Hiding the Lunar Velocities
The Mosul operation is still mainly about taking areas outside the city. This is not yet the Mosul operation, in other words, anymore than a doctor has started an operation if he's only gotten to the stage of wiping the skin clean. Here vis a Mosul-news site:
http://www.iraqinews.com/tag/mosul/As little-by-little continues to take place in the Middle East, I can continue by project of revealing evolutionists as the liars that they are. One of the conspicuous things is that I haven't been able to find much on the moon's slowest velocity. Googling "minimum lunar velocity" finds NASA away on lunch. Google is offering only 5 webpages, in the entire world, with "minimum lunar velocity" in their page. At the Google page for "lowest lunar velocity," there is a link at the bottom for a Google search page for "lunar orbital velocity calculator," but when clicking to the page, there's nothing for calculating the moon's velocity. It would be like a dream-come-true for me to have a lunar-velocity calculator. I had found just one webpage, thus far, telling that the lowest velocity was 2,163 mph.
Ripping NASA for its lunar velocities of less than 2,000 mph needs to have the slowest velocity pinned down. As hard as it is to believe, most pages that Google brings up having "average lunar velocity" are not from NASA or astronomers, but from Muslim pages. One of the pages has: "NASA measured the instantaneous velocity of the moon at various points throughout its orbit. These measurements show that the velocity of the moon varies considerably (from 3470 km/hr up to 3873 km/hr)..." The 3,470 kilometers per hour works out to 2,156 mph. Keep your eye out for 2,156.5 shortly below. We now have a number by which to rip NASA.
https://www.quora.com/Does-the-big-bang-theory-agree-with-Islams-description-of-how-the-universe-formed#!n=12That quote is from a Muslim. Is NASA out to lunch again? Come on, NASA, tell us. Thanks to my son, a NASA "fact sheet" (obtained in 2014 by Wikipedia) was found showing this:
Mean orbital velocity (km/s) 1.022 [= 2,286 mph]
Max. orbital velocity (km/s) 1.076 [= 2,407 mph]
Min. orbital velocity (km/s) 0.964 [= 2,156.40658 mph]http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html
While at it, let me record:
Semimajor axis (km) 384,400 [238,855 miles]
Perigee (km)* 363,300
Apogee (km)* 405,500The asterisks denote mean distances. Note that instead of saying, average distance, the fact sheet has the semimajor axis. Why? Here's from the last chapter: ""A lunar distance, 384,402 km, is the Moon's average distance to Earth." Why isn't it NASA's average figure of 382,500???" Can this explain why NASA uses "semimajor axis" instead of average distance? What does NASA mean by its average distance of 382,500 versus its semimajor axis figure of 384,400? Smoke and mirrors? It is not correct to call the average distance a semimajor axis. The major axis is defined as the distance between the two widest points of an ellipse, making the semi-major axis half of that (half way between apogee and perigee), but this doesn't express the the average distance, according to NASA, anyway.
In Wikipedia's article on Kepler's laws: "The semi-major axis a is the arithmetic mean between r[adius]min and r[radius]max: r max - a = a - r min." If the maximum radius is 20 and the minimum radius is 10, the mean is 15, wherefore the formula above is 20 - 15 = 15 - 10. But this gives the average as per the longest part of an ellipse, and it concerns only one side of the ellipse. It has nothing to do with half the total length of the ellipse. Is this what we want for the average distance of the moon? We are now finding that the semimajor-axis number above is precisely the middle of the mean perigee and mean apogee. But this distance is not necessarily a match with the angular size of the moon midway between the smallest and largest angular sizes. Isn't the latter the representation of the average lunar distance?
I can see someone using, as the average distance, the middle between the greatest-possible apogee, and the least-possible perigee. Is this NASA's average? The footnote at the fact sheet says: "...the distance from the Moon to Earth roughly ranges from 357,000 km to 407,000 km." The middle of those two numbers is 382,000. Yet the numbers are all rounded off.
There are two ways to express midway between apogee and perigee: 1) with a line through the earth's core; 2) along the lunar orbit. I tend to think that midway along the lunar orbit is the way to get at the average distance, for the purpose of finding the orbital circumference. The semi-major axis takes the midway measurement through the earth's core.
