Thread: My bevel angle calculations

So I was bored last night, so I came up with some formulas for the bevel angle given the thickness of the spine, of the width of the razor, and of the thickness of the tape if it is used.

The first picture is of the derivation of the formulas. To clarify terminology, I used:

x = thickness of the spine
y = width of the razor measured from the edge of the spine to the bevel.
t = thickness of a piece of tape. If 2 or more pieces of tape are used, then just use the combined thickness of the pieces of tape for t.
alpha = angle of the bevel with no tape
beta = additional angle to account for the tape (to get the entire angle we need alpha + beta, hence (2) )

(1) is of the bevel angle using no tape.
(2) is the bevel angle accounting for the thickness of the tape.

The second picture I proved that when t=0, then (2) essentially reduces to (1) since the addition angle (called Beta, the funky B) becomes 0.

I know there have been postings about this in the past, but I thought I would just show how I arrived at my formulas.

Enjoy!

Ok to a dyslexic... that looks like alphabet soup...

Excellent stuff, Dave. Saw this thread: http://straightrazorpalace.com/advan...pers-tape.html ? Feel free to amend and enhance.

Regards,
Robin

Originally Posted by StraightRazorDave

So I was bored last night, so I came up with some formulas for the bevel angle given the thickness of the spine, of the width of the razor, and of the thickness of the tape if it is used.

It's been a few decades since I needed to use that stuff, and it's apparent that I did not retain it all that well. Anyway, I'd say about a quarter of the people I communicate with about honing ask for an explanation on taping. I'll be sure to provide a link to your post, because I'm sure it will clearly explain everything!

so is tape good or bad?

why not make it simple?

Let:

x = spine width
w= width of razor
n = number of tape layers
t = thickness of tape
Θ = total bevel angle °


total bevel angle:

Θ = 2{arctan[(.5x + nt)/w]}

this accounts for no tape, one or several layers of tape, etc. (when layers of tape = 0 then nt is 0)

For accuracy, in your diagram you have the spine along the x-axis and the blade along the y-axis. This is fine as long as you position the area of spine wear along the x axis such that the spine wear on the left has coordinates (-.5x, 0) and the spine wear on the right has coordinates (.5x, 0), or make some similar compensation if you choose to not have the spine wear directly on the x axis. The width w would then be accurately measured as the distance from the origin to whatever y value the very tip of the blade occupies, assuming a non-warped blade.

Originally Posted by Ben325e

why not make it simple?

Let:

x = spine width
w= width of razor
n = number of tape layers
t = thickness of tape
Θ = total bevel angle


total bevel angle:

Θ = 2{arctan[(.5x + nt)/w]}

this accounts for no tape, one or several layers of tape, etc. (when layers of tape = 0 then nt is 0)

For accuracy, in your diagram you have the spine along the x-axis and the blade along the y-axis. This is fine as long as you position the area of spine wear along the x axis such that the spine wear on the left has coordinates (-.5x, 0) and the spine wear on the right has coordinates (.5x, 0), or make some similar compensation if you choose to not have the spine wear directly on the x axis. The width w would then be accurately measured as the distance from the origin to whatever y value the very tip of the blade occupies, assuming a non-warped blade.

Ah, I like it. Thanks. I didn't measure the width of the blade directly, but used a diagonal measurement from the bevel to the edge of the spine. I figured this would be easier to measure physically then the actualy width, since you would need to measure the cross-section essentially, which would be impossible if you were measuring the bevel angle in the middle on the blade. Does that make sense? In my diagram, the actual width of the razor is z, but if you actually measure your razor it would be easier to measure y, the hypotenuse.

I like your idea of using an integral multiple of the number of tape layers. I just figured you could just substitute the total thickness in for t, but your way is more eligant.

Honing_Angles_Matrix.xls?

Originally Posted by StraightRazorDave

Ah, I like it. Thanks. I didn't measure the width of the blade directly, but used a diagonal measurement from the bevel to the edge of the spine. I figured this would be easier to measure physically then the actualy width, since you would need to measure the cross-section essentially, which would be impossible if you were measuring the bevel angle in the middle on the blade. Does that make sense? In my diagram, the actual width of the razor is z, but if you actually measure your razor it would be easier to measure y, the hypotenuse.

I like your idea of using an integral multiple of the number of tape layers. I just figured you could just substitute the total thickness in for t, but your way is more eligant.

I can get on board with that - the measurement would be easier. So:

Let:

x = spine width
y= distance from spine/hone contact point to tip of bevel
n = number of tape layers
t = thickness of tape
Θ = total bevel angle °


total bevel angle:

Θ = 2{arcsin[(.5x + nt)/y]}

Originally Posted by Ben325e

I can get on board with that - the measurement would be easier. So:

Let:

x = spine width
y= distance from spine/hone contact point to tip of bevel
n = number of tape layers
t = thickness of tape
Θ = total bevel angle °


total bevel angle:

Θ = 2{arcsin[(.5x + nt)/y]}

I like it! But your new y would have to be measured again if you added tape, since it would be larger if you added tape. But it makes complete sense to me.