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## WHY THE APOPHIS ASTEROID SHOULD HIT EARTH in 2029

(Written 2024)

Back in Texas in the early 2000's, I figured that one could find the sun's distance from earth by following two eclipse lines to its one edge, for the two lines meet at that edge. All I had to do was to discover the angles of both lines, and I could then find how far from the earth they meet.

One line would be a lunar-eclipse line, and the other would need to be a line for specially-chosen solar eclipse that had a similarity with the lunar eclipse. More specifically, I needed to know how large the lunar umbra was upon the earth for a solar eclipse where the sun was the same size, as seen by the eye in the sky, as the sun is during the lunar eclipse. However, I wasn't able to find the data for a solar eclipse, to my satisfaction, to any total lunar eclipse at my disposal from NASA's eclipse pages.

However, I was a novice at this sort of thing at the time. Years later, upon tackling this method again, I realized that the solar-eclipse line wasn't needed at all. It was a beautiful realization than any astronomer would know off-hand in a second. It was at the time of this discovery when the further realization occurred: astronomers have been forced by the establishment not to use eclipse lines to discover the distance to the sun. That's why you have never heard of this idea. The only question is: how has the establishment convinced astronomers not to talk about this?

People in the past were unable to use this method, but with NASA's data online, we now can. This is such a simple and absolute full-proof method. If it doesn't work, then the system would have told you, "by the way, it doesn't work, and here's why." Instead, DEAD SILENCE. It's full-proof because the geometry and math needed is not flexible. The two lines both begin at the earth, and they meet where they meet, full-stop. You cannot change where they meet by one inch. That's what I mean by they meet where they meet. They meet where they meet according to NASA data. If they meet far less than 93 million miles from earth, then NASA is responsible to explain why this is so, because the lines are drawn according to NASA data, and nothing more. All astronomers must explain why they meet far less than 93 million miles away.

One lunar eclipse line goes to the edge of the sun. It doesn't matter which edge because the solar line will go to the same spot exactly. By "edge," I mean the edge we see when we are looking at the sun from earth. The outer perimeter, that is, is the edge. It doesn't matter where on this solar circumference we draw the lunar-eclipse line to, because we are drawing the solar line to exactly the same spot, for the entire circumference is exactly the same distance from the earth.

All we need to do is find the angle of each line, and online calculators are able to tell us how far into space they will meet. Done. Finished. It tells us the true distance to the sun according to NASA lunar-eclipse data. We can even find the solar distance manually, without the calculators.

In order to find the angle of the lunar-eclipse line, we need only the distance to the moon during the eclipse. Everything else is, or can be, known. And that's why NASA does not put the distance to the moon on its lunar-eclipse pages, because it doesn't want anyone to know the true solar distance...which it itself has known all along by this very method.

I had to shut down my quest to find the solar distance, by this method, in the early 2000s, because I didn't know how to find the distance to the moon for any one lunar eclipse. But, later, I found a way, and I'm going to show it to you. Here's the first drawing I've made for you, and for any astronomer willing to talk about it:
http://www.tribwatch.com/photos/eclipseDraw1.jpg

Another thing the NASA pages fail to show plainly is the width of the earth's umbra, in miles/kilometers, during the eclipse. But the pages do offer the umbra size as seen by the eye, and I've found a way to discover the size in miles from that data. Any astronomer knows how to use the so-called "apparent diameter" -- the size as seen by the eye -- to find the size in miles. But I at first did not, back in the early 2000s.

And any astronomer knows how to use the apparent size of the moon to find how far it is from earth. Therefore, as NASA provides the apparent radius of the moon during any eclipse, what is the excuse of astronomers, worldwide, for not using this data to discover the sun's distance by it? If all one needs to find, to find the solar distance, are three things -- the lunar distance from earth, the umbra size in miles, and the angle of the lunar-eclipse line -- why have astronomers not used them and reported back their findings???

I can put this another way: all one needs to know to find the solar distance is one thing, the lunar distance during any eclipse. I can put it this way because NASA pages provide the umbra width, in miles, in two ways, and these two things together are what provide the angle of the lunar-eclipse line. That's all that's needed to be found because the solar line is known from the NASA page without need to find it.

