promhandicam":22qpha7f said:
dchenard":22qpha7f said:
Pardon me for being insistent, but the statement that one will remove more steel on a 50 degree bevel than on a 25 degree bevel just plain doesn't work. . . . Cheers, DC-C
Having read through the hole post, I think that, in your own ways both Derek and Denis are correct.
In the initial set up of converting a blade from straight to cambered then yes more steel has to be removed if you are grinding a 25 degree bevel as opposed to a 50 degree bevel. However once the basic shape has been produced, when it comes to sharpening, less steel will be removed if you add a secondary bevel onto a 25 degree primary bevel than if you are trying to hone the whole of a 50 degree face. Perhaps the following will illustrate what I'm trying to say:
The red lines indicate where the iron would be honed to get a 50 degree angle.
I hope this helps,
Steve
Time for a mea culpa
Looks indeed like I did not interpret Derek's statement properly. As exposed here by Steve, I do not have any argument.
Furthermore, and this is where I must admit being red-faced, in my example where one grinds a camber on the whole face of a 25 degree bevel iron versus the same on a 50 degree iron, I've maintained that the 25 degree iron will require more metal being removed in order to form the same camber. Thanks to Wiley, I must admit that I was wrong #-o
In fact, no matter what the bevel on the iron, grinding a camber on the whole bevel of an iron will result in
the same amount of metal removed, no matter what the bevel angle.
The point that I neglected was that, as the bevel angle goes down, the "effetive thickness" of metal removed to produce a camber goes down at the same time that its length goes up. Think "long and thin" versus "short and fat".
Definitions:
BLT: blade thickness
BET: "effective thickness" of metal removed
BL : bevel length
C : camber of the blade
(theta): bevel angle of the iron
The equations go like this:
BL = BLT / sin (theta)
BET = C sin (theta)
The section of metal removed at any point over the width of the blade is a parallelogram in section, which area can be computed as a rectangle of length BL and thickness BET. Multiplying the two, we get:
Section = BL * BET = BLT * C
Notice that the angle of the equation get cancelled out, or in other words, the bevel angle has no bearing on the amount of metal being removed, only blade thickness and amount of camber matter.
Thanks to Wiley for clarifying this
ccasion5:
DC