Previous Update

(if there are any to speak of)

September 13 - 19, 2016

The United States and Russia have come to an agreement in Syria to coordinate their attacks on terrorist groups, something that Obama was opposed to until now. How long can such a deal last? The situation in Syria and Iraq looks fixed for the foreseeable future. I can't say that any of it is necessarily prophetic fulfillment. In the meantime, I'm putting my live, astronomy-related work in these updates, in case you are interested.

Extra Insight on How Astronomers got the Gravity ConstantKepler is the one who may have originally given the world the idea that a line from the sun to a planet sweeps the same area (square footage) per any unit time, while the planet orbits, no matter what the shape of the ellipse, and no matter how far from the center of the orbit the sun happens to be. I find this hard to believe. More likely, Kepler was wanting to simplify his difficult quests in understanding and predicting the orbits. His third law, "The square of the orbital [time] period of a planet is directly proportional to the cube of the semi-major axis of its orbit," didn't quite work for the moon when I tried it in the last chapter, though it was close. There's little use trying it for the planets as astronomers had ballooned the size of the solar system far beyond the reality.

Wikipedia's article on Kepler's laws, says his third law "has the same value for all the planets in the solar system". In other words, no matter how large the orbit, or the shape, the number will always be the same when one multiplies the period squared by the three times the semi-major axis. That refers to a constant.

The article then says that Newton had it in reverse, that the period squared DIVIDED by three times the semi-major axis = CONSTANT (i.e. for all the planets). And it shows that Newton modified this formula to, 4pi squared / (gravity constant x (sun's mass + planet mass)), to get the same constant. One can begin to glean that the so-called gravity constant was related to Kepler's constant. I can spot that the formula is faulty for more than one reason, for if you go to the article, you can see that the formula was changed to: 4pi squared / (gravity constant x sun's mass), leaving out the planet mass. You can't do that. It thereby becomes clear that the formula gave the wrong number with the planet included, and something closer to the reality if it was left out. The section concludes with, ignoring planetary mass, "Kepler's third law is approximately correct in Newtonian mechanics."

Constants make everything easier. You can imagine these men working into the wee hours of the morning to find some magic constant, and they didn't have electronic calculators in those days. Big big numbers, poor fellows. At Wikipedia's article for Newton: "By deriving Kepler's laws of planetary motion from his mathematical description of gravity..." Newton was looking for a way to turn Kepler's constant into some gravity-related number. Hence, he came up with his formula above. It doesn't mean that his gravity-related constant was devised / measured by any experiment involving gravity; it only means that the gravity acting on the planets was being sought out, and boxed into a constant number while maintaining / reflecting Kepler's third law.

They were able to come up with a number for every planet which came to be called the standard gravitational parameter, and this number was broken down to G x M. It doesn't matter that gravity is not in reality an expression of planetary mass, and it doesn't matter what you name the breaking down of the parameter number, whether you name it gravity constant x planet mass, or oranges x potatoes, that number remains true. So, the particular masses of planets were wrongly assigned as per the Newtonian math, not as per any experiments (impossible, anyway) that could get the true masses. If the G (gravity constant) number changes, so do the masses of the planets. The idea was to get a constant-gravity number even though each planet was under a different level of force from solar gravity. They believed that there could be some magic number that was related to all gravity forces on all planets.

According to the Wikipedia writer, Newton's work was able to claim that multiplying the orbital time period by two, and dividing that by three times the longest radius (or semi-major axis) of an ellipse, got the same number as 4pi squared / the standard gravitational parameter for any planet. But if you just look at the equation, you are able to spot that it does NOT give a constant number. The Wikipedia writer didn't have the normality of politeness to explain why not. We are left with our mouths open going, duh.

Let's test these numbers on their earth distance to discover at least roughly what Newton's constant was. He had period squared / semi-axis cubed, or 365.24 days = 31556736 seconds squared / 93 million miles = 150 million kilometers = 150,000,000,000 meters cubed. We take these numbers to the big online calculator below, to find the answer as: 995827586973696 / 3375000000000000000000000000000000 = .0000000000000000002951. Poor poor fellows.

http://web2.0calc.com/Kepler's law had it in reverse: 995827586973696 x 3375000000000000000000000000000000 = 3360918106036224000000000000000000000000000000000. They didn't even have aspirin in those days.

Using the same formula as Newton's with the moon's period (2354139 seconds) squared and the moon's semi-axis cubed, the result is 5541971094126 / 56800235584000000000000000 = .00000000000009757, not at all the same number as .0000000000000000002951 above. Why does Wikipedia say that this formula = constant? Doing Kepler's formula for the moon: 5541971094126 x 56800235584000000000000000 = 314785263746075038579584000000000000000, not at all the same number as his formula gets for the earth.

It's obvious that when one does the math for 4pi squared / GM, one is not going to get the same answer, simply because GM is different for all planets...wherefore why does Wikipedia say that the answer is constant? I don't know, because, as is typical of writers on these topics, there is no explanation. We are shown the formulas, and often that's all there is. The public is not going to be able to verify anything that astronomers tell us, so long as the leading astronomers can help it, because there are huge black holes in their work from their obstinate attitude. They know the hoax, and they control its dissemination, defending it.

If you take nine different numbers, you can divide each one by an amount that gets the same number, and thus one can call the latter a constant. This is what astronomy did, and the amount-divided-by became the planetary masses. They could have called them the cabbages or the peppers, but they thought it more befitting to call them the planetary masses. They were smart guys, after all.

169 / 13 = 13

457 / 35.15 = 13

1,698 / 1,306.15 = 13The constant is 13. If the left column could be viewed as representing the numbers for the planets obtained in some way by orbital investigations, the trick then is to find some way to discover gravity's role for to assign it a number. Even if it's the wrong number, it doesn't matter, the formula will still work. They would have taken the planet number to divide it by the gravity constant that they chose to use, just as easily as one can take 1,698 and divide it by 13 to find 1,306.15. And, so, they gave one planet the mass of 1,306, so to speak. I have three invented planets above all with a gravity constant of 13, all with their own particular masses, and the left column represents the gravitational parameter that was obtained using the findings related to solar acceleration. The parameter for Mercury was the smallest number, and the parameter for Jupiter was largest, and consequently the masses for all planets worked out fairly appropriate for what one might expect of the planetary masses (i.e. Mercury being light and Jupiter heavy). So if Mercury had a parameter of 169 and earth a parameter of 457, Mercury's mass would appropriately work out smaller than the earth mass. No matter what they chose for a gravity constant, Mercury comes out light and Jupiter the heaviest. It wasn't magic, nor correct, but it was math and playing with numbers.

Solar acceleration is a thing I wanted to look at for this chapter because it accelerates the moon. I wanted to look at the possibility that the moon's orbital / lateral acceleration was due purely to solar gravity and zero to earth gravity. I'm realizing only now that there is an enormous undertaking to check out whether the possibility is correct, and frankly I haven't the need to devote the time to it.

Re-Investigating the Lunar NumbersRight off the bat, one is inclined to view the lateral velocity as a function of the earth's gravity only, because they say that the moon is slowest at apogee, and gains velocity from that point onward. In other words, the report is that the moon accelerates laterally even as it accelerates downward, and then finds its greatest lateral speed when it ceases to fall. The two directions appear completely related. Then the moon decelerates in the lateral direction as soon as it starts to decelerate in the upward direction. It appears that the lateral acceleration and deceleration is wholly due to earth gravity. But I cannot conceive the moon picking up speed laterally just because its falling. I cannot see how gravity can make an object move faster in the lateral direction. In order to have an object move faster in the lateral direction, something needs to push or pull it, and gravity can't do it. I am absolutely sure of this. Either I am absolutely wrong in my absolute certainty, or the astronomers have been bald-faced lying to us. At best, they have been conveying a wrong idea without the integrity to correct themselves.