In fact, as the distance to the moon during an eclipse is hidden on the NASA page, EVERYTHING we need is on the page, and so there are ZERO THINGS we need to hunt down by some other methods. All we need are NASA pages as they are have been showing for many years, to this day.

We cannot know how wide the umbra is, where the moon passes through it during a lunar eclipse, unless we know: 1) how far the moon is at the time, or; 2) how much time goes by for the moon to cross it fully. And we cannot know the angle of a lunar eclipse line to the sun's edge without the diameter of the umbra in miles/kilometers. Here's one of NASA's lunar-eclipse pages:
<https://eclipse.gsfc.nasa.gov/LEplot/LEplot2001/LE2011Jun15T.pdf

Can you spot anything on that page that could indicate to you what the angle is of the umbra line? The umbra line in my drawing is what I mean by "lunar-eclipse line." The line is from the outer edge of the sun to the outer edge of the umbra, and of course it crosses the outer edge of the earth.

In Drawing 2 below, I show how to find the angle to the solar edge from the lunar-eclipse line, by forming a simple, right-hand triangle, shown in red:
http://www.tribwatch.com/photos/eclipseDraw2.jpg

I don't think Drawing 1 needs any further explanation other than what it tells, and so let's begin with some elaboration for Drawing 2. If we can just find the angle of the lunar-eclipse line where the arrow points, we can have the distance to the sun because everything else that's needed is knowable. For example, NASA provides the angular radius of the moon at the peak of the eclipse as .266 degree, though the page doesn't use .266 degree. If it did, it would be easier for anyone to realize that, hey, maybe we can find the distance to the sun with this piece of data.

Instead of .266 degree, it shows: "S.D. = 00° 15' 57.2". You can ask google to convert those numbers to plain degrees, or use this calculator below where you put 00 in the top box, 15 in the middle box, and 57.2 in the bottom box to get .26589 degrees. I'm using .266, good enough.
https://www.rapidtables.com/convert/number/degrees-minutes-seconds-to-degrees.html

As the moon diameter in this eclipse was .532 (twice .266) degrees, all we need to do, in step one, to find the distance to the moon (at the eclipse), is to divide 360 degrees (a full circle) by .532. This gives the number of lunar diameters, back-to-back, in a perfect circle around the earth. That's 360 / .532 = 676.69 moons. As each moon is 2,159 miles wide, that perfect (but imaginary) circle is 1,460,977 miles in circumference. The earth is smack at the center of this circle. How can we use this circle to find the lunar distance? You maybe need a grade 7 education to figure it our from here. That is, every astronomer knows how.

To find the diameter of a circle, it's the circumference divided by 3.14 (pi). So, 1,460,977 / 3.14 = 465,279 miles. The reason we want the diameter is because the moon, at the eclipse, is exactly half as far, or 465,279 / 2 = 232,640 miles. You maybe could have done this in grade 6. That's it, we're done. We have found the distance to the moon at this eclipse. I wish this easy method had dawned on me in the early 2000s. We can now find the angle of the lunar-umbra line to the solar edge.

Here's a drawing as per this lunar-distance situation, if you need it to clarify in your mind what I'm doing:
http://www.tribwatch.com/photos/eclipseDraw3MoonCircle.jpg

Drawing 2 explains that NASA's given apparent size for the umbra radius, .7256 degree, is 2.728 times greater than the moon's .266 degree. From this, we easily find the umbra width, in miles, as: 1,079.5 miles (lunar radius) x 2.728 = 2,945 miles.

Just put the lunar distance, 232640 (miles), into the 'b' box of the calculator below, with 1015 (miles) in box 'a'. The latter number is on Drawing 2 as part of a red, right-angle triangle that I've drawn upon it. This triangle is needed to find the mystery angle of the lunar-eclipse line.
https://www.calculator.net/right-triangle-calculator.html

Upon hitting calculate with the numbers above in the two boxes, and all other boxes cleared even of periods and commas, the mystery angle is given as .25 degree on the alpha-angle line (above the numbers in the boxes). Drawing 2 uses this .25 as .24997 in case the reader thinks .25 is too approximated. This angle is what we want where Drawing 2 asks you to find the angle of the tip of the red triangle. (To get more decimal places than .25, use the calculator's alternative alpha-angle result as 0.0043629 radians. Ask google to convert those radians, and you'll find them converted to .24998 degree.)

