Ohalot, Chapter Twelve, Mishnah Seven



Today’s mishnah deals with a pillar that has fallen onto its side. There is going to be a space on each side of the pillar under which something could lie, because the pillar is round. For this space to be an ohel, the space must be one handbreadth square. Our mishnah calculates how large the pillar must be for the space to be a handbreadth.


Mishnah Seven

1)      [With regard to] a pillar lying [on its side] in the open air,

a)      If its circumference is twenty-four handbreadths, it brings uncleanness to everything under its side;  

b)      But if it is not, the uncleanness cleaves upwards and downwards. 



Okay, ready for some math?

If the circumference of the circle is 24, then according to rabbinic math, the diameter is 8. You could then put a square around this diameter where each side is 8 handbreadths. The diagonal line drawn from one end of this square to another would be 8 + 16/5. This is according to the rule that every handbreadth on the side of a square is equivalent to 1 1/5 handbreadths (we know that the real length is the square root of 128, according to the formula a2 + b2=c2. Still, they were pretty close) At the corners of this square with the circle in it would be a smaller square whose diagonal is 8/5 of a handbreadth (half of the extra 16/5). This is the area which would form the ohel over the uncleanness. Since it is more than a handbreadth, it brings uncleanness to the entire pillar.

If the pillar is smaller than that, then the overhanging part will be less than a handbreadth and it will not convey impurity.