

I've seen several charts showing specs for machine tapers, but the ones for R8 have never been too detailed. Anybody know the exact dimensions or taper angle?

16 deg 57 min 28 sec
16.95776330 deg
or 3.5000 inches/foot
Is that exact enough?
JR

" 16 deg 57 min 28 sec
16.95776330 deg
or 3.5000 inches/foot "
I have some trouble with this formulation of the R8 Taper. Let's assume the 3.5000"/foot is the original definition.
Indeed, 3.5/12 = sin(16.95776330). But the underlying construction implied by this calculation is not a right triangle, so using the sine function seems inappropriate to me. A better mental construction for a taper is a pair of identical, mirrored right triangles. Various trig functions can then be used appropriately.
If one is trying to figure out what angle from the centerline defines the R8 taper based on the 3.5/12 taper, I think the correct approximation is 2 * arctan(1.75/12) or 16.594289... degress.
At least, that's what I think today. If someone sees a flaw in my thinking, I'll appreciate correction.

Actualy the calculation should be arctan(3.5/12)= the angle. I get the angle to be 16.26020470831 degrees.
Ray

Even though it's probably risky to differ with someone whose name includes "carbine", I must. Actually, I'm also trying to judge whether you're in fact just pulling my leg.
A shank taper, such as Morse,B&S,and R8 is a truncated isosceles triangle (2 equal angles). It's geometry is certain not represented by a single right triangle. Using trig functions, whether sine or tangent or arcsin or arctan makes no sense in calculating sides and angles of a single isoceles triangle (implied by the notation arctan(3.5/12)). Thus, one must conceptually cut the isosceles triangle into two mirror image right triangles in order to be using any of these trig functions or to solve the problem correctly.
The "arctan(3.5/12)" solution by definition implies a right triangle with a 3.5" short side and 12" long side. Thus, the hypotenuse is 12.5". I'm unconvinced that this sort of geometry correctly represents a taper shank.

I'm pretty sure you are correct, Bob. I suspected that discrepancy myself, but being poor at math I just indicated a collet I had and then tweaked things until I got full contact with the shank I was turning in the mill spindle using some spotting compound.
The correct numbers wouldn't have done me any good anyway, though if the shank had turned out to be too hard to cut and I ended up sending it out to be ground it might have been a different story.

I turns out my Machinery's Handbook 26 ed. has a nice table on p. 684 for tapers per foot and corresponding angles that puts the 3.5/12 at 16 35 39 (deg min sec). The next page has the equations. The one we've been addressing is cited as angle = 2 arctan (T/24) where T is the taper in inches.
You'd think this was all an abstract wild hair for me. To the contrary, just before seeing this thread I was trying to calculate how much advance of the compound on my lathe (set at 60 degrees) would move the tool into a B&S taper to move the taper how much closer to an arbitrary shoulder.
I'm trying my hand at turning an arbor for my Tom Senior (with a B&S #9 and an unusal dog design). I could buy an arbor but would still have to deal with the dog on the end of the spindle. So I thought I'd see if my skills were up to making something.
Here I've fit the business end into the spindle taper and I'm getting runout of 0.001 to 0.002. I'm thinking I'll tighten up the drawbar and run the spindle for awhile and remeasure.
Here's a shot of the corresponding arbor dog that fits up against a shoulder on the arbor (the reason for the calculations I was exploring).
I haven't made up my mind how to fix the arbor dog to the arbor yet. Counterbored bolts? Pins that I could peen and then turn smooth?
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