I was very happy to see what a simple matter it was to discover the angle of the solar line, shown in Drawing 2. NASA gives "S.D. = 00° 15' 44.72" for its radius, where "S.D" stands for "semi-diameter." Why doesn't the page just use "radius"? Most people on the street don't know what "S.D." means. But if they used "radius" instead, people might figure out how to find the distance to the sun using this piece of data along with the S.D. numbers given for the moon's size. Ya see? NASA is trying to hide these things from you and I. And not a wonder. I found what "S.D." means from a NASA page explaining eclipses, otherwise I would have been in the dark.

We can now go back to the calculator above that converts NASA's S.D. in arcminutes/arcseconds to degrees. We can put the three numbers, 00° 15' 44.72", into its three boxes, then hit "Convert" to find that they are equivalent to .2624 degree, which is the number on Drawing 2.

That is, while the lunar-eclipse line rises toward the sun at an angle of .24997 degree, the solar line rises at .2624 degree. The latter rises more steeply so that it will eventually contact the lunar-eclipse line. We want only to know how far from the earth the solar line contacts the lunar-eclipse line. That's going to be at the sun's edge. We are almost done; we almost know the distance to the sun according to this NASA eclipse page. Aren't you excited?

See Drawings 4 and 5 before going on to see better where I'm going: http://www.tribwatch.com/photos/eclipseDraw4.jpg
http://www.tribwatch.com/photos/eclipseDraw5.jpg

To finish off this calculation, we just put 1 mile into the right-angle-triangle calculator above. Put the number, 1, into the 'b' box. Then, put .0125 into the alpha box because .2624 degree - .2499 degree = .0125 degree. Clear all other boxes but these two. What does this .0125 degree mean in this picture?

It means that the solar line rises .0125 degree more than the lunar-eclipse line, per every mile toward the sun. Therefore, it means that the solar line approaches .0125 degree closer to the lunar-eclipse line for every mile toward the sun. From this information, how many increments of one mile toward the sun will there be before the solar line contacts the lunar-eclipse line? To know the answer, we need to convert .0125 degree to miles. And that's why we have 1 mile in box 'b' of the calculator, and .0125 in the alpha-angle box.

When we hit "Calculate," we find the result for the 'a' box, though the result is shown above that box, at line 'a', as 0.00021817 mile (= about 1.15 feet). In other words, the solar line catches up to the lunar-umbra line by 0.00021817 mile, per every 1 mile forward toward the sun. How can we use this information to discover the number of miles into space before the lines meet? As easy as knowing the earth's radius.

As the two lines begin one earth radius apart, which is 3,960 miles, we know what we need to do here: 3,960 / 0.00021817 = 18,156,800 miles. That is, .00021817 x 18.1568 million miles forward gets 3,960 miles perpendicular to the forward advance. How else can I put it: the two lines close in on each other by 3,960 miles after they both stretch 18.16 million miles into space. After 18.16 million miles forward, the two lines meet.

This 18.16 figure is not from my numbers, but from NASA's. All I did was the math. I checked a dozen or so eclipses very carefully, and they all gave solar distances in this very range, mainly between 17 to 19 million.

One way for science departments to make astronomers ignore this method of finding the solar distance is to have other methods that are pushed as more reliable which get 93 million miles. The trick is to make the eclipse method seem unreliable as per sunlight passing through the atmosphere and thus becoming bent (refracted), which supposedly changes the shape of the umbra as compared with its shape if the earth had no atmosphere at all. However, there's a couple of arguments to make against this claim, starting with the likelihood that gases, including air, cannot refract light. Only liquids and solids can.

