Erik,
the drill sizes you are using are all 0.2 to 0.4 mm too large,
especially with the small values M2 to M6 where it hurts the most. The
resulting theads have got to be too loose, easy to overtighten. Don't
use complicated formulas, especially not percent calculatations in
public, as my math professor said, just subtract the pitch from the
diameter and you are on line.
Peter
Am 23.01.2016 13:31, schrieb Erik Christiansen:
This is what I use:
----------------------------------------------------
Metric Coarse | BS 4168 Clearance hole:
| | Type of fit:
Size Drill E Core | Close Medium Free
----------------------------------------------------
M2 1.70 61% 1.51 |
M2.5 2.15 63% 1.95 | WARNING:
M3 2.60 65% 2.39 | 3.2 3.4 3.6 2.5mm drill -> 82%.
M4 3.40 70% 3.14 | 4.3 4.5 4.8
M5 4.40 61% 4.02 | 5.3 5.5 5.8 With a nut length of
M6 5.20 65% 4.70 | 6.4 6.6 7.0 1.5 * diameter, 50%
M8 6.90 72% 6.47 | 8.4 9.0 10.0 engagement is stronger
M10 8.70 71% 8.16 | 10.5 11.0 12.0 than the external thread.
M12 10.50 70% 9.85 | 13.0 14.0 15.0
M16 14.25 71% 13.55 | For CI and thin sheet,
-------------------------------------------------- stick near 75%.
Or in general: # d is full thread depth.
Drill Size = OD - (E*2d) # E is e.g. 0.65
# d = kP , where P is thread pitch.
Or transposed for E: E = (OD - Drill_Size)/2d
k is: ISO Metric 0.613
UNF/UNC 0.613
# British Association BA 0.600
# BSW, BSF, ME (32 & 40 TPI), Whitworth forms 0.640
# BSB, and BSP parallel.
----------------------------------------
So for M12 x 1.5:
d = 0.613 * 1.5 = 0.9195
E = (12 - 10.5)/(2 * 0.9195) = 81.5%
That's too much, especially for a deep hole. The 10.7 mm hole:
E = (12 - 10.7)/(2 * 0.9195) = 70.6%
would give Tom's spindle a fair chance.
(In my previous post, I forgot that the above table is for Metric
Coarse, i.e. 1.75 mm pitch @ M12. :-(
What would Andy's 10.2 mm hole give with M12 x 1.75?:
d = 0.613 * 1.75 = 1.07275
E = (12 - 10.2)/(2 * 1.07275) = 84%
That's hard on the tap and spindle. Much better is:
E = (12 - 10.5)/(2 * 1.07275) = 70%
Erik
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