I'm Against Sola Scriptura. AND. I'm a Fundie. Unlike a certain Ian Plimer.

This is written* by my friend (with some tensions) Mackey:

I have previously pointed to the ironical - and I think humorous - situation whereby the likes of an anti-fundamentalist professor Plimer can sometimes be clearer about certain principles of biblical exegesis than are those who embrace sola scriptura; whilst the latter can sometimes, here and there, be more scientifically accurate than are the professional scientists.

Ian Plimer will, in the case of the fundamentalists’ global Flood, absolutely and hilariously ridicule - and rightly so - such a notion, using a heady mix of science, common sense, and sailing nous.

He will describe the preposterous situation of a Queen Mary sized Ark being tossed hither and thither in a turbulent global sea, it being overloaded with dinosaurs and other massive animals, not to mention those swarms of irritating insects and pests.

OK, what are Ian Plimer's credentials in shipbuilding or navigation?

Ian Rutherford Plimer (born 12 February 1946) is an Australian geologist and professor emeritus at the University of Melbourne.

Thank you, wiki!

Wait, he has some connection:

Plimer is a fellow of the Australian Academy of Technological Sciences and Engineering

Now, what was he saying again?

the preposterous situation

Pretty please, make it "of living among millions or billions one year, being shut off in an ark, and a year later being lone survivors, a crew of 8" ... that situation is preposterous. But that's how God arranged for our survival.

of a Queen Mary sized Ark

The Ark is actually shorter. Koreans evaluated the Ark to 137 m, while RMS Queen Mary was around 300 m.

being tossed hither and thither

I calculated the rolling period of the Ark, it corresponds to a passenger ship.

in a turbulent global sea,

It would certainly have been turbulent in the bottom streams, touching rock or deposing sediment from hypersaturation. But the turbulence on the surface in a high degree depends on angle of waves. A wave that's 10 m high and 10 or even just 20 m from crest to crest is turbulent. A wave that's equally 10 m high but 100 or 200 m from crest to crest isn't.

Now, the waves can be idealised to circle segments, and the centre of each has a lowest possible placing at the bottom. If a wave of 10 m has a width of 20, probably the sea bottom is sth like 20 m below the crest. But this won't happen if the bottom is about 1 km lower. On the open Pacific, you have winds in which it's unsafe to stand openly on Kon Tiki or Uru, but that's for the risk, of getting swept off the raft. The waves didn't sisk sinking them.

La isla del día siguiente. Crónica de una travesía por el Pacífico** es el relato de esta odisea que lideró Kitín Muñoz y en la que, además de él y de Frattini, ahora con el Reto Pelayo, participaron Pepe de Miguel, Kiko Botana y Juan Ginés García, quienes buscaron emular los pasos del legendario explorador y biólogo noruego Thor Heyerdahl y sus espectaculares viajes con sus naves Kon-Tiki, Ra, Ra II y Tigris.

The Island of the Next Day. Chronicle of a crossing of the Pacific is the story of this odyssey that was lead by Kitín Muñoz and in which, apart from him and Frattini, now with Reto Pelayo***, participated Pepe de Miguel, Kiki Botana and Juan Ginés García, who sought to emulate the steps of the legendary explorer and biologist from Norway, Thor Heyerdahl, and his spectacular voyages with the "ships"° Kon-Tiki, Ra, Ra II y Tigris.

So, Uru and Kon-Tiki were on the Pacific. The Atlantic, where we had Ra and Ra II is less deep. Even there Thor managed. Such waters are not turbulent. Not to an Ark floating sideways in a wave trough where the distance to the crest is safely great.

A large regional Flood would be less safe, since the water would be shallower.

it being overloaded

I don't get that impression, given that one couple of hedgehogs on the Ark easily gave rise to 17 species in 5 genera. There are dog breeds that look more different than Erinaceus Europeaeus does from Hemiechinus auritus.

with dinosaurs and other massive animals,

I'm not a huge fan of Kent Hovind, given his dissertation disses Church Fathers and Alexandrine school, given his insistence "this is not my wife, it's just a picture of her," cute, but is a totally unnecessary polemic against the basics behind Nicaea II in 787, or just his take on alcohol, my grandpa was a distiller. But even Kent Hovind can answer this: if juvenile examples entered the Ark they didn't take up all that much space nor require all that much food. Noah only needed to "take a blue one and a pink one" not necessarily ones that were already ready for reproduction.

not to mention those swarms of irritating insects and pests

I'm not sure how much lice eggs, a k a nits, can survive without a host. But apart from lice, who need human hosts with warm blood, once hatched, I'm not aware of any insect that couldn't theoretically have survived on some flotsam. Genesis 6:20 when mentioning creeping things probably meant reptiles rather than insects. Leviticus 11:20 uses another term for them. So it's not as if the text forced us to believe insects were on the Ark. Those that were (not as passengers) probably were the ones best suited for food (to birds or hedgehogs), perhaps also compostation of waste.

