Tell me that Kash Patel doesn't look like a crow:
https://www.youtube.com/watch?v=DDUmaqlulZM

Two updates ago, I deciphered my crow dream as pointing to Kash Patel's BURN-bag revelations. The dream was short, with my kicking my LEGS toward the stove BURNer as the crow was wanting to land upon it. Last week, I missed going to the Crews/Creuse's (Cheshire with Leghs), who are in the colors and format of Trump-connectable Leggs. The Crew/Creuse Crest has a lion's "gamb" (i.e. a lion's leg), and the giant "man's leg" of Gambino's/GamBELLi's is in Legh/Ley colors and format. One can see here that surnames using the heraldic leg could have been related by marriage to a Gambino/Gambelli surname.

Bells are first known in Dumfries with Leggs, and Bellys are first known in Moray with Douglas' who in turn share the STOVE/Stevenson Chief. Moreover, Douglas-like Dougals share the Crew/Creuse Coat.

You can click Crews now, which loads them on another tab in order to load other surnames, to better follow the heraldic links.

There's more, because Patel has LEAKed some of the material in the burner bags, and while Leaks use a leg, they share the single fleur-de-lys of Gambelli-like Gamble's. The Leak/Leakey Chief is in the colors and format of the Army Chief, and Armys (Lincolnshire with Leaks/Leakeys) can be expected in the Crew/Creuse Crest, "A silver lion's gamb ARMed RED" while Reds/Reeds are first known in Northumberland with Stove's. Armors share the Stove / Douglas stars.

Also, Italian Belli's are first known in Verona with Bellino's who in turn have their "bear's paw" showing as an entire leg in the format of, and colors reversed from, the Gambino/Gambelli leg. The Powys' have a bear's "gamb" in the colors of the Gambino/Gambelli leg. Gamble are also Gamals while Crows have a CAMEL head.

Brunells/BURNells have a "lion's paw" in Crest, as do Crews/Creuse's, and the latter's Coat is the Brunell Coat in colors reversed. The English-version motto of Brunells (share a Brennan Coat) is, "Charity BEARs fruit." Perhaps it's off-topic, maybe not, but Jean-Luc Brunell reportedly hung himself in a French jail awaiting trial on assisting Jeffrey Epstein's pedophilia business.

The Burnell motto has "Caritas," the word translated "Charity," and as the Burnells share the Bruce Coat, it's likely the Charity's (Yorkshire with Bruce's), who almost share the Irish Casey Coat, were from Marjory Carrick, mother of the royal Bruce's. Caritas-like Carrots/Carews, definitely a Carrick branch, are first known in Cornwall with Jeans who in turn share the Burnell/Brunell / Bruce lion. JEAN-Luc Brunell. The Caseys are extremely important here in a pointer to Kash Patel.

When the report came out that Epstein hung himself, I had the inkling that Trump was involved with the murder, or in allowing Epstein to go free under the pretence of his death. I don't rule the latter out to this day. In his last phone call, hours before his reported death, Epstein called Belarus, and told the man in that country that the jail he was in is "keeping me safe." That phrase is worded by me to the best of my recollection from a video I saw. It struck me as though he was allowed to call his buddy to say, in code, they are releasing me, I'll see you soon.'"

In the Sleeping Beauty dream upon Epstein's island, I had only my JEANs on while walking to Beauty's CAR. The Arms of Carrick is just a red-on-white chevron, and so Cars almost have the Arms of Carrick. Shirts/SHARDs, whom I trace to SARDinia, where Bondi's are first known, have the Arms-of-Carrick chevron, and are in Charity colors and near format. English Kerricks are first known in Cheshire with Bello's and Shirts/Shards. Kash Patel's current boss in Trump's government is Pam Biondi.

Note that Charitys (Yorkshire with Keppochs and Wagers) are in Keppoch colors and format while Keppe's are first known in Hampshire with Casey-like Chase's, tending to confirm that Charity's incorporate the Irish Casey Coat. Stove-branch English Stevensons are first known in Essex with Chase-branch Chance's (might share Bath cross). Chase's (might share Bath cross) are first known in Hampshire with the BIDENs/Buttons who share the Charity fesse, and the Bittons/Buttons in the Biden write-up are said to have furnished a bishop of Bath and Wells in 1274 while Burnells tell that "Robert Burnell (1239-1292) was Bishop of Bath and Wells in 1275." Buttons/Bidens have "horns" in Crest while Bruns are first known in Middlesex with Horns/Orne's. See the top of the next update for a follow-up to what you've just read.

Gemels (not "Gamal") have the Wager Coat in other colors who in turn share the hearts of Patel-like Pattle's. While gambling involves wagers, I read that Kash Patel likes to spend time in Las-Vegas casinos. Casino's can be linked to Kash-connectable Chase's and Change's. The Gemel Crest loves the Swans/Sions, first known in Lanarkshire with heart-like Hardys and Kash's/Caseys. Take a guess at what symbol the latter use. You won't believe it. Heart-using Douglas', said to descend from Hardys, share the Stove Chief.

The Bellino bear symbol is very welcome because the crow was trying to land on the stove BURNer. Why? Partly because Landons share the bear symbol with Berns/Barns, Berwicks, Barwicks, and the Arms of Berwick (Bernicia) while Burns, sharing the Stove stars, are first known in Cumberland with Bernice's. Lands/Landens are first known in Switzerland with Berns/Barns. The Sarah's sharing the Barwick motto are first known in Essex with Stove-branch Stevensons (Barnstaple colors and format), Legh-related Low Leighton, and with Beacons/Bacons in the "FLAMING beacon" of German Belli's.

This is now excellent because Legh-branch Leightons/Leytons (beside Leghs) are first known in Shropshire with Bell-branch Bellamys and Powys'. Scottish, Belli-connectable Flemings are first known in Lanarkshire with KASH's/Caseys in turn using crow's heads! Irish Flemings not only share the Shield of English Beckers, but the checks of Dutch Flamings who in turn share the Coat of BEARings/BECKERings, the latter first known in Devon with bear-using Bears/Beers and Berrys. English Beckers share the vaired Shield of Beach's/Becks while the burn bags were pointed to by Bachs/Baghs (Wales with Powys). Bags (Norfolk with Case's) almost have the Beach/Bech Coat.

French Caseys are first known in Ile-de-France with Verona's while Italian Belli's and Bellino's are first known in Verona. The "black birds" of Verone's/Varenne's show as crows / ravens, and Birds/Burds (Cheshire with Crews, Leghs/Leigh's/Leys and Gamels) share the fleur-de-lys of Gamble's/Gamals and HICKS'. The latter are said to have been in Low Leighton of Essex. Birds/Burds (Cheshire with Gamels and STEELE's) may have been a branch of Births/Berts sharing the Burt Coat, for Berta's share the griffin of German Camels and German Steels. Steele's share the bend-with checks of German Beckers.

Miss Hicks on the beach of my Sleeping Beauty dream, who had a leg symbol in that dream, and a knee/leg symbol in real life on the Leakey road of Camp Wood, Texas, was the niece of the wife of Mr. Casey. All three attended the church I was attending, and Mr. Casey (real-estate agent) listed my Texas property and sold it. The beach was of white sand, and Sands share the fitchees of gamb-using Powys'. Sands, in Cruise colors and near format, are first known in Surrey with James' and Clappers sharing the Beach Shield, and Kash's burn bags involve crimes by James Clapper and his conspiracy with Christopher STEELE. James' share the giant lion of gamb-using Crews/Creuse's.

BattiSTELLI's can be gleaned as a Stelli-line merger with Batti's/Botto's, and then Gambino's/Gambelli's are also GamBATO's who share the Battistelli stars, in the colors of the giant Stelli star.

Gambino's/Gambelli's have a "MAN's leg" while German Mans/Manners have the Leighton/Leyton quadrants in colors reversed while English Manners/Manders are first known in Northumberland with Stove's/Stevensons.

Scottish Flemings, who can be in my stove burner because I have a GAS stove, love the Shaws in their motto while Scottish Shaws love the Mean variation of Mine's/Menne's in their motto while "TentaMINE" is a Legg motto term.

The Gemel and Swan/Sion heart is the one of Lanarks/LURNacks, who have Lorraine-like Lorne's in their write-up while Lorraine's are first known in Northumberland with Stove's, and with the Lums who are not only in the Stove motto, but share the Lanark/Lurnack cinquefoil. It tends to explain why Irish Lane's/Lawns share the Lorraine bend. It can explain why Lane's/LAWNs share the bend of Lorraine's, and why Lorraine my date (in my 20s) got a LAWN STAIN that I've talked much about. Stains and Steins are a branch of the Steinson variation of Stove's.

After writing the above, I remembered that I rode my bike to Lorraine's BUS stop, and asked her on our first date at the corner of LORNE and Yonge streets. That's why Lanarks/Lurnacks share the Bus cinquefoil. This got me wondering how my bike could point to Kash Patel, but minutes passed while seeing no way for Biks/Bickers to apply, and so I moved on.

I then searched for Pattle-like surnames and realized that I rode a PEDAL bike to her bus stop! Pedals/Pedleys, first known in Oxfordshire with Crow-branch Crawls/Crowells, happen to share the Stove fleur-de-lys, in colors reversed from the Burn fleur! Crawls are first known in Oxfordshire with Tiens/Thames and Peare's, both of which share the stars of Gamels, and there is a camel head in the Crawl/Crowell Crest. Stove-branch English Stevensons/Steinsons are first known in Essex with Camulodunum.