The average reported velocity number (NASA says 2,286 mi/hr) definitely matches the reported lunar circumference, itself based on the reported average lunar distance of nearly 239,000 miles. However, I had found a way, using the middle-size moon, to get an average distance of about 236,000. Perhaps astronomers sincerely gave themselves over (it's no excuse) to the concept that earth gravity speeds the moon laterally, and from there they entered error in certain ways. In this picture, they really haven't got a clue as to how fast the moon is traveling at any time. Everything they think they know is pure and correct math based on a wrong model...giving wrong "facts." At 236,000 miles, the circumference becomes 1.48283 million miles, and the velocity is that number divided by 655.7 hours = 2,261 mi/hr, not much off from their reported figure.

NASA has no way of knowing the lateral velocity aside from knowing the average distance to the moon. On its lunar fact sheet, if reports a lunar distance, from the earth's equator, of 378,000 kilometers = 234,878 miles. Why is this number rounded off? NASA claims to have put solar deflectors on the moon, and to have used them to get the true lunar distance to virtual precision, yet here we are not shown how amazing that tool is because we get a fat, round number for the distance. NASA doesn't really know the true distance, does it?

http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.htmlBut the middle-size moon is something that can be measured, and NASA's fact sheet has it at 1,896 arc-seconds, which, when multiplied by 3,600 is .52666 degrees. It doesn't say that this is the average, but we assume it to be, for the moon's apparent size is always changing. NASA's number, when divided by 60, gets 31.6 arc-seconds, yet, five chapters ago, when this was the topic, it was found that others report an average lunar size of 31.1 arc-minutes. It seems that NASA, or the leading astronomers, have been changing this figure too, and this wouldn't be based on error, because, if they claim to be able to measure the width of a star accurately, how can they get the moon's size wrong? If they have been changing this middle-size moon figure, something smells. You are not supposed to change that. It shows that they are changing the average lunar distance too.

By the way, there is a question on whether the lunar month of 27.32166 days is of 24 or 23.9244 hours. NASA's fact sheet has the moon spinning on its axis once every 655.728 hours (= the lunar day). The lunar month must match this, otherwise the face of the moon would continually change as we see it from earth. In my opinion, the fact that we always see the same lunar face is a miracle, a thing that God worked into the scheme. NASA claims that there is a perfectly natural explanation for this, but I cannot see it, and, of course, I'm not prone to thinking that astronomers are honest about much. NASA's fact sheet has the lunar period at 27.3217 days, and this works out to 655.7208 days when using 24 hours, nearly the 655,728 hours that they have for the lunar day. But how professional is it to use 24 hours when they say that a true day is some four minutes short of it?

Using the true day of 23.9344 hours, the lunar month of 23.32166 days works out to 653.92 hours, which is not even close to NASA's number. There is a difference of 1.808 hours between the two figures, which is a whopping difference of 24.17 hours annually. That's going to change the face of the moon by an entire rotation every 655.728 / 24.17 = 27.13 years. We would see opposite sides of the moon every 13 years. NASA is not primarily in the business of astronomy, but rather in the business of deliberate deception, explaining its sloppiness and inconsistencies.

We go to the fact sheet again to find that the average perigee is reported at 363,300 kilometers, and the average apogee at 405,500 kilometers, with the semi-major axis at 384,400 kilometers. The latter number is perfectly midway between the perigee and apogee figures, and therefore requires a straight line (from apogee to perigee) through the earth. In most months, there is not a line between apogee and perigee that goes through the earth too, but, I suppose, when dealing with the average scenario, one can assume a line through the earth. Apogee is at times less than 13 days, at times more than 14 days, after perigee. Therefore, while apogee and perigee are always the furthest two points in an ellipse, the earth is not always directly between them. In an oval sideways on the page, and a horizontal line through the middle, the earth is usually going to be either above or below the line, depending on the month.

As the middle-size moon is the average lunar distance, it reveals that the same distance (384,400) will be obtained for midway along the orbit as in half a straight line (i.e. the semi-major axis) between the 405,500 and 363,300 distances used by NASA. It means that 384,400 kilometers (238,607 miles) should be viewed as both the average distance and the semi-major axis, yet NASA uses another average figure of 382,500 kilometers. What is that?

For the lunar-orbit's eccentricity, it gives 0.0549. The eccentricity figure is an expression of how far from the perfect center a planet or moon is. The figure is derived from: (distance of planet from the long side of the ellipse minus distance of planet from the opposite side) / (distance of planet from the long side of the ellipse plus distance of planet from the opposite side). In other words, the closer the earth to the perigee moon, the greater the eccentricity. The closer the earth to the apogee moon, the more circular the orbit. Using NASA's average numbers, the eccentricity is: (405,500 - 363,300) / (405,500 + 363,300) = .0549 (same number as reported above). It's thereby revealed that, when apogee and perigee are painted at opposite ends, with the perfect ellipse center between them, it is the situation used to derive the average eccentricity.

In the last chapter, I realized that half of the apogee and perigee distances combined equaled (237,543 miles) exactly the number I was using for the middle-size moon. In other words, it doesn't matter whether we measure midway along the round orbital path, or measure half the straight line from perigee (222,686 miles) to apogee (252,400 miles), the distance is 237,543 (these numbers are for the second half of July of 2,000). Well, that's not fully true, because the earth was not on a straight line between apogee and perigee. In that particular month, it was established (last chapter) that perigee was a little to the right of o'clock. With the earth on the line between 12 and 6 o'clock, and below the clock's center, the earth will be closer to 6 o'clock than to any point to the right of 6 o'clock. Therefore, when one line from the apogee moon (12 o'clock) to the earth is added to a second line from earth to the perigee moon, at a total distance of 252,400 + 222,686 = 475,086 miles, the major-axis, from 12 to 6 o'clock, will be less distant, meaning also that the semi-major axis will be less than 237,543 miles. It's not a big deal, but it's good to point this out. It's better to find the average lunar distance along the orbital track than to use the semi-major axis.

We still want to know why NASA's average distance is less than its semi-major axis. As the full length of the lunar orbit (ellipse tip to ellipse tip) is very-slightly more than the apogee and perigee distances combined (475,086), we can approximate the plotting of the earth at 252,400 / 475,086 = .531 the way from 12 o'clock. As you can see, the earth will be just .031, of the full 12 to 6 o'clock line, below the center of the clock. With the earth so close to the center, the lunar orbit is not much of an ellipse at all, yet it's officially an ellipse.

In the last chapter, the moon at a height of 237,543 miles was at first put near the 9 o'clock position due to my having both a time circle and a distance circle (time units are not proportional to distance units), but, the distance circle was eliminated when realizing that lateral distance achieved was probably proportional to the drop-distance, and this simplified things (otherwise it was confusing). The height of 237,543 miles could then be plotted between 8 and 7 o'clock based on the height of 239,194 miles coming roughly a couple of minutes after 8 o'clock. With the height of 239,194 miles occurring at 2/3 the time between apogee and perigee, it needs to be a little after 8 o'clock when perigee is placed a little after 6 o'clock.

However, on further thought, the lateral distance achieved is only the reported distance, and I'm now coming nose-to-nose with the question on whether they are based on fact or theory. If it's merely a theory taken as fact that the lateral velocity is directly due to the fall or rise of the moon, then the lateral velocities at any given point are likely concluded / assigned with proportionality to the drop / rise distances. In that case, I wouldn't trust their figures for lateral velocities at any given time, but then I haven't found any such velocities reported. Nor can I find a moon-velocity calculator. I have nothing to work with to derive velocity but the reported distances between the moon and earth. If the latter are incorrect by a significant amount, then it appears that we can't use the eclipse of July, 2000, to find the solar distance with precision.