From the last chapter: "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. The middle of the two extremes is 384,900 kilometers, not even the 384,403 that the writer reports as the average, not even the NASA figure for its average. What a mess. Fred Espenak's page has more-specific numbers: "Earth varies with mean values of 363,396 km at perigee to 405,504 km at apogee." The middle number is now 384,450. His average is almost the same as the furthest-ever in the quote above his. Have they got this rigged to keep us from getting at the truth, that they are lying?
The predicament for NASA remains. Its average distance of 382,500 is about midway (but not quite) between the 357,000 and 407,000. The latter two are supposed to represent the greatest angular size, 33.5 arc-minutes, and the smallest-possible angular size, 29.43 arc-minutes, of the moon. The middle number between the two is 31.465, which should correspond with 382,000 kilometers, but it doesn't. We are not getting the truth.
NASA's mean/average orbital velocity of 2,286 mph gets an average lunar orbit of 2,286 x 24 hours x 27.3217 days (the "revolution period" on the fact sheet; Espenak uses 27.32166) = 1,498,978 miles (close to 1.5 million). To find the diameter of the lunar orbit, we divide by 3.14159265 (pi), and finally we divide by 2 to get the average lunar distance, coming in at 238,570 miles. We can see from this that it's where they get their average lunar distance of roughly that much, yet, as it was shown in the last chapter that this is the wrong average figure, we can now glean that their average lunar velocity is wrong too, for the 238,570 is based on that velocity, or vice-versa.
We can work backward starting from the average distance that I've roughly arrived to: 235,900 x 2 x 3.14159265 / 27.3217 / 24 = 2,260.4 mph. When one uses 2,260 / 2289 to find the difference of .9875, 3, This can call for a decrease in the maximum and/or minimum lunar-orbital speed, but, more likely, their average velocity is the only problem, just as much as they have their average distance too high. It's impossible to work with precision with this mess, but I'm thankful that we can at least have NASA's figures for these matters, especially the 2,156.4 mph, for I've been using 2,163 until now. It means that, for the lunar eclipse of July 16, 2000, they have a lunar velocity at more than 157 mph slower than their own minimum-velocity figure. You can't produce rocket scientists with bad math like that.
There is something I have not yet done for this eclipse. I haven't yet obtained the umbra-path width by adding the time span between U1 and U4, 3.9439 hours according to their figures. We simply multiply the latter by the slowest lunar velocity of 2,156.4 mph, and that gets a minimum of 8,505 miles across (between the moon's position at U1 versus at U4). As we can see in the diagram below, the umbra diameter is virtually one lunar diameter (2,159) less than U1-U4 so that the umbra works out to 8,505 - 2,159 = 6,346 miles minimum. However, I'm going to change the time spans for the eclipse, and this will change the umbra diameter too.
http://eclipse.gsfc.nasa.gov/LEplot1/LEplot1951/LE2000Jul16T.GIFIn case the eclipse disappears from online, here it is from my files:
http://www.tribwatch.com/photos/lunarEclJul2000.gifThere are at least two other methods on the page to find the umbra diameter, one mentioned in the last chapter as 5,738 miles, and the other at 2.656794 x 2,159 = 5,736.6 miles (the small difference may be due to the use of an approximation, 2,159, for the lunar diameter), but these are their numbers based on their faulty, too-low lunar velocity.
When one converts their lunar size of 14'43.2" (on the page) to degrees (.4906666), and divides it by twice .6518 degrees (the number on their page for the umbra diameter), it tells that the umbra, in their claim, was 2.656794 times larger than the moon. This is the umbra diameter that astronomers want us to focus on and believe, while anything above 6,000 is one that no one is to know nor find, yet it's as easy as trying what comes natural when viewing the diagram with the times sitting in front of you. The near-central path of the moon is what makes it irresistible to tinker with their data to see if they have it correct, and they definitely do not. Their lunar velocity is so incorrect that they have got to be lying to the world rather than making an honest mistake. They have reset the "facts" as best they can to maintain their 93-million-mile hoax, but even the umbra diameter that they use gets a sun far less than 93 million miles (more like 15 million).
Now knowing how to find the lunar distance by first knowing the diameter, I can provide the length of the shadow, and therefore its angle to the edges of the sun, according to any shadow diameter. It will provide the true solar distance, but I need to nail down the true umbra diameter first. And this is the sacred cow that NASA wants to keep from us. The math method for this task was in the last chapter; here's an example: 6,346 / 7,918 = .80147 of the shadow length. If we knew the length of the shadow, we wouldn't need to find the lunar velocity nor the umbra diameter to find the true solar distance. But that's why NASA (or Fred Espenak) won't give us the lunar distance of this or any other eclipse.