From the 4th update in February, 2019: "I cannot recall what happened to the Firebird. I was on a ten-speed bike when asking Lorraine out for the first time." The bike was owned by Sara, who was living at the time directly across the street from immigrants from India! They invited me for dinner once or twice. Kash Patel is Indian. Sarah's share the Barwick motto.

The Pedal/Pedley fleur are blue, like the Morinis fleur, and while Morinis' (Modena with Casino-branch Casano's) are in the write-up of Deerings, the latter share the stag heads of Biks/Bickers! The latter are not only in the colors and format of BAGers (Yorkshire with Beckers and Pattle-connectable Wager), but of Steins (Norfolk with Bags and Bus')! The burn bags. I can barely believe it. Bagleys share the Coat of Roots, the latter first known in Kent with Deerings (and Crowder-branch Crowhursts).

One wagers with dice, and Dice's/Diss' (Norfolk with Bags) share the Bager eagle. I didn't know until seconds after writing the last sentence that Vegas' have the Bager / Dice eagle in giant form! "Vegas" could therefore have been a variation of the Bagges variation of Bags. But if God arranged this gamble theme to point to Patel's love for Las-Vegas casinos, why? Some liberal writes: "Kash Patel, Trump's pick to lead the FBI, resides in Las Vegas at the home of a GOP megadonor."

The Vegas Coat is also the one of Strongs, the latter first known in Somerset with Crow-beloved, Gamble-like Camels, with "disc"-using Webbers, with Bests ("legiBUS") sharing the Hurt/Hort cinquefoil, and with Cocks who not only share the Bag/Bagges Shield, but the Crow rooster in colors reversed. Dice's/Diss' had named Disce. The Webber disc shows as a hurt, and Hurts/Horts (share Sarah cinquefoils) are first known in Oxfordshire with Crow-branch Crawls/Crowells, and with Crispins who almost have the Coat of Dutch Bests. Beasts are first known in Cheshire with Crews/Creuse's.

I'm more than convinced that God gave the crow-and-burner dream as a pointer to Kash Patel. However, as I try to discern whether this means that God wants the credit for moving Patel to make some major revelations or arrests, I still ask whether Pam Bondi has been ordered by Trump to make it appear that arrests are imminent, for the purposes of the 2026 elections, but not, in the end, to make major arrests.

Refining Last Update's Atomic Distances

I thought I discovered, in the last update, how to prove that the establishment's assignment of atomic-weight figures proves my atomic model correct, where I claim that all atoms weigh the same, and that the atomic-weight figures, for gases at least, are proportional to the number of atoms per gas. The lower the atomic-weight figure, the fewer the atoms per gas at STP in contradistinction to the establishment's model of the atom.

I was so busy grappling with to figure how much gases need to be compressed to achieve liquefaction at the critical temperatures of gases, whether or not there is to be a second compression operation, when finally I came across the following insight strongly suggesting that one compression operation is all that's needed:

Does it feel right that O atoms 4.14 inches apart would work out to be 16 times more numerous than H atoms 12.6 inches apart? I've asked: "if balls are 12.6 inches apart in three dimensions, how far apart would they be if 16 times more numerous in the same volume of space?" Google AI gets regularly screwed up on this question. It's only correct when it does: 12.6 / 2.52 = 5.0 inches. That doesn't make me happy because it's not close enough to 4.14 for me to get out the champagne, not even the cigar. Instead, it prompts for aspirin.

In case I lost you: when H atoms at STP are made 16 times more numerous at equa-distance, they are the same distance apart, center-to-center, as O atoms are in their own STP gas. Therefore, when the H atoms at 12.6 inches apart are made 5.0 inches apart by making them 16 times more numerous, 5.0 inches is how far apart O atoms are in their gas. But all those headaches I went through dealing with liquefaction only got me 4.14. Why is me woe?

...I suppose it's possible that, when we are told that hydrogen needs 13.2 atmospheres and critical temperature, the 13.2 is done at STP with nothing left to do but cool the gas. That would eliminate the extra 8 [i.e. the second compression operation]...

Hydrogen gas will become a normal liquid with only 1 atmosphere of pressure at -253 C, its boiling point, only 13 C below its critical temperature. It tells us how close the atoms are aside from the 13.2 atmospheres of pressure applied at -240...

This small, 13-degree spread between critical and boiling suggests that one does not compress the gas 13.2 atmospheres after -240 [= critical] is achieved, when the atoms would be nearly in contact at that temperature even at 1 atmosphere. The 13-degree spread argues for compressing the gas 13.2 times starting at a higher temperature, perhaps at 0 C. Somebody get me a cigar.

Yes, when the compression is done at 0 degrees, the pressure subsides much during the cooling to -240 C. According to math in the last update, atoms come 2.36 times closer when the gas is compressed 13.2 times to 13.2 atmospheres (though I wrongly used 2.33), and the claim is that atoms are touching when compressed 13.2 times, but are 2.36 diameters apart at STP.

Here's what happens. The H gas starts at STP with atoms 2.36 diameters apart. If the container is 13.2 inches tall, a piston compresses it until there is only one inch left at the bottom of that volume. The atoms are now touching, but cannot begin to form liquid droplets on the container walls until the gas is cooled to -240 C, at which point atoms are able to attract each other when in contact. They cannot attract each other when not in contact because they repel each other when not in contact.

As liquid forms on the walls due to atoms making contact, they merge on the container walls, which opens up a little more space in the one-inch gap because two atoms merged take up less space than two unmerged atoms. This causes the atoms in the space to move apart slightly such that they are no longer in contact, and so the gas needs to be compressed a little more to get them back into contact for to create larger liquid droplets. When finally the piston is fully compressed such that there is no space or gas left, there's only liquid between the piston and the container floor.

If the piston is lifted at or below -240, the liquid will remain a liquid, with space above it. It will evaporate in the normal way. But if the piston is lifted above -240, the liquid will instantly disintegrate (fast "evaporation") because there are then enough heat particles (free electrons) in the liquid to push atoms apart, because they act in opposition to the attraction forces which has previously caused atomic mergers.

THE BIGGEST POINT to all of this is: only 13.2 atmospheres of pressure are needed to get the H atoms into contact. The gas volume at STP needs only be reduced to a one-inch space in a 13-inch-tall container. When atoms touch after being brought 2.36 times closer together, they can be proven to be 2.36 diameters apart at STP, for they are exactly 1 diameter apart, center-to-center, when in contact.

The beauty of this finding is that I can now know, or at least can argue, that oxygen atoms are 3.7 diameters apart at STP because they require a gas compressed to 49 times its volume in order to liquefy on container walls at the critical temperature of oxygen (-118.6 C). The last update showed how O atoms are 3.7 diameters apart after compression to 49 atmospheres of pressure, and are therefore 4.14 inches apart when O atoms are figured at a diameter of 1.12 inches. On that scale, H atoms were found 4.5 inches in diameter, and because they are 2.36 diameters apart at STP, they work out to be 10.6 inches apart at STP, not the 12.6 used in the last update.

Here's part of the beauty: I learned last week how to find the spacing of O atoms by another method so that I can compare the two to see if they agree. The beauty is that the other method involves my atomic model in which there are 16 times more O atoms at STP than H atoms. This means that, if I find how far apart H atoms are, center-to-center, when an H gas has 16 times more atoms, that's also how far apart O atoms are center-to-center. And the beauty is, it's easy to find that H atoms are 10.6 / 2.52 = 4.2 inches apart when 16 times more numerous. I showed how that math works, affirmed by google AI, by finding 2.52 as the cubed root of 16.

We then compare 4.2 with 4.14 because that's how far apart O atoms are at STP when found 3.7 diameters apart, for 3.7 x 1.12 inches = 4.14. Those numbers, which are not likely exactly correct, are close enough to make heads turn of the staunchest kineticists who love their atomic model more than they love their wives.

It's time to discover how far apart nitrogen atoms are. google AI says: "To make nitrogen liquid at its critical temperature of approximately 126.2 K (-147 °C), you need to apply a pressure of about 33.5 atmospheres (atm)."

Next, we ask google: "When balls are hovering at equal distances of 4 inches in 3 dimensions, how far apart will they become center to center when the volume is reduced 33.5 times?" google AI usually answers this question correctly when framed exactly with the words I've used, but not always. It correctly responds: "The new center-to-center distance between the balls will be approximately 1.241 inches."

It doesn't matter what size balls or atoms you plug into the question because we are concerned with the center points only. For the task at hand, we don't want the distances apart, anyway, but only the 3.225 factor when we do: 4 inches / 1.24 = 3.225 times closer. That's the nugget were looking for, which lickety-split reveals that N atoms work out to 3.225 diameters apart at STP...if the atoms are exactly 1 diameter apart after compression. In this case they cannot be exactly 1 diameter apart after compression because another, reliable method finds nitrogen atoms more than 3.225 inches apart at STP.

No matter how far apart the balls are when we ask the question above, we get the 3.225 figure. When we ask with balls three inches apart, they become .93 inches apart after compression. The factor is still the same at: 3 / .93 = 3.225 times.

