Thread: Tape

Originally Posted by adjustme69

17 is pretty steep.

Nope, that's in the ballpark. Most razors I've measured have been between 15 and 19 degrees.

First three I grabbed at random:

5/8 Clauss Barber's Special (mint):
0.65" by 0.18".
ATAN(0.18/0.65) = 15.5 degrees

6/8 Henckels Friodur (new, honed once):
0.75" by 0.24".
ATAN(0.24/0.75) = 17.74

6/8 Robeson Shuredge (NOS, honed once):
0.73" by 0.23"
ATAN(0.23/0.73) = 17.48

8/8 Wade & Butcher "Celebrated Razor for Barber's Use":
0.97" by 0.31"
ATAN(0.31/0.97) = 17.72

For computing the angle, only the distance from the edge to the wear mark on the spine matters. If you mistakenly measure from the edge to back of spine then most of these wind up in the 12-15 degree range.

In a PM to Lynn I handwaved off any error from the lack of right angles. I apologize, it's been a long time since I've done this for real and I didn't want to calculate chords on the curve the the honing flat makes as you sweep it around the fin of the blade, and I knew that my original calculation was understating the actual angle a bit. I smacked my head this morning when I realized I was being stupid, I simply had to correctly use the blade width measurement as the tangent and half the spine thickness as the sin and everything becomes simple again. The correct calculation is 2 x ASIN((spine thickness/2)/blade width), which gives you the correct right angle needed for the calculation. The asin calculation gives you the angle for one half of the blade. Corrected figures below...

Originally Posted by mparker762

Nope, that's in the ballpark. Most razors I've measured have been between 15 and 19 degrees.

First three I grabbed at random:

5/8 Clauss Barber's Special (mint):
0.65" by 0.18".
ATAN(0.18/0.65) = 15.5 degrees

The correct value is 2 * asin(0.09/0.75) = 15.9 degrees

Originally Posted by mparker762

6/8 Henckels Friodur (new, honed once):
0.75" by 0.24".
ATAN(0.24/0.75) = 17.74

The correct value is 2 * asin(0.12/0.75) = 18.4 degrees

Originally Posted by mparker762

6/8 Robeson Shuredge (NOS, honed once):
0.73" by 0.23"
ATAN(0.23/0.73) = 17.48

The correct value is 2 * asin(0.115/0.73) = 18.1 degrees

Originally Posted by mparker762

8/8 Wade & Butcher "Celebrated Razor for Barber's Use":
0.97" by 0.31"
ATAN(0.31/0.97) = 17.72

The correct value is 2* asin(0.155/0.97) = 18.4 degrees

Originally Posted by mparker762

For computing the angle, only the distance from the edge to the wear mark on the spine matters. If you mistakenly measure from the edge to back of spine then most of these wind up in the 12-15 degree range.

I apologise for the error.

Yea, ummmmmm thats just what I was thinking . . . either that or "holly cow this is getting complicated."

Are you saying you're dividing the Blade thickness by the height measured from where the shoulder hits the hone and then hitting your tangent button??

Lynn

The calculation needs right angles for correctness.

You can think of the blade as an isosceles triangle described by the edge and two honing flats, with the two sides of the blade as equal length, and the base of the triangle being the distance across the flats. If we divide this into two triangles, one for each half of the blade, then we get a right angle between the line down the center of the blade and the line between the two honing flats. We know the distance between the honing flats - that's the thickness of the blade, so half of that gives us the length of the short end of this right triangle. The length of the long side of the triangle is the width of the blade from edge to honing flat. Given the length of two sides of a right triangle we can calculate the third side (using the pythagorean rule) or any angle in that triangle. Let me see if I can draw an ASCII picture:

Code:

     h  /|  (T)
    t  / |   h
   d  /  |   i
  i  /a  |   c
(W) -----|   k (T)
  i  \a  |   n
   d  \  |   e
    t  \ |   s
     h  \|   s

The width of the blade is W, the thickness at the spine is T. The angle that's easy to compute is "a", half the angle of the blade. Since we know W (the width of the blade along the side) and T (thickness of the blade), and the height of this half-blade triangle is half of that (T/2) we can calculate the angle "a" by the inverse sin function ASIN( (T/2) / W). But this only gives us half the angle of the edge, so we multiply that by 2 for the final answer.

The differences between the inverse tangent, inverse sin, and inverse cosine functions are that they calculate the angle "a" using different sides of the triangle. Inverse cosine would be used if you knew the blade width across the side of the blade and the blade width through the center of the blade, and inverse tangent would be used if you knew the blade thickness and the blade width through the center of the blade. In this case we know the thickness of the spine and the width along the side of the blade, so we need the inverse sin function. On a calculator these functions are represented by ASIN, ACOS, and ATAN, or sometimes as SIN-1, COS-1, and TAN-1 (where the "-1" is an exponent).

Edit: It just occurred to me that some of the confusion over the blade angle may be because I'm calculating the full blade angle, whereas knife guys generally talk about the half-angle, since this is what they set on their honing jig if they're using a Lansky, or eyeballing the angle between the knife and stone if they aren't. So if you want to compare these numbers to knife angles, don't multiply the ASIN value by 2....

Mparker,

Sorry, I'm confused:

I'm not completely sure what the definition of the angle of a blade is - is it 2a? From the way your ascii art is drawn, I guess it is.

Another question: if you rotate your picture 90 degrees clockwise, is that the equivalent of resting the blade on its back with the edge in the air? I'm confused about what you call the width of the blade - is it the horizontal distance from spine to point, or is it the hypotenuse of your picture?

If it's the hypotenuse, I get why you use sin. I'm just not sure the width is the hypotenuse - I'd have thought sqrt((t/2)^2 + w^2) should be the divisor in the arcsin calculation.

Or am I getting it all wrong?

James.

It seemed clear to me that what Michael called Width is the hypotenuse and that if you rotate 90 deg clockwise it would be like resting the blade on the spine. And that the blade (edge) angle is 2a.

Unfortunately I remember nothing about sinus, cosinus, etc. - what a shame

Cheers
Ivo