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...
