Thread: blade width vs, thickness question.

Or in the case of a 1 to 4 thickness to height ratio you can just divide 60 by 4 and get 15. That's math for the liberal arts major.

Originally Posted by Hart

Here the example doesn't match our blade geometry but imagine an isosceles triangle with c? being the short side (spine) and gamma () our bevel angle.



Two sides and the included angle given (SAS)
Here the lengths of sides a, b and the angle \gamma between these sides are known. The third side can be determined from the law of cosines:

c = \sqrt{a^2+b^2-2ab\cos\gamma}.

So for a 7/8" razor (.875") to get 16°

\sqrt{0.875^2+0.875^2-2x0.875x0.870xcos\16}

which is:
the square root of (blade hight² + blade hight² - 2 x blade hight x blade hight x the cos of the angle you want (16°)

equals 0.2435529268 or about 1/4"

Cut and paste from Google and calculated on my phone's scientific calculator; I could enter the whole equation in one line and get the answer.

Originally Posted by JDM61

Or in the case of a 1 to 4 thickness to height ratio you can just divide 60 by 4 and get 15. That's math for the liberal arts major.

OK, where did you get 60 from?

My point was for any given blade height you wanted to make, find the width needed for the correct honing angle.

With the Isosceles triangle as you starting point, all three angles are 60 degrees, Divide by the long ratio. 60 divided by 4 equals 15. Not exact science as the angle of each side would actually be slightly more that 8 for a total included angle of like 16.1 on the 1 to 4 ratio but fairly close. Gives you a tiny bit of wiggle room, Like i said. math for liberal arts majors kind of like the "3 pennies tall" sharpening angle for full width chefs knives.

Originally Posted by Hart

OK, where did you get 60 from?

My point was for any given blade height you wanted to make, find the width needed for the correct honing angle.

A 1 to 4 ratio is actually 14.36* inclusive.

I/2 thickness/hyp > A-sin x2

Originally Posted by JDM61

With the Isosceles triangle as you starting point, all three angles are 60 degrees, Divide by the long ratio. 60 divided by 4 equals 15. Not exact science as the angle of each side would actually be slightly more that 8 for a total included angle of like 16.1 on the 1 to 4 ratio but fairly close. Gives you a tiny bit of wiggle room, Like i said. math for liberal arts majors kind of like the "3 pennies tall" sharpening angle for full width chefs knives.

Equilateral triangle is what you mean.

Originally Posted by JDM61

With the Isosceles triangle as you starting point, all three angles are 60 degrees, Divide by the long ratio. 60 divided by 4 equals 15.

The only reason that I can see for this coming close is that a radian is ~ 57.3 degrees

Originally Posted by bluesman7

The only reason that I can see for this coming close is that a radian is ~ 57.3 degrees

You lost me as my only extensive exposure radians was milliradians when trying to get a rough idea of where an artillery shell might fall and even then, we couldn't say the entire word. Just mils. But what happens if you use that "formula" on any of the common 8th widths? 6/8 with a 3/16th spine width should give you pretty much the same result, right? Slightly "scientific" than trying to decide which coins to use and how many when figuring out the angle to use to sharpen your new Shun.