Friday, October 20, 2017

The Weight of the Teiva

If I were such a prolific author that I would have a "magnum opus," I suppose this would be it. To this day, there are still people who identify me as "that guy who wrote the thing on the teiva."

It is told that one year, on a 12th Grade chumash test, Rav Moshe Heinemann שליט"א asked his students how to calculate the weight of the Noah's Ark. He did not ask for an answer, he simply asked how one would go about figuring it out. These are the calculations. And the answers:

Later on in the Parsha, (8:4), Rashi calculates based on the rate at which the waters of the flood receded, that the ark was submerged 11 amos in the water. A variety of commentaries deal with the calculation cited by Rashi and its validity, most notably the Ramban. The Sifsei Chachamim quotes the Nali"t as saying that the figure of 11 amos is only a minimum but it could have been more. There are a number of problems raised with different aspects of the calculation, some of which will be dealt with later on. Nevertheless, if the words of Rashi are taken at face value, they hold within them the key to unlocking this mystery. With the application of a single principle, the weight of the ark can be calculated. The law required for this calculation is Archimedes' Principle which states that the weight of a body floating in water is equal to the weight of the water it displaces. The ark's virtually cubic structure (according to Rashi) makes the measurement of water displacement easy to achieve. The ark was 300x50x30 amos3 in volume (Breishis 6:15). Therefore, the water displaced by the ark was 300x50x11 = 165,000 amos3

The next step, of course, is to convert the figure of cubic amos into conventional measures. Unfotunately, we are unsure as to the exact measure of the amah. There are three primary opinions amongst the contemporary poskim as to the actual length of the amah: Chazon Ish, R' Moshe Feinstein and GRA"CH Noeh. Because of this disagreement, they will differ on the measure of the ark's water displacement and therefore, the final figure for the weight of the ark will be different according to each. The following is a chart calculating the water displacement in cm3 based upon each of the opinions.
Metric to Imperial conversion table below
Chazon IshR' Moshe FeinsteinGRA"CH Noeh
Length of amah57.66 cm.53.98 cm.48 cm.
Volume of cubic amah
0.192 m30.157 m30.111 m3
Calculation= 165000 x 0.192
≈ 31630
= 165000 x 0.157
≈ 25950
= 165000 x 0.111
≈ 18250
Water Displacement31630 m325950 m318250 m3

Now that we have determined the amount of water displaced by the ark, all we have to do is calculate how much that water weighed. Then by Archimedes' Principle we can assume that the ark weighed the same amount. This, however, is not necessarily so simple. The density of sea water is slightly more than that of regular water at approximately 1025 kg/m3. This figure usually remains about the same, without significant deviation, regardless of the exact temperature. The only drastic changes are observed when the water reaches extreme conditions such as freezing or boiling.

The first difficulty encountered is that during the initial 40 days of the flood, the waters were boiling hot (Rosh HaShanah 12a). This would change the density of the water substantially and consequently interfere with the calculation. However, it is important to note that Rashi's calculation is based on the rate at which the water receded after the 150 days which followed the 40 days of destruction. By that time, the waters had calmed down and most probably dropped to a more moderate temperature. Therefore, it can be assumed that the temperature of the water is a negligible factor in the calculation of the water density. However, what does seem problematic is that Rashi brings in the figure of 11 amos in 7:17 when the waters were at their highest intensity. It is almost certain that the density of the water at this point was much less than it was 190 days later. If the ark was calculated to have been submerged 11 amos by a calculation based on cooler waters, that figure should presumably be greater at the time of the actual flood.

The next issue of question in this calculation is the fact that the water was not necessarily pure sea water. It is suggested in Rashi (6:14) that the water contained sulfur. The presence of this sulfur and whatever other solvents in solution with the water could change the density of the water and affect the accuracy of the calculation greatly. This is only a problem, of course, if the words of Rashi are taken literally. The Sifsei Chachamim seem to suggest that what Rashi means is that the sulfur caused the heating of the water. Even if the interpretation is as originally perceived, it is possible that the ratio of solute to solvent was such that it would not have affected the density anyway. Therefore, for the purposes of this calculation I have chosen to ignore whatever effects the sulfur could have had on the water density and thus we are left with approximate figure of 1025 kg/m3. Based on this figure, these are the final calculations of the weight of the ark according to the three aforementioned opinions:

Chazon IshR' Moshe FeinsteinGRA"CH Noeh
31630 m3 25950 m3 18450 m3
x 1025 kg/m3
32420750 kg26598750 kg18706250 kg

In conclusion, considering the relevant opinions, it would appear that the ark weighed somewhere between 18 and 33 thousand metric tons. In comparison with other famous ships, the Queen Mary weighed 73,850 tons. It was 309 m long, about twice as long as the ark. The Titanic weighed approximately 42,000 tons. Of course, this refers to the weight of those vessels without anyone inside whereas the above calculation for the teiva included the inhabitants.

