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  1. #1
    Senior Member khaos's Avatar
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    Quote Originally Posted by pjrage View Post
    I agree with you that a razor edge can only get so sharp. That's where I was going with my light saber comment

    IMHO, polishing the bevel is what makes the razor sharp in the first place. The second the two sides of the bevel meet, EVERYTHING you do from that point forward could be considered polishing the bevel. They can meet at VERY low grits. I think, though, that you are defining "polishing the bevel" as starting the second that you have made some portion of the edge as sharp as it's going to be (0.5 micron assuming that number is correct).

    So, I'll jump on board with that argument then, that around the 4k level I will assume (I have no proof) that you are starting to get into the realm of "some areas of the edge are as sharp as they will get."

    So, think about it like this. At the 4k level, you have two sides of the bevel, both with scratches in them, meeting up to form an edge. As far as I can tell, the scratches in these two edges of the bevel cause the two sides to not meet up as evenly where they meet (the edge). Since each side of the bevel is basically varying in width from having scratches, the very edge is varying in width also. The complete edge would look jaggedy in width (on a microscopic level) if viewed straight on. That is to say, the edge may be 0.5 microns (or whatever the sharpest they can be is) at some points, and, say 2 microns at others. Where the two valleys of the scratches meet perfectly, it might be the perfect 0.5 microns in width, but where two peaks of the scratches meet, it is wider, say 2 microns (just making up that number). Does this make sense? I wish I could draw a picture to describe what I'm saying.

    Anyway, as you take these scratches out further and further (polish), the edge becomes more and more consistent in width. Just for the sake of argument I'm going to pull some numbers out of the air, assuming that your 0.5 micron width is the sharpest a razor can get. Anyway, let's say that at the 4k level, the edge varies between 0.5 microns and 2 microns in width. Some points where the scratches on both bevel sides meet perfectly are as sharp as possible (0.5 micron), but where the scratches do not meet perfectly, you have varying edge widths as wide as 2 microns. Now, you move on from the 4k to a finer stone. Now you are removing the 4k scratches with 8k ones which aren't as deep. Now the edge varies in width from the same perfect 0.5 micron to, let's say, 1.5 micron at the spots where the scratches "meet up" the worst. Then you move on to say a 12k stone, and now you are varying between, say, 1 micron and 0.5 micron. Then you go to a 16k and get to 0.8 micron and 0.5 micron. Then you get to say, a 30k hone and now you much closer to a consistent 0.5 microns because the scratches are so thin that you really can be very consistent.

    Now, these numbers are completely made up to try to illustrate what I'm thinking, but does that make sense?

    If you subscsribe to this making sense, then I would argue that as you polish the edge, you are making it a more consistent width that is closer to as sharp as it can possibly get. When the edge is more consistent and also a smaller width, it will cut better. If it cuts better, I define it as sharper. Therefore, polishing makes it sharper.

    Or I could be completely wrong, but it makes sense in my head anyway.
    This is slightly wrong. The triangular profile where the edge meets will always be a consistant width (if we are assuming it is triangular aka perfect honing, not round tip). However, if you look at it side on, there will be a lateral profile (at the microscopic level) like a very complicated sine wave! JOY!

    Now, as I am learning at college, one can explain this with... FOURIER SERIES! If we assume the scratches on one side are regular, with period x, and on the other side, with slightly different period 1.000001x, but out of phase, then the sum will give you a very different sine function, which will be the edge profile. Now, the scratches on a large grit will not be consistant, they will have underlying frequencies and require fourier transforms of there own, (so imagine doing this with something like 4cosx+3sinx+19cos2x+2sin2x+7cos3x...) and as you can imagine, it would be highly implausible that two sine functions that look like that will add up to give you a straight line. However, as the grit becomes smaller, your scratches are smaller (lower amplitude of variation) and more consistant (simpler transforms, fewer underlying frequencies) and so matching the periods will come closer to a straight line (due to simpler transforms) and will have less deviation along the edge (due to lower amplitude). YAY! I bet none of that made sense to anyone, but at least my thousands of dollars spent on education finally turned up a little bit useful...

    Quote Originally Posted by jendeindustries View Post
    [COLOR=Red]

    I totally agree (and actually follow) your thinking here! I had this very epiphany last week (without reading this thread), and posted it on my blog. To take your thought one step further, if you spend too much time on the final stone, you will again end up with an uneven edge because you have established those high and low points on the blade.

    How many strokes does it take? We'll have to ask the tootsie pop owl (he'll say 3)
    In theory this is right, but would only be true if every stroke were in exactly the same place, which would be statistically implausible (the politically/scientifically correct way of saying impossible). One stroke you may form peaks/valleys along a(x)=ysin(zx), the next stroke would maybe be b(x)=ysin(zx)+.000000001, the next would be c(x)=ysin(zx)+.0000000004, etc. Eventually, if every stroke were very slightly different (which it will be) you will reach an edge equal to the root mean square of the untranslated sine function, which mathematically speaking, is a perfectly straight line equal to Fsum(x)=y*sqrt(2). (Variables: a,b,c, etc are the terms of the fourier series, Fsum(x), which is the overall edge profile. x is the position along the length of the edge, y is the grit of the hone, and z is the fourier transform of a hone's particular profile frequency)
    Think of this like electricity. The wall socket is constantly alternating at Fsum(x)=110sin(30x/pi) (60 hertz period, 110V amplitude). This means you are in theory only recieving electricity some of the time. Sometimes you are at peak voltage of 110V, but 60 times every second that number dips negative. Now it would be a bitch to calculate your energy bill if they had to integrate the function of your power consumption. But they don't have too. Because we use so much power, and not just at 2 second increments, they just charge us the root mean square of 110sin(60) times the amount of power we draw. (the point of this analogy was to demonstrate the root mean square idea.) So yes, you could reestablish valleys, but the world is not perfect, so you can't really polish too much, unless you are overhoning.
    Also, this begs the question, won't sharpening 500000 laps on a 4k do the same thing? No, because of the high grit, it "gouges" rather than gently scrapes. The last pass will always leave a slightly more distinct profile. This profile on a 30k will be a **** ton smoother than on a 4k.