Perhaps I should mention where I did my calculations of number of animal pairs, rolling period, and so on: Baraminological Note · For Sea-Farers .... · Rolling Period of Ark? · Ark : empty weight and freighted weight, number of couples on the Ark. · Small Tidbits on Ark, Especially Mathematical.

With the competence Ian Plimer probably has from the Australian Academy of Technological Sciences and Engineering, he could arguably prove the feasibility of the Ark better than I, if not as well as the Korean team, but he was set on ridiculing Creationists, so, he gave his techno skills a vacation.

Hans Georg Lundahl
Paris
Sts Hermes and Hadrian of Marseille
1.III.2026

Massiliae, in Gallia, sanctorum Martyrum Hermetis et Hadriani.

PS. I was tired this night and didn't attend to the fact that Lord's Days take precedence over most Saints' Days, especially the ones in Lent. It's obviously Second Lord's Day of Lent, also known as Reminiscere, and Sts Hermes and Hadrian are just remembered, not actually celebrated, even in Marseille./HGL

* Genesis, Flood, Ark Mountain (you may need to log in to Academia), despite this passage, the typological readings seem very decent. Edifying. Wish he had left out that non-edifying words, but, but ...

**La isla del día siguiente
http://nauta360.expansion.com/2016/11/03/de_costa_a_costa/1478197268.html

*** I suppose the Spanish means "now in couple with cancer survivor Reto Pelayo"

° Rafts or Egyptian style reed ships.

Small Tidbits on Ark, Especially Mathematical

Baraminological Note · For Sea-Farers .... · Rolling Period of Ark? · Ark : empty weight and freighted weight, number of couples on the Ark. · Small Tidbits on Ark, Especially Mathematical

I

Someone - I think former JW The Truth Hurts - made the argument that all peoples who have a Flood story also have access to big water.

I have already argued, for the peoples in the Altai region, this is not true.

But here is more - nearly all countries in the world are with sea access. On wiki, List of sovereign states states:

The 206 listed states can be divided into three categories based on membership within the United Nations System: 193 member states,[1] 2 observer states, and 11 other states.

And the article Landlocked country states:

In 1990, there were only 30 landlocked countries in the world. The breakup of Yugoslavia, the dissolutions of the Soviet Union and Czechoslovakia, and the independence referendums of Eritrea and South Sudan have created 15 new landlocked countries and 4 partially recognized landlocked states while the former landlocked country of Czechoslovakia ceased to exist on 1 January 1993.

In other words, human populations overall have a tendency to have access to big waters.

II

Dimensions of wood on the ark were calculated in my article Ark : empty weight and freighted weight, number of couples on the Ark as if the "walls" of the container had no overlapping parts. This is not true. You have prisms:

4 * hind wall X horizontals
4 * front wall X horizontals
4 * right wall X horizontals
4 * left wall X horizontals
4 * long X short verticals only

And you have 16 cubes where these prisms intersect.

The proper maths to deal with a correction would be to first subtract the prisms and then add back the cubes. Why so?

Imagine you had only three walls. Three prisms belong to two walls each. These are therefore already accounted for twice, so you subtract them once, the one correct count being left. The cube where all three walls meet also belongs to the three prisms. It is added three times over, once in each wall, and subtracted three times over, once in each prism. It must therefore be added once to have the one correct count of its existence.

III

Executing previous corrections ... in order to get the right total volume (which is not our main concern), one would for each deduct the first three values and add the fourth one, but as deduction from volume of wood ultimately means addition of freight capacity, we well add the three and subtract the fourth (in order to get the volume one would then deduct the result from the volume).

For readability, I will approximate after three decimals and for the final result, we'll consider that the m³ of water = mt (metric ton).