Table of Metric Conversions
57.66 cm=22.7 in.
53.98 cm=21.25 in.
48 cm.=18.9 in.
31630 m3 =1117003 ft3
25950 m3 =916416 ft3
18450 m3=651556 ft3
25o C=77o F
1025 kg/m3=2260 lb/61024 in3(35.3 ft3)
32420750 kg=71475519 lb = 35737.8 tons
26598750 kg=58640206 lb = 29320.1 tons
18706250 kg=41240222 lb = 20620.1 tons
6372500 m=20907152 ft
53 cm=20.87 in.

The Constant Rate of Recession

No, this has nothing to do with the American economy. There is another difficulty with the calculation that Rashi uses to conclude that the teiva was submerged 11 amos. How could Rashi base his calculation on the depth of the water decreasing at a constant rate. One can generally assume that when water decreases, it does so at a constant rate of volume. However, mathematically, if the volume of a sphere decreases at a constant rate, the rate of change of the depth will increase as the waters become shallower. The shallower the water gets, the faster it will decrease depthwise. How then could Rashi assume that the depth decreased at a constant rate? This is the question posed by מהרי"ל דיסקין. He gives his own answers to this question. One, for instance, is that the waters receded, the ground became more saturated which slowed down the overall receding process and hence balanced out the constant rate of change of depth. But a Rebbie of mine from Yeshivas Ohr Yerushalayim posed this question of none other than Nobel Prize winner Yisrael Aumann. He answered simply that mathematically, none of this is needed. True, the rate of change of depth is not directly proportional to the rate of change of volume. However, considering the size of the globe, the difference between the two within the scope with which we are dealing, is negligible and would not affect Rashi's calculation. Is this true? The short answer is "Yes". The longer answer requires a little Calculus.

The radius of Earth is 6372500 m. To make things simple we will convert this to amos. Instead of using three separate measures of the amah, we will keep things neat and use an average figure of 53 cm. (6372500 ÷ 0.53 = 12023585) That translates to 12023585 amos. To make things simpler, we will round it off to 12000000 amos. This will have little effect on the final outcome. This figure will be called rw.

The standard equation for volume:
Vw= 4/3πrw3
Through implicit differentiation:
ΔV= 4πrw2 Δr,
where ΔV is the rate of change of volume and Δr is the rate of change of radius. We have already set rw to be 12000000 and Δr is ¼ (amos/day according to Rashi). Therefore,
ΔV= 4π (12000000)2 ¼
ΔV= 4.524 x 1014 (constant)
The goal of these calculations is to see whether or not Δr changes significantly over the course of the decreasing of the water. To see how much Δr changes, we must switch around the equation to define Δr and instead of using the figure of 12000000 for the radius, we will use the new radius when the top of the mountains became visible, 11999985.
As stated before, ΔV= 4πr2 Δr
Therefore, Δr2 = ΔV/ 4πrnew2
Δr2 = 4.524 x 1014/ 4π(11999985)2
Δr2 = 0.2500006250012
This means that if the waters were receding at a rate of change of depth of 0.25 amos per day when they began receding, then 60 days later they were receding only 0.0000006250012 amos/day faster, a rather negligible amount indeed.

Wednesday, October 4, 2017

How many בקשות in יעלה ויבוא

Since we are about to say יעלה ויבוא numerous times (perhaps as many as 40 times) I figured it would be a good time to explore this:
A friend once approached with an interesting project - to calculate the total number of בקשות in יעלה ויבוא. Now, of course, that's not as simple as it seems. It is not simple addition. It involves a lot of multiplication.
So let's dive into it:

יעלה | ויבוא | ויגיע | ויראה | וירצה | וישמע | ויפקד | ויזכר= 8
(Keep in mind that all the above verbs will apply to all of the following nouns:)
זכרוננו | ופקדוננו | וזיכרון אבותינו | וזיכרון משיח בן דוד עבדך | וזיכרון ירושלים עיר קדשך | וזיכרון כל עמך בית ישראלx 6
= 48
(And then the following modify all of the above:)
|לפניך לפליטה | לטובה | לחן | ולחסד | ולרחמים | לחיים ולשלוםx 7
So the entire first section = 336
זכרנו ה' אלוקינו בו לטובה | ופקדנו בו לברכה | והושיענו בו לחיים טובים + 3
= 339
Now we add the final portion:
ובדבר ישועה | ורחמים= 2
חוס | וחננו | ורחם עלינו | והושיענוx 4
= 8
So the final count is 339 + 8= 347

Wow, 347 בקשות packed into one small תפילה!