Measuring the Solar Distance with Sunrise/Sunset Lines
or
Triangles Don't Lie, Astronomers Do
or
The Magic Number
First of all, I'd need to say that my aging brain is suffering with the astronomical tasks lately. I've caught myself doing some pretty dumb things with a head filled with numbers, and trying to keep track of things. I only have one head, and when it's filled, I can't put it on a shelf and use a new, fresh head. As the hours wear on, it gets tired of doing the math, and probably wants to revolt. I force it to comply with my wishes, but my own head might deceive me into taking short-cuts that are not good for precision. This is what math does to a person. So, hoping that I've cleared out all the mistakes, read with caution, just as though you actually believed that I am not perfect. If anyone has the time and capacity to study what's in this chapter, it is revolutionary, I believe, for distancing the sun.
I wanted to know the angle of a star / planet dead-center on the horizon. It's not exactly 90 degrees with the ground because the ground circles downward to the horizon. I've just realized that the 57-method can find the angle to the horizon, starting with this: "For an observer on the ground with eye level at h = 5 ft 7 in (1.70 m), the horizon is at a distance of 2.9 miles (4.7 km)." This assumes a perfectly flat ground i.e. without hills or valleys.
The following shows that the 93-million mile distance is a hoax. If you would like to know this, take the time to make a drawing. I know how much you love to draw those little planetary circles. Put an earth circle on the left of the page an inch round. Put in a horizontal line through the center of the earth, extending to the right of the page. Choose a spot at the right of the horizontal line for the sun; draw it with a center on the line, and 3 inches round. Draw lines from the top and bottom of the earth to top and bottom of the sun. These are your sunrise and sunset lines. Put a tick at the very bottom of the earth, and mark it M for morning; but a tick at the very top and mark it E for evening. Your two lines can be called line M and line E. Put a circular arrow on the earth to indicate its counter-clockwise spin. Put a dot in the earth circle to indicate the north pole, so that you won't forget that the sunset and sunrise lines take off from the equator. Busting the astronomers is so much fun.
Although you can't make it out on a drawing so small, you can understand that lines M and E contact the earth a little behind (on the dark half of the earth) M and E points; the latter are not to be viewed as the beginning of the lines, but as the very top and bottom of the earth circle. Specifically, M and E are exactly on the equator on both the June-21 and December-22 solstices. You will use their M and E lines to reveal the hoax.
I don't want to put additional lines on your drawing to indicate zero degree. You can imagine zero degree at each of two horizontal lines extending from E and M. Lines E and M are to be given angles in relation to these zero lines.
You now need some background information to answer some questions passing through your mind. When you view sunrise at the equator, you are looking down line M with the sun's EDGE on the horizon, not the center of the sun. Just so you know for sure, the page below verifies: "However, 'sunrise' is defined as the moment the upper EDGE [caps mine] of the sun's disk becomes visible above the horizon - not when the center of the sun is visible. In the same sense, 'sunset' refers to the moment the Sun's upper edge, not the center, disappears below the horizon" The page shows that every latitude has an equal day and night on the two equinox's (March 21 and September 22), except for at the equator. The equator has a longer day, than night, because the sun is bigger than the earth. There is no other reason. Day and night are distorted from geometrical perfection either above or below the equator, wherefore the equator is where we need to have E and M. There is a diagram at this page showing that the June-21 and December-22 dates (called the solstices) have the equator, at both sunrise and sunset, straight across from the center of the sun, which is the case on your drawing.
http://www.timeanddate.com/astronomy/equinox-not-equal.htmlThe chart at the page below has to do with sunrise and sunset times for Quito, Ecuador, virtually on the equator. This page is live on the day you visit it. On July 29 as I write, the Quito page has a chart telling that every day in July had a day 12 hours, 6 minutes and 30-ish seconds long, which predicts the same for June 21. For July 31, the day length is said to be 12 hours, 6 minutes, 23 seconds, and for July 1 it's said to be 12 hours, 6 minutes, 34 seconds. From extrapolation, the day length for June 21 looks to be 12 hours, 6 minutes, 37 seconds i.e. very little change.
http://www.timeanddate.com/sun/ecuador/quitoThe timeanddate.com article above lies to us, and even fails to tell why the day is longer at the equator when claiming to explain that very thing. You can see on your drawing that the day will be longer due to the sun being larger than the earth. The edges of the sun shine a little beyond the top and bottom of the earth circle for that reason.
The page below talks about mirages due to atmospheric refraction, but I've never seen any, and I'm no youngster. One should not misjudge heat-wave ripples in air as due to refraction. Heat-wave phenomena are where light finds it more difficult to transfer straight through speedily-rising heat particles, like off the hot hood of your car. One can witness a shade that this effect causes. The page even says that sailors avoid using stars when they are low to the horizon, which sounds ridiculous because a star a little out of place won't alter course much. I get the impression that the article is hard-up sharing proof of light refraction in air. It has the following that I do not think can be proven, but where the reader is made to think that refraction levels are proportional with the amount of air the light is passing through:
Atmospheric refraction of the light from a star is zero in the zenith, less than 1' (one arc-minute) at 45° apparent altitude, and still only 5.3' at 10° altitude; it quickly increases as altitude decreases, reaching 9.9' at 5° altitude, 18.4' at 2° altitude, and 35.4' at the horizon; all values are for 10 °C and 1013.25 hPa in the visible part of the spectrum.I see a hoax to keep us from finding the solar distance using sunrise/sunset data. While one can be deceived by twice as much refraction between 10 and 5 degrees off the horizon, the claim goes further, nearly doubling refraction between 5 and 2 degrees, when the thickness of air is virtually the same. Or, they double refraction between 2 and 0 degrees when, absolutely, the amount of air that light passes through is the same for both cases. It sounds like a hoax, absolutely.
There was a great need for refraction of light with the evolutionists who were set to feed mankind the huge-universe hoax; to this end, one can expect faked scientific experiments disseminated as respectable science, and this would be fed to the public to this day. It goes on to say that, sometimes, refraction near the horizon is high, then low, depending. Sure, depending. Depending on what? They can make up a reason: "Refraction near the horizon is highly variable, principally because of the variability of the temperature gradient near the Earth's surface and the geometric sensitivity of the nearly horizontal rays to this variability. As early as 1830, Friedrich Bessel had found...blah blah" More than one scientist is quoted to give appearances that refraction is a fact studied by many, and as they all had varied results, we can imagine, that's why the above says "Refraction...s highly variable." It sounds like smoke and mirrors. As you can see, "geometric sensitivity" carries the idea that the refraction is going to be higher than the expectation. Something along our view of the horizon is very sensitive, sure sure. Temperature changes, sure sure, let me be your fool. As we view in the horizontal direction, the temperature variability gets far more effective in refracting light than in the vertical direction, sure sure, a perfect hoax to make the sun appear to be doing what it's not.
