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December 14 - 20, 2021

My Quest For the True Size of the Solar System
It Needs Tweaking
Anyone Can Work It Out

If you're waiting for Jesus to return, see Post-Tribulation Rapture

I didn't do much heraldry this week, for a change. Instead, I posed a way for anyone to discover the true distance of the sun based on work I started more than 15 years ago. I felt in my bones that the solar distance taught to all of us as unquestionable fact was deliberately erroneous. I'll tell you why later. This week, I had a hard-set realization that allowed a calculation of the solar distance in combination with publicized NASA data. I've been thinking not to share this work yet in case I've made a mistake, but on second thought I'll just warn that I may have made a mistake. My cheap pocket calculator at times seems to screw up when it doesn't get enough light. This work has been about ten days full-time in the making.

The realization, which has been like a blind spot for me all these years, is that we don't need an actual solar eclipse, after all, to find solar distance, because we can make up our own, anyone at all, with the moon at any distance from the earth, as long as: 1) the sun is the same distance from earth as per in a lunar eclipse that's used for the math; 2) the lunar-shadow angle in our make-believe / on-paper solar eclipse can be found out.

There is a 99.9-percent chance that most readers will not read deep into this section, and so I'm repeating here the surprise found late in this work, on Saturday of last week:

Below is a lunar eclipse of June 15, 2011, in which the moon passes through the shadow's equator. U1, U2, U3, and U4 are all given, which therefore gives a lunar velocity. The claim on the page is that one lunar diameter of travel, between U1 and U2, took 28 seconds shy of one hour (59.0088 minutes), meaning that the moon is said to be traveling a tad faster than 2,160 miles per hour, or 2,159.1 / 59.0088 x 60 = 2,195.367 mph to be precise. The time between U1 and U3 is given as 2 hours, 39 minutes and 45 seconds, or 2.6625 hours. This route, 2.6625 x 2,195.367 = 5,845 miles, is NOT almost an entire shadow width.

The page gives the shadow radius as .7256", and for the moon radius (SD) of 00 15 57.2, I convert it to .2659 degree at the onlineconversion page above, which makes this shadow .7256 / .2659 = 2.729 times larger than the moon, wherefore the shadow diameter works out to 2.729 x 2,160 = 5,895 miles. In case you didn't catch it, there's about a 50-mile difference between the U1-U3 distance and this umbra diameter. How can that be correct?

[Insert -- Before going on, you should know that I've found a much-better, if not perfect number, to replace the 252,650 in all calculations below. On December 31, this was written in the last update of this month:

The previous method used the formula, .49 / lunar size X 252,650 [it's in the paragraph below this insert]. It wasn't perfect because the 252,650 wasn't likely the correct furthest distance, but we can now try that formula using the lunar distance, 252,464, obtained above by the easy-pi method: .49 / .5418332 x 252,464 = 228,313, exactly the 228,313 obtained in the eclipse calculation!! It proves that the method is accurate.

It means that:

360 / .5418332 x 2159.1 / 2pi) = .49 / .5418332 x 252,464;
360 / .52176 x 2159.1 / 2pi) = .49 / .5176 x 252,464;
etc., using any lunar angle whatsoever between .49 and .558

It's important to note, though, that if the furthest-possible lunar distance has an angular size smaller than .49, the 252,464 above will go up toward 252,650. End insert]

First, let's give the angle of NASA's lunar-eclipse line by first finding the lunar distance at the eclipse. As the moon at that eclipse has an apparent diameter of .2659 x 2 = .5318, the math to find its distance is: .49 / .5318 x 252,650 = 232,791 miles. We put this latter number into box, b, of the triangle calculator. For box, a, we do: 3,960 - 5,895/2 = 1012.5 miles of lunar-eclipse-line spread. With the latter number in box, a, the eclipse line's angle [press "calculate"] is .249 degree. It looks down-to-earth.

Next, we want the angle to the edge of that sun, from the standpoint of a person standing on earth nearest the sun. For this we need the size of the sun at the eclipse, which is said to have a radius of 00 15 44.7, which converts to .2624 degree, and then that's doubled to .5248 degree (about the smallest the sun ever gets) to get the diameter. So, we put a moon of that .5248-degree size into the sky for our make-believe solar eclipse, and to get its angle, we need to know its distance from earth: .49 / .5248 x 252,650 = 235,897 miles. We put the latter into box, b, and for box, a, we use the lunar radius of 1,080, to find an angle of .262 degree, and it happens to be larger than the .249 above for the lunar-eclipse angle, so this is good.

The good news is, we can actually do a solar-distance calculation now that the lunar-eclipse line is smaller than the solar-eclipse line. For the lunar-eclipse line, we found a spread of 1,012.5 miles over a distance of 232,791 miles toward the sun, and to find the spread per 1 mile toward the sun, we do: 1012.5 / 232,791 = .00435 mile. For the solar-eclipse line's 1,080-mile spread per 235,897 miles toward the sun, it's 1080 / 235,897 = .004578 mile of spread per 1 mile toward the sun. It means that the solar-eclipse line catches up to the lunar-eclipse line by .004578 - .00435 = .000228 mile, per one mile toward the sun. As the two lines begin 3,960 miles apart, it will take the following number of miles for the lines to meet: 3,960 / .000228 = 17.37 million miles. If NASA's data for it's lunar eclipse of June, 2011 is accurate, it appears that the sun is 17.37 million miles away.

Science Task For Time on Your Hands

I'm going to show you how you can figure out the distance to the sun, but in order to do this, you need to make it your task. This is not for pleasure reading. This is for the one who wants to get the true data from sources whom I think have long been hiding true numbers so that the true distance cannot be found. I don't think they publicize the correct distance. I was able to glean that they are as much as 10-20 times too far into space.

Even before it dawned on me that the correct distance can be found easily by knowing the true angle of a solar-eclipse line, I had suspected that evolutionists made the stellar system much larger than it is because they had wanted the stars as far distant as possible to play into their scheme of a gigantic universe large enough to fit stars moving explosively apart for some 15 billion years since the "big bang."

We are not talking true scientists here, but anti-God fools who use science to advance any little piece of data that can play to their big-bang view while hiding evidence that plays against it. They are just as insidious, controlling and falsifying as power-hungry anti-Christ politicians. The two are of the same cloth, miseducating the world by the force of their powers over the political arenas. They told us that stars are moving further apart as part of their proof that a big bang took place, but, duh, the big bang was not in the center of our galaxy, and so why should stars of this galaxy be moving apart?

Part or all of the reality of stars moving apart is that they repel one another constantly, and thus from this picture one expects them to be accelerating away from the galactic center, for when the force of an electromagnetic charge is held constantly to something repelled, it accelerates away i.e. stars don't move away constantly at the same speed, but ever-faster with time. It appears that after God created the stars, they sailed away from each other, though in some cases, stars may be headed somewhat toward the center of the galaxy due to some stars clustered in the outer parts so as to drive others inward to a degree. But, generally, the collective stars are expected to participate in expanding the outer boundary of the galactic system.

The reason stars repel one another as a fact is because they are heavily filled with free electrons, all having innate negative charge. To combat this idea, the NASAites "educated" us on the substances found in the solar wind, telling us that it's filled half with free electrons and half with protons, giving the impression that stars are spewing out both, and thus that stars are neutral in their electric charge due to the positive protons off-setting the charge of the free electrons. The goons even have the audacity to claim/teach that the positive charge of one gigantic proton is exactly equal to the negative charge of one tiny electron. STUPID, but this serves their big-bang lie, and their entire system was built on that lie. You need to get this.

This neutral universe was their wish so that no one could argue that stars were spreading out due to inter-repulsion. So, they lied. You need to get this: NASA LIES. There are no positively-charged protons in the solar wind because bare-naked protons naturally attract electrons and thus become atoms. The idea of bare-naked protons in space is an obvious lie. Once protons become atoms, they have only a miniscule amount of electromagnetic charge, if any at all.

Once you understand their cosmic needs in supporting the big bang, you can better guess why they frame the cosmos in ways that support their big bang, and there is every logic that they would have made the solar system much larger than it really is just because they could get away with lying, for nobody else could produce evidence against their establishment that could stand the backlashes of the establishment. They controlled the cosmic "knowledge." I have never heard of them discovering the distance to the sun using eclipse data, and this is the easiest, most-reliable means. I suspect that I have never heard of it because it betrays their claim for a sun 93 million miles away.

If you keep a notion that astronomers and cosmologists are respectable, honest people, you're their very useful idiot. Time to break free from that mold, or you'll grow hairs like those of Einstein, the goofball whom the establishment loves because he was a useful stooge for them, first by inventing the fallacious photon. What better thing could prove a gigantic universe than a particle traveling at the impossible speed of 186,000 miles per hour? Forget it, nothing can be propelled that fast because there's nothing to do the propelling. Even if they say that the big bang did the propelling, it's vanity because the scientific fact is that all objects making collisions with ANYTHING else slow down. It is a fantasy of these lying pigs that photons always move at the same, fantastic speed. Just follow the logic. They are to be despised because they war with God; they are not the innocent by-standers who get damaged by the missiles of war-hogs; they are the war-hogs, and they war against their Creator.

Light does act at roughly 186,000 miles per second, but that's not the same as a particle traveling at that speed. Light waves can act at 186,000 mps where the light particles involved, of the aether, are moving only hundreds or thousands of miles per hour. Einstein did away with the light-wave aether, replacing it with the photon, because he did not understand the aether. He was even working heavily with electrons issuing from atoms, yet did not realize that the aether consists of free electrons filling our atmosphere, and filling the cosmos too. Electrons issue from the atoms of stars, and inter-repel one another into space, filling space, and he of all people should have known this. Yes, these electrons enter our atmosphere, and he should have known this. Don't let the evolutionist swine convince you otherwise. Never respect a God-killer swine.

Mr. Einstein, if you witnessed light from the sun causing electrons to bounce away from atoms on your desk, then don't you think that light waves AT the sun are going to free electrons altogether from atoms far-more powerfully? DUH. The aether you denied the world was under your very eyes in your very own experiments. Mr. Electric Hair, you became one of those wise-in-his-own-eyes fools, and like-minded fools follow your photon science to this day. Orbiting electrons? Are you nuts?

Whenever two particles collide, they give up energy to one another. The energy received from the collision acts exactly to slow them down. It's a fact of science, a law of physics. If two objects, A and B, moving at x force collide head on, A will transfer x force to B so that B's forward motion is wiped out, because x minus x = 0. It's that simple. And B wipes out the energy of motion of A (where bounce physics doesn't exist). Everytime an object strikes another, even if bounce is part of the collisions, it's energy of motion is absorbed by the struck surface and therefore motion diminishes. Energy is not "destroyed," but is used up. They fool you into thinking that the impossibility of destroying energy means that two or more objects retain their total velocities after collisions. Therefore, the idea that all colliding particles, including atoms, remain at non-decelerating speeds, even in pure darkness, is the fantasy of the anti-Christ pigs who rule the scientific establishments. Let's get our facts straight, shall we, before we pose as the most-intelligent wisemen amongst humanity.

I'm going to make a drawing for you to show how to calculate the solar distance. To make it shorter than following my directions, go to the two eclipse drawings on this page, and draw them on your paper but one superimposed on the other, and don't have the sun in your drawing so that the earth-moon region is large.

If they have it correctly pegged at 93 million miles, I'm fine with that. But I'd like to do the calculation myself, though cannot do so because I don't have access to the needed reliable data: the diameter of the earth's shadow/umbra, with certainty, at the point where the moon passes through it. If you can find a way to access that information that's not on some NASA chart made for the public, then that's the ticket.

For your drawing, put your page vertical on the desk. Draw a half-inch circle (the moon) a quarter way up the page from the bottom, followed by a 2-inch circle (the earth) half-way up the page, followed by another half-inch circle (the moon) three quarters up the page. The sun is off the top of the page. The lower moon is under a lunar eclipse, and the top moon causes a solar eclipse about six months later, but you can visualize them taking place simultaneously. The sun should be roughly the same distance for both eclipses, and the solar distance (relative, not actual) is defined here as the sun's apparent size in the sky. We are not striving for perfect accuracy in actual solar distance; we can be two-percent off, no problem.

