velocity of 2,344 mi/hr = .65112 miles per second, which works out to 2.7 miles over 4.2 seconds. In other words, the umbra radius works out to 2.7 miles when using Fred's 4.2 seconds. Changing 2,344 mi/hr by 10 or 20 mi/hr doesn't change the 2.7 by much, all the more reason yet to ask why Fred's umbra-width number isn't matching this math.

If Fred would provide the velocity of the moon at the eclipse, we could test or correct what I'm saying. One thing is certain, his 57.5-kilometer umbra will take a lot more than 4.2 seconds of touch-down. (The umbra starts at zero kilometers wide at initial touch-down.) For the time between U3 (leading edge of umbra no longer contacts earth) and U4 (other edge of umbra no longer contacts), he has only .6 of a second, showing that the umbra path is very small at that point. This eclipse has the shadow tip out in the ocean, where no land elevations exist to explain why Fred had the umbra width at 57.5.

The trick now is to find the angle of the umbra line, then combine it with the sun-size line that we seemed to have nailed down to .268, and discover the true distance to the sun.

Fred's 16'07.4" for the size of the sun converts to decimal format by first dividing 7.4 by 60 seconds to find .12333 of one minute; then add 16 = 16.123333 minutes. We divide it by 60 to find .2687222 degree. The moon's size that he gives is almost identical, 16'07.6" = 16.126666 minutes, but when we divide the latter by 16.12333, it's just 1.000267 larger, a lot smaller of a difference than the eclipse magnitude of 1.0159 that Fred uses on his page. Perhaps his magnitude figure is suited for his overblown 57.5 path.

Fred's pages define eclipse magnitude: "The Eclipse Magnitude is the fraction of the Sun's diameter eclipsed." If it's larger than 1.0, it concerns how much larger the moon is than the sun. The way I interpret 1.0159, the moon's diameter was wider than the sun by as much as .0159 (almost 1/63rd) the sun's diameter. How does one express this mathematically? If he is correct with a moon at 16.126666 arc-minutes, it seems to me that his sun should be 16.126666 - (.0159 x 16.123333), yet that gets 15.87 arc-minutes. I don't see how that can work. The only way for the math to work is: 16.126666 - (.0002067 x 16.123333) = 16.123333. Compare .0159 with .0002067. Why the difference?

With his angular-moon size (.268777) in the angle-A box and the lunar radius in edge-a, the lunar distance is given as 370,512 kilometers. It's almost the 365,361 (core-to-core distance) that the perigee calculator has for perigee, three days after the eclipse. If his 370,512 is the core-to-core distance, it is surely wrong because the moon does most of its falling after the 3/4 mark from apogee, and three days before perigee is exactly 3/4 the way from apogee, wherefore the moon could not possibly be as far down as 370,512 (shy only by about 5,000 from the perigee altitude) kilometers at the eclipse.

We can allow that 370,512 is the surface-to-core distance, meaning that the core-to-core distance was 370,512 + 6,372 = 376,884. In that kinder-to-Fred picture, the moon, over about nine days, had fallen 404,560 - 376,884 = 27,676 kilometers (3,075 daily), and then, with three days to go, it fell 11,523 kilometers (3,841 daily). This combination does not strike me as correct. He has too much drop for the first nine days. The daily drop over the nine days should not so nearly approach the daily drop over the last three days. Falls are such that, after six days of a 12-day fall, the drop is only 1/4 the distance of the entire fall. After nine days of fall, the drop distance is only .5625 the total fall distance. You can check this on the free-fall calculator. But he has .706 the full drop over the first nine days.

http://keisan.casio.com/exec/system/1224835316Try to remember that his 370,512 will be treated here as the lunar distance from the earth's surface. The angle that he gives on his eclipse page conforms to 370,512 kilometers from the earth surface. This angle is not the umbra-line angle, but the angular-moon-size line. The latter should be viewed as two lines beginning from one point on the earth, such as your eye (I need speak of only one line because they are identical in angle). This eye-point is always to be placed in the center of the umbra, which is why you're reading of the umbra

radiusa lot. The moon-size and sun-size lines always start in the center of the umbra, or at the center of the earth circle.So, as the total fall between apogee and perigee was 39,199 kilometers, the approximate, predicted fall distance after nine days is 39,199 x .5625 = 22,049. It's a fall to an altitude of 404,560 - 22,049 = 382,512 kilometers. Earlier, I had 382,501 as the more-precise lunar distance at the eclipse (it shows how close to the three-quarter mark it was). Yet Fred's numbers give either 370,512 or 376,884. Blare the sirens; this is getting criminal. It means that Fred has a false angular-moon size (too large, and at least 5,600 kilometers too close). But why?