What we can do is to prove with the eclipse, using their average distance to the moon, that the sun is not nearly 93 million miles. The problem is, there are several average distances reported online for various reasons that no one explains. I'm apt to suggest that NASA and/or leading astronomers have been changing their average distance over the years. While some are saying that the average distance is 238,700, NASA's semi-major axis is 238,855 miles. It hard to see how one would report 238,700 when viewing 238,855 in some page. NASA's currently-reported average of 382,500 kilometers = 237,674 miles is close to the 237,543 average for the month of the eclipse, which can suggest that all the lunar distances in the apogee and perigee calculator (below) are not necessarily true, but rather geared to NASA's current number.

https://www.fourmilab.ch/earthview/pacalc.html.However, we just saw that the average-size moon must conform to very-near the semi-major axis distance, and so why is NASA's average almost 1,200 miles smaller than its semi-major axis? I had found no one reporting the average-size moon size at 31.6 arc-minutes, but that's the size on NASA's moon fact sheet. I had found 31.09 and 31.1 as the middle-size moon, and we must assume that these numbers were taken from records provided by astronomers, perhaps even NASA itself. I've noted that NASA doesn't have all three sizes (smallest, largest and average) on its fact sheet. As the largest-size moon is roundly said to be 33.5 arc-minutes, let's divide 31.6 by 33.5 to get .94328 the difference. Then, let's divide the average lunar distance by the greatest lunar distance, both claimed by NASA: 382,500 / 407,000 kilometers = .9398. Let's try 31.1 / 33.5 = .9283. The 31.6 claimed by NASA is a closer representation of the numbers obtained from NASA's distances than 31.1, wherefore, where did the latter number come from? Can't astronomers do a simple thing like measure the width of the full moon correctly? Of course they can. But then why are these numbers drastically different?

If the middle-size moon is truly 31.1 arc-seconds wide, then the average distance must be the middle of two lunar distances that, when divided by one another, gets .9283 the difference. It's been a long time since NASA put those reflectors on the moon by which they could get the lunar distance accurately, but then NASA never did put the reflectors on the moon. It was a hoax. NASA wanted to portray itself as mighty during the Apollo missions, but it was all a hoax. NASA is a farce, and the inconsistencies with the numbers at hand tend to support that accusation.

I have just found a page, by using 31.6 in the Google search, that makes matters even worse. The page does not report the middle-size moon at 31.6, but says, "Now that your brain hurts, how about I just tell you that the angular diameter of the sun as seen from the Earth varies between 31.6 and 32.7 arc minutes, and the moon between 29.3 and 34.1 arc minutes." What!? It's not 33.5 anymore? And others say the lower number should be 29.43, not 29.3. This is ridiculous. Can't astronomers measure the width of the full moon??? Of course they can, meaning that they have been lying to us, changing their size numbers according to the changes they have made in their lunar distances. NASA is a fraud. There is a difference between making an honest mistake, and being a fraud by dishing out wrong information as fact.

http://findersfree.com/science-nature/why-do-the-sun-and-the-moon-appear-to-be-the-same-sizeThe webpage above uses NASA's semi-major axis number (384,400 miles) as the average distance. What a mess. It gives the formula for finding the lunar distance, but this formula may be faulty. It's: angular diameter / 206,000 = linear diameter / distance, where 206,000 is said to be the number of arc-seconds in a radian. By linear diameter, the true diameter of 3,476.2 kilometers (= size on NASA's moon fact sheet) is meant. So, let's try it using 31.6 as the angular diameter, but first multiply it by 60 to get 1,896 arc-seconds (the number on NASA's fact sheet), and moreover, let's get the precise number of arc-seconds to one radian.

I'm reading, "one radian is just under 57.3 degrees." I know what this is; it's identical to what I've called the magic number, 57.295 (very useful for finding small angles), equal to 180 / pi (actually, the number should be 57.29577). It's the distance on your protractor, measured in degrees, between the center of the protractor's base-line and the circle on the protractor. In other words, the length of a curved line on a protractor that goes from 0 to 57.29577 degrees (= 1 radian) is equal to the distance of a straight line from 0 degrees to the center of the base-line. Anyway, the point here is that one radian works out to 57.29577 x 3,600 = 206,264.8 arc-seconds. There is a page that converts radians to arc-seconds, and it gives 1 radian as 206,264.8 arc-seconds.

http://www.convertunits.com/from/arcsecond/to/radianThe math: 1,896 / 206,264.8 = .00919207. Therefore, as 3,476.2 x distance is supposed to equal .00919207, we do: 3,476.2 / .00919207 = 378,173.8 kilometers = 234,986 miles. We have a big problem, Houston. How did NASA ever get to the moon feeding their mathematicians the wrong numbers? So, if this formula is accurate, it reveals another way to find the lunar distance without speed-of-light measurements. I have not known of this formula before, but it works directly off of the angular sizes of the moon. It means that it can get for us the greatest, least, and middle distances of the moon. Why haven't I seen this formula elsewhere? What a handy tool.

It's not a wonder that I have never read of speed-of-light measurements getting the same lunar distances obtained by this formula. It's also not a wonder that I have not come across this formula before, because it's not giving the reported average lunar distance. It could appear that the formula is being hidden from all of us, or more writers that treat the angular sizes of the moon would have shared it.

If we re-do the formula with 31.1 degrees (1,866 arc-seconds) as the middle-size moon, it looks like this: 1,866 / 206,264.8 = .0090466. Therefore, we do: 3,476.2 / .0090466 = 384,254.9 kilometers = 238,765 miles. We are now close to 238,855 miles, NASA's semi-major-axis distance, but I'd like to see a near-perfect match to indicate that NASA once used the figures. I read from one person that the number should be 31.09, and with that refinement, the average distance works out to 238,841, close enough for the indication. It appears that some are reporting a middle-size moon of 31.1 based on NASA's semi-major-axis number, but that number can now be proven inconsistent with the formula above, for NASA is required to report a middle-size moon corresponding to 234,986 miles, and the semi-major axis was shown above to be always almost-exactly the middle-size moon. The problem can be put another way, that while 31.09 gets virtually the same semi-axis as reported by NASA, the latter organization now reports a middle-size moon of 31.6. What a mess.

Thus, the distances in the apogee and perigee calculator are not correct. They are based on NASA altering the numbers...for a reason unknown / unsaid, apparently.

With this radian formula, we could actually find the true lunar distances, if only we could have a reliable source for the true angular sizes of the moon. As everyone I've read, prior to finding the page above, has the largest size at 33.5 (2,010 arc-seconds), the radian formula makes the following calculation for the nearest-possible moon: 2,010 / 206,264.8 = .009744755. Therefore, we do: 3,476.2 / .009744755 = 356,725.2 kilometers = 221,659 miles. NASA's fact sheet has 357,000, wherefore 33.5 looks very consistent.

We can do the formula for the furthest moon using the smallest number, 29.3 arc-minutes, that I've seen: 1,758 / 206,264.8 = .008523. Therefore: 3,476.2 / .008523 = 407,861 kilometers. Yet, NASA's footnote has 407,000 as the furthest, meaning that NASA currently does not use 29.3 as the smallest. Where did the latter number come from in order to get nearly 1,000 kilometers more than NASA acknowledges?

If we do the formula using the 29.43 figure that some report, it's: 1,765.8 / 206,264.8 = .0085608. Therefore: 3,476.2 / .0085608 = 406,060 kilometers. Don't the astronomers know the difference between 407,000 and 406,000??? One could excuse a high-school student for getting it that wrong, but isn't astronomy supposed to hire math experts for that field? With all of the inconsistent information on the Internet concerning the angular sizes, might NASA find it within its heart to report the correct figures on its moon fact sheet? But, nope, they are not there, just as though NASA doesn't care about the world reporting false numbers everywhere. Why should NASA care if it itself is the arch-criminal?

As the apogee/perigee calculator below has 406,692 kilometers as the highest altitude (year 2025) between the years, 2000 and 2035, I'd say that it and NASA together must be working off of a lunar size between 29.3 and 29.43. We can do the math backward to find out: .3,476.2 / 406,692 = .0085475 wherefore .0085475 x 206,264.8 = 1,763.048 arc-seconds = 29.384 arc-minutes. This has got to be very-near the furthest distance possible, and it jibes with NASA's 407,000 approximated figure. And it's correct only if the smallest moon measured is 29.384 arc-minutes (.4897 degree). But why is no one reporting that number when we ask Google for lunar sizes?

https://www.fourmilab.ch/earthview/pacalc.htmlI now have the dismal problem of not knowing the correct lunar distance at the apogee of July 15, 2000. There has got to be at least one astronomer in the world that can measure the full moon accurately. Surely, they know when to measure it for finding its furthest distance. Don't they? Regardless, they can measure it hour by hour until they get the smallest size. And, surely, there should be a wide consensus in the world as to what that size is. Yet, NASA, the one represented as superior, doesn't even bother to put the number on its moon fact sheet, just as though no one in all the world would like to have it. Come on NASA, wizen up. Give yourself up. Come out with your hands on your head.

