## Saturday, March 10, 2012

### A Fuller Picture of the Kiyor

The Kiyor [Laver] was a copper vessel which held the water used by the Kohanim to sanctify their hands and feet in preparation for the sacrificial service. It was used in the Tabernacle as well as in the First and Second Temples, although this Kiyor is not described in any great detail in the Scriptural or Talmudic sources.

In the First Temple Solomon constructed ten additional lavers which are described as containing 40 bas (a volume equivalent to 9 cubic amos — Ralbag) of water and standing 4 amos tall (I Kings 7:38). Malbim provides further details of these lavers: the top section of the laver was cylindrical, 1¾ amos in diameter and 2½ amos tall; the bottom section was square, 1¾ amos wide and 1½ amos tall (all the numbers given here are the outer dimensions). Now, Rashi (to II Chronicles 4:6) writes that that these ten lavers were built "in addition to that of Moses," which to me implies that they were exact replicas of the Kiyor that Moses made for the Tabernacle, just as Solomon's ten copies of the Menorah and Shulchan [Table] were exact copies of the originals. I therefore model the Kiyor of the Second Temple after Malbim's description of Solomon's lavers. Here is an image of what the Kiyor would have looked like:

Knowing a) the total volume of water held by the Kiyor, and b) the outer dimensions, allows us to calculate the thickness of its walls. From this we can then figure out the weight of the Kiyor when empty and how much it would have weighed when filled with water.

Thickness of the Walls

The thickness of the Kiyor's walls can be calculated by solving the Volume equation using the known outer dimensions and the known inner volume:

Voutside = Vcylinder + Vcube

Voutside = πr2h + base2h

To obtain the inside capacity of the Kiyor we write:

Vinside = π(r-x)2h + (base-2x)2(h-2x)

where x is the thickness of the wall (a uniform wall thickness is assumed).

Now solve using the known dimensions of the Kiyor:

9 = π(0.875 – x)2(2.5) + (1.75 – 2x)2(1.5-2x)

0 = -8x3 + 27.85x2 – 30.363x + 1.604

x = 0.05562 amos, or approximately 1 inch

Weight of the Kiyor

To find the weight of the empty Kiyor first calculate the volume of the walls using the known dimensions.

Vwalls = Voutside – Vinside

Vwalls = (Vcylinder + Vcube) – Vinside

Vwalls= πr2h + base2h - Vinside

Vwalls = π(0.875 )2(2.5) + (1.75)2(1.5) - 9

Vwalls = 1.604 amos3 (or 10,825 in3)

To find the weight, multiply this volume by the density of the copper walls using a value of 0.31 lbs/in3:

10,825 in3 × 0.31 lbs/in3 = 3,356 lbs

Add to this the 9 cubic amos of water (which weighs 2192 lbs) and the total weight of the Kiyor that was raised each morning via the Muchni was 5,548 lbs.

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