Short cubit, thin walls

8 * hind/front walls X horizontals
8 * 22.86 m * 0.2286 m * 0.2286 m = 9.557 m³
8 * right/left walls X horizontals
8 * 137.16 m * 0.2286 m * 0.2286 m = 57.342 m³
4 * long X short (verticals only)
4 * 13.716 m * 0.2286 m * 0.2286 m = 2.867 m³
16 * 0.2286 m * 0.2286 m * 0.2286 m = 0.191 m³
9.557 m³ + 57.342 m³ + 2.867 m³ - 0.191 m³ = 69.574 m³ / mt

Short cubit, thick walls

8 * hind/front walls X horizontals
8 * 22.86 m * 0.4572 m * 0.4572 m = 38.228 m³
8 * right/left walls X horizontals
8 * 137.16 m * 0.4572 m * 0.4572 m = 229.366 m³
4 * long X short (verticals only)
4 * 13.716 m * 0.4572 m * 0.4572 m = 11.468 m³
16 * 0.4572 m * 0.4572 m * 0.4572 m = 1.529 m³
38.228 m³ + 229.366 m³ + 11.468 m³ - 1.529 m³ = 278.298 m³ / mt

Long cubit, thin walls

8 * hind/front walls X horizontals
8 * 30.48 m * 0.3048 m * 0.3048 m = 22.653 m³
8 * right/left walls X horizontals
8 * 182.88 m * 0.3048 m * 0.3048 m = 135.921 m³
4 * long X short (verticals only)
4 * 18.288 m * 0.3048 m * 0.3048 m = 6.796 m³
16 * 0.3048 m * 0.3048 m * 0.3048 m = 0.453 m³
22.653 m³ + 135.921 m³ + 6.796 m³ - 0.453 m³ = 164.917 m³ / mt

Long cubit, thick walls

8 * hind/front walls X horizontals
8 * 30.48 m * 0.6096 m * 0.6096 m = 90.614 m³
8 * right/left walls X horizontals
8 * 182.88 m * 0.6096 m * 0.6096 m = 543.683 m³
4 * long X short (verticals only)
4 * 18.288 m * 0.6096 m * 0.6096 m = 27.184 m³
16 * 0.6096 m * 0.6096 m * 0.6096 m = 3.625 m³
90.614 m³ + 543.683 m³ + 27.184 m³ - 3.625 m³ = 659.669 m³ / mt

IV

And now let's apply this to the freight capacity ... and let's stay with 14 cubits high water line, meaning we need not just to multiply above corrections with density, but also with 7/15:

Smallest weight available, biggest weight in walls, for small cubit:

20,069.565 - 6,812.184 = 13,257.381 mt
278.298 * 0.88 * 7/15 = 114.288 mt
13,257.381 + 114.288 = 13,371.669 mt

Biggest weight available, smallest weight in walls, for small cubit:

20,069.565 - 1,362.437 = 18,707.128 mt
69.574 * 0.352 * 7/15 = 11.429 mt
18,707.128 + 11.429 = 18,718.557 mt

Smallest weight available, biggest weight in walls, for big cubit:

47,572.302 - 16,147.399 = 31,424.904 mt
659.669 * 0.88 * 7/15 = 270.904 mt
31,424.904 + 270.904 = 31,695.808 mt

Biggest weight available, smallest weight in walls, for big cubit:

47,572.302 - 3,229.48 = 44,342.823 mt
164.917 * 0.352 * 7/15 = 27.09 mt
44,342.823 + 27.09 = 44,369.913 mt.

V

Revisiting two paragraphs:

On the other* post, I had estimated the number of couples to 2032, meaning 4064 individuals, on the average size of a sheep. Now, a sheep eats 7 kg green fodder per day, 365 days. 10,383.52 metric tons for food. How much would 4064 sheep weigh? Tame sheep weigh 45 to 160 kg for the bucks, 45 to 100 kg for the ewes. Let's add the numbers together and divide by four : (45+45+160+100)/4 = 87.5 kg. Let's multiply this by 4064. 355.6 metric tons. Living passengers with crew therefore 355.6 + 10,383.52 = 10,739.12.

So, even the smallest weight available for the animals and food and tools and the walls of the small chambers and so on has 500 tons for tools and temporary water supplies on top of needed animals and food weight. And the 2032 couples I got by reducing an evolutionist's estimate for number of species (taking mammals, birds, reptiles and amphibians separately) by only 16, the number of species I thought there were of hedgehogs, but that's actually 17.

13,371.669 - 10,739.12 = 2632.549 mt
18,718.557 - 10,739.12 = 7979.437 mt

All that for tools and water?

I think there might actually have been more animals than just 2032 couples. Especially since 7 kg food per day is not exactly typical in proportion to the body weight, the ruminants need more food.

And this would mean, hedgehogs are more diversified as a baramin than the medium.

Hans Georg Lundahl
Paris
St. Josaphat
12.XI.2021

Vitepsci, in Polonia, passio sancti Josaphat, e sancti Basilii Ordine, Episcopi Polocensis et Martyris; qui a schismaticis, in odium catholicae unitatis et veritatis, crudeliter interfectus est, et a Pio Papa Nono inter sanctos Martyres adscriptus. Ejus tamen festivitas recolitur decimo octavo Kalendas Decembris.

Ark : empty weight and freighted weight, number of couples on the Ark.