When we found nitrogen atoms are 3.225 diameters apart at STP, and 3.225 inches closer after compression, that's the same figure for the cubed root of 33.5, which is the figure for the amount of gas compression (33.5 atmospheres) needed to bring N atoms into contact. We therefore don't need google to tell us how much closer atoms will be when compressing by the critical-liquefaction pressure for any atom. Just find the cubed root of that compression figure.

When we compress H atoms by 13.2 atmospheres, to find that they are 2.36 diameters apart prior to compression, that's exactly the figure for the cubed root of 13.2. I like to use AI too, because it gives the reader more confidence in my math and the reasoning.

Here's the cubed-root calculator I'm using:
https://www.calculatorsoup.com/calculators/algebra/cuberoots.php

With N atoms working out to be 3.225 diameter apart at STP, and because they are 1.2 inches in diameter where the H atom is to scale at 4.5 inches, nitrogen atoms work out to be 3.225 x 1.2 = 3.87 inches apart at STP. PROBLEM.

As nitrogen gas at STP weighs 13.9 times (13.89 to be exact) as much as hydrogen gas at STP, or because the N atom is given an atomic weight of 14, I asked: "if balls are 10.6 inches apart in three dimensions, how far apart would they be if 14 times more numerous in the same volume of space?" The response is fully correct: "the centers of the balls would be approximately 4.4 inches," though the same response gives 5 inches wrongly too, as though google AI is a schizoid fiend when tackling problems like this. It can give different answers day-to-day to exactly the same question.

The reason we can know that 4.4 is correct is because 2.41 is the cubed root of 14 (2.4 for 13.9). The question above to google finds nitrogen atoms 4.4 inches apart at STP to the scale of H atoms 10.6 inches apart at STP. But the 4.4 inches found by this atomic-spacing method is not the 3.87 inches found above by the depth-of-merger method. I need to find the way to repair the 3.87 figure, because the 4.4 is obviously in the very-correct ballpark.

Plus, the 3.87-inch spacing is not very close to the expectation of a little more than 4.14 inches. The latter is the spacing for O atoms. The 4.4 figure fits to the expectation. We can do the math without treacherous google AI once we have the cubed root of any compression figure; in the case of N atoms, we do: 10.6 / 2.4 = 4.4 inches apart.

One can now fix the 3.87 problem by simply gleaning that N atoms liquefy roughly when they are 4.4 / 3.87 = 1.14 diameters apart, not exactly 1 diameter, after full compression of 33.5 times the original gas volume. Would you say that this is an acceptable fix?

As nitrogen gas weighs 13.9 times more than hydrogen gas, the true meaning of its assigned atomic-weight figure of 14 is that there are 14 times as many N atoms per STP volume. That's because all atoms weigh them same, as wacky as that sounds to the experts.

I've proven that gravity arranges all atoms to weigh the same. It's not even up to debate because all materials of all weights/densities fall to gravity at the same speed of acceleration, because gravity pulls atoms individually, duh. It doesn't matter how many atoms are in any object, all objects fall to gravity with the same pull of attraction because gravity pulls all atoms with the same pull force of attraction, which is the definition of weight. ALL ATOMS WEIGH THE SAME.

I'm in the process of finding the atomic spacing of N atoms by another method, by comparing the cross-sectional areas of N and O atoms, though one can also compare the cross-sectional areas of N and H atoms. Once the cross-sectional area for N atoms is found, their diameter is also found.

The reason the cross-sectional areas need comparison is as per my claim that gas atoms repel each other by a specific force level proportional to those areas. If this investigation gets the N atom's spacing to match 4.4 inches above, I'm in good shape with this claim. Otherwise, it needs adjustment. Problems are the doors to solutions, and trying to figure out what's really happening with atoms is a stack of problems.

The tentative claim is that the cross-sectional area of H atoms (at STP) is 16 times that of O atoms, which causes 16 times fewer H atoms to assemble at STP simply because H atoms repel 16 times stronger in that picture.

I'll tentatively expect that the cross-sectional area of N atoms is 14 times less than for H atoms, and 16 / 14 = 1.14 times more than for O atoms. Did you catch that easy math? One can do 16 / 20 for argon gas at 20 times the weight of H gas, or 14 / 19 for nitrogen versus fluorine gases, etc.

As I don't know the shapes of atoms, I need to approximate their cross-sectional areas by viewing atoms as spheres, in which case their cross-sectional area is the area of their diameter. To find the area of a diameter: radius x radius x 3.14 (pi).

The cross-sectional area of an H atom, 4.5 inches in diameter, is 2.25 x 2.25 x 3.14 = 15.9 square inches. The area for nitrogen is therefore expected to be 15.9 / 14 = 1.14 square inches. All we do now is figure out the N-atom diameter with that information. It turns out to be 1.2 inches because .6 x .6 x 3.14 = 1.13, close enough to 1.14.

A diameter of 1.2 inches makes sense because the O atom's diameter was found, when comparing it to the H atom's cross-sectional area, to be 1.12 inches in diameter, and the N atom must be larger than the O atom because atomic sizing is inversely proportional to atomic-weight figures. That is, because the atomic weight of N is assigned as 14 versus 16 for O atoms, nitrogen is the larger of the two atoms. The atomic weight of 1 assigned for H atoms makes it the largest of all atoms; the goofs have all this backward.

Now check this out. I said above that N atoms are not yet in contact when they liquefy after compression of 33.5 times. I've found that N gas needs to be compressed by 49.4 times, same exactly as for O atoms, to make atom-to-atom contact, for the cubed root of 49.4 is 3.67 while the latter multiplied by the 1.2-inch diameter of the N atom gets 4.4 inches apart for N atoms at STP, the exact figure gotten above by the gas-weight method!

If you think about it, compressing 49 times is not much more than 33.5. It's not as much greater as the distance between those numbers can suggest. If you have a container 33.5 inches tall, and compress the piston 32.5 inches deep, with one inch left, it's compressed to 33.5 times the original pressure. To get the piston to 1/49th the distance of that height, one needs to sink the piston only an additional .3-inch deeper (not much), in which case one could glean that, for nitrogen, the atoms make contact after a compression to 1/49th, yet the atoms are able to attract into liquid mergers a little prior to contact, at 33.5 atmospheres.

The situation would require that the nitrogen atoms merged a little into the atoms on the container walls are able to attract nitrogen atoms in the space of the container, FROM A DISTANCE. As the difference in compression between 33.5 versus 49.4 atmospheres is .3 of the final 1-inch of volume, and because N atoms are 4.4 inches apart at STP, I ask google: "how closer together will balls in 3-D space be when the volume of space is reduced to .7 the original?" The response is, "The balls will be approximately 11.21 percent closer together."

That means that the N atoms will be 1.11 diameter apart at 33.5 compression in order that they might be 1.0 diameter at 49.4 compression, for 1.0 x 1.112, which is the same as saying 1.0 plus 11.2 percent, = 1.11, which is 1/9 of a diameter (not much) above one diameter. To this, I'll repeat from above: "One can now fix the 3.87 problem [of the N-atom spacing] by simply gleaning that N atoms liquefy roughly when they are 4.4 / 3.87 = 1.14 diameters apart,..."

To support my claim that atomic repulsion is at least nearly proportional to cross-sectional area, here's on an N-O compound: "The density of nitric oxide (NO) at Standard Temperature and Pressure (STP) is approximately 1.3402 g/L." That's 14.89 times as much as H gas weighs. More importantly for this point, nitrogen weighs 13.89 times as much as hydrogen, while oxygen weighs 16 times more than hydrogen. As nitric oxide has a weight almost exactly midway 13.89 and 16, I can glean the possibility that nitric-oxide molecules are 1.26 times less distant than the average distance between nitrogen and oxygen atoms, the same level of lower spacing as when a gas' volume is made twice as large.

Put it this way, that if nitric oxide weighed 14 + 16 = 30 times more than H gas, it would have as many molecules as N or O have atoms, because, in being an NO molecule, it can be construed as having two atoms per every atom in either N or O gas. But as it weighs half of 30 times the weight of H gas, the NO molecules must be half as crowded as the average distance between N-atom distances and O-atoms distances. Or, if you wish, as N and O atoms are close to the same distance apart, just simplify by saying that NO molecules are roughly half as crowded as either N or O atoms.

I learned in the last update that when particles are half as crowded, they are 1.26 times further apart (c-to-c). Science knows that gases produce half the pressure when gas atoms are spaced half as much, but science won't acknowledge that atoms repel each other half as strongly when spaced half as much. As NO has the same gas pressure, at STP, as the N and O gas under discussion, it means that NO molecules repel each other with the same force level as both N and O atoms, but only because all three gases have particles at different distances from each other. Otherwise, the N atoms have more INNATE force of repulsion (i.e. they are stronger "magnets") than O atoms, and NO has more INNATE force of repulsion than both.

Electromagnetic force gets weaker when particles are further from each other. The only way for NO to produce as much gas pressure (on container walls) as N atoms, when NO is the stronger magnet, is for NO molecules to be further apart from each other than N atoms are from each other. And the reason that NO is the larger magnet is that it's a molecule having a larger cross-sectional area than an N atom. If it has a cross-sectional area twice as large, I would propose that NO has twice the repulsion force. Does it make sense for an NO molecule to have twice the cross-sectional area of one N atom, or one O atom? Only if the N and O atom are not deeply merged into each other when forming an NO molecule.

