loughnane
Mechanical
- Jan 3, 2010
- 108
I'm writing some TA code and am having a tough time wrapping my mind around the math for the following.
I've got a plate. On this plate I have 3 holes (.130 dia) in a row, with 1" between each hole.
I've got another plate. On this plate I have 3 shafts (.125 dia) in a row, with 1" between each hole.
Tolerances are .001" on every dimension. I assume a CpK of 1.00 (plus/minus 3sigma)
Nominally, the two plates go together. What I'm working on is, if I make 1,000,000 of these parts, how many will fail?
My first crack at it is below. Note that I used only the hole/shaft radii for Hole 1 and Hole 3 (the outer holes), whereas Hole 2 I used the diameter because it is in the center.
Dim Tolerance
Hole - 1 (radius) 0.065 0.003
Hole - 2 (diameter) 0.130 0.005
Hole - 3 (radius) 0.065 0.003
Shaft - 1 (radius) 0.063 0.003
Shaft - 2 (diameter) 0.125 0.005
Shaft - 3 (radius) 0.063 0.003
Distance 1 - 2 1.000 0.005
Distance 2 - 3 1.000 0.005
Nominal Clearance 0.0100
Worst- case tolerance 0.030
RSS Tolerance 0.01118034
Sigma (assuming RSS = 3sigma) 0.00372678
Failures per million 3645.179046
Here is where I say that I don't feel comfortable with this, as I can't quite think of how I would sketch it out on paper.
Does this sound right?
Chris Loughnane - Product Design
I've got a plate. On this plate I have 3 holes (.130 dia) in a row, with 1" between each hole.
I've got another plate. On this plate I have 3 shafts (.125 dia) in a row, with 1" between each hole.
Tolerances are .001" on every dimension. I assume a CpK of 1.00 (plus/minus 3sigma)
Nominally, the two plates go together. What I'm working on is, if I make 1,000,000 of these parts, how many will fail?
My first crack at it is below. Note that I used only the hole/shaft radii for Hole 1 and Hole 3 (the outer holes), whereas Hole 2 I used the diameter because it is in the center.
Dim Tolerance
Hole - 1 (radius) 0.065 0.003
Hole - 2 (diameter) 0.130 0.005
Hole - 3 (radius) 0.065 0.003
Shaft - 1 (radius) 0.063 0.003
Shaft - 2 (diameter) 0.125 0.005
Shaft - 3 (radius) 0.063 0.003
Distance 1 - 2 1.000 0.005
Distance 2 - 3 1.000 0.005
Nominal Clearance 0.0100
Worst- case tolerance 0.030
RSS Tolerance 0.01118034
Sigma (assuming RSS = 3sigma) 0.00372678
Failures per million 3645.179046
Here is where I say that I don't feel comfortable with this, as I can't quite think of how I would sketch it out on paper.
Does this sound right?
Chris Loughnane - Product Design