This distance of .80147 of the shadow length starts at the tip and goes toward the earth, and the figure represents where the moon is passing through the shadow. The rest of the distance, from the moon to the earth, is easily found as 1 - .80147 = .1985 the shadow length. But what is the shadow length? If we knew the lunar distance (from the earth) during the eclipse, we could easily find the shadow length as: lunar distance + (.80147 / .1985 x lunar distance). For example, if the shadow was one mile long, the math for finding the shadow length is: .1985 + (.80147 /.1985 x .1985) = 1. This verifies the math; we can use any other number in place of .1985.
Knowing that the moon was about 252,000 miles at this eclipse, we do this: 252,000 + (.80147 / .1985 x 252,000) = 1,269,000 miles. With that number in the edge-b box and 3,959 in edge-a, the angle of each line from the shadow tip spreads out at .1787 degree (= angle of E or M lines). Providing you are working with an average-size sun exactly (.5331 degree), the formula for finding the solar distance using these numbers is: 3,959 / (.004652 - (3,959 / 1,269,000)) = 2.59 million miles for the distance between earth and sun. The italics are the amount of spreading of the sun's angular-size lines versus the spreading amount of the E-M lines (starting at the tip and reaching the earth over 3,959 miles of spread). At this time, I'm not keen on an umbral-path diameter of 6,346 miles with its 2.59 solar-distance figure. I tend to see it that path more around the 6,000-mile area.
Two chapters ago will tell you why .004578 is used for the solar lines on the day of the eclipse. It was as simple as using the angular solar radius like so: .262275 / 57.295 = 004578 mile. The 57.295 represents the straight line on a protractor (= the zero-degree line). I turned my solar lines into giant protractors using 57.295 as the distance of the protractor's line. No matter how long the cosmic line, it needs to be viewed as 57.295 units long (360 / pi / 2). Actually, the number comes out as 57.296 when using more decimals of pi (3.14159265).
The protractor's line, for example, is the M line in the E-M triangle. The degree rise on a protractor is then the distance between E and M at the open end of the lines (it can be the earth radius or diameter), and the tip of E-M is where the straight line of the protractor starts. One can do the same with the angular lines of the sun; let the bottom line be the protractor line, and the solar diameter or radius becomes the degree of spread. I can't tell you how happy I was to find this 57.295 magic. It put me on the fast track to solving the task I had undertaken without it. It allows people on the street to test what astronomers have lied to us about, and allows normally-educated teenagers to find the solar distance on their own.
Precisely because the time given between U1 and U4 provides 6,346 is what proves the times given as erroneous, for when using the time span between U1 and U3, the umbra comes in at roughly 6,208 miles (I had used 2.87 x 2,163 = 6,208; the precise math should be 2.8664 x 2,163 mph = 6,200.0). No matter which set of times one uses, the umbra must always work out to be the same diameter, obviously. Clearly, they are lying about the times, and I would like to know what the real times were. The most-important task we can have in this regard is the finding of the absolute-correct angle of the E-M lines during this or similar lunar eclipse. It will prove to the teenager the correct solar distance, and alert the world that astronomy is fraudulent in its claim for the volume of the cosmos. Someone in a position of power needs to call on NASA to get the correct times for this eclipse, and the one also for July of 2018.
We have at least three methods of our own for finding the umbra diameter using the times: 1) we can measure the distance of U1-U2 by knowing the correct lunar velocity; 2) we can measure the distance of U1-U3 by knowing the correct lunar velocity; 3) we can measure the distance of U1-U4 by knowing the correct lunar velocity. And that's probably why the eclipse page doesn't mention the lunar velocity. It would be like NASA hanging itself by the neck to include the lunar velocity at that time. We should hang NASA by the neck, anyway.
We first of all correct U1-U2. There are two ways to correct it. They have it incorrectly as 1 hour, 4 minutes, 48 seconds (1.08 hour), yet it should be, about an hour. As one way to correct it (call it scenario A), it may mean that U1 should start at 12:02 instead of the 11:57 that they give, which increases by about 5 minutes the time frame between U1 and U3. It doesn't in itself change the time between U2 and U3. In the meantime, just as they have U1 starting about 5 minutes too soon, they have U3 ending about 5 minutes too late so that the one tinkering with the math gets the same fraudulent lunar velocity between U3 and U4 as between U1 and U2. In scenario A, there are about 10 minutes too many between U1 and U4. Without including the seconds that the page provides for the times, they have U1 to U4 as: 11:57, 13:02, 14:49, 15:53, but an-attempt-to-correct-it scenario A has it at 12:02, 13:02, 14:49, 15:48. In this way, both U1-U2 and U3-U4 become about 12 seconds less than an hour.