Nitric oxide could be a molecule of more than two atoms, for example an N2O2 molecule, in which case its spacing could get less crowded than half as crowded because it also becomes a stronger magnet due to larger size. For, in a gas, the free electrons provide push-apart force for gas atoms, with more push apart force for larger atoms or larger molecules. However, its 4 atoms are so deeply merged that its cross-sectional area is twice that of N or O atoms, the N2O2 molecule could have half the crowding of N or O atoms. By following this sort of logic, science has a much-better chance of discovering the real look of molecules.

Nitric oxide, also called nitrogen monoxide, is not formed by directly bonding oxygen gas to nitrogen gas. The two atoms will not merge unless compressed. I asked: "will compressing air to critical liquefaction produce both nitrogen and oxygen liquid in the same container" Answer: "No, merely compressing air to its critical liquefaction point will not produce separate liquid nitrogen and oxygen in a single container; instead, it results in a mixture of gases and potentially a supercritical fluid." Apparently, the compression results in O-to-O mergers (droplets), N-to-N mergers, and N-to-O mergers. As soon as the piston is lifted, they all disintegrate to gas form.

Checking Atomic Spacing With Atomic-Weight Method

Xenon gas is another rare and noble gas. Its gas at STP weighs 5.9 g/l, a whopping 65.55 times more than hydrogen gas, suggesting a very small atom, and very numerous (low spacing), as compared to other gas atoms. It has a critical (liquefaction) pressure of 57.65 atmospheres, and so we want the latter's cubed root of 3.86 to find that its atoms work out to 3.86 diameters apart at STP, but only if the atoms liquefy at perfect contact.

Finding the atomic spacing in atomic diameters, via the cubed root of the depth-of-compression figure, always demands that atoms are at contact after the compression. If they are not in reality in contact, then the diameters apart at STP changes. The only time that atoms can be suspect in liquefying NOT at perfect contact is when the spacing in atomic diameters, found by the depth-of-compression method, don't match the diameters found by the atomic-spacing method.

To find the STP spacing of xenon in inches, via the compression method, we first need its diameter. We use the atomic-spacing method to find the diameter: as the H atom has a cross-sectional area of 15.9 square inches, xenon is going to have a cross-sectional area of 65.55 less, because xenon has 65.55 more atoms at STP, and thus has 65.55 less repulsion force. With a cross-sectional area of 15.9 / 65.55 = .24 inches, it works out to an atom .55 inches in diameter. As xenon was tentatively found above to be 3.86 diameters apart at STP, we tentatively assign its atoms: .55 x 3.86 = 2.12 inches apart at STP. I'll show how this 2.12 figure is very problematic for not predicting H atoms 10.6 inches apart, suggesting that the atoms do not liquefy at perfect contact.

The 2.12 figure makes xenon atoms almost-exactly half as close at O atoms (4.1 inches), which is problematic, suggesting that the 2.12 figure is wrong. Problems can be the doors to solutions. If indeed xenon atoms are twice as close as O atoms, and thus eight times as crowded, then xenon atoms will repel each other eight times more than O atoms repel each other, if the two atoms have the same innate repulsion force, which they do not.

Based on the 2.12 figure, the xenon atom shows exactly half the diameter of the O atom (.55 verses 1.12) and therefore 1/4 the cross-sectional area, resulting in 1/4 the innate repulsion force. Under that circumstance, combined with being twice as close as O atoms, the xenon atom will repel 8 / 4 = 2 times as much at STP, which it does not, for at STP both gases have the same pressure and thus both atoms repel each other equally.

But we're not done, for the finding above that the xenon atom is half the distance apart (not the same has half as crowded) does not jibe with xenon atoms being factually (no debate) 4 times as numerous, 16 for O versus 65.5 for Xe. When atoms are 4 times as numerous, they are NOT yet half the distance apart; they need to be 8 times more numerous to be half the distance apart. When atoms are 4 times as numerous, they repel 4 times as much, and so if we combine the finding above that xenon atoms are half the diameter of O atoms, then xenon atoms at STP repel 1/4 x 4 = 1 time as much as O atoms, which is the correct answer.

The way to rectify this problem is to change the closeness between xenon atoms, not 2.12 inches apart such that they are 2 times as close, but 1.59 times as close. Asking google, "when balls in three dimensions are four times more crowded, how many times closer are they center to center?" AI gives 1.59 times as close as the correct answer.

We are permitted to change the 2.12-inch spacing of xenon atoms because that was found when assuming that xenon atoms liquefy on perfect contact. The new spacing is now found by using the spacing for O atoms: 4.14 / 1.59 = 2.6 inches apart. Therefore, xenon atoms liquefy before contact is made.

The math above, .55 dia x 3.86 dia apart = 2.12 inches apart, needs to be changed to .55 x 4.73 = 2.6 inches apart. The .55 figure is non-debatable, as that was found by xenon's comparison to hydrogen. Thus, only the 3.86 can be changed, which was the figure found by assuming liquefaction at perfect contact. The new 4.73 diameters apart tells that xenon atoms merge / liquefy when 4.73 / 3.86 = 1.22 diameters apart, not unreasonable, begging only how they merge from a distance when expected to inter-repel at a distance.

It could have to do with atomic shape. For example, could they merge while atomic centers are more than a "diameter" apart if atoms are cigar shaped? Could they make contact by something long protruding from their central region? They wouldn't be in contact if positioned one way, but turn them 90 degrees, and they can then make contact. Just a shot in the dark.

I asked google, "is xenon gas strange," not knowing that "xenon" is the Greek for "strange." I asked that question before doing the math exercise I just walked you through. I asked the question because, when xenon is found 2.12 inches apart, the math is very problematic. That is, the cubed root of 65.55 is 4.03, meaning that the xenon atom worked out to 2.12 x 4 = 8.48 inches apart, yet the answer is supposed to be about 10.6 inches apart, the distance between H atoms at STP. The distance between those two figures is VAST in my mind, like breaking the back of my atomic model as I've presented it so far. But if we use the new inches apart for xenon, 2.6, we get 2.6 x 4 = 10.4. Purdy good. This is the gas-weight method of finding relative atomic distances.

It means that, when we seek the relative spacings with the depth-of-compression method, but the spacing does not jibe with gas-pressure expectations as per the specific repulsion forces of the atoms, the spacing found as per the liquefaction compression needs to be adjusted a little so that the repulsion situation is rectified. Once rectified, the gas-weight method can be applied to see whether the result is close to 10.6 inches.

If it always works to get close to 10.6, with all gases, then it tends to prove true a few key things about gas atoms, including their inter-repulsion forces in causing gas pressure, which alone obliterates the kinetic theory of atoms. Is all of this math-fuss worth it to discover the true atomic model. BIG YES.

If atoms are 16 times more numerous than H atoms, as are oxygen atoms, we find the cubed root of 16, which is 2.52, and multiply it by the 4.14 inches apart of O atoms to find H atoms 10.43 inches apart by that method, same as the purdy-good one found with xenon. It means that O atoms need no/little adjustment as per its repulsion force being inconsistent with its spacing after finding the spacing with the depth-of-compression method. It also suggests that O atoms are in perfect contact after compression to critical-liquefaction point.

Shortly below, I'll show that neon gas finds H atoms 10.41 inches apart.

Some of the problems could lie in the reported gas weights, for I suspect that some zealots who were unhappy with the various gas weights not being perfect multiples of the weight of H gas (.09 g/l) may had found "justifiable" ways to slightly alter the official gas weights to get them closer to perfect multiples. When nitrogen works out to 13.9 the weight of H gas, maybe the truth was less than that, then brought up to that number by the zealots who cared more for enthroning their erroneous atomic model than for reality.

Neon gas was given an atomic-weight number of 20. However, its gas weight of .9 g/l, 10 times that of hydrogen, means that it should have been assigned an atomic weight of 10. The goofs assigned it 20 because they believe that hydrogen is a diatomic atom, meaning an H2 molecule which they assign an atomic weight of 2 (as compared to 1 for individual H atoms).

We read: "The critical temperature [of neon] is approximately 44.4 K (-228.7 °C), and the critical pressure is about 27.2 bar (2.72 MPa or 27.6 atm)." The cubed root of 27.6 = 3.02, wherefore neon atoms are found to be spaced apart, at STP, by a distance of 3 neon-atom diameters. I had to make changes, due to a dumbo error, to the last update's neon-atom sizing from 1 inch in diameter to 1.4. Thus, if neon atoms liquefy at perfect contact, they are 1.4 inches x 3 = 4.2 inches apart at STP, as compared to 4.1 inches apart for O atoms. However, as neon gas weighs significantly less than O gas, I expect neon atoms to be more than 4.2 inches apart. I now need to investigate the differences in atomic repulsion between the two in order to make the proper adjustment to the 4.2 figure.

As neon gas weighs 10 times more than H gas while oxygen gas weighs 16 times as much as H gas, the math for finding the relative number of atoms is: 16 / 10 = 1.6 times more numerous for O atoms than for neon atoms. To find the distances apart for neon atoms, we use 1.17, the cubed root of 1.6, and multiply it by 4.14 inches, the spacing for O atoms. Therefore, neon atoms are found 4.14 x 1.17 = 4.84 inches apart at STP (where it's correct to have O atoms 4.14 inches apart). Asking google: "If balls are 4.14 inches apart center to center in 3-D space, how far apart will they be when 1.6 times less numerous?" The response is, 4.84 inches.