The tip of the lunar shadow is at times above the earth in a solar eclipse, and at times the shadow touches the earth's surface. For our drawing, we can use a shadow tip kissing the earth's surface, with a shadow one foot wide. In other words, draw a line from the top-center of your earth circle, up and out across the left edge of the top-moon circle. This line goes to the edge of the sun. You don't need to draw a second line from the top-center of the earth circle across the right side of the moon to the opposite side of the sun, but if you'd like to draw it for better visualization of the shadow, go ahead.

Your first task is to figure out the angle of one of the lines (both will be at the same angle). We can be approximate. We can make the moon 238,850 miles away, trusting that astronomy has that figure correctly figured for the moon's median distance. We can assume they have the lunar diameter correctly figured at 2,160 miles wide. Go ahead and write those numbers on your drawing if you wish.

The angle of zero degrees is an imaginary, vertical line (draw it if you wish) through the center of the lower moon, through to the center of the upper moon and of course through the center of the sun. All shadow-line angles are taken off of this zero degrees, and as such all shadow-line angles are very small, near zero degree (they are very large, almost 90 degrees, if taken from a horizontal line through the center of your sun).

The tip of the lunar shadow touches the zero-degree line, and the solar-eclipse line both moves horizontally/latterly on the page and vertically/upward toward the sun. As the zero-degree line splits the moon in half, the angle of this line can be expressed as: half of 2160 miles toward the horizontal per 234,620 miles toward the sun. Yes, that is a correct way to figure the angle of a line. You cannot use a protractor unless you draw this drawing to scale, but you can't do it well on a single piece of paper (moon needs to be 30 times the earth-circle away, if to scale). We want more accuracy here than a protractor can offer.

The lunar distance of 238,850 miles is from core to core of both bodies, but as the eclipse/shadow line begins at the earth's surface rather than the core, the line span to the moon is 3,960 miles -- one earth radius -- lower than the core-to-core distance, which is why you see "234,620" in the paragraph above.

You should be able to easily understand that the solar-eclipse line has moved half of 2,160 miles leftward of the zero-degree line by the time the eclipse line reaches the moon 234,620 miles away. t means that we can have the correct angle of this line even without a right-angle calculator. We can find the distance to the sun by the angle expression above that's not expressed in degrees: x miles laterally per y miles toward the sun is the accurate definition of an eclipse-line angle, and more useful than merely having the degrees of the lines.

Next, we work on the lower moon. You need to draw the earth's shadow between the earth-circle and the lower moon. Draw a line from the left edge of the earth circle toward the moon but let it cross above the lunar surface by 1.5 lunar diameters distant from the core of the moon. An identical line from the right edge of the earth will draw your earth shadow in full. Your moon is buried in the shadow. The shadow should be 1.5 inches wide where the moon goes through it. Extend this line past the earth toward the sun, and call it the lunar-eclipse line, not to be confused with "lunar-shadow" line, for the latter belongs to the solar eclipse.

Look at your page, buddy. What do you see? This is explosive, for NOBODY has ever told you what you're looking at because it's all under wraps. You have a solar-eclipse line going to the left edge of the sun, and you have a lunar-eclipse line going to the same edge. Therefore, get ready, here it comes: where the two lines meet, that's the distance to the sun. It can't get any easier.

All we need, to figure the correct solar distance, is the correct angle of both lines when the sun is at the same distance from earth for both eclipses. It can't get any easier because NASA is supposed to have all the data we need to do the math. But why doesn't anyone mention this method of measuring the solar distance? I'm suspicious.

Ho-ho-ho, we have a big surprise here: we don't need NASA to figure the distance to the sun. We only need to find the true diameter of the earth's shadow at the distance of the moon during an eclipse. That will give us the angle of the lunar-eclipse line (it can be called the earth-shadow line, same line), and we need that angle to figure the point in space where it meets the solar-eclipse line (it can be called the lunar-shadow line, same line). We need the correct angle of both lines, that's all we need...that we don't yet have because NASA is, in my opinion, prone to lying about the width of the earth's shadow.

Instead of telling us the shadow facts by observations in the real world, NASA tells what the earth-shadow dimensions work out to by the math only, with a sun already planted, in the math, at 93 million miles away. NASA works backward with the geometry, starting from the sun, to produce the earth-shadow line, on the premise that the sun is 93 million miles away. We therefore CANNOT use their shadow-length figure, or their shadow-width figure at a lunar eclipse, to discover the solar distance.

How can we find the true diameter of the earth shadow at the distance of the moon? Easy. Measure the time that it takes the moon to traverse the shadow, and multiply by the moon's velocity. I've got news for you: we can multiply all on our own, no need for NASA to do it for us. If, for example, it takes three hours to make it through the shadow, and the moon is moving 2,000 miles per hour, the shadow works out to 6,000 miles wide. We don't need NASA to give us the details, but we do need access to a near-full lunar eclipse with all the correct data provided to the public. And that's the kicker: our star lords are corrupt and anti-Christ / anti-Creator.

My hunch is that, instead of measuring the time it takes the moon to pass through the shadow using a stop watch, they will calculate it with a computer program, but based on a sun plugged into the computer program at 93-million miles off. And so this program draws the earth shadow at certain dimensions, and the same program also "measures" the time taken for the moon to pass through it. Naturally, math applied to the shadow numbers will "prove" that the sun is 93-million miles off.

To obtain the correct lunar-eclipse-line angle, we need the true speed of the moon at the time of a near-full lunar eclipse. We can figure this out on our own by a couple of methods, one way being to trust what NASA claims the velocity to be, but my bet is that the lunar velocity will not be included in eclipse data shared with the public.

The furthest and nearest lunar distances has changed over the decades, suggesting that the cosmic peeps have not been the wizards they pretend to be on lunar distances. I have a Pocket Reference book by Thomas J Glover, published in early 1999, having the furthest lunar distance of 252,737 miles, and the nearest at 221,469 miles, and I assume he obtained the numbers from NASA. Going online now to find those distances, they are different. Why?

Glover's book has the moon's orbital velocity at .64 miles per second, and we assume this is the average velocity. It works out to 2,304 miles per hour. Is this correct? It depends on how they calculated the distance to the moon, for that velocity is directly based on the moon's distance from earth. The further the moon, the higher its speed because it needs to fulfill one orbit per 27.32 days. Having the correct lunar velocity at an eclipse is key to finding the solar distance by the eclipse-line method, and don't the evolutionist goons know this? Have they possibly fudged the lunar-distance numbers to protect their crimes?

If we can't trust NASA's lunar-velocity figure at the time of an eclipse, the way to find it is by measuring the time it takes from when the moon first kisses the shadow, to when the moon has completely disappeared into the shadow. The moon has moved one full lunar diameter. This method works only if they are correct with a lunar diameter of 2,160 miles. If it takes an hour to disappear into the shadow, the moon is traveling at 1 x 2160 = 2160 mph. Conversely, we do 2160 / 2160 to get 1 hour, or, if the moon is traveling at 2,300 mph hour, we do 2160 / 2300 = .94 hour per travel of one lunar diameter.

It's hard to believe that astronomers around the world, for well over a century, would allow the establishment to publish wrong lunar-velocity figures for lunar eclipses when any amateur astronomer can measure the time for the moon travel of one lunar-diameter distance. If the establishment cannot hide the true moon velocity for eclipses, then neither should it be able to dish out inaccurate earth-shadow widths for where the moon passes through. This admittedly makes me hard-pressed to explain how NASA and others have publicized the wrong widths in spite of the many eyes who can easily prove them wrong. Still, when I looked into the NASA data to be found online, there were keys missing to allow the finding of the solar distance by the eclipse method, and that omission of data may be a deliberate cover-up. If you take up the same challenge and find resistance to finding the minor details you need, you too might get suspicious. We are talking minor details.

If we know the distance of the moon at any eclipse, we can know its approximate velocity if we have tables that expose the moon's velocities at those same distances. The lunar velocity changes through its orbit, swinging faster around the earth when it's nearest the earth, and slowing down afterward. It's slowest at apogee (furthest from the earth). Therefore, the velocity is nearly the same, every orbit, per same lunar distance, and this knowledge can be helpful when it comes to finding the lunar velocity at an eclipse.

The best way to find lunar velocity is to time your own eclipse. Barring that, you need to find someone who timed it manually, not by computerized computations where 93,000,000 is part of the computations. It doesn't need to be a full eclipse, but it's helpful if more than half the moon goes dark because one can then measure more accurately the time passed per one lunar diameter.

I assume that the reported distances to the moon are accurate because triangulation seems like a sure way to measure. Triangulation would involve two telescopes upon the earth at good distances from one another, and that distance becomes the base of a triangle. They then measure the angle of both telescopes as they are aimed through the center of the moon at exactly the same time, which is enough information to calculate the true distance to the moon. If they don't calculate the lunar distance by this method, should we become suspicious of falsification? Should we try to find the lunar distances as reported before the technology existed to bounce light waves off the moon? Contrary to their assertions, light does not travel in space at the same speed as it travels on earth, for the aether in outer space is not nearly at the density as in a lab.

In fact, this is how we can know that their assertions of bouncing light waves off the moon, and receiving them back again, are bogus. I never did believe they could do this because the light bounced off of the lunar dust does not return as a laser beam, but gets spread into all directions equally, thus making the light very weak from the first inch off the moon. I cannot believe that this weakened light will be registered by sensors on the earth. They cannot shine a laser beam from earth and off the moon that's visible to the eye in a telescope, meaning that even the light of a laser pointed at the moon arrives weakly to it. It's been shown that voice transmissions of astronauts supposedly on the moon were faked, because in at least one instance, the voices responded to Houston too fast for the sound to arrive at the speed of light. They were tricking the world to believe the astronauts were on the moon.

Once the true lunar distance is found, they can find the lunar velocity with simple math. You with a small telescope can do this too: time the duration for the moon to travel across the sky a distance equal to its diameter, and then find how many diameters there are in that one lunar orbit. The average lunar orbit, they say, is about 1.508 million miles...per 27.32 days. I don't know whether 27.32 is an average versus the period for every orbit, but I've never seen the greatest versus shortest orbital periods, and so I assume all orbits are nearly 27.32 days. If the moon is 2,160 miles wide, there will be 1,508,000 / 2160 = 698 lunar diameters per one average lunar orbit. That's how to do the math. Covert 27.32 days into minutes or seconds, and use the time you counted for the travel of one lunar orbit to find the lunar velocity.

Calculator Calculations

Here's how to use the right-angle-triangle calculator to find the angle of lunar-eclipse line. Let's say for example's sake that you find the earth shadow, at a certain eclipse, to be 5,900 miles wide where the moon passes through it. Divide the figure in half to 2,950, and then subtract it from 3,960 (earth's radius) to get 1,010. This means that the lunar-eclipse line spreads / grows 1,010 miles starting from where the moon enters the shadow until the line reaches the edge of the earth. We can now form a right-angle triangle. At this right-angle triangle calculator, enter 1010 in box, a, and, for box, b, enter the moon's distance from earth (in miles of course) during the eclipse, then click the calculate button, and use the angle result on the alpha-angle line because it will match the angle of your lunar-eclipse line (on your drawing) as taken off of your zero-degree line.

The alpha-angle spot is the tip of the lunar shadow, one side of which is your zero-degree line, and the other side of which is the lunar-shadow line going to the sun. The drawing of the triangle at the right-angle-calculator page needs flipping and rotating to match the shadow triangle on your drawing.

If the lunar-shadow tip is above the earth's surface, the sun is larger than the moon, as the eye sees it. The average apparent size, called "angular diameter," of the sun is .533 degree, and the average apparent diameter of the moon is .518 degree, so, yes, when both sun and moon are at their median distances from earth, the sun looks a little larger. So long as the sun is larger than the moon, the lunar shadow will not strike the earth's surface.

