*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:

V

_{outside}= V_{cylinder}+ V_{cube}
V

_{outside}= πr^{2}h + base^{2}h
To obtain the

*inside*capacity of the*Kiyor*we write:
V

_{inside}= π(r-x)^{2}h + (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 = -8x

^{3}+ 27.85x^{2}– 30.363x + 1.604**x = 0.05562**, or approximately

*amos***1 inch**

**Weight of the**

*Kiyor**Kiyor*first calculate the volume of the walls using the known dimensions.

V

_{walls}= V_{outside}– V_{inside}
V

_{walls}= (V_{cylinder}+ V_{cube}) – V_{inside}
V

_{walls}= πr^{2}h + base^{2}h - V_{inside}
V

_{walls}= π(0.875 )^{2}(2.5) + (1.75)^{2}(1.5) - 9
V

_{walls}= 1.604*amos*(or 10,825 in^{3}^{3})To find the weight, multiply this volume by the density of the copper walls using a value of 0.31 lbs/in

^{3}:

10,825 in

^{3}× 0.31 lbs/in^{3}=**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.**

such a boring essay

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