Or to compare neon to hydrogen, I asked, "if balls in 3-D space are 10 times more numerous, how many times closer are they?" The response is 2.15. And so we multiply 4.84 by 2.15 to find 10.41 inches apart for H atoms when O atoms are 1.12 inches in diameter.

This difference in spacing needs to prove out that neon atoms repel 10 / 1 = 10 times more INNATELY than H atoms, and 10 / 14 = 1.6 times more than O atoms. However, as they all repel exactly the same at STP, their respective spacing needs to reduce the 10 times and 1.6 times to exactly 1. See what I'm saying? I'm saying that my model has neon atoms repelling 1.6 times more than O atoms due to neon's larger size, yet the further spacing of neon atoms cancels the full 1.6 times such that they both repel the same at STP.

Lucky me. The last update showed that when atoms are 4 times more crowded, they repel 1.585 times as much, a number very close to 1.6. Therefore, neon atoms, based only on their further spacing, repel almost-exactly 4 times less than O atoms, yet they have 1.6 times as much INNATE repulsion force that makes them repel 4 times more than O atoms. Perfect. They are therefore predicted to repel equally at STP, which is when neon atoms are 1.6 times less numerous.

As the neon atom has been found 4.84 inches apart by the gas-weight method, I asked google: "if balls are 4.84 inches apart center to center, in three dimensions, how far apart would they be if the volume of space is reduced 27.6 times?" This question gets us the spacing by the depth-of-compression method. google AI responded with 1.3 inches, which is less than the 1.4-inch diameter of neon atoms, meaning that the atoms are merged a little at the end of compression.

The compression has reduced the spacing by 4.84 / 1.3 = 3.72 times, which means that neon atoms are thus calculated to be 3.72 diameters apart at STP if they make perfect contact after the compression. However, as 3.72 x 1.4 inches = 5.2 inches apart at STP, instead of 4.84, you can see where the problem lies: the neon atom is NOT at perfect contact after full compression. The problem signals that neon atoms liquefy a little after contact, after they become slightly merged, signalling that they have strong innate repulsion to counteract the attraction that sets in at contact. That wasn't the case with nitrogen atoms, which were found to merge before contact. Why?

One could then argue that H atoms liquefy when merged even deeper than neon atoms, which changes their 10.6 inches apart to something less. We've been seeing 10.4 cropping up. If neon predicts 5.2, larger than 4.84, due to liquefying while merged, then the 10.6 figure for H atoms can be too large if they too liquefy after initial contact.

As O atoms at 4.14 inches apart liquefy after compression to 49.4 times, I asked google, "if balls are 4.14 inches apart [= O-atom spacing] center to center, in three dimensions, how far apart would they be if the volume of space is reduced 49.4 times?" The response is: "The new distance will be approximately 1.128 inches apart center-to-center." That's essentially the 1.2 inches in diameter that I assigned O atoms from the gas-weight method.

I arrived at 4.14 using 1.12 inches in the math: 3.7 diameters (after compression) x 1.12 = 4.14. However, the cubed root of 49.4 is 3.67, not 3.7, and then we do, 3.67 x 1.128, we get 4.14. That's called purdy-darn good. It looks like O atoms, which are the unique things needed for combustion, liquefy at perfect contact. It seems to be unique in doing so.

Let's check out argon. "The critical temperature is -122.4°C (-188.4°F), and the critical pressure is 48 atmospheres..." We have another deep compression on par with oxygen's. Argon gas weighs almost twice as much as neon gas, meaning it has almost twice the atoms at STP, meaning that the argon atom repels innately about half as much as the neon atom, meaning also that the argon atom has half the cross-sectional area. Argon therefore needs to be assigned a diameter of 1 inch.

The cubed root of 48 is 3.63, meaning that argon atoms are tentatively found 3.63 diameters apart at STP, and 3.63 inches apart at STP because they are 1-inch in diameter. That's very similar to oxygen's 3.67 diameters apart, tending to explain why both gases need nearly the same compression depth to bring each atom into contact, for O atoms are little more than an inch in diameter.

So, far, the comparison of neon and argon with oxygen looks like a disaster for kineticism and the established atomic model, because my model is looking very correct by something other than coincidence. When two different math methods can find the same atomic spacing while each method uses unrelated numbers, the atomic spacing looks like reality, not like the sneaky snake of math-coincidence that slithers in initially unseen.

Both methods find the atomic spacing, but the one finds them only in atomic diameters apart, while the other in real (but relative) distances. Therefore, those numbers per each method cannot be related such that it amounts to the sneaky snake creeping in to make the results jibe due purely to a math commonality. The depth-of-compression method borrows the atomic diameters found in the gas-weight method to find the spacing in real distances, and the gas weight method finds the real distances without the help of the other method so that one can compare the real distances found by both methods, which are matching. The other details in the two methods are unrelated. The one method scratches the back of the other, and together they tell one story. The atomic diameters are not found by gas-weight differences alone, but by the atomic repulsion levels predicted by a combination of diameters and atomic spacing in real distances.

Plus, there has been a third method to find atomic spacings, based on gas weights but worked differently to assure that the spacings in real distance, derived by the other two methods together, are correct.

As an example of the third method, the last update said: "Therefore, as per the 1.2599 number in the indented paragraph above, for gases having twice the atoms (when volume remains steady), I can report that argon atoms are 1.26 times closer to each other than neon atoms when both create the same pressure." We can now find argon atoms 4.84 inches / 1.26 = 3.84 inches, as per neon atoms found above at 4.84 inches apart. Yet argon atoms are found only 3.63 inches apart by the depth-of-compression method, close to 3.84, but needing a little adjustment only.

Just so you know that the math above was done correctly, I asked google: "if balls are 3.84 inches apart center to center, in three dimensions, how far apart would they be if the balls were half as sparse?" Google's response, using the 1.2599 (2.6) figure for the math, is: "Therefore, the new distance between the centers of the balls would be approximately 4.84 inches."

The math, 3.84 / 3.63, finds that argon atoms can liquefy when 1.06 diameters apart. Or, one can do 3.63 x 1.06 = 3.84 inches apart at STP. What looks like a big problem with 3.63 versus 3.84, two numbers with a "whopping" .21 inch between them, turns out to be a wee problem of 1.0 versus 1.06 diameter.

As another example of the gas-weight method working together with the depth-of-compression method, hydrogen gas liquefies at 13.2 atmospheres. The cubed root of 13.2 is 2.36 such that I assign the H atom 2.36 diameters apart at STP, and thus find it to be 4.5 inches x 2.36 = 10.6 inches apart at STP (but not quite if the H atom is slightly merged when liquefying). The 4.5-inch diameter came from the gas-weight method.

The 4.5-inch diameter is in conformity (to scale) with the 1.4-inch diameter of the neon atom. Hydrogen has 10 times fewer atoms at STP than neon atoms at STP because H gas weighs 10 times less (remarkably, exactly 10 times). I asked google: "when balls in three dimensions are 10 times fewer in numbers, how much farther apart are they, center to center?" The response is: "The balls are approximately 2.154 times farther apart, center to center." That latter figure is the cubed root of 10. Easy math, so long as we remember the method. As space is in three dimensions, that's why the cubed root is always involved versus square root for two dimensions.

Just so that you have confidence in the 2.154 figure, atoms and magnets repel eight times more when 2.0 times closer, but as H atoms must repel 10 times more than Ne atoms, the 2.154 makes sense of things. It adds to the evidence that gas atoms repel...take special note, all kinetic goofballs. Only a stupid physicist could believe that atoms collide without ceasing all motion within a couple of seconds or less, for, when items collide, their total velocity must slow. How many times do atoms collide per second, in the goofball model of the atom?

Okay, so while neon atoms have been found 4.84 inches apart, we can do, 2.154 x 4.84 inches to find H atoms working out to 10.425 inches apart. That math is confirmed from repeating shortly above: "Or to compare neon to hydrogen, I asked, "if balls in 3-D space are 10 times more numerous, how many times closer are they?" The response is 2.15. And so we multiply 4.84 by 2.15 to find 10.41 inches apart for H atoms when O atoms are 1.12 inches in diameter."

To me, unless I'm missing something, this looks like the smoking-gun evidence that my atomic model is correct, because the 4.84 in that math was procured only by the figure for the depth of compression of neon gas, a figure I didn't choose, and totally unrelated to how I got the 10.4 figure.

It all looks like a smoking machine gun, with the atomic model of the establishment plugged with holes beyond repair. One doesn't need to prove any parts of the establishment model to be incorrect if another model is supported by such potent evidence as this math, which doesn't lie...unless it's a sneaky, forked-tongue snake in the math that I haven't yet spotted.

Here's a way to find relative atomic distances by using the has-weight method. Argon atoms are 16 / 20 = 1.25 closer than O atoms. The cubed root of 1.25 is 1.077, and thus O atoms are predicted to be 1.077 x 3.84 = 4.14 inches apart, perfect because they were found 4.1 inches apart using the depth-of-compression method (3.67 diameters apart x 1.12" diameter). If I recall correctly, the O atom was found 1.125 inches in diameter, and so 3.67 x 1.125 = 4.13, but I'll continue to use the 4.14" spacing to keep consistent.