God placed the moon at just the right size at just the right distance so that the shadow forms on earth at times, and He placed the sun at just the right size at just the right distance so that it can be exactly as large as the moon appears to us. Evolution has no brain to accomplish this. And God made the moon spin on its axis so perfectly that we never see the other side of the moon. One needs Intelligence to accomplish this, don't let the wicked evolutionists fool you. When you work with cosmic topics, have the wisdom to know that it's God's universe. Leave the evolutionists in the dust of their stupidities.

The smallest the sun looks to the eye is .524 degree, which is larger than the moon at its average distance. The shadow tip at that situation, and during an eclipse, is very close to the earth. However, this is not to say that a kiss-ground eclipse must happen when the sun is very distant, because a kiss-ground shadow can be had with a huge sun behind a huge moon too, for example when both are at the apparent size of .5418 (the largest for the sun), for the moon can get as large as .558 degree. The apparent sizes are directly proportional to the distances from earth.

Okay, but if a kiss-ground eclipse can take place when the moon is at many different distances, which lunar distance in particular should be used in the eclipse-method calculation for the solar distance? I don't think it matters. What matters is that, ideally, the sun for both the lunar and solar eclipse of choice is exactly the same distance from earth. That would be like having two identical moons simultaneously, one undergoing a lunar eclipse, and the other causing a solar eclipse. We would thereby get the two sets of eclipse lines with the sun at the same spot. Eureka, it's going to find the true solar distance. For two separate eclipses, "the same spot" translates to the same solar distance regardless of where the earth is in relation to the sun (the earth is at any one distance from the sun twice each orbit, and again twice for the next orbit, etc, because the orbit is elliptical).

The problem is, we aren't likely to find a lunar and solar eclipse with the sun at the same distance, and so there is a little extra work to do, but it can be done. The trick is to find the correct solar-eclipse-line angle by mathematically adjusting the line angle of a solar eclipse having a sun size as close as possible to the sun size for the lunar eclipse. If for example you use a lunar eclipse with a sun size of .528 degree, while the sun size for the solar eclipse is .53, and if we have found the eclipse-line angles for both eclipses, then we need to figure out what the solar-eclipse-line angle would be if the sun were .528 degree in distance rather than .53.

To convert apparent size to distance, a little math. Here's the angular diameters of the moon according to one page: .49, .518 (doubtful), .558. Those sizes represent the furthest (252,650 miles), the average (238,850 doubtful), and the closest (221,580 miles). When we divide .558 by .49, then multiply the result by 221,580, we get 252,330, not quite the 252,650, but close. In reverse, when we divide .49 by .558 and multiply the result by 252,650, we get 221,861. When we divide .558 by .518 and multiply the result by 221,580, we get almost 238,690, not quite 238,850, but close. The diameter numbers are not quite in sync with the distance numbers because different writers give slightly-different sets of numbers. The middle number between .49 and .558 is NOT .518, but more like .524. Someone else says the closest distance is 221,702 (why so "precise"?). ???

So, if you find an eclipse taking place with a moon's apparent size of .51 degree, do: .558 / .51 x 221,580 = a moon 434,470 miles away. This number goes into box, b, at the calculator page above, and then you need to fill box, a, to find the shadow-line angle (= alpha-angle at the page). For a lunar eclipse, you fill box, a, with the distance that the lunar-eclipse line spreads or rises from where the moon enters the shadow to where it meets the earth, and that distance should be in the ballpark of 1,010 to 1,060 miles. You don't want to know the alpha angle in degrees, however, but rather you want to know it, for example, as 1,010 miles of spread per 434,470 miles toward the sun. In degrees, that angle is .14 degree.

There's only two ways to know the length of the earth's shadow: 1) by first knowing the correct size and distance of the sun, or; 2) the eclipse-line method. You only have the latter choice. When they give the size of a shadow, they give it as a calculation based on their pre-perceived size of sun. My hunch has been that their calculations won't pass the eclipse-line test.

Someone says: "When the Moon is the farthest away, it's 252,088 miles away...When it's closest, the Moon is 225,623 miles away." Why don't those numbers match the 252,650 and 221,580 above? If the establishment throws out conflicting numbers, nobody will be able to promote reliable studies, and that's expected to be a tactic of an establishment (NASA especially) trying to hide its lies. Apogee and perigee occur in every lunar orbit, but they change from orbit to orbit, so don't confuse the apogee and perigee of a given orbit as the furthest-possible and nearest-possible lunar distances. I'm talking to amateurs like myself because anyone with a mediocre grade-11 education can find the solar distance using eclipse lines.

Someone says: "The Moon's umbral shadow is at most 267 km [166 miles] across on the Earth". That is, when the sun is furthest from earth and the moon is closest to earth, the shadow cast on the earth's surface is said to be 166 miles in diameter...according to this writer, anyway. At that time, the would-be tip (if the earth were not in the way) of the lunar shadow will be past the earth on its opposite side. We can find the distance between the shadowed surface of the earth and the would-be tip by using the triangle calculator above. To do it, we need the angle of the lunar-shadow line. The distance of the moon when it's nearest is about 221,600, and so we put that into box, b.

But for box, a, we don't want the full radius of the moon, but rather the radius (1080 miles) minus half of 166 miles = 997 miles. You're making a triangle here with the right-angle spot 83 miles away from the lunar core, for this gives you the angle of the shadow line. The shadow line spreads by 997 miles from the edge of the moon to outer edge of the shadow on the earth's surface. With those two numbers in boxes a and b, we find the angle to be .258 degree (if their numbers do not get a number smaller than .258 for an earth-shadow line, when the sun is at it's most distant from earth, they have given out a wrong earth-shadow length).

Now we can have the .258 angle in the alpha-angle box, along with 1080 (lunar radius) in box, a, to find how long this lunar shadow is (if the earth were not in the way). The calculator gives 239,841 miles. To find how far past the earth's surface this goes, we just subtract 239,840 from the 221,600 above, which finds 18,240 miles. I'm not trying to be accurate here; I'm sharing ways and things to help anyone who wants to dig into this task. I don't have much extra time, and with multiple people seeking the same thing, it stands a better chance of getting results. You may need to call up and press astronomy people to direct you to reliable data.

When one looks at a perfect solar eclipse in which the sun and moon are exactly the same size, the tip of the lunar shadow is at the eye i.e. on the earth's surface. The solar-eclipse line (it can be called the lunar-shadow line, same thing) at that time goes from the eye to the surface of the moon, and finally to the surface of the sun. Easy to grasp. But when the sun is larger than the moon during a solar eclipse, the tip of the shadow will be above the earth's surface, and there's no way to know exactly how far up unless you assume the sun is a certain distance at a certain diameter. No matter how far the sun, it's always going to be as much larger in diameter as compared to the distance between sun and moon. If the sun is 100 times further, it's 100 times larger in diameter.

The page below shows how the liars teach the people with a lie. This page tells how to calculate the length of the lunar shadow, but as part of the information for the calculation, the distance to the sun is used, meaning that the calculation is erroneous if the solar distance is erroneous. What we want is a better way to calculate the length of the lunar shadow, more reliable, and I gave you that other method above: using the distance to the moon. Why doesn't the page below use that method?

For the final solar-distance calculation, you will need to access more decimal points because the calculator page definitely rounds off (to three decimal points) so as NOT to give more decimal points that definitely do exist.

On your drawing, the sun can be anywhere between the two (left and right) solar-eclipse lines (you should have them extending out toward the sun) providing that the sun's edges touch the lines; look at your drawing and capture that in your thoughts. You don't need to figure out the size of the sun on a separate task once you've found it's distance; finding the latter provides the diameter. From earth, the look of things will be the same with a sun one million miles away as compared to 100 million miles, because no matter how far or near, the sun's edges are on the lines. That's what those two lines tell you. So, once you find the true solar distance, you can then figure out the solar diameter with one calculation at the calculator page because you will have line b and the alpha-angle figured out by then.

A solar ball 120,000 miles wide is one huge-hot ball, don't let the evolutionists scoff, saying it's too small. They lie all over the place with the sizes of stars and galaxies, for they hate God that much. Don't believe for a second that they operate on truth, for the establishment assures that the only ones who climb their ladders to the top spots are those both agreeable to, and passionate for, their established science. Anyone who starts to buck against their main tenets is demoted or removed on the premise that "everyone disagrees with you." Sure, everyone, because every "great" astronomer was groomed, formed and framed as a paid liar. They set up the minor schools in the same way. If you don't pass the exams in university, you don't get the degrees. You're brainwashed in this way into the error-stacked sciences (which is not to say that ALL sciences are stacked with falsifications).

Lookie: "Since Earth's diameter is 3.7 times the Moon's, the length of the planet's umbra is correspondingly 3.7 times that of the lunar umbra: roughly 1,400,000 km (870,000 mi)...Earth's shadow is as curved as the planet is, and its umbra extends 1,400,000 km (870,000 mi) into outer space." The Wikipedia writer didn't bother to tell what the solar distance is for a shadow that long. With teachers that omissitory, the student walks away with gaping holes in his education. Plus, the writer is absolutely wrong to claim that the lunar-shadow length is proportional to the earth shadow length. He's playing to a 93-million mile sun in which the earth-shadow and moon-shadow lines are close to the same angle, yet they are not at the same angle, and this fact means that the shadows are not proportional, for two proportional shadows must be at the same angle.

The earth is larger than the moon, and the sun is one, wherefore the shadow lines cannot be at the same angle. The wider the sphere/object at the same solar distance, the lower it's shadow angle. The writer is not being lazy to leave out the important details, but is guilty of misleading the reader. The earth's shadow is longer than 3.7 times the moon's shadow (3.7 is more-accurately 3.667).

EXTRA-EXTRA, Easy Calculation Method of Solar Distance

Let's go back to the solar eclipse in which the moon's shadow kisses the earth's surface. The solar-eclipse line spreads out by 1080 miles per lunar distance. Look at your drawing to understand this, because it denotes the angle of the line in non-degree form. It's not likely that you can calculate (I found a way but you might not) the distance to the sun unless you use this other method (not in degrees) to describe the angle of a line. Having the degrees alone of the line won't help you a hoot. I'll give you a clue, however: you need to know exactly how far laterally one degree spreads per unit distance along the protractor's zero-degree line. I once figured out the exact spread per one degree, but it's lost or buried in an old hard drive somewhere. But never mind that, you won't need it. Let me show you all you need to know.

A right-angle-triangle calculator cannot find the distance to the sun even if you have the correct lunar-eclipse angle. The right-angle triangle can be used for a solar-distance calculator only if the right-angle spot is placed at the center of the sun. Line-a then becomes the solar radius, and line-b becomes the distance between sun and shadow tip, but as we don't know either of the two, the right angle is useless for the task of finding solar distance. We need either the solar radius or the solar distance, and we know neither if that's what we're trying to discover.

If you save this page to your computer / thumb drive, you can come back to it if you need it when you take the task at hand seriously. The key to finding the solar distance is here, folks, in this short section, and so you may as well read it to get the secret into your head.

If the lunar-eclipse line likewise spreads out by 1080 miles per lunar distance, then the lunar-eclipse and solar-eclipse lines will be parallel and never meet, meaning that, in the real world, the two lines can never be parallel, for they need to meet at the sun. Nor can the lunar-eclipse line be spreading greater than the lunar-eclipse line, or, once again, they will not meet. Just for your information up-front: if the earth shadow is 5,840 miles wide where the moon passes through it, the earth-shadow line will spread by 1,080 miles per lunar distance regardless of the lunar-distance figure. This tells you roughly what the width of the earth shadow is where the moon passes through it.

Step one of two. We want to know how far from the zero degree line (on your drawing) an eclipse line has spread laterally after its moved one mile vertically toward the sun. For calculation-presenting purposes only, we entertain here an earth-shadow width of 6,000 that the two lunar-eclipse lines "climb" 1,000 miles each, laterally, to the edge of the earth that is itself 8,000 miles in diameter. In other words, the earth shadow is 6,000 miles wide to begin with, but when a line on either side spreads out 1,000 miles, the two lines will be 8,000 miles apart. We concentrate on one of these lines only. The math is as easy as 1000 miles divided by 238,850 = .00419 of a mile. We can call this the lateral number, the distance a line spreads outward laterally. We can now say that, for every mile toward the sun, a lunar-eclipse line spreads outward .00419 mile (this is not the fact, but only the conclusion as per the 6,000 number entertained).