Baraminological Note · For Sea-Farers .... · Rolling Period of Ark? · Ark : empty weight and freighted weight, number of couples on the Ark. · Small Tidbits on Ark, Especially Mathematical

For empty weight of Ark, thickness times density of the wood, times following measures in square cubits:

2*30*50 = 3,000 square cubits
2*30*300 = 18,000 sq cub
4*50*300 = 60,000 sq cub
3,000 + 18,000 + 60,000 = 81,000 sq cub

The problem is, we don't know what cubit length it is, an Egyptian of 18 inches, or a longer one of up to 24 inches.

So, let's suppose we take each of these extremes in metrics ... first multiply with relevant number of inches, then convert to metric system

30*18 = 540 in, 13.716 m
50*18 = 900 in, 22.86 m
300*18 = 5,400 in, 137.16 m

30*24 = 720 in, 18.288 m
50*24 = 1,200 in, 30.48 m
300*24 = 7,200 in, 182.88 m

2 * 13.716 * 22.86 = 627.096 2 * 13.716 * 137.16 = 3,762.573 4 * 22.86 * 137.16 = 12,541.91 627.096 + 3,762.573 + 12,541.91 = 16,931.579

2 * 18.288 * 30.48 = 1,114.836
2 * 18.288 * 182.88 = 6,689.019
4 * 30.48 * 182.88 = 22,296.73
1,114.836 + 6,689.019 + 22,296.73 = 30,100.585

Then let's suppose the thickness of timber of each wall, floor or roof is either half a cubit or an entire cubit:

16,931.579 * 0.229 = 3,870.559
16,931.579 * 0.457 = 7,741.118

30,100.585 * 0.305 = 9,174.658
30,100.585 * 0.61 = 18,349.317

Now take the density of the wood. I'll take a high density hard wood, and a low density soft wood, rose and pine.

Rosewood has up to 880 kg per m³. With pine, we are down to 352 kg/m³. 100 kg = 0.1 metric ton, so it's easy to convert into tons

Small cubit :

3,870.559 * 0.352 = 1,362.437
3,870.559 * 0.88 = 3,406.092

7,741.118 * 0.352 = 2,724.874
7,741.118 * 0.88 = 6,812.184

Big cubit :

9,174.658 * 0.352 = 3,229.48
9,174.658 * 0.88 = 8,073.699

18,349.317 * 0.352 = 6,458.959
18,349.317 * 0.88 = 16,147.399

Now let's talk about the freighted weight - that means empty weight, plus weight of whatever freight is added, both crew and passengers and whatever inert cargo, like food supplies.

We can take it that the water line was 14 or 15 cubits up. If it was 15 cubits up, the moment when water covered the highest mountains (including the one where the Ark was built) the Ark could start floating but still risk scratching the soil. Not too big a problem if it was fairly soft.

If it was 14 cubits up, the moment when water covered the highest mountains, the Ark could float freely. Perhaps a better option.

Volume displaced of water = weight of all of the ship, with any content, so, in this case, freighted weight. This equals either half the volume of the Ark (15 = 15, 2*15 = 30), or 14/30, 7/15 of the volume (14 < 16, 30 * 7/15 = 14).

And the volumes are either from 18 inch cubit or from 24 inch cubit.

13.716 * 7/15 * 22.86 * 137.16 = 20,069.565
13.716 * 1/2 * 22.86 * 137.16 = 21,503.105

18.288 * 7/15 * 30.48 * 182.88 = 47,572.302
18.288 * 1/2 * 30.48 * 182.88 = 50,970.324

So, what weight is then left for crew, passengers and food?

Small cubit:

20,069.565 - 1,362.437 = 18,707.128
20,069.565 - 3,406.092 = 16,663.473
20,069.565 - 2,724.874 = 17,344.692
20,069.565 - 6,812.184 = 13,257.381

21,503.105 - 1,362.437 = 20,140.669
21,503.105 - 3,406.092 = 18,097.013
21,503.105 - 2,724.874 = 18,778.232
21,503.105 - 6,812.184 = 14,690.922

Big cubit:

47,572.302 - 3,229.48 = 44,342.823
47,572.302 - 8,073.699 = 39,498.603
47,572.302 - 6,458.959 = 41,113.343
47,572.302 - 16,147.399 = 31,424.904

50,970.324 - 3,229.48 = 47,740.844
50,970.324 - 8,073.699 = 42,896.625
50,970.324 - 6,458.959 = 44,511.364
50,970.324 - 16,147.399 = 34,822.925

So, for men and beasts and their food and the tools not specified, we have a weight of between 13,257.381 metric tons and 47,740.844 metric tons, if my observation on the waterline holds water (ha ha).