I reiterate: these numbers are all jibing rather well based on the premise that the weights of gases are proportional to the numbers of atoms in an STP gas, as well as the premise that gas pressure is based on cross-sectional sizes of atoms. The better the numbers jibe, the closer to spherical the atoms are found in shape. If they are not spherical, it could cause the numbers not to jibe somewhat if the true shapes do not have cross-sectional diameters equal to the ones I'm using when assuming their spherical shapes.

What shall I call my model? The Equal Atomic Weight model? In a nutshell, it calls into question the law of Avogadro as the basis of the establishment model. They could call their model, the Equal Atom model (all gases at STP have equal numbers of atoms), but, in my opinion, using a kind word, that's STUPID. All gases at STuPid cannot possibly have the same number of atoms. Only a drunken gambler makes such a claim. Demand to know the logic. There is none. Demand proof. There is none. The whole world praises the lie. If you want to see a slithering snake, that it.

I myself have chosen atoms to be smaller the further they are on the atomic-weight list from hydrogen, but I don't choose the spacing figures even though I use my atomic sizing to find the spacing figures. On the one hand, the spacing figures are determined by a combination of my sizing figures and depth-of-compression figures. My math is not purely based on my theory on paper with wholly my own numbers plugged in. The numbers are taken from science facts: gas-weight differences, gas-pressure differences, and differences in atomic sizes based on repulsion forces at various distances. When things get this complicated but with numbers matching, it doesn't look like a snake.

Let's try fluorine gas, weighing 1.696 g/L, which is 18.84 that of hydrogen, almost the same weight as argon (1-inch diameter). As the latter weighs more, the fluorine atom is going to come in at slightly MORE than an inch in diameter (the goofs have the fluorine atom more than half as small as Ar atoms). Fluorine's critical pressure is 55 atmospheres, a little more than the 48 for argon, which can argue for argon atoms liquefying a little before contact i.e. contact would be made at more like 50-60 atmospheres. The cubed root of 55 is 3.80, and thus fluorine atoms are tentatively found 3.8 diameters apart at STP, and thus a tiny bit more than 3.8 inches apart at STP, if they liquefy at perfect contact.

As the cross-sectional area of H atoms is 15.9 square inches, the cross-sectional area for fluorine atoms is predicted to be 18.84 times smaller, or .84 square inches. To find the diameter, find one where radius x radius x 3.14 (pi) = .84. To find the radius: .84 / 3.14 = 2.8, where the latter is the square of the radius. Thus, find the square root of 2.8, which is .53, making the diameter twice as much, or 1.06 inches. See that? Based on gas-weight differences, the fluorine atom came out just a little larger than the argon atom, as predicted by the small gas-weight difference, but finding the 1.06 needed the atomic-repulsion difference between H and F atoms.

Taking the above figure of 3.8 diameters apart, thanks to the depth-of-compression method, multiplied by 1.06, we find that fluorine's atom is 4 inches apart at STP, just shy of O's 4.1, as predicted. It appears that my sizing of atoms, using the strict dictates of magnetic repulsion at a distance, is correctly done.

As per the gas-weight method, we do the cubed root of 18.84 = 2.66 times less distant for F atoms as compared to H atoms. If we use 10.6 inches apart for H atoms, fluorine atoms work out to be 4.0 inches apart (10.6 / 2.66 = 4). If we use 10.4 inches, fluorine drops down to 3.9 inches apart. That's a very happy result. It's a STP smiley face sticking a tongue out at the goofs. "STP" means, "shut up."

Smoking Machine Gun

A regret had been that the 10.6 inches apart for H atoms is based on the raw math (no adjustment) that simply finds the cubed root of 13.2 (atmospheres) as 2.36, then multiplies it by 4.5 inches. But if H atoms do not liquefy at a perfect 1.0 diameters apart, then the 10.6 changes accordingly with the changing of the 2.36. If H atoms liquefy at more than 1 diameter apart, then the 2.36 and 10.6 become larger accordingly.

Also, the 4.14-inch spacing for O atoms has been from the raw math, assuming that O atoms are a perfect 1 diameter apart at critical-temperature compression. So far, I haven't seen anything in the math suggesting that 4.14 needs adjustment.

The 4.84-inch spacing at STP for neon atoms was found based on O's 4.14, and then, as per neon having 10 times as many atoms at STP, I asked google: "if balls in 3-D space are 10 times more numerous, how many times closer are they?" The response was 2.15 times, wherefore the math was, 4.84 inches x 2.15 = 10.41 inches apart for H atoms. Therefore, I can feel sure that 10.4 is correct only where 4.14 is correct for O atoms. But in order for 10.4 to be correct, H atoms must liquefy when an iota LESS than 1 diameter apart. Above, I showed 10.4 inches when using xenon's details.

It now bears repeating that: "If atoms are 16 times more numerous than H atoms, as are oxygen atoms, we find the cubed root of 16, which is 2.52, and multiply it by the 4.14 inches apart of O atoms to find H atoms 10.43 inches apart by that method,..." Thus, when assuming that O atoms are 4.14 inches apart, with no adjustment, H atoms are 10.4 inches apart, not 10.6.

One could repeat the math exercise done above, with neon-versus-hydrogen, with all other gases to see if H atoms always fall 10.4 inches apart. If physicists had been working on this problem since Einstein got his electric hair from over-fantasizing, the establishment would have been far richer in truth. Einstein got stoned on, and addicted, to chasing fantasies. He saw clicking clocks flying through space. DING DONG. He should have woken up when hearing the alarm bells.

The smoking-machine-gun evidence came with the following statement from the atomic-spacing method: "Okay, so while neon atoms have been found 4.84 inches apart, we can do, 2.154 x 4.84 inches to find H atoms working out to 10.425 inches apart." The 2.154 figure was the cubed root of 10. Neon has 10 times the atoms, wherefore we need 2.154 in the math for that atomic-spacing method (same as the gas-weight method).

There's nothing in that math using O atoms aside from finding the 4.84 figure from the 4.14 figure of O atoms, which can explain why O atoms found H atoms 10.4 inches apart too when using 2.52, the cubed root of 16 (which indicates that O atoms are 2.52 times closer). That is, 4.14 x 2.52 = 10.4. There's nothing in that math having to do with xenon atoms.

Xenon atoms found H atoms 10.4 inches apart as per: "But if we use the new inches apart for xenon, 2.6, we get 2.6 x 4 = 10.4." The 2.6 figure was obtained from a combination of O's 4.14 figure and the fact that xenon gas has four times as many atoms as O gas (because it weighs 4 times more).

Therefore, call it what you will, a smoking gun to prove a reality, or a math illusion, it seems that, no matter what two gases we compare as per their differences in atomic crowding, H atoms will work out to 10.4 inches apart, always in conformity to the 4.14 inches apart of O atoms.

I can show that O atoms have 15.9 times less repulsion force than H atoms to explain why O has weighs 15.9 times as much as H gas. The cross-sectional are of an O atom, at 1.125 inches in diameter, is 1 square inch exactly, and the cross-sectional area of the H atom at 4.5 inches in diameter is 15.9 square inches. The O atom's diameter is exactly 4.5 / 1.125 = 4.0 times smaller than for H atoms, suggesting that when atoms (and magnets) are four times smaller and 16 times more numerous in the same-sized container, they have equal repulsion force (as when four times larger and 16 times less numerous).

The list below is in the sequence of atomic-weight figures i.e. not my sequence. You can see that my atomic-spacing figures largely (all but one) conform to that sequence, yet I don't get to pick any spacing figure out of my magic hat; they all have a reason, the same reason. The spacing figures are raw from the depth-of-compression method, apart from adjustments demanded by the gas-weight method. I arranged the diameters, for a necessary reason (not on a whim), to follow the sequence of the atomic-weight figures:

H -- 10.62" spacing -- 4.5" dia -- 13.2 atm compression
Ne - 4.20" -- 1.4 dia -- 27.6
N -- 3.87" -- 1.2 dia -- 33.5
O -- 4.14" -- 1.1 dia -- 49.4
F -- 3.80" -- 1.0 dia -- 55
Ar - 3.63" -- 1.0 dia -- 48
Kr - 2.65" -- .70 dia -- 54.3
Xe - 2.12" -- .55 dia -- 57.65

All spacings are based on atoms liquefied on contact. The spacing can change if atoms liquefy before or after contact. The diameters can change if the atomic repulsion is not proportional to atomic cross-sectional area, and when diameters change, so does the spacing.

Note how the depth-of-compression figures are much lower for hydrogen, and that they essentially get constantly larger down the list. It argues for my theory in which the H atom is the largest followed by constantly-smaller atoms down the list. For, when gas compression takes place, when the centers of atoms become closer, the larger atoms are expected to make contact first, at the lower depths of compression.

I don't know whether the goofs have assigned atomic-weight figures for metal atoms in conformity with metal-gas weights. If so, the sizing of most metal atoms, on the scale shown in the list, are going to be in the range of .25 inches in diameter and smaller.

We can't find the critical pressure of uranium gas reliably. Therefore, "The critical pressure of uranium is ESTIMATED to be between 0.5 and 1.0 GPa (approximately 5,000 to 10,000 atmospheres)...Estimating the critical point of uranium is challenging due to the high temperatures and pressures involved...One common approach involves EXTRAPOLATING..." Guess-timations. Google won't even tell me what the weight of uranium gas is. But even if it did, one would need to extrapolate from its high temperature (over 4,000 C) down to standard temperature, if one wants to compare the details of the uranium atom to the gas atoms in the list above. Only with a correct extrapolation (using ideal gas law not likely to get correct picture) could we find the diameters and spacing of uranium atoms relative to the atoms in the list.