The thing that air and its rising heat are predicted to do is make it more difficult for all the light to get through. That changes the color of the sun, not the solar position. I can't prove my position, and they cannot prove theirs, for they are always in air. A fish can't prove that light refracts in water. The fish needs to get out of the water to see the light bending as it enters water. Can the astronomer / astronaut see light bend when he's in a rocket ship above the air. No. He therefore has no proof that light bends through the atmosphere. Plus, air is not a liquid, and may not be capable of refraction. Wikipedia's article on Moon Illusion, and others, verify that the moon is not larger on the horizon than it is anywhere else. It says, "although the atmosphere does change the perceived color of the Moon, it does not magnify or enlarge it." If heavy refraction were occurring at the horizon, the sun would appear to be moving faster than usual at the sunset horizon, and slower than normal at the sunrise horizon. The article does not mention this, yet it is the best proof of all for solar / lunar refraction.
They give all sorts of examples for refraction, such as twinkling stars, but this is not refraction. It's an effect of moving air. The moving air is shifting the direction of light. Do they twinkle only near the horizon? The astronomers / meteorologists also use mirages, but these are, once again, heat-related disturbances of light. They also point out the flattened sun on the horizon, but why should refraction flatten the sun? Shouldn't it be the other way around, making it oval in the vertical position? If the bottom of the sun is bending more than the top, shouldn't the bottom look longer than the top? Why should both top and bottom flatten out? Why should the top look flat at all due to refraction? Why should the bottom bend light upward, and the top bend light downward? Isn't that the conclusion from a flattened sun due to refraction? Perhaps the top and bottom are doing nothing, while the sides are expanding. Or, if the flattened top and bottom are due to refraction, this may be the fullest extent of it, not a shifting of the entire sun by as much as three degrees.
They say that light changes direction toward the earth circle...so that light reaches around the top and bottom of the earth to a greater distance than would be the case without refraction. THEREFORE, if this is correct, lunar-eclipse lines would be made steeper than they really are, which is to say that the earth's shadow is made narrower than it would otherwise be. The astronomer apparently makes this apply (fraudulently) where he uses a computer program to indicate diameters for the earth shadow that are smaller than the reality. But the umbra diameters due to refraction must then be enlarged to their real diameters, if refraction made them appear smaller in the first place. He can't win even if he claims refraction, because he is forced to provide an amount of refraction, which we can subtract to find the situation in un-refracted reality. If refraction is responsible for part of the sunrise and sunset, please tell us how much, and we will just undo it to find the real sun lines.
Alas, we read above that refraction at the horizon is 35.4' arc-minutes, which is 35.4 / 60 = .59 degree. Yet, he has sunrise and sunset at .9 degree, meaning that the real situation has the true sunrise and sunset at .9 - .59 = .31 degree. I'll re-visit this number to show that it's impossible. It needs to be smaller.
The page below gives the day length for La Concordia on July 30 as 12 hours, 7 minutes, 7 seconds. We can assume, as per what we saw on the Quito page, that the day length of La Concordia for June 21 was likewise 12 hours, 7 minutes, and so I will use this time as the maximum day length possible near the equator (because La Concordia is smack on the equator). Later, I find true sunrise and sunset about .72 minute before the 12-hour face of the earth, meaning that astronomers knock off more than six of these extra seven minutes using their refraction excuse.
http://time.unitarium.com/sunrise/253215In case you need it, the page below gives the day length for any latitude in the world on December 22, which I assume cannot be identical to day lengths for June 21, for the sun is further away (smaller) in December. However, for the equator in December, the day length is given as 12 hours, 7 minutes, same as the equatorial July days above. The page's chart shows La Concordia, Ecuador, at 0.0 degrees, smack on the equator.
http://time.unitarium.com/events/shortest-day.htmlWe need to find the angle of the M and E lines (both at same angle), which were found above to be .31 as per the refraction hoax. To find the angle as per their faulty sunrise/sunset data, we use the 12 hours, 7 minutes at La Concordia. The extra 7 minutes needs to be split in half, with 3.5 minutes on behind E, and 3.5 minutes behind M, thus making the daytime half of the planet lit for 12 hours, 7 minutes. In other words, for the 93-million-mile hoax, lines E and M start 3.5 minutes on the dark side of the earth from the top and bottom of your earth circle.
We can convert 3.5 minutes of earth spin to an angle. Any line extending from the earth's surface, no matter what its angle, will be at 90 degrees to itself after the earth spins a quarter day = 360 minutes, meaning that the key for finding the angle is 90 / 360 = .25 day. That is, the earth circle in degrees divided by the angle between two line positions equals the time between the two line positions. In reverse, .25 day x 360 = 90 degrees, and as 3.5 minutes = .0024 day, our math is like so: .0024 x 360 = .9 degree. We now know that the astronomer's line M and E are .9 degrees (from the invisible zero lines on your page). This is a huge angle that readily shows its impossibility. Why have astronomers fed us this impossibility? Apparently, they view .59 degree as the refracted part of this .9 angle.
Refraction aside, everyone agrees that lines M and E are each less than .267 (half of .533) degree. You can view their respective zero-degree lines as two horizontal lines extending from the M and E spots, or, you can connect lines E and M behind the earth and view them as less than .533 (less than .267) degree. If we knew the E / M angle exactly, we could find the distance to the sun, as you will see in some triangulation below. It's exactly why the astronomical powers are trying to mask the angles of E / M, for these lines give the solar distance in far less than 93 million.
To understand what refraction does to the reality, we need two more lines on your drawing, starting at M and E, which I'll call the 12-hour lines. These are horizon lines seen from the top and bottom of the earth, with the eyes five feet off the ground. We will start the lines at eye level, and go across the horizon 2.9 miles away, and finish at the horizontal line in your drawing (through the center of the sun). But don't put the 12-hour lines on your page yet. These lines will be steeper than the one starting at the toes, but we do not yet know where in relation to the sun is to their meeting spot.
To get the angle of eyesight from eyes five feet off the ground, to the horizon, the math will use 5 / 5,280 (feet per mile) = .000947 mile of line decline over 2.9 miles: 57 / 2.9 x .000947 = .0186 degree (when 57.295, the correct number, is used, it comes out to .0187 degree). We can find the distance from the earth where the 12-hour lines meet by first finding the decline per mile = .000947 / 2.9 = .000327 mile. As either line declines one earth radius upon reaching your center line, we just do: 3,960 / .000327 = 12.1 million miles from the earth. In other words, if two peoples on opposite sides of the planet looked out at the horizon toward the same spot, their line of sight would meet 12.1 million miles away. This has nothing to do with where the sun is located. As triangulation will find the sun less than 12.1 million miles, I would let the 12-hour lines meet beyond the sun. You can draw them on your diagram now, very lightly (they are needed only temporarily) from the E and M spots to beyond the sun's center.