You will find people claiming that air refracts light, but this may be a myth from the astronomy establishment because it needs to nullify the eclipse method of finding the solar distance. Like me, you have probably never heard that the sun moves at a different speed, when at the horizon, as opposed to when it's directly above our heads. But if air refracts light, then the sun on and near the horizon should, as seen by the eye, appear to move faster than it really is, if the refraction is becoming continuously less pronounced with distance from the horizon.

In water or liquid, there is no refraction of light when it strikes dead-on. When the sun is directly above our head, that's a dead-on strike. Therefore, if air refracts light, it does so at a maximum when the sun is on the horizon, with gradually less refraction with height from the horizon until there is zero refraction above our heads.

But as I've never heard that the sun moves faster at the horizon, I take this as evidence of no refraction in air to speak of. Some online: "The Earth rotates 360 degrees in 24 hours or 1 degree in 4 mins." That's equivalent to .25 degree per minute, and as the sun is about .5 degree wide, it takes about two minutes for the sun to show fully, sitting on the horizon, from when it first touches its front edge at/under the horizon. Yet, someone says: "The Sun travels across one-degree longitude in 4 mins." There's nothing said that it does so at a different speed on the horizon versus overhead at noon. If there is no change is speed, then light refraction by the atmosphere is a falsification known by astronomy and other sciences in spite of their claiming that refraction does occur.

Someone at quora:

When the Sun is near the horizon, its movement appears to speed up because of the angle at which we observe it. This phenomenon is known as the "setting sun illusion" and is a result of the Earth's rotation combined with the curvature of the Earth.

There we have it, it's an illusion. In other words, it doesn't really move faster toward the horizon, and therefore does not move faster upwards from the horizon. The astronomy establishment cannot claim, therefore, that the solar distance cannot be found by using a lunar-eclipse line. Even if there is an ever-so-slight alteration of the umbra shape due to wee-wee refraction, it's probably not going to change the 18.16 million figure above by much. The bulk of the air is only 15 miles thick as compared to an 8,000 mile planet. It should be reliable to say that there's no refraction above 15 miles.

The chief astronomers need to come clean here, but as they are by-and-large anti-Christ, globalist pigs, don't expect them to be reasonable or helpful. Pigs will be pigs. Even if the sunlight alters the umbra radius by the full 15 miles, even 20, the math above will need 3,980 miles instead of 3,960, which is not going to make much of a difference in the solar-distance result.

I now implore you to think about something. Astronomy has announced that the Apophis asteroid will be passing the earth by a near-hit of under 20,000 miles, in 2029. As 18 million is 1/5th of 93 million, it means that, according to astronomy's own data, the solar system is five times smaller than they say it is. This means that, when they calculated to velocity of Apophis, they got it five times too fast.

Plus, if they sized this rock by comparing it to the size of a moon or planet, or even to the sun, they also assessed it five times to large, meaning that they assessed it with a total of 5 x 5 = 25 times more momentum than it actually has. For momentum is equal to velocity x mass. If the solar system is five times smaller than they think it is, every moon and planet (aside from earth and its moon) are five times smaller than they think they are.

That's why I want you to think hard, for momentum is the only thing which can keep the asteroid from being pulled in by earth gravity in a near-hit situation. TWENTY-FIVE TIMES less momentum, and that's not all. On top of this, the 20,000 miles they estimated is in accordance with the size that they think the solar system is, meaning that it can be passing by at just 4,000 miles from the ground instead. Five times closer, five times slower, and five times smaller/lighter. If that doesn't give you concern.

This space rock looks like it's going to surprise the world. Four-thousand miles is just one earth radius. I think it's going to hit, because, as the world is, it seems that the timing is right to fulfill the asteroid in the Book of Revelation: the 2nd Trumpet, described as a mountain-like item all ablaze from the sky, and landing in the sea.

I imagine that food will become scarce at that time, explaining the rationing that comes about at the 3rd Seal, which is a plague to take place shortly after the 2nd Trumpet, for the 21 plagues of Revelation have this sequence: 1 Seal, 1 Trumpet, 1 Bowl, 2S, 2T, 2B, 3S, 3T...7 S, 7T, 7B. It may all be a mistake, but I think people ought to begin saving foods now for that time (I'm writing this is early 2024).