Scenario B proposes that they started U2 about five minutes earlier than the reality, wherefore it changes their times to: 11:57, 12:57, 14:53, 15:53. The time between U2 and U3 now changes to about 1.95 hours, and the velocity from U1 to U2 (1-hour span) is now roughly 2,159 mph, wherefore the distance between U1 and U3 (full umbral path) becomes 2,159 x (1.95 + 1) = 6,369. When we use the span between U1 and U4, which does not change from their times given, the math is: 3.9439 hours x 2,159 = 8,515 miles, from which we need to subtract one lunar diameter to find the width of the shadow path, and that finds 6,360 miles instead of 6,356 miles. These numbers are close enough that scenario B is possible; for example, just reduce the 1.95 a little to get a match at 6,360.
Scenario B requires some precision work to discover what the true times were, IF it's the correct scenario. Scenario A works fine, as far as I can see at this time. For scenario B, we need to use hit-and-miss math, guessing at how long U1-U2 really was. The paragraph above makes it one hour exactly, with a lunar velocity of: 2,159 / 60 x 60 = 2,159 mph. Let's try 1 hour and 20 seconds = 1 + (20 / 3600) = 1.005555 hour. The latter multiplied by the lunar diameter gets the lunar velocity of 2,171.0 mph. We then multiply the latter by the time between U1 and U3 to get the diameter of the shadow path. The U1-U3 time span, to the second, works out to their U1-U2 time of 1.78361 hours, plus whatever extra time span we guess that U2-U3 really had, plus whatever the U1-U2 time becomes after guessing at what U2-U3 might be. The trick is to guess until the shadow path calculated from the U1-U3 time span equals the shadow path calculated from the U1-U4 span. In this scenario, U1 always starts at their time of 11:57:17 am.
Let's start off by increasing U2-U3 by 9.5 minutes, half of it at the start of U2, and half at the end of U3. The math: U2-U3 time = (9.5 / 60) + 1.78361 = 1.9419 hours. The time between U1 and U3 becomes their time of 2.8636 hours + (9.5 / 2 / 60) = 2.9428 hours. U1-U2 becomes their time of 1.08 hour - (9.5 / 2 / 60) = 1.00083 hour. The time between U1 and U4 becomes 2.943 + 1.00083 = 3.9439 hours. We are now ready to do the velocity math, starting with 2,159 (lunar diameter) x 1.00083 = 2,160.8 mph:
U1-U3 time of 2.9428 x 2,160.8 = 6,358.7 miles;
U1-U4 time of 3.9439 x 2,160.8 - 2,159 = 6,363.0 milesWe are close to a match. Let's try decreasing U2-U3 a little to 9.4 minutes. The math: The time between U1 and U3 becomes their time of 2.8636 hours + (9.4 / 2 / 60) = 2.9419 hours. U1-U2 and U3-U4 become their time of 1.08 hour - (9.4 / 2 / 60) = 1.0016666 hour. The velocity math starts with 2,159 (lunar diameter) x 1.0016666 = 2,162.6 mph:
U1-U3 time of 2.9419 x 2,162.6 = 6,362.1 miles;
U1-U4 time of 3.9439 x 2,162.6 - 2,159 = 6,370.1 milesThe gap between the U1-U3 and U1-U4 just grew larger, meaning we need to go the other way i.e. increasing U2-U3 by more than 9.5 minutes. Let's try increasing to 9.6 minutes. The math: U2-U3 time = (9.6 / 60) + 1.78361 = 1.9436 hours. The time between U1 and U3 becomes their time of 2.8636 hours + (9.6 / 2 / 60) = 2.9436 hours. U1-U2 and U3-U4 become their time of 1.08 hour - (9.6 / 2 / 60) = 1.0 hour. The velocity math starts with 2,159 (lunar diameter) x 1.0 = 2,159 mph:
U1-U3 time of 2.9436 x 2,159 = 6,355.23 miles;
.U1-U4 time of 3.9439 x 2,159 - 2,159 = 6,355.88 milesIt's almost perfect. At a glance:
U1-U2: 1.0 hr
U2-U3: 1.9436
U1-U3: 2.9436
U1-U4: 3.9439It would be perfect if U1-U3 was 2.9439, for not only would that make U3-U4 exactly 1.0 hour, as it should be, but 2.9439 x 2,159 = 6,355.88, exactly the number you see above for U1-U4. Probably, the reason that 2.9436 isn't 2.9439 has to do with my using time spans rounded off to the second rather than having the seconds in fractions. It is remarkable, if scenario B is the correct one, that there is a perfect 1-hour period for U1-U2. It would mean that they pushed U2 4 minutes, 48 seconds deeper into the umbra than the diagram is showing. Here are the times (seconds rounded off) at a glance for scenario B:
U1: 11:57:17
U2: 12:57:17
U3: 14:53:55
U4: 15:53:55The center of the eclipse (greatest eclipse) then becomes midway between 12:57:17 and 14:53:55, which is 13:55:36. On their page, at the top, they have "Greatest Eclipse = 13:55:35.5".