As you can see, when a person spinning on the earth reaches the bottom of the earth circle, where the 12-hour line applies, that person will be looking into the sun. That is, a 12-hour line runs into the sun; it does not miss the sun so as to point into outer space. It means that the sun is still on the horizon; it has not yet sailed away from the horizon, when a person reaches the M spot. If the day is 12 hours, 2 minutes long, a person will be at the M spot one minute after perfect sunrise. As it takes the sun 2.13 minutes to travel a solar diameter, the 12-hour lines goes almost dead-center into the sun.
Your sun covers the full distance between lines E and M. According to astronomers, the bottom 12-hour line has a starting point 3.5 minutes in time (nothing to do with arc-minute angles) after someone is spinning the globe at the start of line M. To put it another way, line M becomes the bottom 12-hour line 3.5 minutes after the M line.
You're waking up, yawning. You look out the window of your house boat, and you can just see a blip of the orange rim of the sun on the horizon. The sea is like glass this morning. You are spinning on the earth at one revolution per 24 hours = .25 sky degree per minute (360 / 24 hours / 60 minutes = .25), and will be at point M in 3.5 minutes. The sun is crossing the sky at a known speed of 2.13 minutes per solar diameter. Therefore, after you've been looking at the horizon for 2.13 minutes, you see the sun starting to lift off of the horizon. Oh no, it's only 2.13 minutes in, and already you are starting to see outer space between the sun and the horizon. By the time that 3.5 minutes arrive, there is almost a solar diameter of space between the sun and the horizon. This can serve to show how astronomers define refraction.
Again, line M will become a line (call it M2) at the opposite side of the sun in 2.13 minutes (.553 degree / .25 degree per minute = 2.13), but the 12-hour horizon line, which is the view of the horizon more than a minute after the 2.13 minutes, looks midway into the sun. The astronomer will claim that this is not a problem due to light refraction, which is his confusing smoke-and-mirrors to hide the reality. He says that refraction has shifted the whole sun. He will say that it's like someone pointing a flashlight at you while they are in the air, and you are under water; you see the flashlight where it's not really at.
I have made your drawing to reflect the reality, but he says the reality is shifted over. He therefore shifts the sun two-to-three minutes toward the bottom of your page, and he will bring line M down with the sun so that the sun is still rising at the line-M horizon. By noon, the sun is exactly where your drawing has it, but late in the day, the sun shifts toward the top of the page. And he will shift line E accordingly. The only thing he doesn't shift is the position of the earth; its top and bottom remain at the same places on the page, because the earth is not a part of the refraction effect. The earth must always face the real sun just as I have it in your drawing. The 12-hour lines remain the same too. He now has a way to explain why the 12-hour lines look into space, 3.5 minutes after sunrise, because he has shifted the sun down the page and replaced it with outer space.
This trick will work for him if you simply trust that he knows what he's talking about. He needs this refraction theory because, without it, the angles of the lines under discussion will be absolutely naked to anyone playing around with such lines, and, naked, they can be used to find the solar distance by any amateur astronomer. Therefore, astronomers are fed the refraction garbage to keep them from seeing.
Do not get the idea that NASA can assign every earth shadow, per lunar eclipse, its own refraction level based on the time of day of the eclipse. Every eclipse has the sun and earth in the same position as your drawing, meaning that refraction is always at the same level regardless of what time of day the eclipse is at. Umbra sizes change only as per the distance of the earth from the sun. Any refraction that occurs is constant.
One thing that you must add to your drawing now is line M2. I was unable to find an isosceles-triangle calculator that would give its height (by plugging in the base distance and the angles at the base). The triangle height is the distance from the earth to the sun, wherefore it's extremely important. Fortunately, when one splits an isosceles triangle in half, two right-angle triangles are formed, both retaining the same height as the original isosceles. And I did find a right-angle calculator.
You can easily find the angle between M and M2 as .533 degrees because the two lines start at the same place (almost), and cover the entire solar diameter. At the page below, we have a right-angle calculator. In the drawing at the page below, the triangle's corner, C, can act as the center of the sun on your drawing. The triangle corner, B, is where M2 meets E, and triangle corner A is at M upon your page. Get this settled nicely in your mind before proceeding. Line, a, is the solar radius, and line c (your M2) is the solar distance. We need to find the length of line M2.
http://www.cleavebooks.co.uk/scol/calrtri.htmThe line from C to A is not on your drawing; it's the one that splits the M2-M isosceles in half. We can know all three angles because: 1) C is 90 degrees; 2) A and B must add up to 90; and, 3) B is known to be half of .533 in an average-sun situation. But the calculator will not give us the distances of any triangle side if all we know are the angles. We first need to feed it a distance. Where will we get one? From the other triangle between M2 and E. It's an isosceles that can likewise by split into two right angles, and the base of this triangle is the earth's radius i.e. which we know. Once we feed the calculator the distance of line M2, the calculator will also tell us the solar diameter (at line, a). Wonderbar.
This would all be more wonderful if it were straight-forward to find the angles of the smaller isosceles. Until then, we cannot have the length of M2, and if we don't have that, we cannot have the solar diameter. You now know that your drawing is precious. It has not all been in vain. You might even like to re-do it like an expert draftsman out to conquer the world, to release the masses from the clutch of the goons. You can show this drawing to all your friends, if you don't mind losing them all. There is a chance that your drawing will convert the world, but not right away. The enemy has smoke and mirrors to make everyone's vision distorted. Usually, the artist isn't famous until after death. Sorry.
We know that line E is less than .276 degree. But these angles are according to the horizontal, zero-degree line, and do not inform us of the angle for line E in relation to line M2. Do not confuse these two angles of M2. It is probably a good idea that you draw the base line, from the start of M2 to the start of line E. If all I know is that line E is smaller than .267 degree, I don't know the angle, and therefore can't find the angle of the base with it. Will the astronomer please find us the angle of the base of this triangle? He has, already, but he's not telling. It doesn't help his 93-million-mile picture. He's a louse. He's no-good. A jerk. He jerks the figures, and denies you the truth.
If we knew the angle of line M, we could find the angle of M2 in relation to the base line. That's because knowing the angle of line M gives its time of the day (it's all related). Or, the time of day is what defines line M. One knows the 2.13 minutes of time between lines M and M2, and thus one could find the time of M2 if one has the time of line M. With the time known for M, one also has the time of E, and thus one can draw the base line between the two times, one end at the start of line E, and the other end at the start of M2.
Once you have the correct time between M and zero degree, just subtract it from 2.13 minutes to find the time between zero degree and line M2. The only thing left to do is to find the time, on the clock of day, that line M (sunrise) begins.
Any line, at any angle, that grazes a sphere will be 90 degree to the center of the sphere (where it touches the sphere). M2 (and line E) is such a line, but you must not make the mistake of assuming the base line to be 90 degrees to M2, for the base line goes not from M2 to the center of the earth, but from M2 to the start of line E. It's almost 90 degrees, but not quite.
