Some thirty odd years ago I jotted down a formula to calculate the diameters of three dissimilar gage pins that when bundled together with rubber band could be used to gauge a hole of larger diameter. Recently, the need arose again and after the disappointment of searching my memory, I undertook the exercise anew.
In my experience, the most common gage pin sets a machine shop will have run from .061” – .250” and .251” – .500”. These occur in a variety of tolerance classes. The following has been formulated for “minus” gage pin sets. Manipulation of applicable hole tolerances allow “Go” and “No-Go” gages to be built up for holes sizes beyond gage pin availability.
An equilateral triangle where use of corners as centers provides for an arrangement of three circles inscribed within a fourth and mutually tangent. The diameter of the enclosing divided by that of one of the enclosed establishes a constant for convenience in determining an initial gage pin size.
c = D / d = 2.154700515805138
The size of the hole (D) to be gauged divided by the constant (c) decides our starting point.
D / c = d1
The exact value derived will likely not conform to a nominal gage pin size and will require rounded down to the thousandths with subsequent subtraction of the tolerance class value. As a result, a four-place decimal (“tenths” in machinist vernacular, “ten-thousandths” for the layperson), or possibly five depending on tolerance class, now defines the first pin diameter.
D / c = d1 = .XXXX…n
d1 = .XXX (round to thousandths)
d1 = .XXX – .000X (subtract tolerance class value)
d1 = .XXXX (first pin size)
To avoid size overlap (most shops enjoy only a single set of pins), the simple addition of .002” to the value obtained above settles the size of the second pin.
d2 = d1 + .002 (second pin size)
While the included angle for each of the three pins in the configuration described above is approximately 120°, their exact values need be determined to arrive at the third diameter.
∠d1 = 2 ( sin-1 ( d1 / ( D – d1 ) )
∠d2 = 2 ( sin-1 ( d2 / ( D – d2 ) )
∠d3 = 360 – ( ∠d1 + ∠d2 )
Now, with the allowable angle for the third pin, its diameter can be found.
d3 = (D sin ( ∠d3 / 2 ) ) / ( 1 + sin ( ∠d3 / 2 ) ) (third pin size)
Rounding each of three diameters to their nearest thousandths makes for easy selection from the gage pin set.
Noted as a result of using differing pin diameters is the coincident discrepancy between adjacent pin tangencies and the tangencies of pin and their included angle. However, as it is most likely that pin combinations will be created for gauging holes above the largest pin available, this disparity is quite negligible at these common scales. Additionally, the intentional subtraction of tolerance class value in the calculation to avoid line/line fits.
Example:
Gage Pin Set: .251” – .500” Class Z Minus ( .000100” )
Nominal Hole to be Gauged: .7510” ± .0035”
Largest Allowable Hole (High): .7545”
Gage Pins Required:
.350” ( .3499” )
.352” ( .3519” )
.349” ( .3489” )
Smallest Allowable Hole (Low): .7475”
Gage Pins Required:
.346” ( .3459” )
.348” ( .3479” )
.347” ( .3469” )