On the other* post, I had estimated the number of couples to 2032, meaning 4064 individuals, on the average size of a sheep. Now, a sheep eats 7 kg green fodder per day, 365 days. 10,383.52 metric tons for food. How much would 4064 sheep weigh? Tame sheep weigh 45 to 160 kg for the bucks, 45 to 100 kg for the ewes. Let's add the numbers together and divide by four : (45+45+160+100)/4 = 87.5 kg. Let's multiply this by 4064. 355.6 metric tons. Living passengers with crew therefore 355.6 + 10,383.52 = 10,739.12.

So, even the smallest weight available for the animals and food and tools and the walls of the small chambers and so on has 500 tons for tools and temporary water supplies on top of needed animals and food weight. And the 2032 couples I got by reducing an evolutionist's estimate for number of species (taking mammals, birds, reptiles and amphibians separately) by only 16, the number of species I thought there were of hedgehogs, but that's actually 17.

What would we get with a larger Ark (a larger cubit)?

47,740.844 / 13,257.381 = 3.601

2,032 * 3.601 = 7,317.387

So, the number of couples on the Ark would vary between 2,032 and 7,317 according to dimensions and empty weight of the Ark. With less food per individual, even more so.

Hans Georg Lundahl
Paris
Sts Placidus and Brothers
Martyrs of Messina
5.X.2021

Messanae, in Sicilia, natalis sanctorum Martyrum Placidi Monachi, e beati Benedicti Abbatis discipulis, et ejus fratrum Eutychii et Victorini, ac sororis eorum Flaviae Virginis, itemque Donati, Firmati Diaconi, Fausti et aliorum triginta Monachorum, qui omnes a Manucha pirata, pro Christi fide, necati sunt.

* Ex-JW Took On the Flood on my English debate blog.

Water Temperature, K-Ar Dating, Temperatures around the Ark, and Heating

It so happens, I was in a discussion with a fellow Catholic (except she accepts Vatican II and "Pope Francis") where K-Ar dating came up.

I can briefly just start by telling you how pre-Creation dates are obtained in K-Ar (potassium - argon) according to Creationist theory. Extra argon, on top of the one obtained (in theory) from potassium comes either from the air or from the lava. The Creationists I read on it attribute to Uniformitarians the theory, this argon from air or from magma prior to eruption argon is depleted (while the lava is cooling down?) well before the millions of years elapse which we are supposed to measure with the method.

I brought up in that discussion the volcano of Hualalai on Hawaii, known eruption in 1801. And the video in which I had learned about it:

Why i believe in a young earth by ex-evolutionist Dr.Grady McMurtry Part 2
15.V.2016 | Arne Karlsen
https://www.youtube.com/watch?v=f1Qr9ZZ-Y30

In the video, Grady McMurtry, Bachelor and Master of Science, Doctor of Divinity and Literature, hence Dr is the title he uses, but he is not from Med School as some might assume, tells the story of how creationists took tests, with a robot, from different depths where lava had ran from Hualalai into the sea. Now, citing the numbers from that video, I stated in German and retranslate to English:

• 0.8 miles below surface, the date is 0 million years, within error margin.
• 2 miles down, 12 million years.
• 2.6 miles down, 21 million years.

Ester pointed out (that's the name of her) that with lava being 1400 °C, it doesn't much matter for cooling speed whether water is 20 °C or 4 °C.*

Now, this made me look up cooling principles.

One variable deciding the speed of cooling is indeed difference of temperatures. And 1400 °C - 20 °C = 1380 °C, 1400 °C - 4 °C = 1396 °C, not much different. Supposing of course one deals with subtraction in deciding this factor. I am not a specialist, but suppose we have instead division, as in 1400 °C : 20 °C = factor of 70, 1400 °C : 4 °C = factor of 325.

But one point sticks out to me : temperature of the cooling agent compared to the thing to be cooled is not the only factor. You have quantity of cooling agent too. And you have the inverse ratio of cooling speed to the ratio of temperature isolation between two materials. Water cools faster, because isolates less well, than air. And once temperature is out in the water, you have a lot of convection as well - once we are past the stage where water simply evaporates anywhere near the lava. But that precise evaporation allows water vapour to rise and therefore be replaced by fresh cool water. This obviously more so further out into the sea, meaning the lava 2.6 miles down must have cooled faster than the lava 2 miles down, than the lava 0.8 miles down. For each unit of time, less hot lava get flooded by more fresh water from the side which in its turn evaporated and left place for next gush of water.