There really is a way to find the angle of E and M: Gonzales down in La Concordia needs to time the sunrise at the equator, with hands-on work, rather than merely using NASA's computerized wizardry, which is offset like the devil's teeth. Go to the sea at the equator, with a clock, and find the second in which the sun first appears on the horizon. Then, take the exact time of sunset at the same place. Once you have those figures, you divide the excess of 12 hours by two, and figure the angles of lines M and E with it. Once you have those, the length and angle of line M2 will reveal itself.
Here is how this is done. If, for example, the full day at the June-21 equator is 12 hours, two minutes, zero seconds, remove the 12 hours, and divide the remaining two minutes in half. You now have the way to find the angle of E and M (as measured from their zero-degree lines) by doing this: 1.0 minute / 60 / 24 x 360 = .25 degree. That is, view the earth as a circle of 360 degrees, and back up .25 of them, from the M spot at the bottom of your earth circle, into the dark half of the planet. Line M grazes the planet at this .25-degree spot. It just so happens that this solar eclipse was on June 21.
Next, you subtract the answer, .25, from the actual angular size of the sun for this June eclipse, which was 15'44.2" x 2 = .52455 degree. The latter can be used very appropriately as the number of degrees around the earth circle between the start of Line M and where line M2 grazes the planet. Do you understand that? You can actually find the distance around the earth's circumference, and even the amount of time elapsed, between two spots grazed by two lines upon the surface of the sphere. The math, .52455 - .25 = .275 degree, tells you that line M2 is at .275 degree from the zero-degree line (out from the M spot), while line M is at .25 degree from the same zero-degree line.
Line M2 starts at .275 degrees from the M spot around the sunny side, and line E (same angle as line M) starts at .25 degree from the E spot on the dark side. A line connected between their starting places will not go through the center of the planet. If line M2 went through the center of the planet, it would go from its starting point at .275 sunny half to .275 on the dark half. If line E went through the center, it would go from its .25 on the dark half to .25 on the sunny half. It thereby becomes clear that, when the two lines are connected, they will be off by .275 - .25 = .025 degree as opposed to when they go through the center of the planet, for M2 will go from its spot at .275 to .25 at the opposite side, and line E will go from .25 to .275 on the opposite side.
What does this mean? What are the angles of the two corners of the base line? This can be tricky if you don't realize something important. We create the triangle needed. We can imagine a 90-degree line from the start of line E, through the earth's core, and missing the start of M2 by .025 degree. We now raise this line to meet line M2, and thus the core line becomes the base line of an isosceles triangle, and we think, at first glance, that the angle at the start of line E is closing by .025 degree (i.e. 90 - .025), and we can do the same in reverse to find that M2 appears to be closing by .025 degree. Unless we were familiar with angles on spheres, we could conclude wrongly that both corners will be 90 - .025 = 89.975 degree.
I made that mistake, but, fortunately, had evidence to show that the angles on both corners were only half as much as .025. I probably would have left that mistake here had I not had that evidence, which made me try hard to find the problem. I then realized that angles on a sphere are taken from the core. That is, .025 degree is of a line from the core to a spot .025 degree around the circle, but if the line starts twice as far as the core, and goes .025 on the circle (as do lines E and M2 when they meet one another), it will be half .025 degree. Therefore, the corners of the triangle will be 90 - .0125 = 89.9875 degree.
The base line, to my surprise, remains the full earth diameter, essentially, because .025 (14,400 times less than the earth circumference) is a distance of only 1.73 mile around the earth circle.
We can take these numbers to triangle calculator below, and feed the two bottom corners 89.75 while using 7,920 for the length of the base. The calculator gives the distance to the sun as 18.2 million miles. But this is while using, as the basis, the angle obtained from a day of 12 hours, 2 minutes. If sunrise and sunset are less than one minute on both sides, the solar distance is less than 18.2 million miles.
https://ostermiller.org/calc/triangle.htmlAll we need now is to hear from NASA on the actual lengths of the equatorial day on June 21. But even if they insist that refraction is responsible for a certain part of the day length, we would like to hear what part is not due to refraction. We read above that refraction at the horizon is 35.4' arc-minutes, which is 35.4 / 60 = .59 degree. Yet, astronomers have sunrise and sunset at .9 degree into the dark half of the planet, meaning that the real situation has the true sunrise and sunset at .9 - .59 = .31 degree. They (at least some of them) are apparently admitting, round-about (if we can catch them), that lines E and M are really at .31 degree at the summer equator. We can re-do the math above using the average sun, and thus starting with .533 - .31 = .223, wherefore the angles on both corners of the baseline become .223 - .31 = an impossible, negative number. They go over the limit. They go into unreality.
Let me spell out the start of the problem. It's at the size of .31, for this is the angle of line E or M. The math, .533 - .31 is to determine the angle of M2, but in this case line M (.31) is larger than M2 (.223), an impossibility. One can spot on the drawing that line M2 must be at a greater angle than line M because both go to opposite edges of the sun while line M begins closer to its edge than does M2. That's why line M cannot be larger than .267 (half of .533), yet they are apparently hoping to get away with .31 (they might like one larger but this is already pushing the envelope).
Earlier, we saw that the solar size, .52455 degree, minus .25 degree (comparable to .31 above) gave .275 degree for the M2 line. And we then needed to do this: .275 - .25 = .025. The math was doable, but if .25 were larger than the number to be subtracted from, it indicates an impossible situation. That's why .223 - .31 is impossible.
As line E's angle is the same as the one for M, the angle between M2 and E (where they meet at the sun) is a small difference. To put it another way, line M2 leans over more than .267 from its zero-degree line, and line E leans over, in the same direction, less than .267 degree by the same amount, so that the two lines are nearly parallel.
The difference between the angles of E and M2 will be the angle at the tip of the E-M2 isosceles (when the two are .0125 degree, the tip will be .025). Technically, as the two lines rise at different angles, they are not the same length, and the triangle ceases to be an isosceles; at least not a perfect one. However, as both lines go to nearly the same spot at the sun (off by a small arc of solar circumference), and as they both begin on the earth's surface, they are, for all intents, the same length. That tells us how little difference there would be in their angles. The less difference, the further away the sun. The sun will be 93,000,000 miles away if the tip of the isosceles is .00244 degree (ten times smaller than .025), that being the astronomer's solar parallax number. You can verify this at the triangle calculator above, or by feeding the right-angle calculator .00244 in its "angle A" box, and using 3959 (no comma) in its "edge a" box.
http://www.cleavebooks.co.uk/scol/calrtri.htm
The Triangle-Comparison Method
There are two ways to find the solar distance here, one by finding the angle of the base line in relation to M2 and E, and the other by finding the angle at the tip of E-M2. A third way is to find the angle at the tip of lines E and M if by this one is able to find the angle at the tip of M and M2. The right-angle calculator shows that the tip of E-M (this is the earth-shadow cone) is 850,000 miles behind the earth's core if the tip is .533 degrees, which it cannot be. To get 850,000, enter 3960 in "edge a" box, and 89.73 in the "angle B" box (alternatively, use .267 in the "angle A" box). The figure, 89.73, is just 90 - .276.