When we divide NASA's furthest distance (357,000) by its average, we get .9333 the difference, and when we divide 29.384 by NASA's 31.6, we get .92988, virtually the same number. NASA is midway between 29.3 and 29.43, showing conflict with both. It looks like NASA increased the furthest distance from 406,000 to the 407,000 kilometers.

The page we saw above deviated from everyone else's 33.5, having the closest distance at 34.1, a huge leap. How can the full moon be measured so vastly different? How can they get it so far wrong? In reality, astronomers haven't gotten it wrong by measuring, but have been changing the distances, and then adjusting the sizes to reflect the distance changes. But you can't do that. The size always stays the same. You are not allowed to change the size. If we can't trust them with the moon, why should we trust them with the reported sizes and distances of stars? Liars are not to be trusted. They have been lying with the moon's distance. They have made a mess of astronomy. They need to be booted out and replaced by people who fear God enough not to lie. Evolutionists have no fear of God.

Let's do the math for the size of 34.1 = 2,046 arc-seconds: 2,046 / 206,264.8 = .00991923. Therefore: 3,476.2 / .00991923 = 350,450.6 kilometers (217,760 miles). The smallest distance in the apogee/perigee calculator is more than 356,500 kilometers over the years, 2000 - 2035, wherefore, how did we ever get to 350,473? Shouldn't astronomers, who act like your best friends, inform the people when they make changes to lunar distances? No, because we would then not trust them, but they want us to trust them because they are taking the world away from God, and need to be trusted on behalf of that task.

Space.com is claiming: "At perigee — its closest approach — the moon comes as close as 225,623 miles (363,104 kilometers). At apogee — the farthest away it gets — the moon is 252,088 miles (405,696 km) from Earth." Where did space.com get those figures??? They are not as steep as NASA's 407,000 and 357,000. The middle number between 363,104 and 405,696 (precise figures for a change) is exactly 384,400, the one used by NASA for its semi-major axis! Therefore, NASA was at one time likely claiming 363,104 and 405,696.

On the fact sheet, NASA has the average perigee and apogee as 363,300 and 405,500, and the middle number is again 384,400 exactly. The problem now is that the numbers on the fact sheet are not precise. NASA must have made them up, and deliberately so while keeping to 384,400 as the middle. Why would NASA abandon the precise numbers for these approximations? Doesn't NASA have precision light-speed methods to get the lunar distance correct to less than one mile? Probably, NASA is disseminating false information deliberately to make a mess.

Wikipedia shows that NASA's semi-axis number is not much approximated, and it calls it the AVERAGE: "A lunar distance, 384,402 km, is the Moon's average distance to Earth....The time-averaged distance between Earth and Moon centers is 385,000.6 km (239,228.3 mi)." I showed, in the last chapter, how there can be an average based on where the moon spends its time.

https://en.wikipedia.org/wiki/Lunar_distance_%28astronomy%29NASA scores F on its report card for Frankenstein. I don't think we can fix this with a bolt to the skull. I think this monster needs to be put down. The Wikipedia writer is laughable: "Millimeter-precision measurements of the lunar distance are made by measuring the time taken for light to travel between LIDAR stations on the Earth and retroreflectors placed on the Moon." It leads the reader to believe that the speed-of-light measurements got 384,402 for the average, but Creationists know how evolutionists discard all experiments that don't get the "correct" age of earth rocks, until they find an experiment that gets what they were after in the first place, and that's the only figure they report to the masses. So, I know what they did; they respected and published only the light-speed experiments that matched the lunar distance given by the radian formula. That's not science; that's a bum that needs to be booted out.

The radian formula reveals that one doesn't need reflectors on the moon to get the truth. One just needs the apparent / angular size of the middle-distance moon, and the true size of the moon. Just to prove that a moon half way between two distances is exactly midway between the two sizes, NASA's fact sheet on the sun has the maximum size at 1,952 arc-seconds, the minimum at 1,887 arc-seconds, and the average almost midway at 1,919. The problem is, your demented NASA organization (it isn't mine) refused to use decimals for the arc-seconds. It once again refused specifics for anyone that would like them. It can be shown that 1,919 is not exact in that the http://www.cleavebooks.co.uk/scol/calrtri.htm

">triangle calculator gets 92,955,807 miles (the average) using .533112 degrees = 1,919.2 (put 864938 in the edge-a box).

http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.htmlThe sun sheet has the maximum (152.1) and minimum (147.1) distances, as well as the average (149.6). The average number is exactly midway between the minimum and maximum. Why shouldn't the same apply with the moon? Why didn't NASA include all three sizes on the moon sheet? Why did NASA use just one decimal place for all three on the sun sheet? Is NASA anti-educational? Does it not get paid enough to do a proper / acceptable job?

I have been wanting evidence for a certain question using the lunar figures, but NASA has denied me the answer. I've wanted to know, for example, whether dividing the nearest distance by the average distance gets the same number as dividing the largest size by the average size (the two scenarios are directly related). In the case of the sun, the best that the triangle calculator can give for a maximum size, using the limited number for the nearest solar distance, is .542164 degree = 1.951.79 arc-seconds (the NASA sheet gives only 1,952). To get this number, use 864938, as the solar diameter, in the edge-a box, and then convert NASA's 147,100,000-kilometer figure to miles and put it into the edge-b box. Hit "Calculate" to see the angle, and multiply it by 3,600 to get the same thing as the size.

When dividing 1,919.2 by 1,951.79, it gets .98330, a near match with .983288, the latter obtained by the nearest distance divided by the average distance. If there were more decimals in the sizes and distances, I could become more certain than even these numbers allow.

The best that the triangle calculator can give for the furthest distance is an angle of .5243426 = 1.887.63 (the NASA sheet uses only 1,887). The difference between the furthest and nearest distances (.9671) can now be compared to the difference between the minimum and maximum sizes, the latter being 1,887.63 / 1,951.79 = .967127, a match with .9671269.

THE POINT IS, if we can do the above reliably with solar figures, why not the lunar figures? It's a great tool to have to check the details within the lunar-figure mess. I have divided everything I can using all the lunar numbers I could find, and was unable to make any sense of the numbers by that route. There is no consistency between the sizes and the distances, yet, for the solar numbers, as you just saw, there was precision, no inconsistency whatsoever. Moreover, dividing the furthest solar distance by the average gets a difference of .983563, while dividing the counterpart, the smallest solar size with the average size (both with limited decimals), gets .98355. Not bad at all.

Here is what the radian formula gets with a moon of 31.09 = 1.865.4 arc-seconds: 1,865.4 / 206,264.8 = .009043714 the difference; wherefore: 3,476.2 / .009043714 =

384,377.5kilometers. That's not quite the 384,402 that Wikipedia claims for speed-of-light measurements as early as the 1950s, but it's close. On behalf of the 384,402 figure, let's see what the middle-size moon works out to with reversed math: .3,476.2 / 384,402 = .0090431 wherefore .0090431 x 206,264.8 = 1,865.281 arc-seconds = 31.088. One can see how 31.09 is an acceptable number to publish. Yet NASA's fact sheet claims 1,896 arc seconds = 31.6 arc-minutes for the moon's apparent diameter.[Insert: I recently wrote to NASA's representative for its moon fact sheet, Dave Williams, asking why some are reporting an average-size moon at 31.1 arc-minutes, while NASA uses 31.6. He apparently had never heard of the 31.1 figure, for he asked me to provide a source that used it. He wrote a second message telling that he figured out where the 31.1 comes from: it's the size of the moon when one is standing at the core of the earth. I checked it with the triangle calculator, and it proved true so long as one enters 384,400 kilometers in the edge-b box with the lunar diameter (3476.2) in the edge-a box to get an angle of .518122 = 31.0873 arc-seconds.