Fred has mid-eclipse very near the equator so that we can use the lunar distances as they are, without adjusting them, as would be needed if the umbra was far from the equator. Using 370,512 with the lunar radius minus half of 57.5, the umbra-line angle works out to .2643311 (as it starts a little higher than the center of the earth circle, but goes to the same edge of the moon as the moon-size angle, it's less than Fred's moon-size angle). Clearing the calculator and putting .2643311 in the angle-A box along with half of 57.5 in edge-a, we find that his umbra tip would reach, inward from the earth surface, 6,232 kilometers, which does not match an umbra touch-down of only 4.2 seconds. This depth of the tip makes the event almost fully a total eclipse (defined as the umbra touching down clear-across the earth). It means that he cannot have his 370,512 number if he wants to retain his umbra width. You don't need to be an astronomer to figure this out.

Here is what we know: the angular-size line of the moon must meet the umbra line at the edge of the moon. In other words, we are supposed to be able to use Fred's data to find the lunar distance. We'll use his moon-size angle of .268777 degree in angle-A, with 1 in the edge-b box, to find a line spread (in edge-a) of .00469108 miles/kilometers per 1 mile/kilometer toward the sun. That's his angular-size-of-the-moon line taken from the earth surface (i.e. where the umbra is situated). The same method, applied to his umbra line, first needs the line's angle, and since we can't use the moon's distance while we're trying to discover it, we've got to use the sun's. All we have is Fred's information of 92.2 million, and an umbra radius of 28.75 kilometers (the 92.189 was found with his .268777 in the triangle calculator along with half the solar diameter). So, we put 148381516.8 kilometers (= 92.2 million miles) in edge-b with half the solar diameter (695,991.4) minus 28.75 in edge-a (total 695962.7), to find the angle of .26873585, smaller than Fred's moon-size angle (.268777), as it should be. So far, not bad. To help you visualize things, see Figure 2-1 at the page below, and maybe have it on a separate browser as handy when you need it:

http://www.eclipse.aaq.org.au/index.php/eclipse-information/what-are-eclipses[A day after this chapter was put online, I found Fred's page below where he reports the solar distance as 92.22 million miles for November 3 (day of the eclipse). See his section, "Geocentric Ephemeris for Sun : 2013". His distance figure was at a time of day when the angular size was 1934.6 arc-seconds = .2686944 degree (not quite .268777 yet) as the solar radius. To get the 92.22-million solar distance, multiply his November 3 figure of .992086 AU by 92,955,807 (= astronomical unit).

http://www.astropixels.com/ephemeris/sun/sun2013.html] Next, we put .268777 and .26873585 separately into the angle-A box, with 1 in edge-b in both cases, to discover, in edge-a, what the line spread is. We find .00469108 and .00469036 kilometer of spread per 1 kilometer toward the sun. (By the way, although I say that two lines are "spreading" from the earth, they are actually coming closer to one another. The spreading is from their respective and imaginary horizontal lines at 0 degrees.) I've checked and re-checked my method for finding when the lines meet, to assure that it's correct. We subtract the umbra spread (.00469036) from the moon-size spread (.00469108) to find a small difference of .00000072 (the smaller the difference, the further the distance). Finally, we divide Fred's 28.75 by .00000072 to find how much distance toward the sun it will take for the lines to meet, and it turns out to be 39,930,556 kilometers (24.8 million miles), nowhere near the lunar distance.

To be fair, let's try the more-correct 92,468,473 miles instead of Fred's erroneous 92.2. The same math as above (I'll spare you writing it out) using 148813582 kilometers (92,468,473 miles) in edge-b gets a lunar distance of over 2 million. Better, but hardly correct. In other words, Fred's umbra line as calculated from his sun's distance and diameter, put together with his moon-size line, DOES NOT WORK. You can try it using my lunar size and it's angle of .26476241 having a spread of .00462100871, which gets a lunar distance of 515, 852 kilometers, much better, but that's using his 28,75. If we use the smaller umbra width of 2.7 miles (radius = 2.17 kilometers) with my lunar distance, we get under 39,000 as the lunar distance. If we use the 2.7 scenario with his lunar distance and the solar distance of 92,468,473 miles, the lunar distance comes out as 151,346, the best yet, but still not correct. These numbers can't lie, and so it seems, unless I'm doing something wrong in the math, neither umbra scenario from Fred gets the lines to line up correctly.