I'm reading, "...(the length of Earth's shadow is about 1,400,000km", and that's 869,400 miles. Another page has, "length of Earth's shadow = 1392000km", and that's 864,400 miles. I've seen the latter figure from another page. If I feed the calculator the figure, 864,000 in the "edge b" box, and 3960 in the "edge a" box, it gives the angle of .263, when the sun is near its smallest, which they say is in July (but the solar eclipse page for June 21 has a smaller sun, by the smallest amount, than the page for the July-12 lunar eclipse).
What you will see below is that this is not reconcilable with the facts. The following sets up this section's math exercise:
The only reason that they are lying is that it's a dead-center eclipse, capable of revealing their 93-million-mile hoax. The time given between U2 and U4 (15:53:55) is 2 hours, 52 minutes = 2.87 hrs. If we use the velocity figure of 2,000 mph [non-existent lunar velocity], the umbral diameter [where the moon crossed it in the eclipse] works out exactly to where they want it, at 5,740 miles. We have them right where we want them because we can now use the slowest lunar velocity to get: 2,163 mph x 2.87 = 6,208 miles.
http://eclipse.gsfc.nasa.gov/LEplot1/LEplot1951/LE2000Jul16T.GIFIn case the eclipse dissappears from online, here it is from my files:
http://www.tribwatch.com/photos/lunarEclJul2000.gifThey have a major problem that, thanks to the 57-method, can be revealed. To show this, I've got to first set out the information in this paragraph. My solar-distance figure is based on the lunar distance from earth in the lunar eclipse, and finding that distance is somewhat unreliable. To get it as best I can, I use the 406,199 kilometers shown for June 15 (apogee) at 15:35 pm, and the 358,378 kilometers shown for the following perigee at 7:45 am on June 30. There are 352 hours between the two full moons, and the full drop of the moon from apogee was 406,199 - 358,378 = 47,821 kilometers = 29,716 miles. We then divide the latter by 352 hours to find that the moon dropped toward earth 84.4 miles per hour (average) from its apogee position, which position was 22.33 hours from 13:55 pm the next day, the time of mid-eclipse. Therefore, if the lunar drop was constant, by which I mean to say that it was 84.4 mph on the first day (it probably wasn't), the lunar distance at this eclipse was 252,412 - (22.33 x 84.4) = 250,527 miles. However, it seems to me that the moon would fall to earth at an accelerating pace, meaning it would fall less than the 84.4 mph average in the first day. I therefore have no idea how much higher the figure should be than 250,527, but I set it at 252,050, though I think it can be higher still.
The following is an example of how to work out the angle for any shadow diameter whatsoever, and I'm using it as per the 5,740 miles above: 57.3 / 252,050 x (3,960 - 5,740/2) = .248 degree. I caution, however, that while the 57 method is fast, it's approximate only for these sorts of distances. If you want precision, add the distance being spread (3,960 - 5,740/2) in the edge-a box, and the distance across (252,050) in the edge-b box to find the angle in the Angle-A box; in this case, it's .2477 degree.
http://www.cleavebooks.co.uk/scol/calrtri.htmAs the sun, on the day of the eclipse, had a radius of .262 degree, the difference between the E-M and M2-M triangles, if we take this picture further, is .262 / .2477 = 1.0577. That is, NASA needs to claim that the E-M triangle was 1.0577 times larger than the M2-M triangle at the eclipse. With the shadow diameter at 5,740 miles, there are 1,090 miles from the edge of the shadow to the edge of the earth (i.e. lines E and M). According to their own data, the sun, on the day of the eclipse, was roughly 3,960 / (.00457 - (1,090 / 252,050)) = 16.1 million miles away. That's the math for working out any solar distance based on using the miles between shadow edge and the earth edge. Note that I've changed the .0047 mentioned earlier to .00457. I'll get to the reason shortly.
Whereas I had the earth shadow at 6,208 miles for the eclipse, it was changed to 6,000. I saw a possibility in the minimum of five minutes NASA added artificially to U1-U2 (path outside the umbra), which were perhaps removed from U2-U3 (path inside the umbra). This made a lot of sense because it enabled them to keep the correct duration between U1 and U3 (a full umbra diameter); the only wickedness was in sticking five or more minutes of the umbral part on the outside of the umbra. It's possible that they took the extra minutes from P1, however. The extra minutes were needed to hide the true diameter of the umbra. These extra minutes slowed the moon and thus worked to reduce the distance between U1 and U3. "Between U1 (11:57:17) and U2 (13:02:05) there is 1 hour, five minutes, which is 1.08 hours...The lunar velocity works out to 2,000 mph (2160 / 1.08)." But if there were 5 minutes less, the velocity would work out to 2160 / 1 = 2160 mph. Chances are, the moon was moving faster than 2,160, and that, therefore, more than 5 minutes were involved.
By removing 6 minutes (.1 hour) from U2-U3, the time between U1 and U3 changed from 2.97 to 2.87 hours. However, if they did not borrow 5 or 6 minutes from U2, but from P1 instead (more-likely scenario), then the time for U2, and consequently, from U2 - U3, is accurate. The only thing needed is to remove 5 or 6 minutes from U1, and therefore from U1 - U3.
The central-path lunar eclipse of July, 2018, will reportedly have a U1 - U2 time of 1 hour, six minutes (1.1 hour), which makes the lunar velocity very little more than 2,160 / 1.1 = 1,963 mph, impossible. NASA is definitely lying about this one too, and has the lunar shadow 5,624 miles wide, according to its own shared data (see "2018" two chapters ago) in which none of it is open to my personal interpretation. This eclipse is a virtual replica of the one discussed above, with a reported lunar distance of 406,222 kilometers as compared to 406,199 with the one above. The one in 2018 will have a slightly smaller moon, but a slightly larger sun, suggesting very-closely the same shadow size between them.
If we remove 5 minutes from U1 - U3 (2000 eclipse), the time span is 2.78, which, when multiplied by 2,163 mph gets an umbra diameter of 6,015 miles. For being conservative, a 6,000-mile shadow is preferred, in which case the math was redone like so (the following is still not correct due to .0047):
The 3.05 million figure was obtained with 3960 - 6,208/2 = 856, but with a shadow 6,000 miles instead of 6,208, the 856 becomes 960. The latter number gets an E-M angle of 57.3 / 252,050 x 960 = .2182 degree, and with the solar radius on that day at .52455 / 2 = .2622 degree, it's a difference of .2622 / .2182 = 1.202 times. The solar distance using the 960 figure is like this: 3,960 / (.0047 - (960 / 252,050)) = 4.44 million miles. We multiply the latter by 1.202 above to find that the E-M triangle is now 5.34 million miles long, giving an earth shadow of 5.34 - 4.4 = .94 million, and the right-angle calculator has it at 1.04 million when at .2182 degree.
http://www.cleavebooks.co.uk/scol/calrtri.htm(I apologize for the inconvenience, but the 6,000 miles should be 5,998 because 6,000 is obtained wrongly when using the 7,920-approximation for the earth radius. With the stated 7,918 figure, 5,998 is the correct number to use. However, to make it easier to recall what I'm talking about when reflecting back on this, I'll stick to calling it the 6,000 scenario.)