We do the same math for the solar-eclipse line which was found to move 1080 miles from the zero-degree line per 234,850 miles toward the sun. The math is 1080 divided by 234,850 = .0046 of a mile. We can call this a lateral number too. The two lateral numbers, .00419 and .0046 are similar, but not identical. The solar-eclipse lateral number is larger, and for this reason a solar-eclipse line moves further away from the zero-degree line, per mile toward the sun, than does a lunar-eclipse line...meaning that the solar-eclipse line will catch up to the lunar-eclipse line eventually, and when they cross paths, bonanza, that's at the edge of the sun.

Step two of two. To find how much more the solar-eclipse line moves away from the zero line than the lunar-eclipse line; the math is grade six level once again: .0046 - .00419 = .00041 of a mile. This means that, per every one mile toward the sun, the solar-eclipse line moves toward the lunar-eclipse line by .00041 mile. Are you impressed?

We need to know only one thing more: how far apart the two lines are at the earth. The answer is, one half the earth diameter = 4,000 miles. Wow, can it really be this easy? Yes, we need only to calculate how many times .00041 mile goes into 4,000 miles, which is 4,000 divided by .00041 = 975,600. That it: after the lines go out toward the sun by 975,600 miles, they will have arrived to the sun together. The sun is under a million miles away using these numbers, meaning that the earth diameter ought to be significantly less than 6,000 miles wide...but more than 5,840 miles.

BIG POINT: the lunar-eclipse line's lateral number must be smaller than the lateral number of the solar-eclipse line when the sun is at the same distance from earth for both eclipses.

[Insert -- this section was written before realizing that we don't need a solar eclipse to find the solar distance. All we need is the would-be shadow length of the moon that undergoes the lunar eclipse, and so spare yourself the hassle of finding the most-appropriate solar eclipse: use the angle of the lunar eclipse's would-be lunar-shadow line for this calculation instead of the solar-eclipse line. The angle of a lunar-shadow line is simply the length of the shadow divided by 1,080 miles. That is, the angle is equal to a 1,080-miles lateral spread per length of the shadow. End insert]


I think I have grasped an important key that wasn't settled in my mind in the past...because I do not have astronomy smarts, which only goes to show how guilty astronomers are for not presenting to the public what I'm presenting here. If I can do this, why haven't they? The realization is, we do not need to find a suitable solar eclipse at all to find the solar distance. Once we have the lunar-eclipse-line angle, we take the size of the sun during that eclipse. We then create a make-believe / on-paper situation in which the sun is the same size in the sky. There you stand, looking out at this sun on a beautiful sunny day. You want to know the angle of a line from your eye to the edge of the sun. You can find it by holding a coin out to block the sun where the coin and sun are exactly the same size. The line through the center of the coin through to the center of the sun is zero degrees, and the line from your pupil to the edge of the coin to the edge of the sun is identical to a solar-eclipse line in which the moon and sun are of the same size. The moon is now your "coin."

You can find the coin teaching online, it's nothing new. They use it to show how the angle to the moon can be found, yet why don't they use it to express the angle to the sun too? Because, it's dangerous to them, I'm sure that's the reason. Why have I never seen anyone using the moon as the coin to find the solar angle? It's just such a natural teaching to be shared far and wide, and yet I have never seen it. Have you? So, yes, I think I'm correct, we can use the moon to find the angle to the edge of the sun for any-size sun. And it means we can use any lunar eclipse, total or otherwise, where NASA gives the correct width of the shadow, at the point where the moon passes through it, to find the true solar distance with easy-as-pie math that even you and I can do.

Once you know the distance of the coin from your eye, and the diameter of the coin, voila, you can have the lunar-eclipse-line angle to the sun's edge. But don't rush out to do this coin experiment, because we would do much better to place a larger object between our eye and the sun. We could use anything, including the moon because we know its diameter, and we can know its distance whenever its the same size as the sun! ZOWZERS. Why aren't drawings of this situation plastered all over the school textbooks? Hum?

If I had been an astronomer by profession, I would have known this a long time ago, yet I've never read from anyone this method of finding the solar distance, even if laid out just for fun with geometry. What are they hiding? Let me repeat this: when we find a lunar eclipse for which we can find its shadow length, we take the size of the sun at that eclipse, and that will alone be enough to give us a solar-eclipse-line angle to the sun even though no solar-eclipse data is being used for the angle measurement. We just stick a moon between our eyes and the sun at that distance. We then have both a lunar-eclipse line and a solar-eclipse line to the "same" sun (i.e. at the same distance). It coughs up the solar distance accurately, VERY ACCURATELY because the moon is a huge body, unlike a coin.

I didn't know until now (18th) that there was a near-total lunar eclipse on the 19th of last month. It was visible in all of North America. Maybe it's because I'm in canada, but when asking google for the particular angular diameter of the sun for the current time, or for last month, I get general angular-diameter pages for the high-schooler, nothing of what I ask for. google has become an anti-educational piece of trash. I could at one time find such things lickety-split. No more. No one is allowed to find the answer to, "what is the current angular diameter of the sun?" or "what is current distance to the sun?"...but I did find a page title: "Earth closest to sun on January 3-4, 2022 - EarthSky". It means that, at the eclipse last month, the sun was on the large size.

My bet is that google- and NASA-goons don't want people to know and research when the sun is closest / furthest in case it rubs global-warming claims, of the time, the wrong way.

I found this: "Perihelion in Montreal, Quebec, Canada is on Tuesday, January 4, 2022 at 1:52 am EST (Change city). Distance from the Sun's center to Earth's center will be 147,105,052 km (91,406,842 mi)". Notice that Montreal is in canada, as if, maybe, the canadian government is not allowing google canada to have access to google as fully as Americans have access to it, in which case canada is a piece of anti-educational trash too, no surprise at all. Globalist pigs don't want the general population mucking around with knowledge; we are to repeat only what they teach. Hell-to-pay is almost here for dark souls. I no longer can do extensive google searches, and this has been the case for years.

Take a look at the eclipse on the page below, but for me it looks all wrong. The image makes it appear as though the shadow is moving, when in reality it's the moon moving into the shadow. Moreover, the moon moves east to west, yet they have the shadow coming down upon the moon as though the moon were moving straight up into the sky. This looks like disinformation. I'm starting to wonder whether the globalists had recently decided to feed the world false things in EVERYTHING to make people go crazy, lose touch with reality, get confused, have less power of accomplishment. It explains why everything being taught is suddenly backward in the past year or two.

The page above is as per Montreal again. I'm starting to think that canada is blocking a myriad of pages from all over the world so that canadians get more canadian news and information. This is betrayal of the people by the liberal quacks. They are turning everyone into quacks.

On NASA's Lunar Eclipse page for 2012-2030, we find data on last month's eclipse. At the link below, there is a drawing where you can see how the earth shadow does not strike the moon from its top:

Here's a page you will need to explain the lunar-eclipse pages:

The "SD" on the eclipse pages stands for "apparent semi-diameter," and the SD for the sun and moon are given on, the eclipse page for last month, as 00 16' 11.0" and 00 14' 44.5" respectively. Those numbers can be converted to angles at this page, where, for example, you put "00 16 11" into the first box and hit the Convert button. It works out to .2697 degrees, and we double that to get the full solar diameter of .5394 degrees (larger than it's median size). In the same way, the moon's semi-diameter works out to, .245694, for a full diameter of .4914 degrees, almost as small as the moon can get.

To find the angle of a make-believe / on-paper solar-eclipse line for the eclipse above, start with the sun .5394 degrees, and find the lunar distance when the moon is .5394 degrees too (that's a kiss-earth solar eclipse). Earlier, it was said: ".49, .518, .558. Those sizes represent the furthest (252,650 miles), the average (238,850), and the closest (221,580 miles). When we divide .49 by .558 and multiply the result by 252,650, we get 221,861." So, to find the lunar distance for the size of .5394, do: .49 / .5394 x 252,650 = 229,511 miles. We can thus have the angle of the solar-eclipse line to go with the angle of a lunar-eclipse line for last month's eclipse. Are you excited?

Just put 229,511 in box, b, of the right-angle-triangle calculator, with the lunar diameter of 1079.5 in box, a, and the calculator spits out the solar-eclipse-line angle of .269 degree. The earth's shadow line, for the eclipse of last month, must be less than .269 degree.

This eclipse does not have the moon passing close to the shadow equator, and consequently there are no U2 and U3 times. The closer to the shadow equator the moon passes, the more fully/reliably one can have the shadow diameter by timing the moon's crossing. Otherwise, we are left to trust the shadow diameter as given by NASA. The page gives the shadow diameter via "U. Radius," and in this case this radius is given as .6434". it's the apparent size in the sky, and this number is of course at the distance of the moon at the time of eclipse.

We double .6434 to 1.2868 to find the shadow's diameter, and then, to find how much larger it is than the .4914 moon at the time, we divide the two numbers to find 2.6186 times larger. The claim for the shadow diameter is: 2.6186 x 2,160 = 5,656 miles (radius = 2,828).

I can tell you what the angle of this eclipse is, but I've always got to show my math. It first involves the earth radius of 3,960 miles minus the shadow radius of 2,828 miles = 1,132 miles. That is, the lunar-eclipse line spreads 1,132 miles from the point where the moon enters the shadow to the edge of the earth. The lunar distance for this eclipse was figured above as 229,511 miles, and so we put that in the box, b, of the right-angle triangle calculator, and for box, a, we put 1,132 to find an angle of ERROR BY NASA because the line spread of 1,132 is too high.

I showed above that the sun at this eclipse has an apparent size of .5394 degree. I showed all we need to do is to find the angle to the edge of that sun, from the standpoint of a person standing on earth nearest the sun, is to stick a moon in the sky the size of .5394 degree. That angle turned out to be .269 degree, and so the earth-shadow line (same as lunar-eclipse line) must be smaller than .269, yet NASA claims .282 degree. The latter was obtained, not by my numbers, not by the numbers of an amateur astronomer, but by the size NASA claims for the earth-shadow diameter. They have the shadow too thin where the moon passes through it (and therefore they have the shadow too short for the reality).

Below is a lunar eclipse of June 15, 2011, in which the moon passes through the shadow's equator. U1, U2, U3, and U4 are all given, which therefore gives a lunar velocity. The claim on the page is that one lunar diameter of travel, between U1 and U2, took 28 seconds shy of one hour (59.0088 minutes), meaning that the moon is said to be traveling a tad faster than 2,160 miles per hour, or 2,159.1 / 59.0088 x 60 = 2,195.367 mph to be precise. The time between U1 and U3 is given as 2 hours, 39 minutes and 45 seconds, or 2.6625 hours. This route, 2.6625 x 2,195.367 = 5,845 miles, is NOT almost an entire shadow width.

The page gives the shadow radius as .7256", and for the moon radius (SD) of 00 15 57.2, I convert it to .2659 degree at the onlineconversion page above, which makes this shadow .7256 / .2659 = 2.729 times larger than the moon, wherefore the shadow diameter works out to 2.729 x 2,160 = 5,895 miles. In case you didn't catch it, there's about a 50-mile difference between the U1-U3 distance and this umbra diameter. How can that be correct?

First, let's give the angle of NASA's lunar-eclipse line by first finding the lunar distance at the eclipse. As the moon at that eclipse has an apparent diameter of .2659 x 2 = .5318, the math to find its distance is: .49 / .5318 x 252,650 = 232,791 miles. We put this latter number into box, b, of the triangle calculator. For box, a, we need: 3,960 - 5,895/2 = 1012.5 miles of lunar-eclipse-line spread. With the latter number in box, a, the eclipse line's angle is .249 degree. It looks down-to-earth.