One could of course also argue, some would, that the lava collected from the depth of 2.6 miles down could be from earlier than the eruption in 1801. Well, volcanos don't usually go extinct for 21 million years between two eruptions, so, arguably, the lava 2.6 miles down would be if not from the 1801 eruption, at least not too far before that. And it would still be a case of the recent eruption being dated much older depending on how far down it was in the water when the volcano erupted.

Which brings us to the Flood.

If lots of volcanos were let loose along with subterranean waters, the year of the Flood, from Noah entering the Ark to when he stepped out again, would have seen record many eruptions with record much water from all around to cool it down.

So, in a way, consistent lava dates of millions of years in a series of layers in what looks like inland areas, would indicate that the lava is from the Flood.

Also, such a setting would also provide for lower layers dating older and higher layers dating younger, the way they do in Laetoli, for which the other explanation is, the lower layers are very much older than the younger ones. Which is one I find unacceptable and feel obliged to find an alternative to:

The first lava layer has no heat from earlier lava layers to support it and deeper water above it. The second lava layer has the heat from the first lava layer (if any remains) to support it and the hight of that layer plus some intervening sediment later makes it more shallow water. And for the third time the water is even shallower and the lava heat supported from two lava layers below it. And meanwhile, eruptions world wide are also gradually adding heat to the waters overall, so the incoming water is also less cool, this obviously supposing that the temperature difference is one by division rather than by subtraction.

This brings us to the question : how cool was the water before eruptions warmed it up?

Some who had miscalculated the rain needed for the Flood in 40 days, by presupposing all water of the Flood came from rain, and that covering the highest mountains with 15 cubits of water meant covering Mount Everest, had calculated that the rain falling would produce lots of friction heat and therefore not be able to reach the ground without burning everything in its way (including Ark?) instead of watering. I disagree on the conclusion, but the idea of friction heat indicates that the waters need not have been too glacial at first around the Ark.

We can consider the Ark was well insulated, but even so some periods, between the initial friction heat and heat from waters coming from near magma and the new heat from magma after volcanic eruptions, may have been a bit chilly.

I suggest a solution which was preserved in India was tried out or repeated on the Ark : shit from animals was dried and then burned for heat and light.

Hans Georg Lundahl
Val d'Europe
St. Gavin or Gabinus
Priest and Martyr
19.II.2020

Romae natalis sancti Gabini, Presbyteri et Martyris, qui fuit frater beati Caji Papae, atque, a Diocletiano diu in custodia vinculis afflictus, pretiosa morte sibi caeli gaudia comparavit.

PS, I corrected the spelling of Ester, since she uses this German spelling./HGL

PPS, it can be added that some consider Hawaii was created by volcanic eruptions some time comparable to the 21 million years cited./HGL

* 1400 °C = 2552 °F;
20 °C = 68 °F;
4 °C = 39.2 °F. Google's own converter.

Rolling Period of Ark?

Baraminological Note · For Sea-Farers .... · Rolling Period of Ark? · Ark : empty weight and freighted weight, number of couples on the Ark. · Small Tidbits on Ark, Especially Mathematical

A passenger ship will typically have a long rolling period for comfort, perhaps 12 seconds while a tanker or freighter might have a rolling period of 6 to 8 seconds.

From My Sea Time: Q.What is roll period of a ship and on what factors does it depends ?

Roll Period Time (T) = (K * Beam^2) / sqrt (GM)

Or, from wiki:

T = 2pik / sqrt(gGM)
g = gravitational acceleration
k = radius of gyration about longitudinal axis
GM = stability index

Mathematically the radius of gyration is the root mean square distance of the object's parts from either its center of mass or a given axis, depending on the relevant application. It is actually the perpendicular distance from point mass to the axis of rotation.

Now, I had written in the previous stability related article that ...

"I was out, I calculated that total weight of Ark with load when waterline was 15 cubits up was 50,970 metric tons. I took into account that there were three storeys on Ark, and considering foot tons and calculating for even distribution of weight over three storeys, I got it to a centre of gravitation of either 21.37 feet above keel/bottom, if lowest storey count as ten feet up, or if the foot tons are zero because the height is zero, 18 feet above bottom."

18 - 21 feet up = centre of gravitation.



Problem is, what distance values do I take into account for root mean square distance to the point marked 18/21?

I'll go with three points on the height times three points on the sides. Nine points overall in their distance to the centre of gravitation.

In the middle, where diagonality is not an issue, the three points are high, middle and low. Not marked.

60 - 21 = 39	60 - 18 = 42
30 - 21 = 9	30 - 18 = 12
21	18

Each side has same values, so each following value is counted twice in the mean. High, middle, low on each side is abbreviated as h, m, l.