There was no match for the shadow length. The culprit, I reasoned, had to be the .0047, which you see in the math above. That's when realizing that .0047 had to match line M2 (.00457), and it solved the problem. The angle, .0047 mile spread per mile toward the sun, is at 57.3 / 1 x .0047 = .269 degree, which is not the same as the .262 degree that is the M2-M triangle split in half. In other words, we need to change .0047 to whatever it would be when it's at the .262 degree (should be .262275), wherefore we do this: .262275 / 57.3 = .004578 mile (57.3 should be 57.295), the figure given in the right-hand calculator too. It means that I'm now using, to find the solar distance, the M2-M lines in conjunction with my conservative figure for the E-M lines where the latter is worked out from the 6,000-mile shadow above.
At first, I re-did the scenario above by replacing .0047 with .00457 to find that the solar distance at an exciting 5.2008 million, but after seeking better precision, I used .004578, which lowered the 5.2 somewhat. Tentatively convinced that the math had to find 5.2, I changed the 252,050 to 251,500 in order to bring the number up to 5.2, like so:
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.
http://www.cleavebooks.co.uk/scol/calrtri.htmYou can imagine how happy I was to see the match, with the resolution that .0047 should have been .004578. The match simply means that the math and method is correct, and the reason that it's correct is that .0045728 mile is a length-related expression of .262 degree, which is itself used in the math. If we use any other number besides .262, it won't work with .00457 in the math. If you are working on the same problem using a different-size sun, for example with a radius of .265 degree, you need to change the .00457 accordingly. You would need to do: .265 / 57.3 = .004625. It's the 57-magic. You can do the same with .265 in the angle-A box and 1 in the edge-b box.
Lest you get the wrong idea, the match doesn't necessarily mean that the moon passed through the shadow where it was 5,998 miles wide, yet this method is no small item for the solar-distance task at hand. In the math, you saw two things: 1) M2-M angle / M-E angle to find the difference between the triangle lengths; 2) the multiplication of that difference by the length of M2-M to find the M-E length. Why should the angles of two triangles be related to their lengths? Because they all want to discover astronomers as goons seeking respect for fraudulent math. An isosceles triangle has half the angle when twice as long (providing the base lines of the two are equal). We can be thankful that it's so simply, yet, caution, the math can get confusing. Find the angle difference between two triangles is an expression of their lengths. For example, when the M2-M triangle has a tip angle 2 times (or any number) larger than the tip angle of the E-M triangle, the situation (finds an unknown length) is: M2-M length x 2 = E-M length. Easy enough. In reverse, it's M-E length / 2 = M2-M length. BUT we must not do M2-M length / 2, or M-E length x 2.
If we are working with the angles alone to find the lengths, the best we can do is find the relative lengths, and, for the math, E-M now becomes the smaller of the two numbers: E-M angle x 2 = M2-M angle, or we can do M2-M / 2 = M-E angle. Where we have only the angles and wish to find the difference between them, it's M2-M angle / E-M. Once you have the difference between them by that method, you can find the length of one if you have the length of the other. Your drawing can't be of help until we first find the length of line M2. Once we have that, we can find the length of E-M providing that we have the latter's correct angle. NASA has the information to acquire the correct angle, yet it's clearly lying to the world to cover it up because it's as easy as you just saw it to be to find the true solar distance with the correct angles.
There is even a question on whether NASA's lunar and solar sizes are correctly entered on the eclipse pages with a central-path moon, but for every erroneous thing they add, more inconsistency results when one starts to fiddle with the numbers. A central-path moon through the shadow is their worst enemy because one can figure both a lunar velocity and a shadow diameter with it...unless NASA enters false data. Expect NASA and other goons to enter the wrong data. These people need to be jailed for a long time. NASA needs to be revamped, or excommunicated altogether.
Earlier, we saw some pretty neat things (in some whole numbers) when finding the solar diameter with a summer sun 5.2 million miles away. I had not yet done the following at the time. There is nothing written in stone above to prove a summer sun at 5.2 million miles (it all depends on what the true figure should be for the 6,000 scenario), but what follows makes it more compelling.
It's time to convert to the average sun size. First, the average sun (.533) is 5.11 million miles away when the .5246-degree sun is at 5.2 million. Putting these numbers in the right-angle calculator gets a diameter of 47,612.5 miles. From the solar radius, 23,806 miles, we knock off 3,959 miles to get 19,847 miles of rise (for line E-M) per distance of 5.11 million miles. That works out to 57.3 / 5.11 million x 19,847 = .2225 degree rise. Be on the look-out for a .2221-degree rise as per reducing the solar diameter to the magic number, 47,508.
When we feed the angle-A box .533 degree, with 47,612.5 in the edge-a box, the average solar distance of nearly 5.12 million is found. I don't have the brain power at the moment to find why it's not 5.11. This is an easy way to find any solar distance as per the angle that you use in the angle-A box (along with the solar diameter in the edge-a box).
http://www.cleavebooks.co.uk/scol/calrtri.htmAmazingly, 47,612 miles = 6.013 earth diameters of 7,918 miles, like Someone knew what He was doing. Plus, 47,612.5 miles = 22.06 (i.e. a virtual round number) moon diameters. Or, when the sun's diameter is exactly 6 earth diameters (= 7,918 x 6), it's 47,508, itself 22.002 times the lunar diameter of 2,159.2 miles. I've never seen a double-whammy like this from the same number. It could appear that the atheists actually have the diameters of the moon and earth both correctly figured, thank you. Besides, what are the chances that the moon and sun will look almost exactly the same size to us on earth? It looks like He set this up to reflect the Father and the Son.
Astronomy says that the moon is 400 times closer to us than the sun because it's 400 times smaller. They have the sun at 864,938 miles wide, which is 400.62 units of 2159 miles, but if, by their lunar-distance scheme, they have the moon 92.9 million / 238,700 = 389.2 times closer than the sun, this is wrong when the middle number between their smallest and largest sky moon is the average. The middle is .524417 degree, as compared to the average sky sun at .533112, making the sky moon .98369 the size of the sun (to get the full .533112, click the page's "significant figures" box). This difference suggests that the lunar distance should be based on .98369 of 400.62 = 394 times, not 389. If we divide their solar distance, their refined 92,955,807 miles, by 394.08, we find 236,633 miles as what they should be reporting as their average lunar distance (they may have a different definition of "average"). "The AU [distance to sun] has been defined as 149,597,870,700 meters (92,955,807 miles)."