Next, we want the angle to the edge of that sun, from the standpoint of a person standing on earth nearest the sun. For this we need the size of the sun at the eclipse, which is said to have a radius of 00 15 44.7, which converts to .2624 degree, and then that's doubled to .5248 degree (about the smallest the sun ever gets) to get the diameter. So, we put a moon of that .5248-degree size into the sky for our make-believe solar eclipse, and to get its angle, we need to know its distance from earth, which is: .49 / .5248 x 252,650 = 235,897 miles. We put the latter into box, b, and for box, a, we use the lunar radius of 1,080, to find an angle of .262 degree, and it happens to be larger than the .249 above for the lunar-eclipse angle, so this is good.

The good news is, we can actually do a solar-distance calculation now that the lunar-eclipse line is smaller than the solar-eclipse line. For the lunar-eclipse line, we found a spread of 1,012.5 miles over a distance of 232,791 miles toward the sun, and to find the spread per 1 mile toward the sun, we do: 1012.5 / 232,791 = .0043494 mile. For the solar-eclipse line's 1,080-mile spread per 235,897 miles toward the sun, it's equivalent to: 1,080 / 235,897 = .004578 mile of spread per 1 mile toward the sun. It means that the solar-eclipse line catches up to the lunar-eclipse line by: .004578 - .0043494 = .0002286 mile per 1 mile toward the sun. As the two lines begin 3,960 miles apart, it will take the following number of miles for the lines to meet: 3,960 / .0002286 = 17.32 million miles. If NASA's data for it's lunar eclipse of June, 2011 is accurate, it appears that the sun is 17.32 million miles away. You can scratch your head all you want, but don't call me a nut because this math, which I've done lots of over the years, is full-proof.

[Insert -- If the precise lunar diameter of 1,079.5 is used in the calculations, the solar distance ends up at 17.48 million]

Before claiming that the sun is 17.3 million, the eclipse data at the page above needs to be verified.

If interested, I've found my previous discussions on this and related quests starting in the 3rd update of July, 2016, with the 4th update in August having a total lunar eclipse I worked on heavily:

...Let's look again at the data for the July 16, 2000, lunar eclipse, telling us [inadvertently] that the sun is 15.673 million miles away:
With a shadow path [i.e. shadow diameter] of 5,738 miles, the distance from shadow edge to earth edge is (7,918 - 5,738) / 2 = 1,090 miles. The solar distance using the 1,090 figure is like this: 3,959 / (.004578 - (1,090 / 252,000)) = 15.673 million miles.

None of the figures in the math are my own aside from guessing at 252,000.

I'm going to check that 15.67-million-mile figure today, right before your eyes. It's not likely the correct solar distance because I was rounding off the lunar distance. Here's the address in print to this eclipse of July 15, 2000, eclipse:

The math used in that 2016 update is a little different than the math used here. Plus, as I re-do things here, I'm changing the 1,090 to 1,091, for this is the lunar-eclipse spread when the earth diameter is reckoned at 3,960 miles. I'll also use 1,079.5 miles for the lunar radius.

Where you see "(.004578 - (1,090 / 252,000)," the .004578 is the solar-eclipse-line spread per mile toward the sun, and the "1,090 / 252,000" is the lunar-eclipse-line spread per mile toward the sun, assuming that the moon was 252,000 miles from the earth at the time. Apparently, I may not have been sure at the time how to nail down the lunar distance perfectly, but will do the math here, from the NASA page having the moon at the following size: 00 14' 43.2" = .24533-degree radius = .49066 diameter. So, we do .49 / .49066 x 252,650 = 252,310 miles. Putting the latter in box, b, with 1,091 in box, a, calculates the lunar-shadow line at .248 degree.

I don't care what solar-eclipse angle I used in the 2016 update, because I'm re-doing the full math here with my new realization: I can use a moon in an on-paper solar eclipse that's the same size as the sun at the eclipse, said to be 15'44.2," or 15.7367 arcminutes, which is then divided by 60 to find it equal to .26227 degree (or you can use the onlineconversion page above to find the same). We double that to find a solar diameter of .52454 degree. With a moon at that size, we find it to be the following distance from earth: .49 / .52454 x 252,650 = 236,013 miles, and so we put the latter into box, b, with 1,079.5 in box, a, to find a solar-eclipse-line angle of .262.

Okay, we have both eclipse-line angles, .262 versus .248 for the lunar-eclipse angle, and I can readily see that the spread between these angles is going to give a sun far less than 93-million miles. At the risk of confusing you, the 2016 update had found a way to use the right-angle calculator, and angle degrees, to find the amount of line spread per 1 mile toward the sun (I had forgotten that this was found when writing a EXTRA-EXTRA section above). Put .262 into the alpha-angle box, with 1 the box, b, and pressing the Calculate button, we find a spread of .0045728 mile per one mile toward the sun. For the lunar-line spread, put .248 instead into the alpha-angle box, with 1 again in box, b, to find a spread of .0043284. mile per 1 mile toward the sun.

The two lines begin 3,960 miles apart, and as the solar-eclipse line catches up to the lunar-eclipse line by .0045728 - .0043284 = .000244 mile per 1 mile toward the sun, we do: 3,960 / .000244 = 16.23 million miles. But why doesn't it jibe with the 17.3 million miles that was obtained from the 2011 NASA eclipse? The sun at this 2000 eclipse is said by NASA to be smaller than the one that got the 17.3 million distance, meaning that a smaller sun should be more than 17.3 million away, yet the math gets 16.23 million.

Let's do the math again by the EXTRA-EXTRA method (no angle degrees needed) to see if we get the same numbers (we won't): the solar-eclipse-line spread is 1,079.5 miles per 236,013 miles toward the sun, and the lunar-eclipse-line spread is 1,091 miles per 252,310 miles toward the sun. Refining, the solar-eclipse line spread is 1,079.5 / 236,013 = .0045739 mile per 1 mile toward the sun, and the lunar-eclipse-line spread is 1,091 / 252,310 = .004324 mile per 1 mile. The math now gets a catch-up speed of .0045739 - .004324 = .0002499...versus the .000244 obtained by the calculator in the paragraph above. That's how you can know that the degree-less method is reliable. The final math: 3,960 / .0002499 = 15.85 million.

It looks like the 16.23 million isn't going to stick, and 15.85 million is even further from 17.3 million. What's wrong?

We can now do the geometric math starting at the sun to see if things check out. According to the astronoLIARs, the sun is 94.5 million miles away when it's at its furthest, and it's essentially at its furthest for this eclipse. A sun 15.85 million miles is therefore 94.5 / 15.85 = 5.96 times nearer than NASA claims, in which case the solar diameter, which is proportional to the solar distance, works out to 865,370 miles (their figure) divided by 5.96 = 145,196 miles. The radius is 72,598, and because we want to draw a line from the edge of the sun to the edge of the earth, we knock off 3,960 from the radius, meaning we want about 68,638 in box, a (just to show how to do the math and check the shadow-line angle). This represents line a in the calculator page's triangle, and that line extends from the edge of the sun toward the solar core, but ending one earth radius before reaching the core, for which reason line b is a straight, zero-degree line to the edge of the earth.

Then, we want 15,850,000 in box, b, and when we hit the calculate button, we get the angle of the earth-shadow line (line c) at .248 degree...exactly the lunar-shadow-line angle obtained above with NASA's data.

We can now figure out the ratio between the lunar shadow and the earth shadow by pretending that the moon in the earth shadow is in direct sunlight. With the core of the earth 15.85 million miles from the core of the sun, the core of the moon undergoing eclipse is a distance from the sun of 15.85 million + 252,310 = 16.10 million miles. So, we now put 16,100,000 in box, b, and for box, a, we put the solar radius, 72,598, minus the lunar radius = 71,518 (latter into box, a). This gives a lunar-shadow angle of .255 as compared to the earth-shadow angle of .248. We can now find the length of the lunar shadow of the moon that's in the middle of its eclipse, with 1079.5 in box, a, and .255 in the alpha-angle box, it's this easy (as long as the calculator's working). The lunar shadow is thus found to be 242,551 miles long (it's not the true length; I'm just showing the math method).

For the finding of the length of the earth shadow, put 3,950 in box, a, and .248 in the alpha-angle box, to find it 914,879 miles long. The earth shadow works out to 914,879 / 242,551 = 3.77 times longer in this particular case. If left to astroliars, some will say about 3.667 times, in which case they lie.

The 15.85-million-roughly distance depends on whether NASA entered the correct data on its eclipse pages. If you can either verify or expose NASA's numbers, be my guest. I will accept 93 million if the eclipse lines proves it, but as you saw with the eclipse of last month, the NASA number for the shadow diameter was far-too small in the 5600s.

It's possible that, since my work on these matters back in about 2005, someone(s) complained to NASA concerning their too-small earth shadows, but I'm not making an accusation, for I just can't remember things well enough. If you wish, you can figure out, for this eclipse under discussion, what the width of the earth shadow needs to be in order to get a sun 94.5 million miles off.

In fact, I'll do it for you. The lunar-eclipse line needs to be at an angle of about .2623 in order to get a sun 94.5 million miles away with a diameter of about 865,000 miles. So, using .2623 (in alpha-angle box) for the lunar-eclipse line, with a moon 252,310 miles (in box, b) from earth undergoing the eclipse, the shadow width at the lunar-passage point works out to 3,960 - 1,155 (in calculator a-line result) x 2 = 5,610 miles.

Let me show another way to find the shadow width at the point where the moon passes through it, using the total eclipse of April 15, 2014. The umbra radius given is .6952 degree (diameter = 1.3904). The moon at the time is said to have had a radius of .25858 degree (diameter = .51716), and so just do .6952 / .25858 x 2,160 = 5,807 miles for the width of the shadow at lunar passage. Dare we estimate the solar distance using this eclipse? I don't remember NASA having earth shadows that large so routinely, and the kicker now is, this eclipse has a sun at a near-median size of .5314 degree.

Okay, first we find the lunar-eclipse-line angle by first finding the lunar distance at this eclipse: .49 / .51716 x 252,650 = 239,381 miles. Into box, b, goes 239,381, with 1,056.5 in box, a, as per this: 3,960 - 5,807/2 = 1,056.5. The latter is the lunar-eclipse-line's spread when climbing to the earth's edge. The calculator tells us that this line is on an angle of .253 degree.

Next, we need a solar-eclipse line, or just a lunar-shadow line, no eclipse needed at all. The important thing is, we need a moon, the size of the sun in the eclipse, to go between you standing on the earth and the sun, and that size is .53144 degree. It's a shadow-kiss-earth situation, making it EASY-EASY-EASY to find the angle. A moon of the size above is at the following distance from the earth: .49 / .53144 x 252,650 = 232,949 miles, and so we plug that latter into box, b, with the lunar radius of 1,079.5 in box, a, and the solar-eclipse-line angle is found out at .266 degree (rounded off by the calculator). It's good, because the lunar-eclipse-line angle is smaller at .253 degree.

We can take a short-cut here by plugging .253 and .266 separately into the alpha-angle box, with 1 in the b box in both cases, and we then find that the lunar-eclipse line spreads at .0044157 mile laterally per one mile toward the sun, while the solar-eclipse line spreads .0046426 mile laterally per mile toward the sun. However, as both angles are rounded off by the calculator, this doesn't get a precise solar-distance result. This method needs more decimal points on the angles. The final math is: 3,960 / (.0046426 - .0044157) = 3,960 - .0002269 = 17.45 million miles. The July-2000 eclipse with a smaller = more-distant sun than the sun in this 2014 eclipse worked out to 15.85 million miles. Something seems out of wack a little.

