tag:blogger.com,1999:blog-1241254853273428596.post3956301627981778610..comments2020-09-25T08:31:37.139-04:00Comments on Al Pi Cheshbon: טומאה under a beamShtiklerhttp://www.blogger.com/profile/07498936768989355610noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-1241254853273428596.post-80590048936063354622009-09-11T13:57:21.791-04:002009-09-11T13:57:21.791-04:00I plan to continue by discussing the minimum size ...I plan to continue by discussing the minimum size of a sphere to have טומאה spread underneath it.<br /><br />Yes, a 1x1x1 cube is definitely required. The volume of a cubic טפח is not sufficient.Shtiklerhttps://www.blogger.com/profile/07498936768989355610noreply@blogger.comtag:blogger.com,1999:blog-1241254853273428596.post-20851805839460058942009-09-07T03:19:04.369-04:002009-09-07T03:19:04.369-04:00Where is the continuation?
There's another in...Where is the continuation?<br /><br />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?<br /><br />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.3.141592653589793238462643383279502884197169399375https://www.blogger.com/profile/12059887722695403992noreply@blogger.com