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 amos^{3} in volume (Breishis 6:15). Therefore, the water displaced by the ark was 300x50x11 = 165,000 amos^{3}.
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 cm^{3} based upon each of the opinions.
Metric to Imperial conversion table below
Chazon Ish | R' Moshe Feinstein | GRA"CH Noeh | |
Length of amah | 57.66 cm. | 53.98 cm. | 48 cm. |
Volume of cubic amah (length/100)^{3} | 0.192 m^{3} | 0.157 m^{3} | 0.111 m^{3} |
Calculation | = 165000 x 0.192 ≈ 31630 | = 165000 x 0.157 ≈ 25950 | = 165000 x 0.111 ≈ 18250 |
Water Displacement | 31630 m^{3} | 25950 m^{3} | 18250 m^{3} |
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/m^{3}. 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/m^{3}. Based on this figure, these are the final calculations of the weight of the ark according to the three aforementioned opinions:
Chazon Ish | R' Moshe Feinstein | GRA"CH Noeh |
31630 m^{3} | 25950 m^{3} | 18450 m^{3} |
x 1025 kg/m^{3} | ||
32420750 kg | 26598750 kg | ^{}18706250 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 m^{3} | = | 1117003 ft^{3} |
25950 m^{3} | = | 916416 ft^{3} |
18450 m^{3} | = | 651556 ft^{3} |
25^{o} C | = | 77^{o} F |
1025 kg/m^{3} | = | 2260 lb/61024 in^{3}(35.3 ft^{3}) |
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. |
I discussed this over Shabbos about 5 or 6 times with different people or groups. I think it makes sense intuitively to calculate the volume in cubic amos before converting to modern measurements, especially when prompting people to work it out mentally on Shabbos. Here's what I do: The submerged volume is 165,000 cubic amos. If we estimate an amma at half a metre, then a cubic amma is 1/8 of a cubic metre. We can then approximate the volume as 160,000 cubic amos, divide by 8 and get about 20,000 cubic metres. A cubic metre of water weighs a (metric) tonne, so the weight of the displaced water, and therefore of the box, was about 20,000 tonnes. (Chazon Ish-niks can then multiply by about 1.75 as they would for any volume measurement.)
ReplyDeleteI forgot to mention last week - once again I went through this calculation with a few people over Shabbos. A couple of people asked the same question: How do we know the ark's shape was a rectangular prism? How do we know it wasn't boat-shaped, with a hull at the bottom? Actually you imply above that Rashi answers this question, so maybe you can relieve me of my ignorance and remind me where.
ReplyDeleteIt seems that what you say "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." can be answered by what you stated earlier "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."
ReplyDeleteHere's the big question I don't think I've asked you before: Did anyone in R' Heinemann's Chumash class answer the question correctly on the test? Or did any of them come up with any other creative answers?
ReplyDelete