Remove an earth radius from 384,400, and therefore enter 378,028.6 in edge-b with the lunar radius in edge-a to find an angle of .526248 degrees = 31.575 arc-minutes, not exactly 31.6, but close. If NASA is ball-parking 31.6, I would not be happy. Fact sheets are not for ballpark figures. High-school students doing minor reports on the moon are not the only people using NASA's fact sheet. Mr. Williams says, "31.6 arcminutes is the value adopted by the IAU [astronomical unit = official solar distance] as the mean apparent diameter of the Moon and is used in all relevant calculations." Yet no one else online could be found reporting 31.6. Compare 378,028.6 above with 378,000 below. End Insert]

The moon fact sheet has the average distance at 378,000 (big fat round number hides the precision) from the earth's equator, which works out to less than 382,000 as the average. NASA: "The average distance to the Moon is 382,500 km." There is nothing here for NASA to feel proud about. Instead, shame. The tax payers paid for those lunar reflectors on the moon, and yet the people do not get the benefit of seeing what specific numbers the instruments have produced. Shame, NASA. What are you hiding? And stop with approximations, already. As 382,500 kilometers is 237,674 miles, we find the angle like so: 57.2957 / 237,674 x 2160 = .5207 degree = 31.24 arc-minutes. That's not 31.09, and it's not 31.6. Doesn't NASA know the difference between 382,000 and 382,500? These figures are regarding online data pages, not random conversations over a beer. Get it right and tight, NASA, or ship out.

https://www.nasa.gov/pdf/180561main_ETM.Distance.Moon.pdfObviously, speed-of-light measurements on the moon are bogus. NASA has apparently held to those "experiments" for its major-axis version of the lunar average distance, and yet NASA has changed its nearest and furthest distances, showing that speed-of-light experiments are bogus, for if they had merit, the distances obtained would have been clinched in the 1950s or 1960s, with no need to ever change them. Wikipedia gives an example of such an experiment in the 1950s that got 384,402 miles for the average. It needs to be stressed that this figure is correct only if they were using the correct apparent size for the average-distance moon. NASA's fact sheet doesn't even bother to give the average-distance size. Shame. It's such a basic part of lunar astronomy, but NASA deliberately keeps the number off the page.

If the speed-of-light measurements are so accurate, down to a centimeter, how can there be such a difference of almost 200 miles between 363,104 and 363,300, or between 405,696 and 405,500? They can know exactly when the moon is the furthest by when it's the smallest, and so they can do their speed-of-light measurements at that time. What's the problem? I've got to assume that the rounded-off distance numbers (for average perigee and average-apogee) on the fact sheet were, at one time (perhaps originally perhaps not) 363,104 and 405,696 (these are the two that got 384,400 as the middle number). Let's do the math in reverse to see what arc-minutes we find for these distances:

3,476.2 / 363,104 = .0095736, wherefore we do: .0095736 x 206,264.8 = 1,974.69 arc-seconds = 32.91 arc-minutes. It's not close to either 33.5 or 34.1.While others give the diameter of the moon as 2,159 miles, and others 2,159.2, NASA's moon fact sheet gives 1738.1 kilometers as the moon's radius, which is equal to 1080.0053 miles, or 2,160.01 miles for the diameter. One can use the triangle calculator below to find the size of the moon depending on what angle they use. For example, 31.088 arc-minutes (divided by 60) is .51813333 degrees; we take half of that (.25906666) and put it into the angle-A box with 1080.0053 (no comma) in the edge-a box. The latter represents half the lunar diameter in order that the calculator gets a lunar distance of 238,854.8 in the edge-b box. It's nearly 238,856 miles, equivalent to their light-speed-begotten distance of 384,402 kilometers (the latter is the number corresponding to 31.088 arc-minutes). It means that the triangle calculator works as well as the radian formula. In this picture, the earth and one's eye is at the corner of the triangle named, A, and the eyes are looking at one edge of the moon, at corner B, at an angle of .259066 degrees. Corner C is the center of the moon:3,476.2 / 405,696 = .0085685, wherefore we do: .0085685 x 206,264.8 = 1,767.38 arc-seconds = 29.456 arc-minutes, not close to 29.3 but close to 29.43.

http://www.cleavebooks.co.uk/scol/calrtri.htmClearly, since the 1950s at least, the middle-distance moon was rendered as 31.088 arc-minutes. This was not due to light-speed measurements, in my guess-timate, but rather due to a high-level decision that was supported by phony / manipulated light-speed experiments. If we put half of .526666 (equivalent of 31.6 arc-minutes) in the angle box with 1080.0053 in the edge-a box, the middle-size moon thus equates with a lunar distance of 234,985 miles. The radian formula had obtained 234,986 (I don't know why the triangle calculator is off by about 1 mile in each of the two cases above). However, the math done above for the 31.6 scenario only had 234,485 miles, suggesting that the numbers used to get it were wrong to begin with. It's just one problem after the other. After weeks of having it on-mind to resolve the issue of true lunar distances -- furthest, nearest and middle -- I'm not seeing the light at the end of the tunnel yet.

Wikipedia's Moon article quotes NASA's middle-size number as the average-distance moon: "...at an Earth-equator to Moon-centre distance of 378 000 km, the angular size is 1896 arcseconds [31.6 arc-minutes]." There are many pages quoting this sentence, and, apparently, none did the investigation to see whether 1,896 arc-seconds is representing 378,000 kilometers. It's not. I've just written to the email address on the fact sheet to inquire concerning whether their 1,896 number is accurate. Waiting for a response. It looks to be accurate, yet it's not the middle-distance moon, as the Wikipedia writer assumed. NASA needs to make clear what its 31.6 refers to, that it is not representing the average lunar distance. NASA should include the average-distance moon size. NASA should put precise distance figures on the fact sheet. Otherwise, NASA needs to switch to the garbage business.

I now have two reliable methods to get the lunar distance just by having the correct angular lunar sizes, and one of them, the triangle calculator, verifies the correctness of the other method, the radian formula (though the numbers are not a perfect match). I hope this word can get around because the radian-method tool is not easy to come by, and use of a triangle calculator may not come to mind.

http://www.cleavebooks.co.uk/scol/calrtri.htmSo, the average distance of 31.088 jibes with the light-speed measurements while not being the middle number between 29.43 and 33.5 (middle number = 31.465), nor the middle number between 29.43 and 34.1 (31.765), nor the middle number between 29.3 and 33.5 (middle = 31.4), nor 29.3 and 34.1 (31.7). It's a Frankenstein Failure.

Let's try the very lowest and very highest numbers I could find for lunar size. Let's put 29.3 into the angle-A box as .244166 degree (29.3 / 60 / 2) along with 1080.0053 in the edge-a box, to find a furthest-lunar distance of 253,431.5 miles (407,858.5 km). Let's do the same with 34.1 / 60 / 2 = .284166 degree to find a nearest-lunar distance of 217, 757.4 miles. Let's subtract the two distances to find 35,674, and then multiply it by .5, and add the product to 217,757 to get 235,594 as the middle-distance number (it's an alternative way to get the middle number, as opposed to (253,431.5 + 217,757.4) / 2 = 235,594). As you can see, the astronomers may secretly have the average moon way down at 235,600, approximately.

Or, if we go with the second-highest and second-lowest, it's 29.43 and 33.5, the middle of which is 31.465 = .524417 degree. Here in italics is what I wrote five chapters ago on this number: "

When we put .524417 degree in the angle-A box and their lunar diameter of 2,159 in the edge-a box, the lunar distance comes out at 235,877 miles, close to the 235,896 that I used in the last chapter. But if we put 2,159.2 in the box, the lunar distance works out to 235,899." It's almost 235,896, important because that was the distance related to a sun having a diameter (at 5 million miles away) of about 47,508 miles, the latter number exactly 6 earth diameters and almost-exactly 22 moon diameters (47,508 / 22 = 2,159.45). For all we know, it may be exactly 22 lunar diameters. Evolution can't provide that situation; God can. Six chapters ago, I shared: "With 47,508 in the edge-a box and .533112 as the angle [astronomy's average solar angular size], the solar distance becomes 5,105,736, or 5.106 million miles. We divide the latter by 21.644 to get 235,896 miles as the tentative [lunar-distance] average."Here is a perplexing problem to note and avoid. The angle of 29.456 corresponds to the reported distance of 405,696 kilometers. When we get the middle number between 33.5 and 29.456, and enter half of it in degrees, or .26231665 degrees, along with 1080.0053 in the edge-as box, the lunar distance is said to be 235,895.5, shy of my magic number (for lack of a better term) by only half a mile. The curiosity is: if we convert 33.5 and 29.456 to degrees (.2804166 and .24546666), and enter half of their values separately into the triangle calculator to get both distances (220,669 and 252,088.7 respectively), the middle number is