To get more decimal points on the angles, and one may get them by using the 00 15' 56" that we find beside the .266 degree, for example, when doing the operation with 232,949 and 1079.5. We then enter 00 15' 56" in the first box at the onlineconversion page above, and hitting the Convert button to find, .26555 instead of .266. We do the same exercise with information that derived the .253, to find it more-precisely as .25277 degree. Re-doing the math now by plugging those new angles into the calculator with 1 in box, b, we find: 3,960 / (.0046348 - .0044117 = .0002231) = 17.75 million, a substantial difference from the 17.45 above. However, 17.75 is still wrong, because the accurate way to do it is like this:

1,056.5 / 239,381 = .0044135 (.49 / .51716 x 252,650 = 239,381) 1,079.5 / 232,949 = .004634 (.49 / .53144 x 252,650 = 232,949) .004634 - .0044135 = .0002205, then 3,960 / .0002205 = 17.9

The moon size on the second line, .53144, is also the sun size for the eclipse. The moon size in the first line is for the lunar eclipse. You can do any eclipse by copy-and-paste of these numbers, changing the moon-size numbers and re-doing the math. The first number, in this case, 1,056.5, is the eclipse-shadow radius subtracted from the earth radius. The reason for the difference between 17.9 and 17.75 is, probably, that the 00 15' 56", for example, has 56 rounded off.

The total length of this shadow is with .253 in the alpha-angle box and 3,960 in the box, a, which gives the shadow length on line, b, of 896,797 miles. The sun is nearly at it's median distance here, yet we saw the Wikipedia writer claim that the earth shadow is 870,000 miles long. He's wrong / misleading, isn't he?

If NASA's eclipse data is correct (at this time, anyway), and there's nothing wrong with my math methods, it appears that the sun is, on average, 16 million miles off.

To show a neat thing, we will use my prize-winning, 2-foot round watermelon hanging on a rope from a helicopter. Hold her steady over the moon. Up a little more, left a little more, er, a little back to the right and...hold it right there, don't move. You are standing on the ground, and the watermelon is up in the sky a ways covering the moon perfectly. A line from your eye through the center of the watermelon, and through to the core of the moon, is the zero-degree line, and as the moon at the time is 235,900 miles away, we know it's angle of .262 degrees, this being the line from the edge of the moon across the edge of the watermelon and into your eye.

Okay, so now we are able to measure how far the watermelon is, either with a tape measure, or with the moon. As we know the angle from the edge of the watermelon to be .262 degree, we put 90 - .262 = 89.738 into the alpha-angle box, and we put 1 into the box, a, because it's the radius of the watermelon, and when we press the calculate button, it tells us that the distance to the watermelon is 218.7 feet. This number depends squarely / wholly on the diameter of, and distance to, the moon, because it was those two lunar things that got us the .262 angle. Therefore, if the watermelon is not nearly 218 feet away, when measured with a measuring tape, then neither do the astronomers give us the lunar diameter and distances correctly. Put your watermelons to work today, and find the true distance to the moon.

If they haven't got the lunar distance correct, though I reckon they do, then neither can we use it to find solar distance. They know about the watermelon test to lunar distance, though they don't use watermelons, I must confess. Or helicopters.

Asking google how people of the past determined the lunar distance, it gave a video at the top of the results featuring my watermelon method but with a tiny coin instead. At the 1:10 point of the video, note that someone in the past used the solar-eclipse method of finding solar distance, with the conclusion that the sun was about 20 times the distance of the moon, or about five million miles. How did he come to that conclusion? And why doesn't the video tell of it? Might he have done better with a watermelon?

The video seems to imply that the solar distance was estimated anciently by a measurement of the moon's time travel through the earth shadow during a near-total eclipse. Why did the solar distance come out to about 5 million miles by that method??? The ancients didn't have the accuracy of lunar-eclipse lines that we have today, but then again, I don't think we have accuracy of lunar-eclipse lines today because the astronomical establishment is a fink of satan. If someone today measures the real time for a lunar passage through a shadow, the establishment is ready for such as person, and has measures in place to block him/her. This establishment assured that it owned the Internet, it's no coincidence that Internet giants are all anti-Christ. The childish google will bury his/her video / article before long if he/she tries to publicize the correct times. That's how the establishments have always worked, for a century and counting. One could buck against the establishment's main tenets only by ruining his/her career / reputation / worse.

If you'd like to figure out the solar distance for yet another good total lunar eclipse, see the one of July 27, 2018. It has one of the smallest lunar sizes in combination with one of the smallest solar sizes, meaning that it's one of the longest earth shadows with a moon as far down along it as possible. The moon in this case is .49038 degree (smallest possible = .49). The sun size is .525 (smallest possible is .524). The umbra width, where the moon passes through, is 5,713 miles. The umbra-time data on the page, in case it changes, is now like so:

U1 = 18:24:28.4 (UT)
U2 = 19:30:16.8
U3 = 20:21:45.4

[Update the week after this update -- the same eclipse on the page below has U1 to U2 from 18:24:27 to 19:30:15, and there's a 2.9-second difference between the two records for total passage through the entire shadow, showing that eclipse times depend on who's giving them. We may ask if there are larger errors in any eclipse.>

End insert]

We find the lunar-eclipse-line angle by first finding the lunar distance at this eclipse: .49 / .49038 x 252,650 = 252,454 miles. Into box, b, goes 252,454, with 1,103.5 in box, a, as per this: 3,960 - 5,713/2 = 1,103.5 (or, you can do 7,920 - 5,713 / 2 = 1.103.5). The latter is the lunar-eclipse-line's spread when climbing to the earth's edge. The calculator tells us that this line is on an angle of .25 degree.

Next, we need a solar-eclipse line, so we stick a moon the size of the eclipse's sun, .525 degree, between us and the sun, and thus, for box, b, we put the lunar distance when the moon is .525 degree, and that's found with: .49 / .525 x 252,650 = 235,806. The lunar radius of 1,079.5 goes into box, a, and the solar-eclipse-line angle is then given as .262 (verses the smaller lunar-eclipse line of .25).

We can put .25 and .262 separately into the alpha-angle box, with 1 in the b box in both cases, and we then find that the lunar-eclipse line spreads at .0043634 mile laterally per one mile toward the sun, while the solar-eclipse line spreads .0045728 mile laterally per mile toward the sun. The final math is: 3,960 / (.0045728 - .0043634) = 3,960 / .0002094 = 18.91 million miles. Not bad, but something's still out of wack a little.

You might find it a good challenge to discover why these numbers are not quite adding up expectedly (try different lunar distances, or ask whether NASA is being dishonest with its numbers). If the smallest-size sun on their 93-million scale is 94.5 million miles away while their middle-size sun is about 93 million, then the 1.5-million difference is going to be almost six times smaller between a medium size sun at about 16 million versus the smallest sun on that scale i.e. the smallest can't be as high as 18.91 million unless the 16-million is wrong. The total spread from largest to smallest sun should be about .5 million miles.

I asked google for NASA's angular diameters of the moon, but google pretends not to know what I want. So, I was left to go to Wikipedia's Moon article to find that the lowest angular diameter is 49.3 arcminutes, which converts to .48833 degree. Will .48833 make much of a difference, as compared to my use of .49, when tabulating lunar distances? The Wikipedia page gives the largest moon as 34.1 arcminutes, which works out (at google's arcminute converter) to .56833 degree. WHAT? I've never seen such a high number for the moon's largest size. How did we get from the .558 claimed largest by others to a whopping .56833?

The total length of this shadow is with .25 in the alpha-angle box and 3,960 in the box, a, which gives the shadow length on line, b, of 907,559 miles.

In the video below, note the discussion on the distance between mars and the sun, according to Kepler's third law:

The reason I'm showing the video above is to suggest that astroLIARs are nuts. Kepler worked out a law for finding planetary distances based on how fast planets circled the sun. Achem, stupids, the orbital period has NOTHING to do with distance from the sun. We could excuse Kepler of centuries ago, but by now, stupids should have become wiser. Orbital periods are based on the mass of planets versus their distances from solar gravity. Hey, STUPIDS, if one were to put a planet at the same distance as mars, only heavier, the orbital speed needs to be faster, and so the period of this make-believe planet is less time than for mars, yet they are at the same distance. Astronomers, you got rocks for brains if you have a law claiming that both planets must be at different distances from the sun.

In other words, if the planetary distances are based on Kepler to this day, the astronomers have made fools of themselves in yet another way. God could put ten planets at the same distance as mars, each with a different period. God could have made a lead-core planet circling the sun faster than one year but at the orbit of mars. There's no reason for the impossibility of such a scenario. Therefore, orbital periods cannot disclose planetary distances, sorry.

Once you get to number 3 in the video below, he's on Kepler's law under discussion. You will see his erroneous formula in which a semi-area of the orbit of a planet, when cubed, is equal to the period of the plant squared. I think that when a man stretches the brain for decades too far, he becomes dangerous to his own thoughts, more like a magician than a scientist. When you get to the dots cutting the orbital area in half, that's a good time to pause the video to ask: if a make-believe planet were taking the same orbital path, or a similar path having an identical semi-area, and if it was traveling the same orbit at a different, average velocity than the real planet, out the window goes Kepler's law.

So why do astronomers keep Kepler's law if it's so obviously wrong? Apparently, they lie here for a need(s), and, chances are, the need(s) serves their big-bang view of cosmological creation.

When the video above gets to Newton's law of gravity, which modern astroliars hold to, they make fools of themselves yet again by assuming that the amount of planetary mass is proportional to the level gravitational force/power. That is, the gravity force between the sun and a planet is equal to / caused by the masses of both planets because the fools see a gravity particle(s) in every atom. FOOLS, they like to play games, and care not if they fill you with trash, as long as they can derive honor as wisemen. Newton had this wrong, but it plays wunderbar for cosmic evolution, and that's why modern science adopted it. It's why modern evolutionists traffic in this illicit mind-bending trash. You can gather that, if Newton was wrong, so are the shapes of planetary orbits as modernists claim as facts. I know what I'm talking about here. I know I'm correct. I did not allow myself to be brainwashed, step-by-step, school degree by school degree.

If you didn't understand a thing the wiseman said on the gravity section of the video, consider yourself blessed. Move on, move along, it's trash. He's acting like he knows his stuff to boggle the minds of his students, to act like a wiseman, even while he's a fool. Be logical. There are no gravity particles in atoms. Be wise, and realize that gravity is from the negative charge in electrons, not from electrons bound to atoms, but electrons freed from atoms. Any pool of free electrons in a planetary body constitutes the gravitational force of a planet. No gravity particles can be proven to exist, but we do know that electromagnetism exists, and it does act exactly as gravity acts.

The only "problem" is, a negatively-charged definition for gravity obliterates cosmic evolution as the fools teach it. They need all atoms to possess an innate pull so that atoms attract atoms, and thus the cosmic debris from the big-bang could attract and form stars as the particles sailed along through space. We could say, okay, protons attract electrons to form atoms, but unless atoms attract one another, they cannot form stars, which is why they welcomed and embraced Newton's gravitational fantasy like the whore of Revelation, for they are the whore of Revelation. They are the killers of God and His people.

If stellar gravity is from the negative charge of free electrons within stars, then, oh no, stars repel one another, and galaxies cannot form. Ouch, that ruins cosmological evolution. So, absolutely not, gravity cannot be defined as negative charge, even if it's true.

My discovery: gravitational force is proportional to the heat within a planet. It's not proportional to the mass of a planet. The God-haters claim to know the masses of all the planets from their orbital speeds and solar distances, yet they don't even know the true distances of the planets. It's that bad. They are that ruinous of the facts of reality. They live the lie and pose cosmological fantasies, and the people who praise them are the sleepers. But we are to be awake, to understand what God is doing in exposing His enemies, for God does not need to expose them so that He wakes up, but that we wake up. The willfully asleep can never enter the Kingdom of God. Stay awake, stay wise, know the liars and know the lies.

Heat is the flow of free electrons into and out of materials. The kinetic theory of heat is bogus. Electrons stream from the sun because the sun is hot. The sun is hot because it consists of many free electrons. The heat in the sun is the solar gravity because electrons have negative charge. Stars repel stars because they are all net-negative. Stars attract planets because negative charge attracts atoms. Negative charge attracts atoms because negative charge repels outer electrons from atoms, making atoms net-positive. The closer atoms come to gravity, the more atomic electrons get blown away, the stronger the force of gravity on the atoms. That's logic. It works, and it's provable.