    Honemeisters please correct me as experience trumps explanation, but from what I've read, Glen agrees. It can only get so sharp, but you can polish it to be smooth which translates interestingly enough into literally a comfortable smooth shave.
    Last edited by khaos; 06-16-2009 at 08:36 PM.

  2. #2
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    Quote Originally Posted by khaos View Post
    This is slightly wrong. The triangular profile where the edge meets will always be a consistant width (if we are assuming it is triangular aka perfect honing, not round tip). However, if you look at it side on, there will be a lateral profile (at the microscopic level) like a very complicated sine wave! JOY!

    Now, as I am learning at college, one can explain this with... FOURIER SERIES! If we assume the scratches on one side are regular, with period x, and on the other side, with slightly different period 1.000001x, but out of phase, then the sum will give you a very different sine function, which will be the edge profile. Now, the scratches on a large grit will not be consistant, they will have underlying frequencies and require fourier transforms of there own, (so imagine doing this with something like 4cosx+3sinx+19cos2x+2sin2x+7cos3x...) and as you can imagine, it would be highly implausible that two sine functions that look like that will add up to give you a straight line. However, as the grit becomes smaller, your scratches are smaller (lower amplitude of variation) and more consistant (simpler transforms, fewer underlying frequencies) and so matching the periods will come closer to a straight line (due to simpler transforms) and will have less deviation along the edge (due to lower amplitude). YAY! I bet none of that made sense to anyone, but at least my thousands of dollars spent on education finally turned up a little bit useful...



    In theory this is right, but would only be true if every stroke were in exactly the same place, which would be statistically implausible (the politically/scientifically correct way of saying impossible). One stroke you may form peaks/valleys along a(x)=ysin(zx), the next stroke would maybe be b(x)=ysin(zx)+.000000001, the next would be c(x)=ysin(zx)+.0000000004, etc. Eventually, if every stroke were very slightly different (which it will be) you will reach an edge equal to the root mean square of the untranslated sine function, which mathematically speaking, is a perfectly straight line equal to Fsum(x)=y*sqrt(2). (Variables: a,b,c, etc are the terms of the fourier series, Fsum(x), which is the overall edge profile. x is the position along the length of the edge, y is the grit of the hone, and z is the fourier transform of a hone's particular profile frequency)
    Think of this like electricity. The wall socket is constantly alternating at Fsum(x)=110sin(30x/pi) (60 hertz period, 110V amplitude). This means you are in theory only recieving electricity some of the time. Sometimes you are at peak voltage of 110V, but 60 times every second that number dips negative. Now it would be a bitch to calculate your energy bill if they had to integrate the function of your power consumption. But they don't have too. Because we use so much power, and not just at 2 second increments, they just charge us the root mean square of 110sin(60) times the amount of power we draw. (the point of this analogy was to demonstrate the root mean square idea.) So yes, you could reestablish valleys, but the world is not perfect, so you can't really polish too much, unless you are overhoning.
    Also, this begs the question, won't sharpening 500000 laps on a 4k do the same thing? No, because of the high grit, it "gouges" rather than gently scrapes. The last pass will always leave a slightly more distinct profile. This profile on a 30k will be a **** ton smoother than on a 4k.

    Honemeisters please correct me as experience trumps explanation, but from what I've read, Glen agrees. It can only get so sharp, but you can polish it to be smooth which translates interestingly enough into literally a comfortable smooth shave.
    Khaos,

    I actually completely understand what you are saying (but I didn't break out the calculator to double check the math)
    It makes complete sense, though, and is in line with the same thinking.

    I tried to look at the edge straight on with my microscope, but couldn't get a good focus to see if the edge was even, or how even it is. Someone else may be able to...

  3. #3
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    Geez I hate math---LOL!

  4. #4
    Senior Member blabbermouth hi_bud_gl's Avatar
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    Default This is the blade i think you were talking about it.

    Jaguar really great razor. will get scary sharp
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  5. #5
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    Quote Originally Posted by hi_bud_gl View Post
    Jaguar really great razor. will get scary sharp
    Wow, that is a gorgeous razor. Even better that it gets nice and sharp!

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    Sham that is a beautiful razor! I've never heard of that brand before. German I guess?

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    I have a few razors which are crazy sharp, but not all shave equally good.

    1. FWE Special, this blade makes hair disappear no noise their just gone when doing hht, and shaves great.
    2. Joseph Rodgers another scary sharp edge and shaves great.

    3rd Johan Engstrom frameback wedge. Probably the sharpest edge I've seen
    Sadly one of the worst shavers imho, feels like a cheese grader on my face.

    I am not sure how this applies to this topic other then to me sharpness is only worthwhile when coupled with smoothness or shaveability.

    Cheers
    D

  9. #8
    Senior Member khaos's Avatar
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    Quote Originally Posted by Gunner777 View Post
    German I guess?
    I have a French razor just like that except Spanish point, and I have seen a few TI's exactly like that, with several different points. My guess then, though Sham has to say, that it is French usign Sheffield steel (all the TI's were stamped with something like "finest Sheffield Silver Steel". Plus the blade marking is French. Am I right?
    PS: I still think this is more evidence that there was some connection between all the companies... a common supplier, them buying eachother out, a common grinder, etc...

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