236,379. The average between them thus ceases to be the magic 235,895.5 above. Why should this be, since the magic number was obtained from the middle number between 33.5 and 29.456 to degrees?. What can be the difference between the two scenarios, and which one is the correct one to use? The middle number between .2804166 and .24546666 is .26294 for the one scenario, not quite the .2623166 above that's from the middle number between 33.5 and 29.456.I can show this again by converting 32.91 and 29.456 to degrees and entering half of them separately into the triangle calculator (with 1080.0053 in the edge-a box) to get both distances (225,631 and 252,088.7); the middle number is

238,860. However, when we get the middle number between 32.91 and 29.456, and enter it alone as half of .5197166 degrees, the calculator has a distance is only238,127miles. The radian formula for 32.91 and 29.456 gets 225,622.4 and 252,084 miles respectively.Finally, I think I solved the problem. When we enter the two angles to get their middle distances, we use the same lunar diameter. The C line for the lower of the two angles needs to go a little further in distance before it achieves the lunar diameter, and thus it throws off the calculation by getting a lunar distance higher than it should be. It means that the magic number-scenario is the correct one, and that we should always use only the middle number, between the smallest and largest apparent sizes, when seeking the average distance.

We are now touching upon the reported nearest/furthest distances of 363,104 and 405,696 (these are the two that got 384,400 as the middle number). We saw that 363,104 unacceptably got a number too low at 32.91 arc-minutes, while, on the other hand, 405,696 looked acceptable because it got 29.456 arc-minutes, reasonable beside 29.43. My possible view there is that while the 29.456 distance is correct, they falsely made the nearest number 32.91 so that the average got their speed-of-light distance.

The Solar Distance Still EvasiveAs idea came to me that two telescopes, one earth diameter apart, should aim at either edge of the moon so as to make the tip of a triangle at a certain, unknown distance behind the moon. I wanted to use the triangle calculator for this, but didn't know what angle the telescopes would be aimed at when pointing to the edges of the moon. I decided to enter .51816666 in the angle-A box, this number being 31.09 arc-minutes expressed in degrees. Half the earth diameter, or 3,959 miles, is to be entered in the edge-a box. The edge-b box then tells us that the tip of the triangle, A in the diagram below, is 437,751 miles from the center of the earth (at corner C). There is only one telescope in the drawing, at corner B; the drawing would need a second triangle (just as though the first were being flipped) to get the second telescope below corner C.

http://www.cleavebooks.co.uk/scol/calrtri.htmThe trick now is to find how far the center of the moon is. With the distance, from the earth to tip A, of 437,751 miles, and with the base of the triangle at 3,959 miles, the distance to where the two lines are 2,160 miles apart (= lunar diameter) should be obtainable even by a dummy like myself. Well, the full triangle (double the one in the diagram) we are making is an isosceles (defined as having two equal sides = pyramid shape), and the decreasing distance between the two, rising lines of an isosceles is proportional to the distance away from the base = earth. For example, half the distance between the lines is at half the distance from the base, wherefore the math would simply be 7,918 (double 3,959) x .5 to get the distance between the lines at the halfway point. That's the first step, and the second step is that, where the moon is half way, we use 437,751 x .5 to find its distance.

How much further is the moon than half way? The only way to do the first step is 2,160 / 7,918 = .2728 the distance from the tip. We then do 437,751 miles x .2728 = 119,418.5 miles from the tip, and subtract that from 437,751 to find 318,332.5 as the distance from earth. While this is not the correct lunar distance, one can adjust the angle until we get to the ballpark of the average lunar distance. I have found that an angle of .7 gets a distance to tip of 324,032 miles, which, when multiplied by .2728 gets a moon 88,396 miles from the tip, which, when subtracted by 324,032, gets a lunar distance of 235,636 from the earth.

We can repeat this method for a size of 5,738 miles instead of 2,160, as this is the diameter of the earth umbra, claimed by NASA, for the eclipse of July, 2000. That is, NASA claims that the moon passed through the umbra where it was 5,738 miles wide at a distance from earth of 252,280 miles from earth. Let's see what the angle of that umbra was using the method above, for this will tell us where the umbra lines point in the opposite direction, when going to the edges of the sun. In this case, we divide 5,738 by 7918 to find .724678.

Next, we find the angle of .24755 getting a distance to the tip of 916,310.16 miles (this represents the length of the umbra in this scenario), which we multiply by .724678 to get the moon 664,030 miles from the tip, which is 252,280 miles from earth. Perfect. We now enter the angle of .24755 in the angle-A box with half the reported solar diameter (432469) in the edge-a box to find the calculator telling us that such a line will travel out 100 million miles before it finds a sun at 864,938 miles wide. So, you see, that's how the NASA goons have the eclipse rigged, to get their solar distance, roughly. But anyone can play that game with just one line (i.e. all lines to the sun but all at one angle). You can put the sun at any distance with just one line. But if you are not a goon, you quickly realize that two lines (i.e. at different angles) are necessary, and you find the sun that way, one with a lunar-eclipse line, and two with the line from the angular size of the sun on the day of the same eclipse. Where the two lines meet, that is where the sun is. As NASA, nor anyone else I know of, has shown such a method, they are all falsifying goons who control astronomy. The two-line method would become obvious to any astronomer, wherefore fear of repercussions, or spiritual corruption, keeps them from speaking out.

This is how close the human race is from knowing the true distance of the sun: if NASA would only report the true times of the eclipse, we would have the means to a very-close approximation of the umbra width where the moon passed through it, and thus we could find the angle of the eclipse line. I implore astronomers worth their salt to study the eclipse of July 16, 2000, on NASA's page by Fred Espenak, and use the information provided there to find the lunar velocity. I am sure that you would find, as I did, a velocity of 2,000 miles per hour, which is impossible. It means that NASA gave false times for the eclipse that are in the neighborhood of five minutes too many for the U1 to U2 period. The umbra would become more than 6,000 miles wide if the moon's velocity were properly more than 2,156.4 mi/hr, which is NASA's own slowest moon velocity. The velocity of the moon for the total eclipse for July of 2018 has a moon even slower than 2,000 mi/hr.

Solar Eclipse Data Offers No HopeIn hopes of breaking the NASA roadblock to these lunar eclipses, I've been neglecting the possibility of finding the solar distance by using a solar-eclipse line in conjunction with the solar size on the day of the eclipse. Possibly, details for solar eclipses are given more factually rather than by a computer program with the 93-million-mile hoax built-in. It dawned on me that the reported width of the moon shadow on the earth's surface can be verified / checked by various satellites not friendly to NASA, which should keep NASA talking straight about such details. On the other hand, no one (and no satellite) can see the earth shadow in order to keep NASA straight about its details.

When one has the diameter of the moon shadow on the earth, and the distance to the moon, one also has the angle of that line to the edge of the sun...which is why I don't expect NASA to give both needs on solar-eclipse pages. The solar eclipse of August 21, 2017, is passing through the north-central United States, and Fred Espenak is being held responsible, by his own admission, for "all eclipse calculations." Place your bets on whether Fred will publicize the width of the lunar path?

For the total solar eclipse of March 9, 2016 (2 am Universal Time), the so-called "Path Width" at the greatest eclipse is given as 155.1 kilometers. It should be true that this phrase represents the literal width of the shadow as well as the width within which people can see the full eclipse (as opposed to partially). The angular sizes of the sun and moon are also given, but not the distance to the moon. This eclipse happens to be the day before the moon was at perigee, and the perigee calculator says that the moon was 359,508 kilometers away at 7 am on the 10th. The question now is whether I can venture a good guess on what the lunar distance was when comparing the lunar size at the eclipse to the expected lunar size at perigee the day after. If we knew the moon size with certainty for the closest-ever or average perigee, this would be more reliable, but NASA has failed us badly on this thing. It doesn't even give the largest and smallest moon sizes even on its moon fact sheet. NASA seems to leave out a critical parts of information needed to draw a correct line to the sun's edges. It looks deliberate.

