#### spydercofan

##### Plastic

- Joined
- Feb 25, 2015

I am trying to calculate the minor diameter limits for an

**external**thread via ASME B1.1-2003.

Let me preface, I understand the practicality of the minor diameter is of less concern during the manufacture of the thread. However, this is for theoretical purposes. Therefore, my question can be asked:

*How do i solve for the*

**theoretical**minor diameter limits of an external threadFor example purposes, let us use the thread series

*1/2 - 13 UNC 2A*

M. Handbook provides the following specifications of size:

Allowance = 0.0015

Major = 0.4985 - 0.4876

Pitch = 0.4485 0.4435

**UNR Minor Dia. Max (Ref.) = 0.4069**

My assumption is that the .4069 column, indicates the maximum the minor diameter can be an ensure the thread's internal 'perfect' counterpart can successfully mate. Anything below this value would simply be clearance.

Moving on, I attempt to use the UTS design profile to solve for this 'maximum'. And my math looks like this:

- pitch (p) = .0769
- fundamental height of triangle (H) = .866p = .0665
- True depth of thread (D) = .625H = .0415
- Depth of thread to bottom of fundamental triangle (d) = .875H = .0576
- minimum minor diameter (Dmin) =
**.4876**- (2 * .0576) = .3724 - maximum minor diameter (Dmax) =
**.4985**- (2 * 0415) = .4155 - Tolerance of minor diameter (T) = .4155-.3721 = .0434

Second, the tolerance range exceeds that of which the standard states.

Let us look at ASME B1.1-2003, chapter 5.8.1 which discusses how to calculate the minor diameter tolerance range of an external UN thread:

*(2) UN Classes 1A, 2A, and 3A. To intersection of flat root with flanks of threads (see Figs. 2 and 3), equals pitch diameter tolerance for class of thread specified, plus 0.21650635P*

Using this formula, T = .0216

But the standard fails to remind us to multiply by two, to find a diameter value? T = .0433

So, as you can see, I am confused in trying to compute this range. What is the proper way? And what does .4069 mean?

Thank you