Forming a burr by no means guarantees a sharp knife. Leading edge strokes will sharpen without forming a burr.
Impossible! Even if a burr is formed, the actual sharpness of the finished edge does not rely on burr formation.
Raising a burr has nothing to do with sharpness.
If the edge isn't chipped or deformed, raising a burr only wastes metal.
A few leading edge strokes are all that's needed the vast majority of the time.
Why not use a hatchet? Batoning is merely a bogus excuse to test the toughness of a knife. Why risk injury in the middle of the woods for no reason? I've yet to see the head of a hatchet break into pieces and go flying.
If you must baton your D-2 Bog Dog, wood shouldn't be a problem...
If this was done for a purpose, why is it being resolved?
This isn't done by hand. It's merely a matter of inputing the design and letting the CNC do the work.
To compare the prices of each fairly, an additional $200.00 has to be added to the Hinderer XMs. While Sebenzas have a full...
Come on now. That's beyond a broad statement. That's the equivalent of stating your address is Earth. While Earth is correct, it's a bit vague.
This must be huge. I can't open this without my computer hanging.
Atomic decay? Are you forming carbides with elements heavier than lead...
How are they tied in?
Nor does it mean it will not. It depends on which elements are present.
It has nothing to do with 420 HC at all, actually.
The atomic mass of each element will determine carbide concentrations. Heavier elements are more stable than lighter elements. As atomic mass...
Then why is C able to form carbides at a percentage concentration of .46%?
What are these percentages based on?
It's more than simply the percentages of Cr and C which are present that determine the percentage of Cr carbide formation.
I suppose a guesstimation can possibly apply if only Fe...
That's an impossible statement to make with 100% certainty. How can you be sure no V is bonding with the C? Granted, there may not be enough to notice a difference, but that doesn't automatically equate to no vanadium carbides being formed.
What is the basis for the percentage of Cr...
420 HC may not be the highest carbide steel in the drawer, but it certainly does contain carbides; a small amount of V carbides, with the majority being Cr carbides.
These are the only comparisons I pay any attention to.
I'd like to see independent Charpy tests conducted on finished knives...
As a general rule of thumb, things that are tough are usally strong.
However, something can be strong, and lack toughness.
A two cubic foot block of ice is strong. It'll easily support quite a bit of weight without failing. Tap it with a hammer, and you'll see it isn't tough at all...
No, it's just that, even. Powder steel, despite its evenly dispersed carbides, doesn't automatically provide greater wear resistance than ingot steel.
Because based on that fact alone, on paper, they should; but that's not the case, as proved by the ease required to sharpen powder steels as...
1. Even carbide dispersion equates to even wear.
2. The fact that the carbides are dispersed evenly, and the possibility of clusters is eliminated, should result in greater wear resistance.
3. If more effort is needed to hog off steel, it has to be more wear resistant. Less effort would...