

is there some formula to determine the relief angle necessary for a given feed rate to successfully turn a given diameter of round stock?

Probably but 99.999% of home ground tool users eyeball it. 7 degrees is about right. In the case of screw threads having large helix angles, the 7 degrees applies with respect to the helix angle.
A multiple lead screw thread or worm having a 30 degree helix angle has a 23 degree relief on one side and a negative 23 degree on the other. Since large helix angles are usually cut with the tool profile normal to the pitchline helix angle, shops that speciaize in this kind of work have tool holders tilt the tool by one of several means to suit.
Single point tools used in tangential heads on gear hobbers when nicely ground are works of art.
[This message has been edited by Forrest Addy (edited 07152004).]

Just what Forrest said!
But if you must know
Cot helix angle= (PI* dia)/lead (feed rate)
That gives a number that is relative to the axis of feed. IOW, if you have a 10" dia part that you feed at .005"/rev, the helix angle is 89.99 deg.
If I remember, the best explanation that I ever saw was the setup directions for a Landis thread grinder.
JR

I am the one who usually advocates heavy feed at lower rotative speeds.
Even under conditions of very aggressive feed rates for turning in the lathe, you will not run into clearance problems at from 5 to 7 degrees.
You want greater clearance for softer materials, not because of feed rate but because of the springback of the softer materials that makes them want to rub behind the cutting edge.

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 lessused 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 mistakeprone 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 moreeasilyunderstood 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 greaterthanHelixAngle 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 straightforward:
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 longlead screwthread to drag the flank on a toolbit with 5 to 7 degrees of Side Clearance.
John


Geez John, you sound just like my brotherthe aerospace engineer (John too). Everytime I ask him for help in calculating stresses, weldments or colinear supports, he gives me 5 pages of the formula and expects me to understand it.
I sure am glad he gives me the answer. I'm not sure I could find it on the paper.

FWIW, CCMT inserts have a 7 degree relief. Kennametal's CPMT have 11 (or is it 12?) degree relief.
In my experience, both work fine. Kennametal's relief seems to be a slight bit freer cutting. But is possibly offset by the extra cost of the Kennametal inserts.
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