The following is a "revoked" standard (which is probably why you can find it freely available here http://www.chinafastener.biz/gb/am/file_list.aspx?action=download&id=215&ext=pdf). I am going through the academic exercise on page 49 of the standard (Appendix B), and I am having a hard time understanding how they calculated the pitch diameter tolerance for tolerance grade 5. If I use table 10, I find that for the basic major diameter of 30mm, I have two pitch values which I interpolate between. For tolerance grade five, I interpolate between 100 and 118, to get 109 microns. If I use the formula in 6.12.5 (page 28), I calculate Td2(6) = 90*(1.25)^0.4*(30)^0.1 = 138.267 microns, which when converting to tolerance grade 5, I get Td2(5) = 0.8*Td2(6) = 110.613. If I round that value to the nearest R40 preferred number, I get 112 microns. If I round Td2(6) to the nearest R40 preferred number before converting to tolerance grade 5, I get 140 microns, which when converted, is 112 microns. I'm quite confused on how they arrived at 108 microns in the example. I really want to make sure I get this correct for no reason other than that this shouldn't be too hard to calculate otherwise it's usefulness would be limited.
Hopefully this was the right section to post this in, but if not please move it accordingly.
Cheers,
Matt
Hopefully this was the right section to post this in, but if not please move it accordingly.
Cheers,
Matt