Now, for scenario A, which took dogged work only to come to nothing proved. First, here is their scenario, showing a match, with U1-U2 at 64.8 min (1.08 hrs), giving a lunar velocity of: 2159 / 64.8 x 60 = 1,999.074 mph:
U1-U4 = 3.9439 hrs x 1,999.074 - lunar diameter = 5,725.148 miles:
U1-U3 = 2.8636 hours x 1,999.074 = 5,724.54 milesThere is a virtual match when comparing U1-U4 with U1-U3, but there is a big sore spot where U1 is much too slow for reality. They probably arranged a shadow path of 5,725 because they had a full umbra diameter of 5,738 miles, which tends to suggest to us that the path was 13 miles less than the full diameter. If one had the method, one could probably disprove this because the moon missed the center of the umbra by only 119 miles, and I do not think the circle of the umbra would drop as much as 6.5 miles over a distance across of 119 miles. They say that a pair of eyes 5 feet off level ground (imagine the earth as a smooth billiard ball) can see the horizon as far as 2.9 miles away. A drop of 5 feet is .000947 miles, and the difference between the two numbers 2.9 / .000947 = 3,062 times, whereas the difference between 119 and 6.5 is only 18 times. Something is wrong with their numbers again.
Below is my scenario A with U1 shortened by 4 minutes, 49 seconds to 59 minutes, 59 seconds (59.98333 min) and U4 shortened by the same, so that the lunar velocity is 2,159 / 59.98333 x 60 = 2,159.6 mph:
U1-U4 = 3.783345 hrs x 2,159.6 mph - lunar diameter = 6,011.51 miles:
U1-U3 = 2.78333 hours x 2,159.6 = 6,010.88 milesIt's roughly the same sort of match as their scenario, but with a velocity of 2,159.6 mph, exactly what should be expected if 2,156 is the slowest velocity. I've chosen 59 minutes, 59 seconds (instead of 60.0 minutes) to show you better how to do the math if you'd like to tackle this problem. The figure above, 3.783345, is 9 minutes, 38 seconds (2 x 4 minutes, 49 seconds) removed, and 2.78333 is 4 minutes, 49 seconds removed. For example, from their time of 3.9439 hours (for U1-U4), we remove 9 minutes, 38 seconds = .160555 hour, to get 3.783345.
One can try different time-removed periods other than 4 minutes, 49 seconds, but the virtual match will remain roughly the same, which is not good news because we can't arrive to the reality by this method. In other words, one scenario is as good as the other so long as the velocity is in the correct ballpark. Here are some examples.
1) With 5 minutes, 48 seconds removed, we're left with 59 minutes, 0 seconds (59.0 min) from their 64 minutes, 48 seconds, wherefore the lunar velocity is 2,159 / 59 min x 60 min = 2,195.593mph:
U1-U4 = 3.750567 hrs x 2,195.593 mph - lunar diameter = 6,075.72 miles:
U1-U3 = 2.766933 hours x 2,195.593 = 6,075.059 miles.
Difference = .661 mile.2) With 4 minutes, 55 seconds removed, we're left with 59 minutes, 53 seconds (59.8833) from their 64 minutes, 48 seconds, wherefore the lunar velocity is 2,159 / 59.8833 min x 60 min = 2,163.2 mph:
U1-U4 = 3.780012 hrs x 2,163.2 mph - lunar diameter = 6,017.923 miles:
U1-U3 = 2.78166 hours x 2,163.2 = 6,017.27 miles.