Next, we do a method that gets roughly the same lunar distance. Let's read from Wikipedia: "the Moon's angular diameter can vary from 29.43 arc minutes at apogee to 33.5 arc minutes at perigee". When divided by 60, 29.43 arc-minutes is found as .4905 degree. According to these figures, we find the average moon size with, (29.43 + 33.5) / 2 = 31.465 = .524417 degree, and finally we divide .4905 by .524417 to find a difference of .9353 times between the two distances. The moon at the eclipse is said to have been 29.43 arc-minutes (same number quoted above, supposedly the smallest it could be = furthest distance it could be), and to show that NASA has wrong numbers for lunar distances, it has 406,199 kilometers = 252,400.4 miles the day before the eclipse. As the day before was on apogee, the only day that the moon can be as small as 29.43, it already suggests that something is wrong, one way or the other, with 29.43. We can multiply 252,400 by .9353 to find 236,070 miles as their (not mine) average lunar distance (albeit it depends on how they define "average"). But this is not their stated average lunar distance, and it's not the same as the 236,633 above, suggesting something a little out of place with their stated lunar distance for the moon at the eclipse.
If we use the 251,500 miles (instead of 252,400) that I'm using for the day after the eclipse, the math goes even smaller to 235,228 miles. Clearly, they were either lying to the masses on this matter of lunar distances for eclipses, or they were being deceptive about the average lunar distance.
They can correctly figure, with certainty, that the average sky moon is .98369 times the size of the average sky sun. Now, as there was reason to size the sun 22 times the lunar diameter, one can start with the assumption that the average lunar distance ought to be .98369 of 22 times closer than the sun. In other words, instead of dividing the solar distance by 22, we need to divide it by .98369 of 22, which is by 21.64. Which solar distance should I use in going forward? Should we use the 47,612.5 that was derived from the 6,000 figure? Or, as NASA's eclipse data won't let us know whether 6,000 is the correct figure, should we use the magic solar diameter, 47,508, 22.0026 times the lunar diameter of 2,159.2? Should we gamble on the idea that God used a solar diameter closer to 47,508 than 47,612.5? Let's go for it. In going for more precision, we may as well use .98369 x 22.0026 = 21.644. Instead of 57.3, we can use 57.295.
With 47,508 in the edge-a box and .533112 as the angle, the solar distance becomes 5,105,736, or 5.106 million miles. We divide the latter by 21.644 to get 235,896 miles as the tentative average. What follows is an example of a lunar eclipse (my creation) with the moon and sun at the two distances above, and with the lunar-eclipse lines spreading at .22211 degree, the number needed to get the 5.106 million distance. When we put half of 47,508 into the edge-a box along with .22211 degree in the angle-A box, the E-M triangle is 6.12758 million miles long, wherefore I've arranged the math to provide 6.12749 (close enough for today). It shows that the scenario (and all the numbers) corresponds to a solar diameter of 47,508 miles. I apologize for the large decimal places, but I wanted to get much precision. The .266556 figure is half of .533112, and is used to find .266556 / 57.295 = .004652 mile spread per mile toward the sun (you can verify this with .266556 in the angle-A box and 1 in the edge-b box). (I don't recall how I got the .533112, but one can use .533 and .2665 alone.) We don't need to find the angle of the solar lines from a solar eclipse; we can just use the size of the sky sun in that way, using whatever size the sun was at any lunar eclipse. The (916 / 235,896) below is the spread of the lunar-eclipse lines, and the (.004652 - (916 / 235,896)) is part of finding when the two sets of lines meet at the sun:
With a shadow path of 6,089.05 miles, 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 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)I'll need to assume that my 1.0217 million figure for the shadow length is wrong due to decimal shortfalls, and go with the calculator's 1.0213, which is exactly 1.021263.6. When we divide the latter by my 5.10575 million (th M2-M triangle length) it turns out to be 4.9995. Look at that fat, round number. It's saying that the earth shadow from an average-distanced sun is 5 times longer than the distance between earth and the average-distanced sun, no matter where the moon is going through the shadow. Aside from the moon in the picture, the paragraph above makes the shadow lines at .2211 degree, and they reach the edge of the sun of 47,508 miles wide and 5.10575 million distant. Thus, with the magic number in use, not only is it 6.0 times the earth diameter, and virtually 22.0 times the moon's diameter, but the shadow is now found to be 5 times the distance from the earth and the sun. It's as though a Designer done it. Here's the calculator if you'd like it:
http://www.cleavebooks.co.uk/scol/calrtri.htm I would suggest that the true lunar diameter is 47,508 / 22.0 = 2,159.45 miles. I cannot find more than one decimalized lunar diameter, reporting in at 2,159.2.The 235,896 was not merely in the scenario, which I'll call the 6,089 scenario, just because it fit the rest of the math. It was in there as per a comparison between the sky moon and the sky sun. The 235,896 number is what determined the 6,089 number, and the two together created the .22211 degree angle, which was exactly the angle needed for the 47,508 scenario. Even before the 6,089 scenario was conceived, it was apparent that the flip side of the 235,896 coin was a solar diameter 22 times the lunar diameter.
This all started with 6,000 miles for the path of the moon in the eclipse; it's what got us on the 5.2 million in the first place, which is what gave the 5.11 as the average solar distance. If 5.2 million is the furthest the sun can get, the average is expected to be 5.2 x .98366 = 5.115 million, because 92,955,807 / 94,500,000 = .98366. The math to find the average distance is simply this: true solar distance at its furthest x (92.955 / 94.5)...correct if the astronomers have their distances to scale with their angular sizes, and I think they must.
I have no basis for a sun 5.11 million at its average, aside from the 6,000 figure, or whatever else it might be. To put it another way, we can find the solar distance by first finding the average lunar distance of 235,896. Secondly, we need to know the diameter of the umbra at the eclipse, and thus far my reasoning is that it should be near 6,000 miles. I was viewing the 6,000 as the minimum, expecting it to be larger. If it's made larger, the 6,089 figure needs to be larger too, spoiling the magic-number scenario eventually. Here are the two scenario's side-by-side for a comparison:
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. It means my math checks out.
http://www.cleavebooks.co.uk/scol/calrtri.htm
With a shadow path of 6,089.05 miles, the M-E shadow-line spread is 914.475 miles over 235,896 miles. With 914.475 in the edge-a box along with 235,896 in the edge-b box, it gets .222112 degree. With the solar radius at .266556 degree on its average, it's a difference (between the two triangles) of .266556 / .222112 = 1.20009. 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 1.20009 above to find that the E-M triangle is now 6.1274 million miles long, giving an earth shadow of 6.1274 - 5.10575 = 1.0216 million miles, and the right-angle calculator has it at 1.0225 million with .222112 degree in the angle-A box (use 3959 in the edge-a box)I apologize for using .5246 in the first scenario. I had that sun mixed up with one in a solar eclipse. The number should be .52455 for better accuracy. It might explain why the shadow lengths do not match.