The umbra width is dictated by the solar distance combined with the lunar distance, and Fred's lunar size is too large, giving an umbra width too large too. Using math alone (without earth-surface experiments), there is no way to calculate the umbra width unless the true solar distance is known.

In case that you don't trust my method, it can be verified. For example, take Fred's moon-size line with its spread, (.004691076). Then take his umbra line found with his 370512 in edge-b along and the lunar radius minus 28.75 in edge-a to find an angle of .26433110 with spread is .0046134808. Subtract the latter from his moon-size spread to find .000077595, and divide 28.75 by .000077595 to find 369,959 kilometers. It should have come out to 370,512 if all the decimals had been more precise. It doesn't matter what the umbra width is; so long as the umbra line and moon-size line are using the same lunar distance, this method will get that lunar distance too.

What we need to do is to toss out his moon-size line, and then try one that conforms to 376,130 kilometers. We put that (with no comma) in edge-b with 1738.1 in edge-a to find the angle of .26476241. The spread for this angle (put 1 in edge-a) is .0046210087. Next, we need to find the angle for an umbra line, which requires an umbra width first, and so we'll use my 2.7 miles = 4.34 kilometers. We put 1,738.1 - 2.17 (half of 4.34) in edge-a (I'm using 1735.93) with 376,130 in edge-b to find the angle at .264431862. It's spread comes to .00461523941 kilometers, and so we subtract it from .0046210087 to find a difference of .00000576929, and finally we divide 2.17 by .00000576929 to find a lunar distance of 376,148 kilometers (i.e. which should be 376,130 if all the decimal places were more precise).

If I trusted Fred's sun line, I could calculate the width of the umbra according to my lunar distance. However, in this chapter, I'm realizing that I can calculate my own sun size, and this, if it's being done correctly, can actually tell us the true distance and diameter of the sun. There were two reasons for claiming that Fred's sun is too large, one being his wrong figure, 92.2 million, and the other one where I worked out the more-correct solar distance in which this eclipse ceases to be a hybrid.

By the way, when I say "more correct" number, it means more in tune with what NASA's numbers are. Fred is required to stick to his own system of lunar distances. Plus, he can't use a computer program, for finding eclipse-line angles, based only on his view of the solar distance and diameter. The moon lines need to agree with the suns lines; otherwise the solar lines are erroneous. Here is what Fred did. He first used an angle of .2687222 degree for the angular size of the sun, which is in tune with a sun 92.2 million miles away and 864,938 miles in diameter, i.e. two numbers used by NASA. Next, believing that this was a hybrid eclipse, or for some reason trying to twist it into being a hybrid, he (or his calculator program) had no choice but to make the moon's angular size a wee larger than .2687222, and so he has it at .268777. I'd like to know which came first, his enlarged sun or moon.

One thing that Fred seems to have neglected was to change the eclipse's times to conform to his over-blown umbra. It might be lucky for us if he has actually recorded the true times for U1 to U2, for this can get a good approximation of the solar distance...because the true times allow one to figure the real umbra size if one has the velocity of the moon at the time. However, we can't get lucky in this regard if the umbra didn't even reach the earth.

My sun-size and moon-size angles are .268 and .26476241 degree. With the sun larger, it ceases to be a hybrid. We can test the angular-sun data from Fred, which is his sun-size line said to be at .2687222 degree. He doesn't give his email address on the eclipse pages, or I'd ask him whether this angle is taken from the earth core versus surface. We can start by assuming it from the surface. If it's from the core, we need to find what the sun-size angle would be from the surface (it will naturally be higher at the surface). For simplicity, I always want to start two sets of lines from the surface (i.e. the same place) when checking their meeting place, though this is not necessary.

With the sun line starting at the earth's core, it goes to an imaginary sun a tad smaller than the moon in the case of a hybrid eclipse. We can imagine the sun superimposed upon (or on the backside of) the moon. It's a correct visualization. Therefore, to get an earth-surface angular-sun line (keep in mind that we may not need it in the first place), we need to plot the moon-end of this line, and for this we need to know the apparent width, in kilometers, of the sun that is superimposed on the moon. As this paragraph has the purpose of proving Fred wrong, we need to divide Fred's sun size (16.123333 arc-minutes) by Fred's moon size (16.126666 arc-minutes) to get .999793323 the difference, and then multiply it by the true lunar diameter of 1,738.1 kilometers, to get 1,737.74 kilometers for the sun size. It's as though the sun were that large and stationed smack on the backside of the moon. My findings had found the sun to be 1,759.35 kilometers when superimposed on the moon.