Difference = .653 mile.3) With 4 minutes, 45 seconds (4.75 min) removed, we're left with 60 minutes, 3 seconds (60.05) from their 64 minutes, 48 seconds, wherefore the lunar velocity is 2,159 / 60.05 min x 60 min = 2,157.2 mph:
U1-U4 = 3.785568 hrs x 2,157.2 mph - lunar diameter = 6,007.227 miles:
U1-U3 = 2.78443 hours x 2,157.2 = 6006.58 miles.
Difference = .647 mile.4) With 3 minutes, 48 seconds removed, we're left with 61 minutes, 0 seconds from their 64 minutes, 48 seconds, wherefore the lunar velocity is 2,159 / 61 min x 60 min = 2,123.6 mph:
U1-U4 = 3.81723 hrs x 2,123.6 mph - lunar diameter = 5,947.277 miles:
U1-U3 = 2.80027 hours x 2,123.6 = 5,946.653 miles.
Difference = .624 mileThe difference is dropping with the time-removed dropping, but it'll never be enough to get perfect match. Even when we get down to their scenario, with zero time removed, there's still a difference of .608 mile. I don't think we are supposed to have a match because it wasn't a perfectly-central lunar path. The miles in italics are not the literal diameter of the shadow, but the width of the moon's path. The difference of .65 of a mile can be due to the shadow path being .65 x 2 miles less wide than the umbra diameter. But, like I said, one scenario is as good as the other so far as matches go. I've spent days on this, and, in the end, no luck on using it to find the real framework of the time spans for this eclipse.
I'm left without a knowable umbra diameter. I've learned, however, that scenario B is a viable option, as far as I can see. Scenario B has a sun much closer to the earth than the 5-6 million of scenario A. The shadow lengths can be calculated for both scenarios. Scenario B had a lunar path of 6,355 miles, and we can use this as the full umbra diameter in this case. The math to find how far the moon was (last chapter) from the earth using this scenario is: 6,355 / 7,918 = .8026 from the tip, and therefore 1 - .8026 = .1974 from the earth. As the latter figure needs to represent 252,000 miles from earth, the .8026 figure is 252,000 x (.8026 / .1974) = 1.0246 million miles, making the umbra a total of 1.246 million + 252,000 = 1.2766 million miles. I don't necessarily see any problem with this scenario. The angle of the umbra tip can be found with that total distance in the edge-b box, and the earth diameter in the edge-a box: it's .35537 degree.
The same math above can be done for scenario A having a lunar-path width of 6,010 miles. But before doing this, we need to see whether scenario B checks out. With the tip at .355 degree, each line, E and M, goes to the solar edges at half that angle (.177685). The distance to the sun in this scenario is found with the following:
With a shadow diameter of 6,355 miles, the distance from shadow edge to earth edge is (7,918 - 6,355) / 2 = 781.5 miles. The latter number in the edge-a box along with 252,000 in the edge-b box gets .177685 degree, and with the solar radius on that day at .524555 / 2 = .2622778 degree, it's a difference of .2622778 / .177685 = 1.476 times. The solar distance using the 781.5 figure is like this: 3,959 / (.004578 - (781.5 / 252,000)) = 2.68 million miles. We multiply the latter by 1.476 above to find that the E-M triangle is now 3.956 million miles long, giving an earth-shadow length of 3.956 - 2.68 = 1.276 million, and the right-angle calculator has it at 1,2766 million when fed .177685 degree [put 3959 in edge-a box].There is nothing wrong with that math. It gets an umbra length matching one that was obtained by another method above the paragraph above. We can now get the distance from the tip of the shadow to the sun, and check whether the E-M triangle has an angle matching with the shadow-alone triangle. If there is no match, scenario B cannot be true. Before we can get the E-M triangle angle, we need the solar diameter with a sun 2.68 million miles away. The latter number goes to edge-b and .2622778 into the angle-A box, to find the solar radius as 12,268 miles. We therefore put the latter number in the edge-a box with 3,956,000 in edge-b to find each line, E and M, at .177681 degree, a match with the .177685 above. Finally, we find the umbra-alone angle, expecting it to be .17768 degree, and so we put the earth radius (3959) in edge-a with 1,276,000 million in edge-b, and the angle is said to be .177769 degree. It seems like a match (all decimals were not included in all of this).
