## Thursday, May 7, 2015

### טומאה under a beam

In the משנה יומית, we recently encountered a משנה I had long been awaiting. Just as a background, טומאה which is inside of a room will generally make anything else in the room טמא. In order for the טומאה to spread, however, it needs to be inside a space which is at least a cubic טפח. So, the משנה אהלות י"ב:ז discusses how big a cylindrical beam would have to be in order to be certain that there is a cubic טפח underneath it. The משנה gives the figure of a 24 טפח circumference. Let us explore the math behind this:

The key to figuring this out is actually simple geometric fact mentioned in the .גמרא סוכה ח
כל אמתא בריבועא אמתא ותרי חומשא באלכסונא
An isosceles, right-angle triangle whose two sides are a טפח will have a diagonal of 1.4 טפחים. Or, more simply, if you draw a line from one corner of a 1x1 טפח square to the other, the length will be 1.4 טפחים. This is actually a simple (and rounded) version of the Pythagorean Theorem. The square of the hypotenuse is equal to the sum of the squares of the other two sides. 12+12=2 so the hypotenuse should be 2 which is more like 1.414. Let's be as precise as we can for this.

Above you will find a rough sketch of the problem we are trying to solve. We are going to solve for the radius of the beam. So in order for the space to be a cubic טפח, the diagonal has to be 2 טפח. If we draw a line from the centre of the beam to the corner of the square which surrounds the beam we can form a triangle which I have highlighted in faint green. It is a right-angle triangle where the two sides are of length x and the hypotenuse would be x+√ 2. But using our rule from סוכה (and Pythagoras) we know that the length of the hypotenuse can also be expressed as √ 2x. Therefore, we have
 √ 2x = √ 2 + x √ 2x - x = √ 2 (√ 2 - 1)x = √ 2 x = √ 2 / (√ 2 - 1) x ≈ 3.41 Double that to get the diameter and multiply by π to get the circumference c ≈ 21.45 טפחים
Now we know that the תלמוד does not necessarily use 100% precise figures. The aforementioned גמרא in סוכה clearly uses a round figure of 3 for π. But how would the משנה have come to a figure of 24 for this calculation?

רע"ב explains: A circumference of 24 means a diameter of 8 (when using the talmudic π of 3) If the sides of the large square above are 8, the diagonal, from one corner of that square to the other, which includes the diameter and two cubes, would be 8 * 1 2/5. Basically, that's 16/5 extra, 8/5 diagonal on either side. But for a טפח x טפח cube, we were looking for a diagonal of just 7/5. רע"ב explains that משנה was not concerned with the minuscule margin of the extra 1/5. The troubling issue is that if you go through all those calculations with a circumference of 21, a diameter of 7, you get exactly 7/5 on the dot! I saw explained according to one source that since a beam would likely be sunken into the ground slightly, the משנה did not want to give the exact measurement which would end up being overly stringent.

To be continued...

1. Where is the continuation?

There's another interesting issue here. Your calculations are based on the assumption that the cubic טפח required for טומאה to spread is not just the volume of a cubic טפח, but actually a cubic space with each side measuring one טפח. Is this clearly the הלכה?

If we don't use this assumption, but rather we simply require a cubic טפח of volume, then of course the question as stated is unsolvable because the volume would depend also on the length of the beam. (This suggests that in fact your original assumption is probably correct.) But if we assume the beam is one טפח long, then how big should the circumference be? Let's assume the one-cubic-טפח volume can't be split into two parts. If x is the radius (as in your picture), we need x squared = pi x squared / 4 + 1, giving x = 2.2 (approx) for a circumference of 13.6. Doesn't help.

2. I plan to continue by discussing the minimum size of a sphere to have טומאה spread underneath it.

Yes, a 1x1x1 cube is definitely required. The volume of a cubic טפח is not sufficient.