If we put 1,734.14 into edge-a with 376,130 in edge-b, the earth-surface angular-sun line is at .264159198 degree (spread of .00461048042), smaller, as it should be, than my moon-size line of .26476241 (spread = .00462101) obtained with a moon at 376,130 kilometers. However, the spread difference here is .00001053, and multiplying it by 376,130 gets only .396, meaning that the moon-size line has spread by .396 kilometers more than the sun-size line by the time they get to the moon, and that makes the moon-size line only 1,734.14 + .396 = 1,734.54 kilometers when it needs to be 1,738.1. My conclusion here is that this sun is too big. It's still too big.

It was determined that his moon-size line could not have been from the core, but from the earth surface, suggesting that the modified, earth-surface line that I'm giving him here is not a correct picture. But in case you'd like to know, we can use it anyway, with 1,734.14 in edge-a and his 370,512 in edge-b, and finally, going through all the proper ropes, the spread difference turns out to be .00001069. The math is 370,512 x .00001069 = 1.14 so that his sun size now works out to 1734.14 + 1.14 = 1,735.28. No good.

If we didn't modify his sun line, but started it at the earth surface to begin with, at his angle of .2687222 degree having a spread of .00469012, the math becomes: his .00469012 - my .00462101 = .00006911, and finally .00006911 x 376,310 = 25.99 kilometers, that being the distance from the edge of his sun to the edge of the moon when both are superimposed at my lunar distance. To put it another way, the sun works out to be 25.99 kilometers larger than 1,738.1, or 1,764.1 kilometers. It means that Fred could not get a hybrid using the more-correct lunar distance. It recalls (from above) that the solar size, at the same lunar distance, worked out to 1,759.35 kilometers (when using none of Fred's data), which is 4.75 kilometers less than 1,764.1. The latter number is Fred's sun size with my lunar distance versus 1,759.35 (my sun size) and my lunar distance. With Fred's angle in the picture, the sun works out to be larger, once again, than it should be.

The 4.75 checks out. My sun size had worked out to .268 degree with a spread of .0046775165, and his solar size of .2687222 has a spread of .00469012, the difference between the two being .000012604, which, when multiplied by my 376,130 gets 4.74074 kilometers (essentially 4.75). Therefore, using the correct lunar distance, he can't get his hybrid. He needed to make his moon larger to get the hybrid. The distance between the superimposed sun and the moon, in his picture fully, uses .268777 with .00469108 and .2687222 with .00469012 = a difference of .00000096, which we multiply by his 370,512 to get .3557, meaning that he's claiming a sun size of: 1738.1 - .3557 = 1,737.74 kilometers = a hybrid, but barely.

We can work his two lines in reverse, from the moon to the earth, to see what his umbra width will work out to. To do this, we get the more-correct solar distance of 148,813,582 kilometers (92,468,473 miles), sun core to earth core, minus 1 earth radius (6372) to get the span to the earth edge (i.e. where the umbra is), minus the more-correct lunar distance (376,130), = 148431080 into the edge-b box. In edge-a, we put half his solar radius (695,991) minus a lunar radius, = 694253 kilometers. We now have a line from the edge of NASA's sun to the edge of NASA's moon, and the triangle calculator (below) says that it's at an angle of .26798617. Of course, this line (the umbra line) arrives to earth at that same angle, and to find how wide the umbra will be at the surface, we clear the calculator, and put 376,130 in edge b with .26798617 in the angle-A box, finding 1,759.26 in edge-a, 21.16 kilometers higher than the lunar radius (1.738.1). It means that the line will not reach the earth surface. It will instead reach the umbra tip in the air after is has come down 1,738.1 kilometers.

http://www.cleavebooks.co.uk/scol/calrtri.htmRepeat from earlier in this chapter: "

My findings had found the sun to be 1,759.35 kilometers when superimposed on the moon." Look at how close it is to 1,759.26. That's because the .26798617 angle in the paragraph above is nearly my sun-size angle of .268 degree. Yet Fred's numbers do not reflect this non-hybrid picture. Where does Fred get justification, therefore, for an umbra 57.5 kilometers wide? In order to get his 28,75, we would need to do the exercise above with about 150,500,000 kilometers (instead of 148,813,582) and his lunar distance. We can't do that, however.His sun at 16.12333 arc-minutes = .2687222 degree (sun radius), when multiplied by 2, gets an angular diameter of .537444 degree, which seems about right against the average-size sun of .533 degree, for, on November 3 (day of eclipse), the sun is larger than .533 degree. But let's test this to see if his sun is at least close to accurate. Here's what was said above:

