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The Holy Ark was 2 1/2 Cubits, by 1 1/2 Cubits, by 1 1/2 Cubits (Shemot 25:10).

The Talmud discusses (Baba Basra 14A-B) how much space the Luchot occupied in the Ark. The Talmud says that each tablet was 6 handbreadths, by 6 handbreadths, by 3 handbreadths. The Talmud goes on to discuss exactly how much room the Luchot occupied in the Ark, depending on whether the Cubits used to measure the Ark were made of 5 or 6 handbreadths.

The Talmud goes say there was some empty space left over and discusses what was in the empty space.

My Question:

According to the opinion that the Whole Luchot and Broken Luchot were placed in the Ark (Baba Basra 14B), how did they both fit?

I'm assuming that both sets of Luchot were the same size, since G-d tells Moshe to hew a new set of Luchot like the first set (Shemot 34:1). No matter how you calculate the Handbreadths/Cubit ratio used in the Ark, there isn't enough room for a second set of Luchos.

There are opinions that Moshe made a second Ark which were eventually used for the broken Tablets (see for example Rashi to Devarim 10:1), but:

  • Others (such as the Ramban 10:5) Disagree with this.

  • The opinion in the Talmud that says both sets of Tablets were placed in the Ark seems to be saying that they were both placed in the same Ark.

  • Tosafot (Eruvin 63B) is of the opinion that there were two Arks, but says that when King Shlomo built the Beit Hamikdash both Luchot were placed in one Ark (according to the opinion that the broken Luchot were also in the Ark).

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2 Answers 2

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I haven't looked at the relevant g'mara excerpts, but, off the bat, I don't see the problem. If the second set was broken, then it doesn't have to fit upright or the like: it was in pieces. So as long as the total interior volume of the aron is (slightly more than) the total volume of the two sets of luchos, that's enough. The interior volume of the aron would seem to be more than 2*1*1 square amos, while the total volume of the two sets of luchos would seem to be 1*1*.5*2*2 amos (if there are six t'fachim to an ama), so what's the problem? (Even if there are five t'fachim to an ama — and, again, I haven't checked the sources — you need only (6/5)^3 times that volume, or 3.456 cubic ama, which the interior of the aron should be fine with (2.5*1.5*1.5=5.625, so even once we subtract the width of the walls (which I don't know offhand) I imagine it's fine).

Again, this is all without recourse to resources, but it seems right.

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I see my mistake. I figured that there were 4 luchot, so I multiplied every dimension by 4 (i.e. 24*24*12). Instead, I should have just multiplied the volume of one tablet by 4. That's embarrassing. –  Menachem Aug 23 '11 at 18:53
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I made these from Baba Basra 14. It shows the difference between how R. Yehuda and R. Meir calculate the dimensions of Aron:
enter image description here

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