There is no reason to expect that scenario A (6,010-mile shadow width) won't have matching triangles. But let me check, anyway. 6,010 / 7,918 = .759 from the tip, and therefore 1 - .759 = .241 from the earth. As the latter figure needs to represent 252,000 miles from earth, the .759 figure is 252,000 x (.759 / .241) = 793,643 miles, making the umbra a total of 793,643 + 252,000 = 1.0456 million miles. The angle of each line, E and M, can be found with that total distance in the edge-b box, and the earth radius in the edge-a box: it's .21694 degree. Here is the next exercise:
With a shadow diameter of 6,010 miles, the distance from shadow edge to earth edge is (7,918 - 6,010) / 2 = 954 miles. The latter number in the edge-a box along with 252,000 in the edge-b box gets .2169 degree, and with the solar radius on that day at .524555 / 2 = .2622778 degree, it's a difference of .2622778 / .2169 = 1.209 times. The solar distance using the 954 figure is like this: 3,959 / (.004578 - (954 / 252,000)) = 4.9968 million miles. We multiply the latter by 1.209 above to find that the E-M triangle is 6.04 million miles long, giving an earth-shadow length of 6.04 - 4.9968 = 1.0432 million miles, and the right-angle calculator has it at 1.0458 million when fed .2169 degree. It means my math checks out.A couple of chapters ago, I ruled out a shadow-path width of anything in the ballpark of 6,208 as impossible due to the fact that the eclipse of concern was with a moon nearly as far from earth as possible, where the shadow diameter is smallest. But until I can discover some method of determining the ultimate length of the umbra, I should probably not rule out a shadow path 6,200 miles wide or more for the 2000 eclipse. The longer the umbra, the thicker it is in the ballpark of 252,000 miles away. Judging from the sizes of various shadow-path widths as reported by NASA, it is more important, for decreasing the widths, that the moon is further from the earth as opposed to the sun being closer to the earth. All of NASA's shadow-path widths are computations, and therefore erroneous. Instead of deriving the shadow width based on a combination of time spans, the lunar diameter and lunar velocities, they do it with the solar lines pre-programmed in their computers as being between 91 and 94.5 million miles off. They feed the program the solar and lunar distances (or angular sizes) at the time, and these two things are all they need to provide for themselves the shadow-path width. However, even the shadow-path widths they use do not jibe with their own solar distances. Below is my math using their shadow diameter for the 2000 eclipse, telling that their number gets a shadow 915,291 miles long, which is longer for it that they claim, and the number (5,738) moreover places the sun closer than they have it:
With a shadow diameter of 5,738 miles, the distance from shadow edge to earth edge is (7,918 - 5,738) / 2 = 1,090 miles. The latter number in the edge-a box along with 252,000 in the edge-b box gets .247825 degree, and with the solar radius on that day at .524555 / 2 = .2622778 degree, it's a difference of .2622778 / .247825 = 1.0583 times. The solar distance using the 1,090 figure is like this: 3,959 / (.004578 - (1,090 / 252,000)) = 15.673 million miles. We multiply the latter by 1.0583 above to find that the E-M triangle is 16.587 million miles long, giving an earth-shadow length of 16.587 million - 15.673 million = 914,000 miles, and the right-angle calculator has it at 915,293 million when fed .247825 degree [put 3959 in edge-a box].So, their shadow-diameter figure gets a sun only 15.7 million miles off. Don't you think someone has noticed this and complained to NASA? Or did they notice and not complain? Is this problem only in the most central-path eclipses, where they tweaked their computer program to provide their best-scenario garbage? To have the math above get 91 million, as it should be, their shadow diameter needs to be smaller than 5,738. Perhaps they thought it to be too risky to go smaller, because it makes the lunar velocity even smaller than the 2,000 mph that they have it at right now.
The up-coming central-path eclipse of July 27, 2018, has a lunar velocity of significantly less than 2,000 mph. It reports an umbral radius of .6488 degree, as compared to .6518 for the 2000 eclipse (their umbral radius figure does not reflect the path width). The diameter is smaller because the sun is a little closer, and the moon is a little further. They do not provide a path width, for, in order to do so, they would need to make it match what I did to find it...and they would find that their computerized numbers for the umbra diameter are erroneous. Of course, the ones handling these things must know that their numbers are erroneous. How could they not? They must be deliberately lying to protect their 93-million-mile hoax.