The sun's radius for the March eclipse is given as 16'6.05", or 16 minutes and 6.05 seconds, which I interpret as 16.10083 minutes because 6.05 divided by 60 shows that it's .10083 of one minute. The size of 16.10083, because it's in arc-minutes, can be multiplied by 60 to find its equivalent in degrees, which is .268347 degree. Multiplied by 2, the reported solar diameter at mid-eclipse is found as .536694 degree.

The moon's radius at mid-eclipse is given as 16'33.5", or 16.55833 arc-minutes = .275972 degree, or .551944 degree for the diameter.

The so-called gamma for the solar eclipse, which I've assumed refers to where the tip of the moon shadow reaches in the interior of the earth, is given as .2609. I'm leaving open the possibility that my interpretation of gamma, as it relates to solar eclipses, is not correct. Here is NASA's definition: "Gamma is the distance of the Moon's shadow axis from Earth's center in units of equatorial Earth radii. It is defined at the instant of greatest eclipse when its absolute value is at a minimum." I've never read this definition before, having the second sentence (that I don't understand). I had read (elsewhere from NASA) only the first sentence, and assumed that "Earth's center" is the core. Can there be another, surface-related center that is meant? The solar eclipse below has a gamma of .3271, which is nearly a third of an earth radius, yet the spot on the earth surface for the "Greatest Eclipse" shows almost smack at the center of the earth circle. I don't see any logic in viewing this spot one third of an earth radius from any "center" point upon the surface.

http://eclipse.gsfc.nasa.gov/OH/OHfigures/OH2013-Fig05.pdfWhat could it mean that gamma is greater until a minimum occurs at greatest eclipse? The definition as given by the NASA page stinks. They need to re-write it. It does not help as it now sits. If I have gamma interpreted correctly, it alone can be used to get an angle to the sun's edge, yet one cannot trust whether the gamma figure is a factual calculation, or a false one generated by a computer having the 93-million-mile distance built-in. As 1.0 unit of gamma is made one earth radius, the .2609 gamma figure on the eclipse page (for the March-2016 eclipse) can mean that the tip was supposed to be 1,662 kilometers from a center of the earth. What center? It's best to get the angle from the size of the shadow on the earth surface, and then compare with a line as per the gamma number, to test the gamma figure and umbra width both. If both angles point to a sun 93 million miles away, neither of the reported values (not even as per the width of the shadow) are factual.

We can do an approximation based on a moon's reported distance (359,508 km) the day after the eclipse, guessing that it was about 359,750 at mid-eclipse. As this is the distance between lunar core and earth core, remove one earth radius for the distance from the moon shadow (on the moon-side earth surface) to the center of the moon. In other words, view the lunar distance as 353,379 kilometers. As the shadow is reported at 155.1 kilometers, we half it to 77.55, and now figure that the eclipse line spreads out by a lunar radius minus 77.55 kilometers over 353,379 miles toward the sun. That is, 3,476.2 / 2 - 77.55 = 1,660.55 kilometers of line spread per 353,379 kilometers across. We just feed the figures into the edge-a and edge-b boxes to find the angle of line C (it goes to the edge of the sun) as .269234 degree. In the diagram at the page below for this scenario, the edge of the moon is at corner B, and corner C is 77.55 kilometers from the center of the moon. Corner A is where the moon's shadow line begins on earth:

http://www.cleavebooks.co.uk/scol/calrtri.htmTo test whether Fred's numbers are nearly correct, we put 77.55 in the edge-a box with .269234 in the angle-A box to find the moon-shadow tip at a distance of 16,503 kilometers behind the earth surface (facing the moon). But the gamma figure from Fred has the tip at only 1,662 + 6,371.4 (earth radius) = 8,033.4 kilometers from a center of the earth. No matter how wrong the lunar-distance figure that I'm using, it cannot make Fred's number correct (if my interpretation of solar gamma is correct). For, when we put 8,033 (1,662 + 6,371) in the edge-b box along with 77.55 in edge-a, it gets an angle of .5531 degree, more than double the angle that we got for a moon at a distance (353,379 km) that NASA would tend to confirm. For me, this is like a Frankenstein failure from the NASA people. I am once again denied a way to find the solar distance on my own. Of course, NASA would never invite us to seek that distance on our own, using eclipse lines. Go ahead, write to NASA, asking how you can find it on your own using eclipse data, and notify them that it doesn't work when the angular-size line is compared with a moon-shadow line.

To find how far the line will go at an angle of .269234 before it's spread by as much as a solar radius, enter it, along with 695,914 kilometers (= solar radius - 77.55) in the edge-a box, to find 148.1 million kilometers = 92.03 million miles. Yes, it appears that Fred has his numbers based on a computer program having 93 million built-in. One can glean here that they have no correction, in their computer program, for a refraction element that bends sunlight. If there really was a refraction problem, as astronomers sometimes claim (probably in an attempt to explain the problems with their lines), the eclipse line under discussion would not find 92 million miles, as it just did. In March, the sun is always between 91.4 and 92.96 million miles (because it's always 91.4 in early January and 92.96 six months later).

In fact, when they measure the size of the sun at .533 degree, six months after early January, the two lines go to the edges of the sun exactly where they claim the sun is. To prove this, put .533 in the angle-A box with 864,938 miles in the edge-a box (the latter is their number for the sun's diameter), and you will find 92,975,361 miles in the edge-b box. If any significant refraction were in effect, the number in edge-b would not come so close to what they claim for the average-distance sun. I have never read anyone saying that the size of .533 is after refraction's effect.

As we saw, Fred gives .268347 degree as the solar diameter at mid-eclipse. We can now draw two lines, one at that angle, and the other at .269234 degree. The rest of this paragraph is futile, but I'll include it because the method shows how to find the solar distance. To find the spread per kilometer toward the sun for each line, put .268347 and .269234 in the angle-A box with 1 in the edge-a box to find 213.512208 and 212.8088 kilometers of distance toward the sun, respectively, per 1 kilometer of spread. Divide 1 by 213.512208 and 212.8088 both, to find .00468357 and .004699 kilometers of spread per 1 kilometer toward the sun. We now need to subtract .004699 from .00468357, but the answer is less than zero, an impossibility. To put it another way, the lunar eclipse line at .004699 degree spreads more than the solar-size line at .00468357 degree so that the two lines will never meet, neither at the sun where they are required to meet, nor anywhere. This paragraph is useful if the eclipse line is less steep than the solar-size line. If the full core-to-core distance is used (instead of 353,379), the eclipse line comes out at a smaller angle than the solar-size line, but the two lines meet at just 1.145 million miles away from earth.

So, the sun works out to less than zero kilometers away thanks be to the angle of the eclipse line that Fred has given us to work with, albeit he never hoped that we would discover the angle of this line. Fred, the world of astronomy implores you to provide the correct figures.

It looks like Fred is willing to risk being found as a fraud by the satellites of nations unfriendly to the United States. If these nations pipe up to say that a certain moon shadow is smaller or larger than what Fred reports, they can be ignored. But what about honest astronomers in nations friendly to the United States? Can't they be ignored too? On behalf of the holy-cow hoax, yes, by all means.

One can find the line from the sun's edge, at 92 million miles (148.1 million kilometers) away, to the moon's edge when the moon is 359,750 kilometers from earth, and finally bring the line to the surface of the earth to see how large the moon shadow would be from that method. One would subtract 359,750 kilometers from 148.1 million kilometers (or whatever refined number you like) to find that the line decreases in spread by 695,991 - 1,738 (solar radius - lunar radius) = 694,253 kilometers over 147,740,250 miles. Dividing one from the other, it works out to 212.805 kilometers (virtually the impossible number above) across per one kilometer of decreased spread. To find the decrease in spread over the 353,379 kilometers from center of moon to the earth's surface, divide the latter number by 212.805 6.51 to get 1,660.06 and therefore multiple the latter by 1 to find that the line at the earth's surface would be 1738.1 (lunar radius) - 1,660.06 = 78 kilometers, which is close enough to Fred's 77.55 to show that the method is correct. Unfortunately, we can't use the method to find the real shadow size unless we know the true solar distance.