392 + 502 = h2	422 + 502 = h2
92 + 502 = m2	122 + 502 = m2
212 + 502 = l2	182 + 502 = l2
1521 + 2500 = 4021 = h2	1764 + 2500 = 4264 = h2
81 + 2500 = 2581 = m2	144 + 2500 = 2644 = m2
441 + 2500 = 2941 = l2	324 + 2500 = 2824 = l2
1521 + 81 + 441 + 2(4021 + 2581 + 2941) = 21129	1764 + 144 + 324 + 2(4264 + 2644 + 2824) = 21696
21129 / 9 = 2347.67	21696 / 9 = 2410.67
sqrt(2347.67)= 48.45	sqrt(2410.67) = 49.10
2 * 48.45 * pi = 304.42	2 * 49.10 * pi = 308.50
32.174 ft/s2 * 21 = 675.654	32.174 ft/s2 * 18 = 579.132
sqrt(675.654) = 26	sqrt(579.132) = 24.07
304.42 / 26 = 11.71	308.50 / 24.07 = 12.82

So, unless I totally got the way of calculating radius of gyration wrong, the rolling period of the Ark, according to the formula given in wiki, would have been between 11.71 and 12.82 seconds. Recall that first sentence?

A passenger ship will typically have a long rolling period for comfort, perhaps 12 seconds while a tanker or freighter might have a rolling period of 6 to 8 seconds.

In other words, God saw to it, they were fairly comfy on the Ark, whenever the natural rolling period prevailed!

On the road to St James, on a day when I was crying, a man trying to comfort me told me "God gives his what they need, but not too soon / just a little less, so they don't feel spoiled". God measly? Er, no.

If I got the calculation right, the rolling period of the Ark was that of a passenger ship, or even somewhat slower.

Of course, I am not a ship captain and can have got it wrong, but if you are into ships, see for yourself!

Hans Georg Lundahl
Nanterre UL
St. Narnus of Bergamo
27.VIII.2018

Bergomi sancti Narni, qui, a beato Barnaba baptizatus, primus ab ipso ejusdem civitatis Episcopus ordinatus est.

For Sea-Farers ....

Baraminological Note · For Sea-Farers .... · Rolling Period of Ark? · Ark : empty weight and freighted weight, number of couples on the Ark. · Small Tidbits on Ark, Especially Mathematical

I have been looking at a book about "stability and trim", editor William E. George.*

Transverse Metacentric is an imagined but localisable point in the middle of the vertical axis, affecting the radius of a circle segment in which smaller variations of centre of buoyancy occur, around the longitudinal axis.

It is calculated as BM (radius of B, centre of buoyancy, around/below M, Transverse Metacentre) = I/V.

I is calculated in relevance to water plane as Length * Breadth³ / 12, for a rectangular water plane, which Ark arguably had.

Assume 1 cubit = 2 feet.

300 cubits * (50 cubits) ³ / 12 =

600 ft * (100 ft) ³ / 12 = 50 000 000 ft⁴.

V involves Length, Breadth and Draft, and I assume the latter is 15 cubits = 30 feet.

600*100*30 = 1 800 000 ft³.

Now, I/V = BM = 27.777 ft. Or 27 ft 9' 4".

If Draft was 30 feet, then centre of buoyancy "initial" (before any rolling) would be 15 feet above "keel" (if any), meaning, Height of Transverse Metacentre (KM) is 15 ft + 27 ft 9' 4" = 42 ft 9' and 4".

So much for transverse stability.

Now, longitudinal hull strength ... how do we avoid hull breaks ...

First, how do we not avoid hull breaks? A ship progresses through waves and especially waves that are the length of the ship are a strain on the trim. When waves have crests near two ends of ship, a sagging tendency if there in structure is enforced. When one wave has the crest midship, a hogging tendency is enforced. The periodic enforcing of either a sagging or a hogging one can case a break of hull.

Now, how long were waves in the Flood?

I would imagine, a very typical seascape during the Flood would have been long dunes.

If they were very much longer than 600 feet, like 1200 feet, ark would not have had to deal either with hogging or with sagging, even if progressing through the waves. But more importantly, it was arguably not even doing that. The main motion except rolling would have been swaying - moving with waves and having the length of the Ark confortably between two of them, so moving sideways.

Progress would have sunk the Ark, fortunately there was none.

Indeed, one Swedish dialect would translate progress as "framstega" - and in block letters and ASCII values, that adds up to 666.

But, there was no progress, no surge, for the Ark. No progress, no hull break, no sink.

