JRIowa and Company --
There's a nit in the arithmetic used to illustrate JRIowa's algorithm, which I'll rephrase slightly:
Cotangent (Helix Angle) = (Pi x Diameter) / (Axial Feed per Revolution)
Cotangent (Helix Angle) = (Pi x 10 inch) / (0.005 inch)
Cotangent (Helix Angle) = 31.4159 inch / 0.005 inch
Cotangent (Helix Angle) = 6283.18
Since the Cotangent is the reciprocal of the Tangent, the above can be restated
Tangent (Helix Angle) = (1 / 6283.18)
(Helix Angle) = Arctangent (1 / 6283.18)
(Helix Angle) = Arctangent (0.000159155+)
(Helix Angle) = 0.009+ degree.
[Sound of climbing onto soap box]
I firmly believe that the less-used trigonometric functions -- Secant, Cosecant, Cotangent, Versine, Haversine, and so forth -- should today be considered only historical curiosities, and that it is the duty of those teaching applied math today to define and dismiss those functions.
A hundred years ago the use of these functions served a practical purpose, avoiding the relatively mistake-prone arithmetic operation of dividing . . . instead of "dividing by the Sine / Cosine / Tangent of the Angle" we said "multiply by the Cosecant / Secant / Cotangent of the Angle".
Today scientific calculators and spreadsheet programs are almost ubiquitous, and these tools make division easier and less subject to error than multiplication by the reciprocal.
Until the calculator makers and the spreadsheet programmers add Secant, Cosecant, and Cotangent keys or commands to their products, using those functions impairs the practicality of the algorithm.
The arithmetic error inspiring my rant probably would not have happened had the algorthm been expressed as this more-easily-understood way:
Tangent (Helix Angle) = [(Axial Feed per Revolution) / (Pi x Workpiece Diameter)]
[Sound of jumping down from soap box]
Some textbooks say that the Side Clearance Angle on a toolbit should be greater than the Helix Angle on the workpiece by 3 to 5 degrees.
Let's say that we take the midpoint of this range, 4 degrees, as an idealized greater-than-Helix-Angle clearance and take the midpoint of JimK's toolbit Side Clearance Angle range as an idealized toolbit angle. This would mean that we would want the workpiece Helix Angle to be (6 degrees - 4 degrees) = 2 degrees.
How much Axial Feed would be necessary to generate a 2 degree Helix Angle? The algorithm is pretty straight-forward:
Axial Feed per Revolution = Tangent (Helix Angle) x Workpiece Diameter x Pi.
Doing the arithmetic for a 1 inch workpiece, we find that Axial Feed per Revolution = Tangent (2 degrees) x 1 inch x 3.1416 = 0.109+ inch.
For a 6 inch diameter workpiece, Tangent (2 degrees) x 6 inches = 0.658 inch.
Bottom line, you'd almost have to be cutting an oddball long-lead screwthread to drag the flank on a toolbit with 5 to 7 degrees of Side Clearance.
John