The solar eclipse of one year from now will cross the United States fully, and so we need American people to verify, or not, Fred's numbers.

Do you think it's coincidental that no one is given sufficient information to find the true distance to the sun either by lunar-eclipse or solar-eclipse data? How can we find the true moon-shadow width unless someone with a satellite measures it? In fact, after writing that sentence, I googled "satellite photo of moon shadow" to find NASA offering a photo for the eclipse under discussion! And it's huge!

http://www.space.com/32215-total-solar-eclipse-2016-satellite-photos.htmlThe page says, "It provides a bird's-eye view of the total solar eclipse, which was visible from the ground only along a 90-mile-wide (145 kilometers) path that stretched through Sumatra, Sulawesi, Borneo and other islands eastward into the Pacific northeast of Hawaii." I have slow Internet on dial-up service only. It takes about ten minutes to load the page. There were two satellite photographs while the page was loading, the first showing the entire earth and shadow, and the other not. When the page was fully loaded, the one showing the entire shadow disappeared and turned into a link at "photo." But when I clicked the photo link, nothing happened because the page had bumped me offline. I tried to re-load three more times, but each time I get disconnected from the Internet before it's fully loaded. The caption for the photo is still on the page (this is 3 am September 20): "The shadow of the moon is clearly visible on Earth's Pacific Ocean in this spectacular full-planet view of the total solar eclipse of March 8/9, 2016 from space by the NASA-NOAA Suomi-NPP satellite. Credit: NASA/NOAA/GSFC/Suomi NPP/VIIRS." Inside the photo it says that it was imaged by Suomi NPP/VIIRS alone.

I often see things on a page that others do not due to slow loading. If you have instant loading, you probably will not see this photo if it was programmed not to show. The second photo, which stays on the page, shows blackened ocean; it's about as wide as only 3.5 centimeters of the full-planet shot (that alone makes it far larger than 145 kilometers; Fred says 155). Upon re-loading the page, the first photo came up again; the full planet measures 14.5 centimeters on my screen, and the shadow is at least 5 of those, or over 2,600 miles / 4,100 kilometers...bigger than the moon itself! That measurement is in the vertical direction. Horizontally, the shadow is about 3.5 centimeters (of the 14.5), though the implication is that it's wider in that direction too (though the photo doesn't show it. Why not?).

I saved it while I had it up; see it here:

http://www.tribwatch.com/photos/eclipseMoonShadow.jpgAs the shadow shows wider than the moon, the penumbra is showing, making it impossible to know the size of the umbra. At least, one cannot make out the umbra from the penumbra, and perhaps NASA has tampered with the photo to boot. The shadow does not show round, as one would expect, making the photo suspect. It gives the impression that there is no way at all to define the umbra. The penumbra is always larger than the moon, and the umbra is always smaller than the moon.

If the sun is larger than the moon (i.e. not technically a total eclipse), the umbra does not reach the surface of the earth. In this case, the line situation will be reversed: the solar-eclipse line will be steeper than the angular-size line. Yet, the two lines must still meet at the sun. The problem is, there is no evidence on how high the tip of the umbra is above the earth's surface, and NASA cannot be trusted for this height.

Can we figure out the solar distance from an eclipse where the sun and moon are identical in size? Actually, no, because the solar eclipse line will be identical to the angular-sun line. The two lines will both start at the earth's surface. When moon and sun are identical, an umbra merely kisses the earth's surface so that it can be taken as zero miles wide.

No one can be within the umbra if a piece of the sun can be seen. No sunlight shines in the umbra, whereas sun does shine in the penumbra (see drawing here). The penumbra is supposed to become ever darker with nearness, from its rim, to the umbra rim. If satellite photos fail us, it seems to me that people can record the towns / cities where totality exists in order to find the true width of the umbra. Shouldn't this have already been done in previous eclipses? The lunar-eclipse drawing at the page above can give you a great idea on why a lunar-eclipse line can tell us very-closely the distance of the sun as per where the line meets a solar-eclipse line. You can see that a lunar-eclipse line is always at a less-steep angle so that the solar-eclipse line will catch up to the lunar-eclipse line eventually. You only need to know by how much the one line will catch up to the other per mile / kilometer toward the sun. I've created the math method for finding out, though astronomers have the same method yet never share this information with us.

You will never here of an evolutionist astronomer using eclipse lines to find the solar distance because it is dangerous for them to give-out even the mere idea. The first thing one will discover by looking into this method of finding solar distance is that NASA's lunar-eclipse data is erroneous; the moon is far too slow according to the size of their umbra diameter. The world will nail these goons as frauds right away, and, as I showed above, their solar-eclipse data can be shown to be fraudulent too.

The angular-sun line is not on the drawings at the page above, but is a line from the middle of the earth surface to the edge of the sun. It's exactly equivalent to the line shown on the annular-eclipse drawing. "Annular" is probably related to "annulet" = solar ring around the moon.

As you can see, one can find the solar distance by using the umbra and penumbra lines together, but, for this, one needs to know exactly where the umbra begins and where the umbra ends. NASA will never give these two items up truthfully until its leadership is made devoid of evolutionists.

As we saw, an eclipse line spreading out at .269234 degree spreads one kilometer per 212.8088 kilometers toward the moon or sun. It will therefore spread by 93,000,000 / 212.8088 = 437,012 kilometers (271,547 miles) over 93 million miles, and that's supposed to be the sun's radius, yet it is just .628 of what they report for it. Nothing much is working for them when they insist on a sun at 93 million miles. Reminder: the angle of .269234 was obtained from Fred's 155-kilometer shadow path, a number that cannot be trusted. I'm only showing here that his lunar-path number is fraudulent, not even getting the reported solar diameter or distance. It begs the question of why Fred didn't use a number that gets the reported solar distance. The answer is: you can get that distance dishonestly in one way, but the rest of the math is thrown out of whack because it's not the correct distance. We saw how Fred had the distance at 92 million miles by one way of looking at it.

The best I can do now is with a set of true satellite pictures on solar-eclipses, with the penumbras showing well enough to get their diameters. As you can see in NASA's photo above, the penumbra is not round, not at all well-defined, and this may be due to tampering, for one can draw a penumbra line to the edge of the sun, if the penumbra is well defined, to find where it meets the angular-sun line. In the satellite photo, one can glean that the darkest / central area (ignoring the dark areas in the vertical direction) is at least three centimeters from the edge of the penumbra. As 14.5 centimeters is the width of the earth at 12,743 kilometers, the 3 centimeters works out to 3 / 14.5 x 12,743 = 2,637 kilometers.

We now have a line that starts on the earth surface 2,637 kilometers from the center of the umbra. The penumbra line thus closes by 2,637 kilometers minus the lunar radius (1,738), or by 899 kilometers, per 353,379 kilometers (= lunar distance to earth surface), which works out to 899 / 353,379 = .002544 kilometers of spread per kilometer toward the sun. On the other hand, the angular-sun line spreads by .00468357 kilometer (math shown above) per kilometer outward. These two lines will never meet. The penumbra line must go toward the sun at a greater angle than the angular-size line in order for the two to meet, wherefore the penumbra radius needs to be much larger than 2,637 kilometers. You can see in the drawing that the penumbra line must be the steeper of the two.

I've just realized that we can actually figure out the minimum size of the penumbra using the solar size on the day of the eclipse. We just do the math in reverse until we find .00468357 (instead of .0025028), in which picture the two lines will be at identical angles. We can then say that the umbra needs to be larger yet. The math in reverse is: .00468357 x 353,379 + 1,738 = 3,393 kilometers for the umbra radius. An umbra thus having a diameter of 6,786 kilometers would be: 6,786 / 12,743 x 14.5 centimeters = 7.7 centimeters wide on the satellite photo, or more than half the total earth diameter. Just so you know.

So, there are two ways to get the solar distance using a solar eclipse: 1) by finding where the umbra line meets the angular-size line; 2) by finding where the penumbra line meets the angular-size line. Comparing the one with the other, if they do not get the same solar distance, proves them to be the darkened goons that I accuse them of.

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