Apophis is too small for to use triangulation to find its distance from earth. The stated distance to Venus, which is an erroneous distance, was supposedly "discovered" using the "transit of Venus" across the sun, which tends to indicate that Venus is too small for triangulation calculations. In that case, the same should be true for Apophis.

Triangulation uses two telescopes at a known distance apart on the earth, with each telescope measuring the angle to a rock at the same time. This forms the three sides of a triangle, and the distance to the rock is the height of the triangle...which is figured easily when the two angles to the rock are known from both sides of the triangle's base line on the earth. But Apophis is so small that it wasn't discovered until 2004. It nears the earth every 8 years, roughly, and so it's come nearest only two times after initial discovery.

An asteroid, in case you don't know, circles the sun near-circular like a planet, versus the long oval of a comet. Apophis has an orbit very near the earth orbit. It circles the sun once in .89 earth year. The calculated velocity of the asteroid is based on a sun 93 million miles away, but if the sun is 5 times closer, then the orbital distance is 5 times smaller, making the velocity 5 times smaller.

Plus, as the asteroid circles the sun in the same direction as the earth, it approaches the earth from behind, wherefore it's not moving very fast as compared to the earth. Without doing any articulate thinking about it, it seems correct that, as the orbital period of Apophis is .89 year, it's moving about 13 percent faster than the earth. That is, after the earth has done one perfect orbit, the asteroid has done 1.13 orbits, or one orbit plus 13-percent of its next orbit. This well explains why the comets returns to nearness with earth every 7.5 years on average, for 13 percent x 7.5 is almost 100 percent. That is, every 7.5 earth orbits, Apophis does an extra one, or 8.5, bringing the two close to each other.

It seems to me that these numbers reveal the speed of Apophis to be 13 percent faster than the earth. But these numbers then conflict with the astronomy's numbers as follows. With the earth moving around the sun at 66,600 mph (note the 666), as astronomy claims, Apophis at 13-percent faster works out to 75,260 mph. As the two orbit the sun in the same direction, they would collide at 75,260 - 67,000 = 8,660 mph, ouch, that is very fast, predicting huge devastation, but it only gets worse with other estimates from astronomy.

Am I right? Yes, if the two move in the same direction with Apophis sneaking up from behind, the asteroid would enter the atmosphere at about 8,700 mph, if we use their numbers here. But, as they think the earth is moving 66,600 mph, we must divide it by 5 to get the truer speed of about 13,320 mph, and then add 13 percent (multiply by 1.13) of it for Apophis, or 15,050 mph. Now, the fly-by or impact speed becomes only 15,050 - 13,320 = 1,730 mph.

Let me show you an amazing thing I've just found, because the claim from astronomy is that Apophis will fly-by at 2,075 mph, almost the 1,730 I calculated with a solar system 5 times smaller. As I can't see how they justify 2,075 using the size of their solar system, I'm getting the impression that they secretly know the solar system to be about 5 times smaller. I'm also inclined to accuse that chief astronomers in-the-know will not reveal how they arrive to the 2,075 number.

There are reports of a speed of 30.7 kilometers per second = 68,675 mph for Apophis. But how did they arrive to this number? It doesn't work with a sun 93 million miles away. We now get an approximate, relative speed, for the asteroid, of 68,675 - 66,600 = 2,075 mph. They have Apophis moving only .3 percent faster than earth. How can that be?

If we imagine the asteroid moving 100-percent faster (= twice as fast), it does two solar orbits for every one of earth. And if it moves 50-percent faster, it does 1.5 orbits per 1.0 of earth. See the relationship 50 percent versus 1.5? They both have 5s. Or, if it moves 25-percent faster, it does 1.25 orbits per 1.0 of earth. Now do you see the relationship? Good. Therefore, if it moves 3-percent faster, it does only 1.03 orbits for every earth orbit, in which case it would arrive near to the earth only once per every 33 years roughly. The reality is that it approaches once every 7.5 years on average.