None of this knowledge is necessary for life with Jesus, but life with Jesus in the end times behooves us to identify the enemy and his tricks, and I'm telling you, the enemy is a fool in more ways than his rejection of Jesus. Do not respect the educators, do not let them lure you to themselves with their aura of professionalism, with their claim that they have solved great mysteries. Rather, they have denied the plain facts before their noses if those facts served to demolish the big-bang farce.


Here's Polly with another new find, that the Nazi's appear to be in charge of vaccine passes in Britain. Keep watching, don't give up, and by the 12th minute you'll see why. This is a mark-of-the-beast look-alike for the time being; they are pushing this without sign of defeat, and it can become the mark of the beast, so don't take chances: start preparing for it just in case. The Western leaders are not nice people, as you might think; they will force you to vaccinate continually to get your food.

In the 24th minute, meet the richest woman in Germany, Susanne Klatten, daughter of the Nazi family (QUANDT by surname) that owns the vaccine-pass company (Entrust) under discussion. She is part-owner of SGL Group, a graphite company i.e. a product that makes graphene-oxide. It was the Graff family that lived across the street from me, where the mother and father of the family still had a GERMAN accent, that pointed to graphene oxide. I think I now know why they were German.

The Klatten surname looks like a possible branch of Klassens, a significant topic over my last two updates. Klassens use "LADY FORTUNE" as their symbol, and Ladys/Laudymans have a version of the Anchor/Annacker Coat while Graffs/Graffens happen to have an anchor as code for Anchors/Annackers. Jewish Glass' share red wings with German Quands/Quandts. English Quands are listed with Quints, a branch of Quince's/Quincys (Northamptonshire, same as Ladys/Laudymans) who are in the write-up as Faucets, the latter first found in East Lothian with Fortune's. (Load Quand link now to have access, on another tab, to other Coats of Arms.)

Adolf Hitler's father had married Miss GLASSl. Klassens/Class' are a branch of Glass'. Stuarts and Houstons lived in the GLASgow area with Pollocks, and while Hitler's mother was Miss POLZl, Hitler's nephew, William Hitler, changed his surname to Stuart-Houston. Klatts are listed with Glatz's while Glaze's share the pheons of the Lords/LAUDs (like the Laudyman variation of Ladys) in the Glasgow-surname motto.

I had found evidence that president Franklin Delano Roosevelt helped Hitler escape alive to Montana / Idaho (the latter became home to Aryan Nation). CARNeys share the Glaze / Lord/Laud pheon, and CARNagys share the giant Roosevelt eagle. Scottish Watts were first found in Carnagy, and Pollocks of Glasgow, who built Rothes castle, were proto-Rothschilds. These Pollocks married Watt-branch Watsons who in turn married Leslie's, earls of Rothes because Pollock titles came down to them through these marriages. Leslie's are from Lesce (Sava river), home of a CARNI people group who named an area there. Rothes castle was right-near the Rose clan to which Roosevelts may be traced. Jewish Rothchilds use roses. Dutch Roosevelts share roses on stems with SCHERfs/Schere's (compare with Walkers), and one George Herbert Scherff, Nazi, was the father, I believe, of George Herbert Walker Bush. The latter was adopted as a son, for spy purposes, by Nazi-supporter, Prescott Bush, banker.

President DeLANO's surname was first found in Brescia, beside lake Garda to which the "Garde" motto term of Carricks traces, and Carricks, first found in Ayrshire with ALDa's, share the ALTAN/ALDaM dog while Susanne Klatten is the sole own of ALTANa, a drug company. English Lane's/Lawns, very provable as a Delano/Lane branch, look like they use a version of the Glasser/Glazier Coat.

Back to the Ladys/Laudymans (probably a Lord/Laud branch) with a Coat like that of Anchors, for the latter's lozenges are colors reversed from the same of Leiths (Edinburgh, beside Fortune's and Seatons) who in turn share the Seaton crescents. The Let's/Late's/Leits are in the GLASgow motto with the phrase, "LORD, let," and Leith's use a "Trustie to the end" motto to go with "True to the end" of Home's/Hume's. The Trusty surname is listed with English Tristans (Devon, same as Seaton location) while the other English Tristans were first found in Cornwall with Trusty-like Trysts (rose). The English Delano's are listed with Lannoys while French Lannoys (and Lyons) have the Home/Hume lion in colors reversed.

The two mottoes above both have Tooth-like "to the," and the Tooths share the giant griffin of Leith-like Letters and Lauders (Berwickshire, near Edinburgh).

Fox news totally out of touch with what matters: "'The Five' rips Biden for insulting Americans in explaining bad poll numbers". Fox is a loser game for a loser audience. Fox takes you nowhere. It wastes your time, Christian. It uses you for profits. It's in the game of hooking you so that it has ownership of your eyes and mind every day.

The greatest in the kingdom of satan is not the one who can kill a man with heart, but the one who can kill the most men without heart. That's how small satan and his thugs are. The more you insult them, the more they want to kill you. That's how small they are. But Jesus makes us strong enough to turn the other cheek, to take insults without fighting back. Be strong, be the real man. The goons are cowards if not for their possessing military might in many nations. The anti-Christ will be glorified for his beastly military, but that's just indication of the wee-wee that he is. The one who considers himself great for his killing machines becomes a lunatic. We all know the face of a lunatic when we watch Hitler speak. Where's he now in regrets? Same place as the most leaders in the West are going.

God can be likened unto a giant heart. The entire Bible, when God comments and directs, is full of heart, the stuff that humanity needs. God allows heartless evil to fester and act, temporarily, to make a show of evil on this stage of history. But God is the Clean God too. When Jesus said that our whole body is clean if our feet are clean, I'm sure He wasn't referring to soil and sweat. I think He meant that if where our feet take the body is not to violence, corruption and other human degradation, then we will be remain clean.

Globalists testify against themselves when they fake goodness for their globalism because, in the end, they will do an about-face and act openly sinful, outrageous, wacko, murderous. They will testify against themselves because they show they KNOW the right things...because they are feigning the right things. They can't claim ignorance, therefore.

In the 8-9th minute of this Stew Peters show, the lawyer shows a book by Alan Dershowitz, "The Case For Vaccine Mandates." That's right, Mike Lindell the Christian hired this wicked man to bring election-integrity to the supreme court of the United States:

I follow the news daily, though not from big media, but I don't recall hearing until this week that Netanyahu of Israel called publicly for Biden's presidency over Trump's after the ballot-box whore houses with Fox news betrayed Trump at the election. How do we explain this? If we suggest that world leaders found out about Trump's deep involvement with something like Jeffrey Epstein, it still doesn't justify Netanyahu's endorsing and accepting Biden as president. I suggest that the Israeli's play every American president and leader for all they can get from them, and so once Netanyahu saw that Biden was going to become the president in spite of the cheating, he decided he had best act like Biden's friend, because this is how Israelis play American leaders, even if they are anti-Israel, such as Obama was.

This video has a nurse before a Louisiana legislature revealing that most medical people don't even know what VAERS is, meaning that the reporting system (VAERS) for "adverse" events" from vaccines is not even nearly telling of all the deaths and maimings, and the politicians, for the most part, don't care. This defines love gone cold, heartless people willing to see humanity die off to make more room in society for the survivors. EVIL.

I don't know who this guy is, but he sounds like he knows what he's talking about on aluminum poisoning the brain with vaccines:

The U.S. supreme court decided lately, in a 6-3 decision, that New York's mandatory-vaccine law should stand without a court hearing to hear why it should be knocked down. Why did they bother to become top-of-the-line judges if they don't want to decide top-of-the-world disputes? Two of Trump's supreme-court picks -- Kavanaugh and Coney-Barrett -- in-effect supported mandatory vaccinations, and the winning side even disallowed religious-reason exemptions. How can this be? If the top court allows New York to do it, it will allow all the liberal states to do it. Now what? Should all Christians move out of liberal states? That may exactly God's direction at this time. Stack the Christian-leaning states with more Christians coming out of wicked-empowered states, and that will bode better during the 666 program. Let the wicked people in the wicked-empowered states suffer the lockdowns, the vaccinations, the high taxes, the demented politicians, the sleazy, the violent streets, and the rest of the corruption. We pay government workers to get corruption off the streets, not to facilitate corruption in the streets. They are baboons swinging in the tree of evil.

Gateway Pundit: "CDC COVID-19 Response Team published a weekly report on Friday where...The U.S. Centers for Disease Control and Prevention (CDC) reported that of the 43 COVID-19 cases caused by the Omicron variants, 34 people were fully vaccinated. Of those fully vaccinated, 14 people had received their booster shots but five of those received their additional shots less than 14 days. Did the CDC capitulate to the reality that vaccines have been a total bust? Or did the CDC put this out there to play to Fauci's turning up the vaccine intensity for coming boosters?

You see, the CDC is making it appear as though vaccinated people are being invaded by the omicron when the likely reality is that the vaccines are causing it. If we ask how that can be possible when even the unvaccinated are getting omicron, that question explains why the CDC put this data out there, because it includes a few unvaccinated who have omicron. But nope, this is erroneous. It is not possible to test anyone for omicron, or any other COVID variant. The claim by the medical establishment that it can differentiate variations is a part of their scam. Always remember that. This scam is perpetuated by those medical people -- especially hospital owners, managers and paid-off doctors and nurses -- who are swimming in the money due to the "pandemic," and they want this to last forever. They have been guilty of taking the lure into gross sin, but the money won't buy them out of their mental instabilities that come knocking with sin.

Demons cause "mental illness." That's because demons can't affect us physically, and so they go for the thoughts, to screw them up, drive people insane, if possible. Drip-drip-drip, they can affect our mental operation, and then attitudes, over time, and when our actions become destructive to us, when we're fooled into sinful adventures, mental illnesses, or perhaps better described as a fearful mental dimness, sets in. The key word is "set" in. When cements sets, it's hardened. When Jesus said, "the Truth will set you free," my bet is that He was talking about new-found and exhilarating mental health. Free from demons = yahoo mental health, and love for Jesus in abundant appreciation for being the one who sets us free.

Go ahead, take a Jesus pill today, but never forget it if He takes your begging seriously, and frees your mind from demons, because if we forget Him, He can stop protecting us from mental ills until we are swamped all over again with "darkness." Free from demons = Light in the mind. White clouds feel beautiful again. The sky is thrilling again, and new life calls. But when one's mind is "down," a catch-all term for all sorts of spiritual debilitations, the sky may as well be an outcast, or, at best, a riddle. When God is in the mind, we can see Him in the sky.

To have God in your mind, you've got to talk to Him. He will show Himself when our talk is right. Get your God-talk right. Try-try again until you get it right. God is not cheap, so cheap-talk won't cut it for Him. He is magnificent, more capable than all the world's computers put together. His "brain" in unimaginable. I love you, Great One. Can I make you happy tonight with my talk? Rather than me ask You for something, how about You ask me for something. How can I bless Your heart, Sir? I'm not very capable, and I'm not very Light-ful, but if I could just be part with You, if I could just be counted as one of your children, your loyal children, I would really appreciate that. Thank you, Jesus, for the day of pain you suffered for the family of God, won't you please accept me by the forgiveness you offer freely for those who chose to become your brothers. Be with me to remind me of all you want me to be. If we can't be one on the same team, then I won't be happy. I want what you promise. I want that life you spoke of. You are the only thing that matters.

There you have an example of intimate talk that get's God's attention. See what's in your heart that God might appreciate. Then watch for the signs that God gives you secretly to show that He heard your prayer, and likes what you said. This is our secret victory over the fools of this planet. They remain in darkness because they don't want to ask God for anything...except maybe to bail them out of unexpected troubles.