Hans Georg Lundahl
Paris
St Fidelis of Sigmaringen**
24.IV.2018

PS : I was out, I calculated that total weight of Ark with load when waterline was 15 cubits up was 50,970 metric tons. I took into account that there were three storeys on Ark, and considering foot tons and calculating for even distribution of weight over three storeys, I got it to a centre of gravitation of either 21.37 feet above keel/bottom, if lowest storey count as ten feet up, or if the foot tons are zero because the height is zero, 18 feet above bottom.

I also considered that lifting a ton in pulleys from its resting on floor of one storey to its hanging from its roof, twenty feet higher, ten cubits higher, would add twenty foot tons on that side. Note, with pulleys, even one man can lift a ton, especially if he only needs to lift it one inch.

And this would imply a possibility of leaning a floor (with all "ship", or if it has independent slant from that of whole Ark), when cleaning or food distribution can be facilitated by that.

Supposing the lowest floor counts as ten feet for weight position, and lifting is done at one side between second and third floor, the sides would then have a momentum of 594,150 foot tons on one, and 594,170 foot tons on other side - for all of Ark. Independent movement of a floor would add agility to occasional slanting and a floor could be secured to the side except when this is done./HGL

* Stability and Trim for the Ship's Officer, Third Edition, based on original edition by John La Dage and Lee Van Gemert, Third Edition, ed. by William E. George Cornell Maritime Press, last copyright 1983. ** Sevisii, in Rhaetia, sancti Fidelis a Sigmaringa, Sacerdotis ex Ordine Minorum Capuccinorum et Martyris; qui, illuc ad praedicandam catholicam fidem missus, ibidem, ab haereticis interemptus, martyrium consummavit; et a Benedicto Decimo quarto, Pontifice Maximo, inter sanctos Martyres relatus est.

Baraminological Note

Baraminological Note · For Sea-Farers .... · Rolling Period of Ark? · Ark : empty weight and freighted weight, number of couples on the Ark. · Small Tidbits on Ark, Especially Mathematical

Hedgehogs come in 5 genera and 16 species. Erinaceus, Atelerix, Mesechinus, Hemiechinus, Paraechinus. Together these are known as the "subfamily" Erinaceinae.

These are considered as belonging to the family Erinaceidae together with the subfamily Galericinae a k a Hylomyinae.

Echinosorex gymnura, Neotetracus sinensis, Neohylomys hainanensis, three species of Hylomys and two species of Podogymnura.

At least the 5 genera of hedgehogs, on my view would be from a single couple on Noah's Ark.

The 5 genera of Hylomyinae? Same couple, with loss of spines? One other couple? A separate one for Neotetracus sinensis too?

So, Erinaceinae would on my strictly amateur estimate be 1 to 3 couples of fairly small animals on the ark. Deinogalerix might be a relative of Hylomyinae which did not get on the Ark, because Noah wanted a smaller one (or got a smaller one from God).

If someone were to consider each species separately as a candidate for the Ark, we would get 24 species, 24 couples, not counting the extinct Deinogalerix, which is also another species, clearly, which would bring us to 25 couples, one of which is rather large.

When however Evolutionists say that these are part of Order Eulipotyphla together with "non-African" shrews, Soricidae*, with Talpidae, that is Desmans and Moles ... I feel like drawing a line.

On the other hand, Nesophontes being extinct is not a parallel to Deinogalerix, since Nesophontes went extinct in early 20th or perhaps 16th C. This means we cannot point to this as a separate pre-Flood entity, as far as I could gather about their fossils. Also, solenodons, considered to be relatives of them ("diverged 40 million years ago" said about a Nesophontes species in relation to solenodons sounds like evolutionary ideology, but take a look at the features) have a face clearly reminiscent of hedgehogs.

So, if all Eulipotyphla come from one couple on the Ark, that makes it real spacy. It is still spacy enough if for instance moles and shrews are all from one couple each.

Another Ark was not too crowded on Good Friday : the Church. Christ, the Blessed Virgin, St John, St Dismas, some women, it seems people weren't quite crowding up for salvation that day on Calvary.** However, Heaven is much spacier than Noah's Ark, so don't fear it will be too crowded up there if you make your eternal salvation. If visible universe is two light days across and the Heaven where blessed souls and angels are adoring God is outside that, this means it has really very much space. While a Hell which is about 6000 km below our feet, or less than 4000 miles down, risks getting very much more crowded than Scoffers consider that the Ark would have been. And it seems secular scientists agree it is very hot down there too.

Hans Georg Lundahl
Nanterre UL
Good Friday
14.IV.2017

PS, CMI (one of authors Jonathan Sarfati, as yesterday, other Lita Cosner) did a better job today!

* One of Soricidae is African, but here I use "non-African" to distinguish from Afrosoricida. Considered an order on its own. ** Those who came uninvited by Romans were probably however crowded before arriving to the "one holier hill than Rome".