Therefore, how did they choose an asteroid speed of 30.7 kilometers per second??? It's incorrect according to the size of their solar system. The astronomers are indeed lying to us about the size of the solar system, and they do not wish to inform the people on how they derived the speed of 68,675 mph for Apophis. The reason they use this speed, I assume, is to provide the truer impact force so that science departments all over the globe can prepare more properly for the fall-out should the asteroid ever strike.

However, there is yet the question on whether they have properly sized this asteroid. If they have calculated the truer velocity in accordance with a solar system 5 times smaller, perhaps they do have its truer size too, which is not good news because it would have been better, for impact reasons, had it been 5 times smaller. On the other hand, the larger it is, the less chance that earth gravity can pull it in for impact. On the other hand, if this is the rock of the 2nd Trumpet, it will strike.

Few of those who report on this asteroid know the true size of the solar system. Therefore, we would expect them to report a different velocity than the 68,675 mph above. This figure is wrong, not only as compared to the reality, but as compared to the velocity predicted by a 93-million-mile solar system. This number has been reduced only slightly from that prediction (should be over 75,000), yet it's been reduced such that the relative speed between asteroid and earth is nearly what I predict with an 18-million mile solar system.

I have not yet concluded that the sun is 18 million miles away, on average, but rather this is a tentative number based on NASA data, which cannot be trusted for perfection. I was able to find that NASA's eclipse data (as reported to the public) is at least slightly imperfect.

The webpage below has a graph (see purple line) showing the "relative" velocity at 5.9 kilometers per second, or 13,200 mph, early on the day that the asteroid crosses the earth. Is this nuts or what? How possibly can the number be that high? It then predicts a speed of 7.4 k/s as the asteroid is closest to the earth, or 16,550 mph.

There are several webpages reporting that Apophis will achieve 12.6 kilometers per second, or 28,190 mph, relative to the earth: "Apophis has a mass of 2.7*10^10 kg and is expected to be traveling at a speed of 12.6 km/s relative to the Earth at the time of closet approach."

The page above tells what others do, that the predicted path will be almost central, above and along the earth's equator, that is. That's the worst-possible scenario for allowing gravity to pull it in. The prediction is that it will be closest above western Africa while headed toward the Americas. But, if could strike before reaching Africa altogether.

Claim: "Based on Goldstone and Arecibo radar images taken in 2012–2013, Brozovic et al. have estimated that Apophis is an elongated object 450 × 170 metres in size,..." I have no knowledge whatsoever on this radar technique, but I do have cause to be suspicious of it, because astronomy establishments report what they choose to report, usually erroneous. Others say from NASA that it's 340 meters wide, and the page above: "...a 250 to 300 meter wide asteroid". If they all can't agree, that's how we know they really don't know how to measure it accurately.

Radar, I assume, is shot to the asteroid, then bounced back to earth, or to a spacecraft. The sizing of the rock is then partly in accordance to the distance between it and the radar source. But if they are working on the basis of a 93-million mile sun, then this asteroid is being sized 5 times too large. That's good news for its much-reduced impact force, but bad news for the globalists because it's 5 times easier for earth gravity to pull it in for a strike.

The silver lining in this asteroid is that it will close the giant gap between rich and poor. Many companies will go bust, but not until the stock market is totally destroyed will God be happy. Not until corporations treat the workers and buyers with charity, rather then maximized prices for the sake of greedy shareholders, will God be happy.

Wikipedia:

Based upon the observed brightness [is that all?], Apophis's diameter was initially estimated at 450 metres (1,480 ft); a more refined estimate based on spectroscopic observations at NASA's Infrared Telescope Facility in Hawaii by Binzel, Rivkin, Bus, and Tokunaga (2005) is 350 metres (1,150 ft).

When one sizes a rock based on the observed brightness, the size of the solar system is a factor in the result. Therefore, this rock ought to be less than 450 / 5 = 90 meters wide.

Wikipedia: "The maximum apparent angular diameter will be approximately 2 arcseconds." That apparent diameter is equivalent to .00056 degrees, and where the moon's apparent diameter (as seen by the eye) is about .5 degree, this rock, as they report it, will be visible to the eye as: .5 / .00056 = 890 times smaller than the moon's diameter.