The scenario where the moon passes through a 5,998 mile stretch has both triangles significantly longer, yet the M-E versus M2-M triangles share a difference of 1.2 times (1.199 versus 1.2)...showing that the 1.2 ratio maintains itself on both sides of summer i.e. spring and fall (the 6,089 scenario is at the average solar distance, which occurs in both April and October). If the 5,998 figure was never used, the math would have used the 6,208-mile shadow diameter, and the 1.2 ratio would have been 1.35 instead (longer shadow). If the winter ratio is likewise around 1.2, then it seems certain that the 6,208 figure is not correct. I haven't done any winter calculations, but it seems that, if the 1.2 ratio is maintained from April through to October, it should stay roughly the same through winter.
The difference in shadow size between 1.05 and 1.02 million is 30,000 miles. It means that while the shadow grew 30,000 miles between the two scenario's, yet the ratio stayed the same, meaning that the M2-M triangle, extending from earth to sun, grew by 30,000 x 5 = 150,000 miles, because that triangle is five times longer. You see, if both grew by the same percentage, that's when the ratio maintains itself. Can we prove that the triangle grew by 150,000 miles? There is something wrong because it was established that the one scenario was at 5.2 million miles, and the other at 5.1, a difference of only 100,000 miles. If the shadow size were 1.4 million, this would be a match (100,000 on both sides). I'll deal with the need for a change in the 6,000 scenario in the next chapter, where you will see that 1.04 is the correct figure. The difference in shadow size between 1.04 and 1.02 million is 20,000 miles, meaning also that the M2-M triangle grew by 20,000 x 5 = 100,000 miles.
One thing seems certain, that if this discussion were about 6,086 verses 6,208, a large problem develops because 6,208 is in the other way (larger shadow) from 6,089, than is 5,998. It seems impossible, you see. The 5,998 mile width is of course further from the earth than the 6,089 spot, but a width 6,208 miles is closer to the earth than the 6,089 width. The 6,089 scenario is with a sun and moon closer to earth than the 6,000 scenario, and as the sun becomes further for the 6,000 scenario, the shadow grows wider so that the shadow at 5,998 miles of width must definitely be further than where the 6,089 width will be. However, as the 6,089 figure was not from an actual eclipse, I have decided to remain open to the possibility of a larger umbra diameter for the 2000 eclipse, for as yet I have not nailed down the length of the umbra for this eclipse.
The difference of 91.05 miles between the two shadow paths (of 5,998 versus 6,089.05 miles) is quite expected seeing that the two paths were about 16,000 miles apart. The difference is 252,050 - 235,896 = 15,604 miles. A decrease of 91.05 miles of shadow width is split 45.525 for the line E and 45.525 for line M. Each shadow line (of the cone) must drop, as they extend toward the tip, by 45.525 miles after the moon has changed positions along the shadow,15,604 miles closer to the tip than where it started in the 6,089-mile scenario. The line E at .22211 degree drops by .22211 / 57.295 = .003877 mile per mile away from the earth, which we multiply by 15,604 to find a drop of 60.496 miles over that span. However, as the situation evolves to the 6,000-mile scenario, the shadow grows wider because the sun gets further away; this is the process that changes line E from .22211 degree to .218 degree. The latter angle is less steep, and therefore increases the shadow width (outward direction). In the competition between the two opposite factors, the inward direction (full capacity 60.5) wins by 45.525 miles, suggesting that the sun's contribution to the competition is 14.972 (60.496 - 45.525) in the opposing direction.
Can it be proven that the sun's part was a mere increase of the shadow by 14.975 miles on each side? In this process of increasing the shadow by 30,000 (or is it 20,000?) miles in length, we are to view the lines at .22211 degree to begin with (sun at average distance and moon at 235,896 miles), but evolving to .2187 degree (to be exact, using the right-angle calculator) -- a difference of .00341 degree -- when the moon reaches 251,500 miles from earth. How far up will the end of a line go, 251,000 miles out, if it increases by .00341 degree? Ask the 57 method: .00341 x (251,000 / 57.295) = 14.968 miles (the right-angle calculator gets 14.968), the figure (virtually) at the start of this paragraph).
The next chapter will show the shadow distances of their 93-million-mile scheme, with the longest less than 885,000 miles. The reason that my scheme gets shadows more than a million miles long is that I use E-M angles smaller than they do (allows a longer shadow cone). In order for them to have larger angles for E-M, they need to artificially increase the solar diameter beyond the lunar-eclipse lines.
To find my average distance using astronomy's angular sizes for the sun, we need to find what they use as the smallest versus average angular sizes. Google doesn't make a search for this easy to find. The magic number (in and around 47,508) requires best in the range of 5.1 for average distance. We could use the furthest distance of the sun, but this is not easy to find with Google unless you can remember the word aphelion: "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)." These numbers are supposed to be directly proportional with the angular sizes.
To check their figures: see if they match their angular sizes. I don't recall coming across an article giving these sizes. Usually, we only get the .533 angular size. On the day of the July 16 eclipse, the sun, as reported by the eclipse page, was 15'42.2" = .5246 degree, and nearly at aphelion. Believe it or not, by googling "aphelion angular size," Google brings up the Flat Earth Society first of all. I can see no one on the page giving the angular size for aphelion aside from the Flat Earth Society, which has it at 31'36" = .5227 degree. Should we trust the flat earthers? I'm just wondering where they got that figure.
There is an angular-size calculator. Angular size is taken from one spot, such as the eye, to the outer edges of a body. When feeding 94.5 million along with 864,938 (their solar diameter), the angular size is given as .5244, and so this must be their figure for smallest sky sun. The .5246 for the June solar eclipse, slightly larger, seems about right, since it occurred two weeks before aphelion. We can use .5244 in the angular-size calculator along with my solar diameter of 47,508 miles, finding a solar distance of 5.19 million, meaning that there is correlation between my system and theirs. At .533 and 47,508, the calculator gives 5.106 million.
It may seem unscientific to go with a solar distance that corresponds to a solar diameter 22 times the lunar diameter while also being 6 times the earth diameter, but the rest of the math outlook, which is scientific, came very close to the magic number. That's how it was found. This makes the magic number compelling. It also tends to verify that they have the lunar diameter very-closely correct, which enables us to size lunar shadows obtained from eclipse data. Having the lunar diameter correct also allows us to find the true solar diameter. If their average lunar distance is not the way in which we would understand "average," it doesn't necessarily mean that their lunar distances, as reported in eclipse data, is wrong.
One way or the other, and in more ways than one, they are lying about this particular eclipse to cover something. Their July-of-2018 eclipse, at apogee, has a moon a tiny-bit further than 252,400 miles, and the moon there is said to be 29.423 arc-minutes, smaller than the smallest one at the Wikipedia statement above. They need to get their facts straight; if they keep changing things, people are bound to notice. If they keep changing things, maybe they should never call them facts. They have 406,222 kilometers for the lunar distance of this 2018 eclipse, which are 406,222 x .62137 = 252,415 miles. This online kilometer converter says that 1 kilometer = .612137 mile.
http://www.unitconversion.org/length/kilometers-to-miles-conversion.html
