Newer Tables, Preliminaries

Newer Tables: Preliminaries · Flood to Joseph in Egypt · Joseph in Egypt to Fall of Troy.

A) The anchor points.

In order to calibrate carbon 14, you need anchor points, where an object of known age is also carbon dated. To me, the Biblical events are (except for I/II and VIII, a physical turning point in the early post-Flood era, and the fall of Troy) are of a known age, at least with approximation about what text is the correct one. Originally there were nine anchor points, but I dropped point IX and I introduced points I/II and VI/VII.

I, the Flood.

2958 BC. Dates as 37 000 BC. 0.0162771032469586

I/II, a midpoint.

2738 BC. Dates as 20 933 BC. 0.1106908081611022

II, death of Noah, beginning of Babel.

2608 BC. Dates as 9500 BC. 0.4344336256776197

III, birth of Peleg, end of Babel.

2557 BC. Dates as 8000 BC. 0.5176637713737836

IV, Genesis 14.

1936 BC. Dates as 3500 BC. 0.8276258493022752

V, c. 1700 BC, death of Djoser.

1700 BC. Dates as 2800 BC. 0.8754082729655837

VI, c. 1590 BC, death of Sesostris.

1590 BC. Dates as 1839 BC. 0.9703280628990107

VI/VII, Exodus.

1511 BC. Dates as 1609 BC. 0.9882151189089242

VII, fall of Jericho.

1471 BC. Dates as 1550 BC. 0.9904890385842258

VIII, fall of Troy.

1179 BC. Dates as 1179 BC. 1 (100 pmC)

B) Distances and speeds.

Between two anchor points, there is a specific distance in time.

That one will involve the carbon 14 level in the atmosphere present at the beginning per se decaying corresponding to the time, a factor of for instance 0.5^(220/5730) for the first of these distances.

It will also involve this atmosphere, which is a sample but an open sample, not lowering the carbon 14 level, but in each of the cases I studied so far actually raising it, because while the original carbon 14 content decays, new carbon 14 is added.

The normal addition over such time as it is today is 1 - that decimal fraction or if it's translated into pmC (a percentage fraction), it is 100 - so many pmC points.

However, these occasions, what's getting added will be added faster, and we well calculate how much faster. The level at the end of the table's time span will consist of a part that's the remainder of the level at the beginning and a part that's the addition over this time. Once you have determined the actual addition, you divide that by the normal replacement. That will mean how much faster the carbon 14 is added over these periods.

The level was not low because the carbon 14 was (after the Flood) added slower than now, which lowered it over time, it was low because it started out low after the Flood and got high this fast since then by adding carbon 14 faster than now.

I—I/II

2958-2738 = 220 years

0.9737380239639717 * 0.0162771032469586 = 0.0158496344 remainder

(0.1106908081611022 - 0.0158496344) / (1-0.9737380239639717) = 3.6113494917 replacement speed

I/II—II

2738-2608 = 130 years

0.9843971533174005 * 0.1106908081611022 = 0.1089637165

(0.4344336256776197 - 0.1089637165) / (1 - 0.9843971533174005) = 20.859649258 replacement speed

II—III

2608-2557 = 51 years

0.9938496186938799 * 0.4344336256776197 = 0.4317616932 remainder

(0.5176637713737836 - 0.4317616932) / (1 - 0.9938496186938799) = 13.9669516199 replacement speed

III—IV

2557-1936 = 621 years

0.9276310630766214 * 0.5176637713737836 = 0.4802009946 remainder

(0.8276258493022752 - 0.4802009946) / (1 - 0.9276310630766214) = 4.8007455894 replacement speed

IV—V

1936-1700 = 236 years

0.9718551869271131 * 0.8276258493022752 = 0.8043324745 remainder

(0.8754082729655837 - 0.8043324745) / (1 - 0.9718551869271131) = 2.5253604741 replacement speed

V—VI

1700-1590 = 110 years

0.986781649588181 * 0.8754082729655837 = 0.8638368197 remainder

(0.9703280628990107 - 0.8638368197) / (1 - 0.986781649588181) = 8.0563186722 replacement speed

VI—VI/VII

1590-1511 = 79 years

0.9904890385842258 * 0.9703280628990107 = 0.9610993101 remainder

(0.9882151189089242 - 0.9610993101) / (1 - 0.9904890385842258) = 2.8510060751 replacement speed

VI/VII—VII

1511-1471 = 40 years

0.9951729639914483 * 0.9882151189089242 = 0.9834449689 remainder

(0.9904890385842258 - 0.9834449689) / (1-0.9951729639914483) = 1.4592950262 replacement speed

VII—VIII

1471-1179 = 292 years

0.9652938815432484 * 0.9904890385842258 = 0.9561130087 remainder

(1 - 0.95611300868100765329074873108872) / (1 - 0.9652938815432484) = 1.2645318252 replacement speed

C) For the calibration, I hack up the timespans to pieces.