http://eclipse.gsfc.nasa.gov/LEplot/LEplot2001/LE2018Jul27T.pdfIn order to find the path width, one needs only two things: 1) add the time span between U1 and U4 and remove the time span for one lunar diameter of travel; 2) multiply the time span by the lunar velocity. But upon having the path width by this method, the one doing the math will quickly see that the velocity used makes their umbral radius figure glaringly wrong. Therefore, they do not report the width of lunar paths. But there is a way to find the path length by noting their Greatest Eclipse figure at the top of the page. In the eclipse below (September 26, 1931), you can see that the greatest eclipse position is exactly midway along the lunar path regardless of whether or not the path is central. The greatest eclipse position is midway between U1 and U4; the time they provide for Greatest Eclipse will verify it.
http://eclipse.gsfc.nasa.gov/LEplot/LEplot1901/LE1931Sep26T.pdfSo, in the eclipse above, they have the center of the lunar path at 19:48:04.9 Universal Time, and they have U1 at 17:54:37 Universal Time. The span between the two is 1 hour, 53 minutes, 31.9 seconds. The math to decimalize this number is: 1 + (53 / 60) + (31.9 / 3600) = 1.8922 hours. By finding the lunar velocity (a thing they do not report on the eclipse page because the survival of their hoax depends upon not showing it), one multiplies it by 1.8922 to find half the lunar-path length plus half a lunar diameter. By removing half a lunar diameter, one has the distance for half the lunar path. .
Let's go back to the 2000 eclipse and find the time between U1 (11:57:17 UT) and Greatest Eclipse (13:55:35.5), which is 1 hour, 58 minutes, 18.5 seconds = 1 + (58 / 60) + (18.5 / 3600) = 1.9718 hours. Their velocity figure was found at 1,999.074 mph (lunar diameter / U1-U2 time), wherefore their figure for the lunar path is 1,999.074 x 1.9718 x - (2,159 / 2) x 2= 5,724.55 miles, which matches the math done earlier in this chapter:
U1-U4 = 3.9439 hrs x 1,999.074 - lunar diameter = 5,725.148 miles:
U1-U3 = 2.8636 hours x 1,999.074 = 5,724.54 milesWe now know that they do give a shadow-path distance, round-about, but it's not based on the true times (which provide the faulty lunar velocity). They have it based in a calculation-program with their faulty velocity number built-in. If anyone looks into this in a non-central path eclipse, they wouldn't likely notice the problem, but with a central-path eclipse, the velocity is way too low from their own minimum-velocity figure.
My scenario A and B had similar lunar velocities. The difference was that scenario B had 9.6 minutes longer for the similar velocity to increase the shadow-path width. The faster the moon during any time period, the longer the path, or, the longer the time period, the longer the path for any one velocity. NASA and Fred Espenak need to be called upon to fix the lunar velocities for all the lunar eclipses, for as their umbral radii are all in conformity with the umbral radii in the July-2,000 eclipse, all of their umbra-radius figures are incorrect, meaning that all of the lunar velocities implied by the figures will be erroneous. This is not the way to lead astronomy, by lying to the people. I see jail time as appropriate. Falsely teaching a larger solar system than is the reality, when it's taught knowingly wrong, deserves jail time.
Update September 26,2016: I recently wrote to NASA's representative for its moon fact sheet, Dave Williams, asking why some are reporting an average-size moon at 31.1 arc-minutes, while NASA uses 31.6. He apparently had never heard of the 31.1 figure, for he asked me to provide a source that used it. He wrote a second message telling that he figured out where the 31.1 comes from: it's the size of the moon when one is standing at the core of the earth. I checked it with the triangle calculator, and it proved true so long as one enters 384,400 kilometers in the edge-b box with the lunar diameter (3476.2) in the edge-a box to get an angle of .518122 = 31.0873 arc-seconds.
Remove an earth radius from 384,400, and therefore enter 378,028.6 in edge-b with the lunar radius in edge-a to find an angle of .526248 degrees = 31.575 arc-minutes, not exactly 31.6, but close. If NASA is ball-parking 31.6, I would not be happy. Fact sheets are not for ballpark figures. High-school students doing minor reports on the moon are not the only people using NASA's fact sheet. Mr. Williams says, "31.6 arcminutes is the value adopted by the IAU [astronomical unit = official solar distance] as the mean apparent diameter of the Moon and is used in all relevant calculations." Yet no one else online could be found reporting 31.6. Compare 378,028.6 above with 378,000, the latter being the average lunar distance, according to NASA's fact sheet on the moon, from the earth's surface.