As they have it 5 times larger as well as 5 times further away than the reality, then the reality remains at 2 arcseconds of width (at its closest approach). However, if it's trajectory is directly toward earth, this rock will grow in size/brightness and put the fear of God in all inhabitants of the earth. Right? One would think.

The reckoning time will arrive, yet, ironically, the Bible tells that the earth will yet refuse to repent, right up to the very end of the 1,260 days. If it were otherwise, I don't think God would send the 2nd-Trumpet rock in the first place.

The .pdf page above: "Apophis will spend only 2 hours at an altitude below 40,000 km from the Earth surface." Not so where it's moving 5 times slower. That ramps it up to 10 hours on a near-miss situation, giving gravity 5 times longer to suck it in. Plus, there are many more hours, as it approaches from 10s of 1,000s of miles off, for gravity to bring it down.

The same page tells that there is a plan to land men on Apophis in 2029, but, for me, this sounds as though NASA is plotting computer animations and/or staged productions to feign such a landing. It is not possible to land a craft backward on a planet, or they would be doing the same on the earth in full view of us all. I do not believe that SpaceX rockets land backward on earth, but believe that Musk is working with NASA to deceive the world. When I see a planetary / moon lander being landed on earth successfully, I will change my tune, but, without doubt, sufficient technology did not exist when they pretended to land men on the moon.

Same .pdf page: "Twenty four hours before the closest approach, Apophis velocity relative to the Earth is still under 6 km/s, its distance to the Earth being 500,000 km." That's a speed of earth approach of 13,400 mph.

Same page: "The orbit of Apophis with respect to the Earth during its April 2029 flyby will be...over the Earth equator and a perigee altitude between 29,500 and 33,800 kilometers over the Atlantic Ocean (42.9°W, 29.0°N)." When we convert the 29,500 kilometers, it's 18,330 miles, considerable less than 20,000. Cutting 18,330 by 5 gets us significantly less than one earth radius, meaning that the people telling us that this rock DEFINITELY won't hit the earth are lunatics. They should be striving to get the people to store foods, starting now (2024), but, instead, the globalists are trying to destroy farms, starting a year or two ago. Do you see the devil in the details?

The statement above claims that the asteroid will be closest to the earth when it's also closest to the sun for this particular orbit, for "perigee" is defined as the orbit's closest distance to the sun. It's a wild "coincidence," but then maybe not in God eyes. There may be a reason that it strikes at perigee. Solar gravity is strongest then, upon the asteroid. A planet or asteroid is moving at its fastest at perigee. If this is the fulfillment of the 2nd Trumpet, what does this say for the mercy of God? Nothing. Instead, it appears that God wants it as fast as possible for maximum destruction.

I've read that the asteroid will cross over Australia, then Africa, then over the Atlantic. The video below shows this path, and even has it going over Washington DC. However, the video is confusing me because the earth is known to orbit the sun counter-clockwise when viewed from above its north pole, yet the asteroid is seemingly shown moving either clockwise, or even TOWARD the sun; the latter direction is impossible because the asteroid never moves toward the sun.

With the earth orbiting counter-clockwise, we are looking at its backside as it's shown on the screen above. This video looks all wrong. There's no such thing as the asteroid motion "relative to earth" if the asteroid is moving in a direction opposite the earth. Plus, for what looks like another error, the video has the closest approach at 37,000 kilometers (keep eye on bottom of screen) when the rock is above the mid-Atlantic.

However, when we see the video below, the asteroid is going counter-clockwise with the earth (so far so good), and the sun is again off the left side of the screen. Only this time, the asteroid is crossing over Africa FROM the Atlantic ocean rather than over Africa INTO the Atlantic. Is there nobody we can trust on astronomical topics? Is there disinformation deliberately to confuse, to keep us from realizing the deliberate errors of astronomy?

Scroll down this NASA page to see virtually the same scenario as the one directly above:
https://science.nasa.gov/solar-system/asteroids/apophis/

I now need to adjust my thinking, because I was told that the path would be over Australia toward Africa. NASA's page has its closest approach over mid-latitude Africa. Let's work with this.