Or, if you think that you don't like God because you think He amounts to a lame (unexciting) life for you, then you have the choice of feeding before the serpent of modern drugs...that drain you deeper into mental self-butchery:

People who depend on drugs for mental stability are taking a dagger to their own minds by rejecting the Healer and opening doors to demons. That's the bottom line. Once a person starts to believe that they have a "mental disorder," demons screw them up further with all sorts of mental gymnastics. My understanding is that evil spirits can cause both thoughts and mental visions / dreams to influence our beliefs on many matters. Give them an inch, and they'll put a foot into the door so you can't close it back up.

The video above, by the way, is a high-caliber documentary, The Marketing of Madness, of over two hours. It mentions Robert SPITZER (died 2015), an up-and-coming, influential psychologist pimp at the time of my spider event that pointed to graphene-oxide in vaccines. The Spitzer surname is like "spider," and I still seek a person with Spitzer surname who can apply to the spider event. The documentary is a great backdrop to Fauci-ism. Is a form of graphene-oxide in the body a mind-controlling facilitator? Just asking. Are sinister psychology departments secretly pushing graphene-oxide? Just asking.

The spider event took place on Doner street in Gormley, Ontario. I've just looked up Doners to find a green "serpent" entwined around a sword in the way that medical fields use an Asclepios rod (snake entwined around a rod). The Spitzer unicorn, on a red Shield, is much like the horse of Jewish Rothschilds (not "Rothschild), for both are rare in that they face the viewer's right side. The Spitzer unicorn is "facing the sinister" (left side of the Shield is viewer's right). The Hall-of-Names description of Jewish Rothschilds mentions "sinister twice" (once the German Rothchilds pop up, click the arrow beside "German" to get to the Jewish Rothchilds).

The uniCORN is part-code for the Ceraunii Illyrians (probably of Greek origin), from the entity represented by mythical Coronis (Greek myth), mother of Asclepios. Is it coincidental, therefore, that Spitzers use a unicorn? Consider that myth made the Illyrians children of CADmus and HARMONia, when they left Greece, who were turned into two snakes at that time. These two snakes have been deciphered by myself as the two snakes of the HERMes CADuceus, a symbol almost identical to the Asclepios rod. Hermes was code for ARMENians, and so was HARMONia, but more in particular, Harmonia can be gleaned as code for CADUSii-branch Armenians at mount HERMON in Phoenicia, for "CadMUS," a king of Phoenicia, can be deciphered as part-code for Cadusii, and part-code for Mus in Armenia. The ARAS river in Armenia named Ares, a detestable mythical figure, father of Harmonia above. The Phoenicians around mount Hermon migrated to the land of Taphians at Calydon, where there was a mythical Oeneus, and then the Ceraunii above were at or beside the Oeneus river of Illyrium. The Cadusii look like they named Kadesh in Syria, in Amorite territory.

Crime minister trudeau thinks it's better if the youth do marijuana freely, without feeling as though they are doing evil. He doesn't care what the parents want for the kids; he's made it legal for kids to smoke pot in public, and parents can have less say because word goes around that it's just dandy to smoke pot. But drugs do not allow the youth to deal with their problems so as to become strong adults. Instead, drugs are a cop-out, a waste of humanity along with a waste of time. Adding a pot evil in the midst of youth who tend toward alcoholic bashes cannot pose promise for a better world in the next generation. God gives life difficulties to make us grow into strong-stemmed plants, useful for producing fruit, but trudeau is contributing to teenage wasteland all over again, even though he knows that marijuana is, for some, a stepping stone to harder drugs.

People, especially teens who have got very-little knowledge of adult life and its many difficulties and challenges, do not need drugs to obstruct them, to trip them up, to make them less capable of overcoming their challenges. trudeau is guilty of being a reckless fool. Let Jesus be your sure crutch, and don't feel inferior for having a Crutch, if you need One, because Jesus as your crutch will get you to the point where you are no longer crippled. And it could be very fast for you. Lean on Jesus, and you will not be disappointed. Get under His feathers, and be safe. Learn from Him, and know the good life. Experience God in your heart, mind and soul. Let God be your fuel as you become accustomed to the cleansing and transforming words of Jesus. Drop your old life today, and start a new one in Jesus, that's what it's all about, which the powers in this world do not want you to know.

Beautiful on the mountains are the feet of the One who brings good news. The beauty about God's plan is that while demons in our midst seek to frustrate and pain humans, and to leave them in the dark without God's healing powers, they inadvertently cause some crippled souls to throw themselves upon Jesus, in a moment, and then a flash goes off, and they see what they've been missing, and they give up the old lifestyle for Jesus, and the demons thus contribute to their salvation, how perfect for paining them in return. Pain your demons today by turning things over to the Jesus Way. Make them screech and hiss.

This is a great but sad video featuring a courageous, caring and persecuted doctor:

The doctor above says that more than 50-percent of his vaccinated patients had blood clots, suggesting that people will die sooner one day than would otherwise be the case, the near "perfect crime" for the population-control fiends. There is no doubt that murder is going on deliberately, the signs are everywhere, and the fiends need to strategize continually to keep the cat in the bag as much as possible, yet they had been sloppy / disorganized partially due to over-confidence in having lots of "right" people in power positions, and partially because this is a huge task that can't be kept fully under control as desired. A federal court has re-instated Biden's vaccine mandates this week, a very horrible ruling that will pain many who wish not to risk the shots.

They are going in for the kill of children. The CDC panel this week put out an "admission" that some children have died / gotten sick already, but only a handful in numbers to give the impression that a few have suffered. Even so, it is necessary for Fauci and his bosses to cease vaccinations the moment three of four children suffer severely. But they are not stopping, which is how you know they have orders from the top to continue the slaughter, but also how you know that billions of "trusted" doctors, police, judges, national leaders, and people from all fields, are evil at the bottom line. About half the world seems to be tolerating this slaughter without a care, and supporting it even.

I personally don't know what I can do to help. The ignorant seem to be cemented in their ways, and the only thing that will change their minds is the killing and miming of people close to them, afterwhich the damage has already been done. The anti-Christ media is responsible for hiding the slaughter as best it can, but, later, this slaughter will ease up on Christian persecution, I think, when the perpetrators receive a backlash that will last years. During this backlash, they will ease up on such measures in spite of enforcing the mark of the beast, or so goes my hope. In other words, my hope is that while they continue forward with denying the people purchasing power, they will not go about full-throttle persecuting those who resist the vaccines after learning that the people will bite them back for trying it. Many persecuted people are family members of the ignorant ones.

It is one thing altogether different if our side would wish the other side dead because it is they who are part of the murder-machine. Having such sentiments is not equal to their side wishing our side dead for doing no wrong but speaking out against the slaughter. I wish Biden dead if he does not call to an end of vaccinations. This sentiment is a natural reaction to the mass-murder reality. I do not wish him in prison only, but dead by a terrible disease that pains him, and makes an example of him. That's how I see the mind of God working, and if we have the mind of Christ, we will feel the same way. They would kill us for thinking in this way, but we are the innocent ones, a big difference. Warning to trudeau; God is a killer, and will kill painfully as the crime deserves, after much patience has run out. During His patience, He is measuring-out your "reward" by how deeply you commit the crimes.

I watched a video from a wacky pre-tribulationist who pretends to care for his Christian audience, who claims as a sure fact that forced vaccinations are paving the way to the mark of the beast. I'm not willing to say that. It sure looks like it can be, but I can't know for sure. I suggest we prepare our foods and other needs as much as doesn't risk too much, unless you want to take the risks, just in case the situation now is indeed the lead-up to the mark in the 2020's. Storing food poses no risk at all to finances as long as foods are stored correctly. Storing food is just a wise thing at this time. I continue to dry foods, with carrots and squash over the stove burner as I speak.

The election-fraud events since last year have exposed further to us that the enemy is strong enough to commit crimes openly and get away with them. It's a warning to us as to what the political situation is that brings on Christian-targeted persecution. This is not to be treated lightly, and pastors need to talk about it to their people. Some are, we can be sure, but even the pre-trib man I refer to above is shocked that churches aren't openly talking about vaccine mandates leading to a mark-of-the-beast-like situation. But the wise amongst us who watch prophecy know. All I can do to help is to say: store up your needs. Forget vacations and restaurants; save all your money and store the oil in your jars, so to speak, so that when the night arrives, you will be more ready. If you don't have money, try drying foods for others who don't have time, for a little food in return for your labor. A stay-at-home mom home-schooling kids will find it overwhelming to dry foods too; ask if she'd like you to dry them for her.

Gateway Pundit headline: "NFL Revises Covid-19 Protocols, Will End Mandatory Testing For Asymptomatic Fully Vaccinated Players". SURE, because vaccinated people are testing positive like rabies spreading through the pack of wolves, and the NFL feels responsible because it forced them all to become vaccinated. So, stop testing because it's incriminating as well as embarrassing. The CDC even said months ago that it was going to end testing, and now we know why.

So far, the biggest COVID losers by far are those who ran to get vaccinated, the ones who loved the idea of wearing masks to rub things into the faces of normal, Christian-leaning people. These people now stand the chance of dying by complications due to blood clots and who knows what more, as they continue to get shots. Some of the useful idiots will have changes of heart when they see the many complications, the failed promises, and the escalation of insanity, but some, even though they themselves fall to permanent injury, are so deluded of soul they insist on vaccinating even their own children in order to raw-raw-raw the team forward with this scheme. Souls of pollution, they drink from the fountain of the cesspool. trudeau turns the tap on, and canadians swallow the sewage, praising father government like the mothers of all idiots.

There is little hope here, folks, truly, there are too many demon-obeying souls by now. canadian churches kept their mouths shut as demons took hold of people all around. Church leaders were hoping things would just stay calm forever, until the pre-tribulation rapture. They are going to feel the heat of their folly, I hate to say it. By the looks of things now, they will be responsible for delivering their congregants to the fangs of the mark of the beast. I wouldn't want to be such a pastor.

Christians who say that the USA is one nation under God are delusional. Are they blind? There's a big difference between wanting something and the reality. The reality is, the USA is the most-spiritually polluted, the most-wicked nation, in the world. God wouldn't touch it with a billion-mile pole. Get real, Q-Christians, you are deceiving your fellow believers. God's not interested in a country ruled by Washington masons, get real, DOPES. Brothers you are not if you continue in this vain vein. God does not put his throne in end-time Sodom, get real. Beware the Kingdom-Now believers who think that they are going to defeat the devil so that Jesus can return. Beware the phonies who make you think that things are about to get better again at any instant. These are the Q-Christians who hold out a promise not from God. Prepare your heart and means for persecution, in case it arrives in a few years, or earlier. There will be NO REVIVAL in the last seven years, probably not even in the last 10 years.

The attempt by Nova Scotia to ban people from grocery stores a couple of weeks ago has already died a quick death, making it appear that there's not going to be a vaccine ban on groceries this winter, but don't celebrate because there's always next winter, and there's going to be more pilot projects conducted by the goons this year so that they can engineer a workable plot for denying food in efforts to vaccinate the entire globe.

There is more we will need besides food. When they disallow entry into building centers, who will you get to go buy you a water pump, or shingles for the roof? If appliance stores lock you out, who will buy you a new stove? These are the things we need to consider starting now. We would do well to spread the news that wealthy Christians ought to get back-up items purchased soon, the sooner the better. If you know you're going to need a pump, a fridge, a toilet tank, a washing machine, in the next ten years, it won't hurt to buy those things on-the-cheap side now/soon. If societal conditions get better in the future, you can give those things away to your kids, if you want better quality, or even to your aged parents who don't need better quality.

I've seem Mark Dice do this before with another issue, to show how liberals are very happy to see us in jail or worse, for resisting faked globalist plots:

Here's a list of videos / articles that anyone can find useful if ever one needs to take a government / corporate body to court due to de-facto enforced vaccinations. One can file a suit before or after the vaccination, whether the vaccination does harm or not, because vaccinations are not approvable, and may do long-term damage, especially via the increase of artery blockages. There has been a long-standing law saying that a person cannot be forced to take an injection. Blocking entry into vital stores is a defacto forced injection:


Here's all four Gospels wrapped into one story.

For Some Prophetic Proof for Jesus as the Predicted Son of God.
Also, you might like this related video:

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