The more-correct line to the sun should therefore have been at .268 degree [= angular diameter of .535933]. When we use that in the triangle calculator along with 376,130 (above), we find 1,759.35 in edge-a..which more than 21 kilometers larger than the lunar diameter. That's not a hybrid eclipse. A hybrid needs a sun visually smaller than the moon.Anything larger than the earth radius assures that the umbra does not reach the earth surface because there is a ring of sun around the moon due to a sun larger than the moon. There is nowhere you can go on earth to be in the umbra when you can see a ring of sun around the moon. Some sun will always shine on you; you will be only in the penumbra.

So, Fred has a sun diameter at .53744 degree, and I get a smaller sun at .53593 using a math and method (and their angular-sun numbers) that I can't see anything wrong with. Why would he artificially increase the size of his sun, if that's what he did? Was he trying to make a hybrid out of an eclipse that was really an annular? You would naturally be trusting Fred over me; after all, he really knows his stuff. Yes, and because he really knows it, that's how you can know that he's guilty of passing false information. The more he claims to be an expert, the worse his guilt. And the buck stops with NASA.

When using Fred's sun-size angle instead of mine along with my moon distance, the superimposed sun works out to 1,759 kilometers (unacceptable for Fred), but when using Fred's sun size with his moon distance (in angle-A and edge-b), the superimposed sun works out to 1,737.75, which only-just-barely makes it a hybrid. He then used this small sun-moon difference for an umbra width of 57.5 kilometers.

Any number under 1.738.1 is falsely pushing a hybrid, and can explain why I've found Fred's sun too big, and almost as large as the moon. An annular is defined where the sun is bigger than the moon, but if Fred artificially makes the moon larger than the sun, he turns an annular into a hybrid. If this is what he did, he naturally needs to give times for U1 and U2 to reflect a hybrid. (It's still a mystery to me as to why Fred has U1 to U4 times for all annular eclipses, where there is no umbra-tip contact with the earth).

The shadow path for this November eclipse was in the ocean so that there were no witnesses to any umbra touch-down. It means that Fred could lie or make an honest mistake, and no one would know.

Again, so long as Fred insists that the umbra touched down, he cannot use a number greater than 1,738.1 for the edge-a box. It can explain why he has an over-blown moon size, as compensation. In other words, when a larger moon is used for the edge-b box, the angle goes up i.e. closer to what it should be for accommodating a November sun of virtually the same size. If he didn't enlarge his moon (from the reality), he wouldn't have had a hybrid. But why would he enlarge his sun too, if his aim was a hybrid, where the reality was a sun already larger than the moon?

In the next update, I'm going to try to see whether I can find the angle of the umbra line even though the umbra tip didn't reach the surface. Also, I'm going to use the method of this chapter, for figuring the true solar size, in conjunction with a total

lunareclipse's umbra line. That should spell the end of my emphasis on eclipses, unless something surprising pops up.I also plan to have a section that returns to bloodline topics and heraldry, as per Julius Bassianus' bloodline.

By the way, Dave Williams at NASA's fact sheet didn't write back after I asked him for "NASA's smallest and largest angular sizes of the moon, and their precise distances." He promptly wrote back after my first two mails to him. This makes me more suspicious yet. I plan to re-write him with the same question.

Update -- Dave wrote back October 13, but still did not provide the three angular sizes of the moon. He didn't even mention why not. He's totally ignoring this request, making me more suspicious of NASA. He gave me a page (below) that gives the lunar distance for any second of any day over a three year period only, now showing 2015-2017 (should show 2016-2018 starting in January 2017), and this allows us to figure the angular sizes ourselves for any point in time. The page has a link where one can purchase a MICA program for obtaining those distances in all other years. Very apparently, NASA is not providing this information, at least not freely. MICA is from the Naval Observatory, and the latter's distances and other information may be from NASA to begin with. If so, it suggests that NASA is making money off the people while permitting the Observatory to sell it. It is not moral (and may not be legal) for NASA, paid for by the people, to be selling the people its information. The information is already owned by tax payers.

http://aa.usno.navy.mil/data/docs/geocentric.phpThe page now providing lunar distances for 2015-2017 gives them to within a couple of kilometers of the distances in the apogee-perigee calculator. As soon as lunar distances are available for 2018, one can have NASA's lunar distance for the central-path, total eclipse of July 27, 2018. I will elaborate on this in the 3rd or 4th update of November.

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