Each piece is of equal length. It has its own level of decay, a multiplication with a decimal fraction starting with 0.99 ...

It obviously also has it's own level of normal replacement. One of the numbers, 22 years, is used twice over these tables.

22 years (I—I/II, 10 times, V—VI, 5 times)

0.9973422400389199 * decay

(1 - 0.9973422400389199) = 0.0026577599610801 normal replacement

13 years (I/II—II, 10 times)

0.998428650637827 * decay

(1 - 0.998428650637827) = 0.001571349362173 normal replacement

17 years (II—III, 3 times)

0.9979456554564614 * decay

(1 - 0.9979456554564614) = 0.0020543445435386 normal replacement

23 years (III—IV, 27 times)

0.9972216007456389 * decay

(1 - 0.9972216007456389) = 0.0027783992543611 normal replacement

19.6666666666666667 years (IV—V, 12 times)

0.9976237884822478 * decay

(1 - 0.9976237884822478) = 0.0023762115177522 normal replacement

15.8 years (VI—VI/VII, 5 times)

0.99809052947232 * decay

(1 - 0.99809052947232) = 0.00190947052768 normal replacement

20 years (VI/VII—VII, twice)

0.9975835624104119 * decay

(1 - 0.9975835624104119) = 0.0024164375895881 normal replacement

24.3333333333333333 years (VII—VIII, 12 times)

0.9970607710539279 * decay

(1 - 0.9970607710539279) = 0.0029392289460721 normal replacement

D) Putting it all together

We begin by putting the B and C parts together. I'll take an example from table VII—VIII.

VII—VIII

1471-1179 = 292
0.9652938815432484 * 0.9904890385842258 = 0.9561130087
(1 - 0.9561130087) / (1 - 0.9652938815432484) = 1.2645318252 !!

24.3333333333333333 years (VII—VIII, 12 times)

0.9970607710539279 *
(1 - 0.9970607710539279) = 0.0029392289460721 !!

The two quantities marked with !! will enter a multiplication. It's the speed times the normal replacement.

1.264 531 8252 * 0.0029392289460721
= 0.0037167485 !!

Here I have marked the specific replacement.

Note, while tables I—I/II and V—VI have the same piece length, the same decay, the same normal replacement, the different speeds mean they have different specific replacements.

So, after the specific replacement is calculated, for each new year after the first anchor point, the following calculation module is added, where x = first or any other immediately previous year's carbon level.

0.9970607710539279 * x (decay)
+ 0.0037167485 (specific replacement)

For each new stop, you calculate the year first, like the first or immediately previous year in the table minus the piece. In this case 24 and 1/3 years. The result is not yet rounded to an integer, OK.

Then you insert x, and do that calculation.

1471-24.3333333333333333 = 1446.6666666666666667

0.9970607710539279 * 0.9904890385842258
+ 0.0037167485

You can actually put them on the same line, easier to copy-paste and put it into the calculator.

0.9970607710539279 * 0.9904890385842258 + 0.0037167485

Giving as result the new carbon 14 level:

0.9912945131

As this is put below the year, you continue to next and so on, and in each case I added a formula for obtaining the year.

As honour where honour is due, I have already noted in With a Little Help from MathQueen! how her method has helped me to not despair on getting this perfect. Carbon 14 calculators are not bad, but they will round the result. Her video is Can you crack this beautiful equation? – University exam question. Below the link, on the post, I derive the formula I need. When you have calculated all years (without rounding) and carbon 14 levels, below them, you temporarily give each this formula. The first line obtains the number of instant age extra years from the carbon 14 level (must be expressed as decimal fraction). The second line adds the unrounded year of the real chronology.

1446.6666666666666667
0.9912945131

5730 * log() / log(0.5)
+

Put the variables into the appropriate slots:

1446.6666666666666667
0.9912945131

5730 * log(0.9912945131) / log(0.5)
+ 1446.6666666666666667

Put the whole formula on a single line:

5730 * log(0.9912945131) / log(0.5) + 1446.6666666666666667

Get it through the calculator:

1518.9468913523

The result is only now rounded.

1447
0.9912945131
1519

And some readers may appreciate what you put the decimal fraction into the standard form of a % faction, as pcM go.

1447 BC
99.129 pmC
1519 BC

This is how each of the stops within a table was constructed. And including the last one to check it lines up with the first of the next or with 100 pmC at Troy.

Now, I'm making these posts a bit early, but they are out on Christmas Eve, after First Vespers./HGL

PS, in case you wonder about my Biblical chronology, I use the Martyrology reading for Christmas Day:

Φιλολoγικά/Philologica: What Martyrology, by the way?
Saturday, November 21, 2020 | Posted by Hans Georg Lundahl at 5:40 AM
https://filolohika.blogspot.com/2020/11/what-martyrology-by-way.html