The predicted angle of strike, back when they thought it could strike, was at a 45-degree angle. I don't recall reading where they thought it might land. But this 45-degree angle is as per their sizing of the rock, and their velocity deductions. If it's five times slower and smaller, it could be more like a dead-on (90-degree) impact, sending tsunamis in all directions instead of one main direction as per a 45-degree strike.

A dead-on strike would be directly in the rear of the planet (backside according to orbital direction). Wikipedia echoes somebody when saying: "The close approach will be visible from Europe, Africa, and western Asia." This view tends to make the rock come closest at a longitude through Libya and Sudan. But if it doesn't fly over that area, and instead strikes dead-on, where might it land according to the predicted path as NASA shows it?

Well, let's assume it's to be directly over Sudan at its closest approach (which Wikipedia has at 21:46 universal (British) time). They say the asteroid is slated to enter south of the equator, and so the backside of the earth from Sudan, when the rock is in the southern hemisphere, is 3,960 miles west, or in the Atlantic between nearer to South America than to Africa. The backside of the earth is exactly one earth radius behind the closest-approach spot, or 3,960 miles back.

But we need more fine-tuning. As the asteroid is predicted to be moving 16,550 miles per hour at the closest approach, I cut that by 5 times to 3,300 mph, suggesting that it will strike the earth's backside 3,960 / 3,300 = 1.2 hours before it's slated to be over the longitudinal line through Sudan. We now need to turn the earth 1.2 hours backward on its axis, so that the asteroid lands 1.2 hours closer to Africa. That keeps it in the Atlantic ocean. But, I don't know for sure whether the closest-approach spot is at Sudan. And it may not be a dead-on strike. And the data I'm using from astronomy may be wrong to boot.

Still, this proposed landing is smack at the equator, not only in the mid-Atlantic east-to-west, but in the mid-Atlantic north-to-south. Tsunami's can go from there to many parts of the Atlantic. There's nothing to stop the waves en route to Washington DC, and there may also be a clear path to much of Britain. The Gulf of Mexico would be protected by Cuba, Haiti and the American Virgin Islands, and Texas would be protected by Florida.

If you stop the NASA video above at the middle of the eighth second, you will see that the asteroid has reached the earth line. It's going to go past the earth after the 8th second. At the 8th second, the back of the earth, which faces you, shows the mid-Atlantic facing you. You can see north-west Africa as the light patch on the globe's right side. Just re-position the gold path of Apophis so that it's striking the earth, and you can see little chance of it striking the Pacific ocean, but a huge chance of striking the Atlantic. But none of this matters if NASA wrongly positioned the earth for the fly-by. s

This page shows that the average closest-earth approach of this asteroid is over 10 million miles, as the goofs measure it, yet some claim that it will be as close as 18,000 miles in 2029, yet the goofs say, no worries, our gang has this pin-pointed as sailing right by without incidence. Yaa-yaa, sure thing.

They even pretend to know exactly how many minutes the asteroid will be from perigee as it reaches closest approach, even while it's still years away. Do you think maybe they could show some humility instead of trying to make themselves appear so expertly? Not a chance. Astronomy is filled with deceptive egos and lunatics, haven't you seen the signs?

In the beginning, shortly after it was discovered 20 years ago, some in astronomy were saying that impact looked likely. Of course, this set off all sorts of internal, behind-doors discussions. And if they agreed there that impact was likely, they might also agree to lie to us to keep the world from panic. But, is this a wise thing to do, to keep the world from preparing? Or, perhaps they view this as a God-send in the sense that they desire "population control," and what better way to kill the masses off but by an asteroid? So, best thing: keep people from preparing.

In this way, God and his enemies are "working together" to destroy many of the ungodly, the unrepentant, the anti-Christs...if Apophis is the 2nd Trumpet.

I don't want to be guilty of not warning people concerning this asteroid in the face of a solar-earth distance of only 18 million miles, which is not my figure, but NASA's (though NASA won't own up to it). Some of my main work on the eclipse data is at 5th update in December, 2021.

NEXT CHAPTER

Brass Tacks of Tribulation Preparation
Some points on starting your own wilderness retreat,
even if you don't have the money.