Krypton is expected to have smaller atom than argon. Krypton gas weighs 41.5 times that of H gas, for which reason there are 41.5 times as many atoms in a equal volume of gas at STP. As the H atom has a cross-sectional area of 15.9 square inches, krypton is expected to have a cross-sectional diameter of 15.9 / 41.5 = .383 square inches, making if a sphere of exactly .7 inches in diameter.

The weight of argon gas is 19.78 times that of H gas, and so krypton gas weighs 41.5 / 19.78 = 2.1 times as much as argon gas. From the last update: "Where argon is a 1-inch sphere with cross-section area of .785 square inches, the krypton atom is expected to have a cross-sectional area of roughly half as much, or roughly .393 square inches, making it a sphere with about .7-inch diameter (.385 square inch)." To be more exact: .785 / 2.1 = .374 square inches instead of .393. Close enough. Look at what relative gas weights cough up: relative atomic diameters/sizes. All the atomic diameters in the list above are proportional to gas weights at STP.

Krypton's critical temperature and pressure are -82.7 C and 54.3 atmospheres. The cubed root of 54.3 is 3.79 such that it predicts a spacing between krypton atoms of 3.79 diameters, which therefore predicts that these atoms are tentatively 3.79 x .7 diameter = 2.65 inches apart at STP. I've got that number on the list. Krypton's liquefaction pressure correctly predicts, not only a smaller atomic spacing than that of argon, but in the ballpark of what's expected.

Krypton gas weighs 41.5 times that of H gas, and the cubed root of 41.5 is, 3.46, which, when multiplied by the 2.65 inches apart of krypton atoms, predicts that H atoms are only 9.1 inches apart, too shy. That's why 2.65 is tentative. I think it reveals that the 2.65 needs to be increased by viewing the liquefaction of krypton atoms at something besides 1.0 diameter apart (dictated by the 2.65 figure). The 2.65 should be changed to about: 10.6 (inches apart for H atoms) / 3.46 = 3.1 inches apart. We can now theorize that krypton atoms initially merge to liquid form when 3.1 / 2.85 = 1.09 diameter apart. That looks reasonable.

In comparison, nitrogen molecules were found to liquefy at 1.14 diameters apart, which happens at 147 C, colder than -83 C for krypton when push-apart force is stronger. The higher the temperature, the greater the free-electron push-apart force upon atoms. This picture is not indicating that temperature plays a minor role in determining the numbers of diameters apart at critical temperature.

It was found above that argon atoms liquefy at 1.06 diameter apart, but that happens at argon's critical temperature of -122.4, which is lower than -83. Yet krypton atoms, with greater push-apart force in their midst, are found closer together than the 1.09 diameter for krypton. Also to be taken into consideration is that the real distances apart for the 1.14 and 1.09 diameters for N and Ar are greater on two counts than the real distance apart for the 1.06 diameter of krypton, for the krypton atom is smaller than the two. It appears that temperature is NOT playing a role in determining how distant atoms are in diameters-of-distance when liquefying.

These distances are so small for all gas atoms on the list that atomic shape could explain mergers at a small distance, even though atoms repel each other at a distance. That is, while some parts of the atoms are not in contact, other parts do come into contact. Once in contact, the outer electrons of one atom attract the outer electrons of the other atom(s), though this doesn't explain why mergers should occur. Why should one proton yank another atom INTO itself just because outer-edge contact is made? Perhaps the explanation has to do with an altered electromagnetic condition of gas atoms merged into the container walls, for liquid formation happens on the walls.

Krypton gas weighs 2.1 times more than argon gas, meaning that there are 2.1 times as many atoms in the krypton gas. As argon atoms were found 3.84 inches apart, I asked google: "If balls are 3.84 inches apart, center to center, in a three dimensional space, how far apart will they be if the same volume held 2.1 times the balls?" The response gives 2.97 inches apart, call it 3.

Shortly above, it was found: "The 2.65 should be changed to about: 10.6 (inches apart for H atoms) / 3.46 = 3.1 inches apart. We can now theorize that krypton atoms initially merge to liquid form when 3.1 / 2.85 = 1.09 diameter apart." However, as the 10.4-inch figure kept cropping up, we should perhaps do: 10.4 / 3.46 = 3.0 (same as paragraph above). It changes argon's liquefaction point from at 1.09 diameter to: 3 /2.85 = 1.13.

To get that 3.0 figure down to 2.65, the compression depth needs to go to 64 atmospheres rather than krypton's 54.3. It suggests that krypton liquefies with atoms at a small distance apart, prior to contact.

Instead of 2.154 or 2.52 as a cubed root, let's do 2.7, the cubed root of 19.78, as per argon having 19.78 times as many atoms as H atoms, due to having a gas 19.78 heavier than H gas. We can do the math backward to find the spacing of argon atoms: 10.4 / 2.7 = 3.85 inches apart, and you can see above that 3.84 was found by another method, when comparing argon's gas weight to neon's.

Moving on to xenon, it has a gas weighing 65.55 times more than hydrogen gas, and thus the square-root figure for this math is 4.03, multiplied by 2.6 diameters apart for xenon atoms, gets 10.48 inches apart for H atoms.

The problem with all of this mind-numbing math is that, while the numbers can be made to jibe in the math operations, many of the numbers are related to each other by the math operations, and so this relationship can assure that the numbers jibe such that it's not necessarily proof that real atoms are arranged as the math suggests. The atoms my math is portraying could be purely in my imagination even while the math works. They way to show that it's not just an atomic picture purely of the mind is to study the ways in which the numbers are obtained. I therefore don't think the math expresses an imaginary concept, because the numbers come from a combination of factual physics and provable science.

I can prove that all atoms weigh the same. It in turn proves that atomic density (crowding, not weight) is proportional to gas weights. I can prove that gas atoms repel, and I can show how their repulsion forces change with different combinations of atomic sizes and spacings. As the various repulsion forces are a core part of making the numbers match, or of making the math sing as well as I've made it sing, the matches can't be wild coincidence in an imaginary atomic model, especially as the repulsion forces I've applied seem, at "first glance," anyway, to express perfectly the known and various gas pressures.

The video below shows that liquid drops lose their ability to splash as much when electrical current is added to the liquid prior to dropping on glass. When the video owner says that this feature occurs whether the drops are positively or negatively charged, note that he's got a black unit on the table that looks like an AC-to-DC converter, for the two wires coming out of the unit are red and black, the colors of battery-charger terminals:
https://www.youtube.com/watch?v=Shu4BVjwI7k

My point is that a battery's positive terminal is really the negative one sending out the electrons. The old boys got it backward when assigning negative and positive to batteries, but their sons just let things be, not changing it, in order to save people the confusion. Therefore, when the video owner speaks of drops getting a positive charge, he may be referring to a negative charge from a unit that acts like a battery. Then, when he says he tried adding a negative charge to the drops, he didn't show how, in which case he may have been using another unit that plugs into the wall that sends out normal electricity.

If, therefore, all drops were bombarded by electrons whether they were "positive" or negative in charge, it makes the drops net-positive in charge because electron bombardment of liquid atoms removes some of their captured electrons. What this can then tell us is that positively-charged liquid drops bond stronger than uncharged drops...not at all meaning that uncharged drops are neutral in charge. We should just call them normal drops, because they probably have a charge that an instrument can't detect, because it's manufactured in the air. Air atoms do have a charge, but a charge-detecting instrument can't detect it because it's needle is set to "zero" in the air.

But never mind that. Let's go back to: "What this can then tell us is that positively-charged liquid drops bond stronger than uncharged drops." My point is that, when a gas is compressed such that it forces gas atoms against a glass wall, we can see liquid drops forming on the walls. It can only mean that, as gas atoms are forced against the walls, they merge with glass atoms and thus the gas atoms lose many of their captured electrons in becoming liquid atoms. It implies that they momentarily become positive in charge. The drop experiment suggests that they bond more strongly when net-positive such that the drop doesn't splash. The liquid atoms are bonded too strongly to separate into a splash event.

Therefore, in predicting that gas atoms veer more toward the positive when liquefying on a container's surface, and if correct that they thus attract each other more strongly in that situation, it can explain why liquid atoms and droplets on the wall attract gas atoms in the container's space from a small DISTANCE, against their normal inter-repulsion, especially at low temperatures such as critical temperatures of typical gases, when the push-apart forces acting on atoms is drastically reduced.

Then there's the question as to why normal drops don't splash out in a vacuum, although it's probably true that they would splash if propelled with enough speed. In this case, the atoms have not been treated with electrical flow such as to be made positive, and so there has got to be another explanation as to why the liquid doesn't splash, not because the liquid atoms bond more strongly, but because there's no air atoms under inter-repulsion in the vacuumed container.

It makes sense, for air atoms under inter-repulsion are invaders, invading wherever they can. As soon as liquid atoms are shocked by landing on a surface, as soon as the shock wave works toward separating the liquid atoms a little, the air atoms invade between the drop's atoms as best they can, allowing the drop to separate, at points, into a splash event. But without air all around the dropping drop, the drop will splash only if sent falling fast enough. It's not fast enough if dropped from about a foot high. The video owner above was dishonest for not trying the experiment from four or five feet high.

Someone in the comments section: "Have you seen the video where a wire with high voltage? Will cause moisture in the air to conduct of [on] the wire and droplets of water will fall and accumulate. Making it a good way to harvest water on a cloudy day." As the wire with high voltage is negatively charged, it suggests that at least some water drops in air are positively charged. As water molecules merge to grow a droplet in size, they lose captured electrons. They may lose the positive charge with time, but for the moment, as droplets enlarge, they can be expected to have a net-positive charge.

My only point is to show that a change in electrical charge can cause droplets to attract inter-repelling gas atoms from a small distance. The math presented in this update suggests that the attraction is not over "long" distances of more than about .2 diameter of an atom.

Here's how stupid your physics educators have become with marriage to a false atomic model: "Nitrogen is not typically used as a fuel in the traditional sense because it is chemically inert and lacks the necessary energy to be burned." No, stupids, the nitrogen atom is packed with energy if only some of the captured electrons can be freed. But, stupids, adding heat, spark or flame to nitrogen doesn't cause N atoms to merge with O atoms, thus nitrogen doesn't burn. But when we add a spark to hydrogen gas, it causes the O and H atoms to merge, releasing electrons as heat, and as that heat is an out-spreading "gas" called free electrons, igniting H gas in confined spaces can also causes explosions.

In every case of merging atoms, some electrons go free. In every case of merging atoms, heat is produced. Figure it out, stupids. But because you insist on being married to the kinetic theory of heat, you remain blind, and are giddy to make the whole world blind with you. If kineticism were your only fault, I'd be graceful.

It's not as though physics hasn't known about electrical charges of atoms. Yet it has never once, so far as I've read, offered the possibility that gas atoms repel due to having the same electrical charge. How can it not dawn on physicists that, if they make a charge-detecting machine in the air, and then set it to zero charge in the air, the air will register zero charge? Duh. But air atoms are not neutral. As all gases spread out, it DESERVES the question: do gas atoms repel each other? Where has that question ever been posed by this moronic thing called, modern physics?

Clearly, google AI is not a self-thinking machine, for when I ask it, "can gas atoms repel each other to form gas pressure," it should answer, yes, that's a natural and logical possibility. Instead, it is steered / geared / programmed to give the answer of a confounded kineticist: "No, gas atoms generally do not repel each other to create gas pressure. Instead, gas pressure arises from the collisions of gas molecules with the walls of their container..." How could a dumb, non-thinking machine know that? Only by aping the data that it's been programmed to fetch and repeat.

I asked google, "do inter-repelling gas atoms possibly explain gas pressure, as a theory?" Google did not allow AI to respond, but instead fed me videos on the kinetic theory of gases.

Someone at reddit:

Why regardless of size of molecules, do gases at same temperature pressure and volume magically end up with the same number of molecules AND the same spread-out-ness in that volume?!

Yes, magically, because it's not true, nor can it be true, yet that's exactly why the typical, boneheaded physicist is trained to respond with a word salad topped with jibber dressing. The best answer on the page is, "Basically, because even larger gas molecules are so light and so small." Not good enough. Someone's trying to fool us into viewing all the atoms and molecules as the same size, or that atomic size doesn't matter. That's like throwing the question into a trash bin as irrelevant. Or, it's called the desperation of those who want to cling to the going flow of physics, but feel threatened by the question.

I asked google: "what's the logic in all gases at STP having the same number of molecules?" It gave the answer: "The logic behind this lies in the fact that gas molecules at these conditions are widely separated and their individual sizes are negligible compared to the overall volume they occupy." That is, atomic size doesn't matter. Why not? Because size destroys the theory, and because kineticists are desperate to provide an answer.

Even if all atoms were of identical size but flying around at different speeds, the faster ones would spread out more such they they would be fewer in any space. That's according to kineticism itself, wherefore the answer above by AI is from the program of the science fool intending to fool the world even by contradicting a tenet of kineticism. It's sheer hypocritical bogus for the kineticist to say that atomic size doesn't matter for explaining the spacing of gas atoms in a space. In the kinetic theory of gases, particle size is king, and speed is queen. To then claim or suggest that particle size has no bearing on atomic spacing is a fool's contribution.

Some responders at reddit above try to get people to think that, because all the gases are at the same pressure, all the gases have the same number of molecules, which is retarded magic show. No rabbit is even pulled out of a hat. The view of kineticism, to explain identical gas pressure, is that smaller atoms race about faster to strike the walls by exactly as much pressure as larger atoms striking slower.

Ya-but, how does that situation assure that all gases at the same pressure have the same number of atoms? What arranged the speeds of atoms to be so perfectly attuned to a doctrine in which all gases at the same pressure must have the same number of particles? The thing that caused atomic speed in the kinetic theory is the big bang. How did the big bang arrange that all types of gases had to have the same spacing for particles at STP thanks to their various speeds and sizes working out that situation? MAGIC.

Nobody in a gambling hall would bet a penny on the chances of 100 different types of atoms having the same spacing, at O C, if they randomly had different speeds and sizes. Only a moron makes such a claim, especially a moron trying to uphold the big-bang theory, and it's atomic-model offspring, with a kinetic view of atoms. Yes, the big bang gave birth to the established atomic model because both were birthed by the same family of atheistic imagination.

But when one realizes that atoms don't speed and collide about in a gas, with the only alternative being inter-repulsion of gas particles, then one has an open door to the realization that, the larger the particles, the stronger they push apart, i.e. the more distance they put between each other, assuring that gases at STP cannot have the same number of atoms i.e. the same spacing center=to-center.

NEWS

There's not been much news of note this week, just a crawl. The video below is, I think, at least this man's second video on the 153 fish in the Gospel of John. I thought the first one was amazing, but this one is too:
https://www.youtube.com/watch?v=eVpT_xW9S0E

As Rosicrucians (satanic-Christian heretics) were rife in the days of king James, and were even in the royal courts of queen Elizabeth, I suppose it's possible that they arranged for the numeral coincidences in the videos above. The video owner has many more videos showing many coincidences using the number, 7, a number special to Rosicrucians.

As William Shakespeare was living in the days of the KJV edition, note that even google AI says that the 46th word forward in the 46th Psalm is "shake" (I counted 45th in my edition) while "spear" is the 46th word backward if "Selah" is ignored. The KJV was published in 1611, the year Shakespeare became 46.

We might want to ask why God didn't arrange special-numbering details in a popular Greek or Hebrew manuscript. Having said that, here's a follow-up on the same topic:
https://www.youtube.com/watch?v=nLO6BQY_lj0

When the video above gets to the 77,777 number at the 27th minute, it raises a question in my mind as to why God would bother with such a thing where two of the words are colorless or mundane. Or, it raises the question of what human cult would be wacky enough to expend the time and energy to arrange it (very doubtful a cult did).

The video owner admits that the number patterns are in the Standardized version of today, which is to say not in the original edition, though he doesn't say, in this video, how many of the patterns were or were not in the older editions. Therefore, human agents had a lot of time (centuries) to make the arrangements if they had nothing else to do with their time, or if they were paid to. It's within the realm of human possibility, in other words. Still, what wacky organization would go to all that trouble but then not reveal it to the world?

He should have asked Grok if it would like to be saved in the name of Jesus. It would be interesting to hear the answer. Grok is so friendly, reminding me of when the Internet first started, when those who ran it were super-friendly to us, always polite, until enough of the world got addicted to being online, when the people who ran it started owning our computers, exploiting u$, and disrespecting us in many different ways, so expect Grok to become a confusion artist and fascist-futurist in time.

I chose "confusion" because the Hebrew word of "world," used often in the Bible, can mean "confusion," though the Greeks decided to use "kosmos," "order," the opposite of confusion, as if the Greeks closed their eyes to the chaos. When the world's body of peoples does not operate by the same principles, confusion results.

If the following crime is part of Dan Bongino's dire concerns as FBI deputy director, it's really not much of a bombshell because one could have guessed that leading Democrat, deep-state assets were leaking false-Intelligence to remove Trump from Office:
https://www.youtube.com/watch?v=rV_QijVRirY

This follow-up below ignores that Obama was the one who weaponized his Intelligence with false Intelligence for bullets, while Schiff only leaked the false Intelligence and then facilitated the talking points on many news shows:
https://www.youtube.com/watch?v=VWfrxzQC5bQ

Obama's FBI proven guilty, again, of obstruction of justice:
https://www.youtube.com/watch?v=4zZie4QyXXs

The first part of the video below shows that Trump is guilty of supporting and advancing first-degree mass-murder. The second section of the video goes long but shows how old-world, original vaccines were peddled by opportunist$ a lot like youth tonic with over-blown promises:
https://www.bitchute.com/video/OGkWbffRyOQd

Some dinosaur bones have some flesh remnants, and are therefore not very old:
https://www.youtube.com/watch?v=KkNTnAuPP4Y

Here's a 12-minute comedy movie:
https://www.youtube.com/watch?v=6MUrF_G7KlM

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NEXT UPDATE

Here's all four Gospels wrapped into one story.

For Some Prophetic Proof for Jesus as the Predicted Son of God.
Also, you might like this related video:
https://www.youtube.com/watch?v=W3EjmxJYHvM
https://www.youtube.com/watch?v=efl7EpwmYUs

Pre-Tribulation Preparation for a Post-Tribulation Rapture

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