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Surface Area of Coarse vs Fine Stone

Posted by jasonstone20 
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Surface Area of Coarse vs Fine Stone
October 09, 2015 03:09PM
I saw this on a sharpening group on Facebook, an thought it was interesting, but maybe not so practical. Any ideas? Does the math here work? I am not sure how to go about verifying it.



Quote

"
Ken Schwartz‎ EDGE SNOBS™"


Yesterday at 3:55am · 

So the question came up - do you get more surface area using a coarse stone or a fine stone on an edge? I went through a theoretical discussion about this and wanted to share it here as well. Let me know your thoughts on what I wrotetongue sticking out smileyythagoras grit levels and surface area - edge theory.So let's lay the foundation for what I hope will be an interesting theoretical discussion. I'm going to start off without pictures and add some if it gets too confusing. Don't be shy asking questions )OK so what is the pythagorean theorem? If you know this skip ahead.A triangle has three sides. A right triangle has one 90 degree angle. To either side of the right angle are two sides which we will call side A and side B. Each side will have some length. Knowing the length of these two sides we can figure out the length of the side opposite the right angle from this using the Pythagorean theorem. This third side - side C - is called the hypotenuse.If we take the A side's length and multiply that length by itself we get what we call A SQUARED. So if the length is 3 inches long 3 squared is 9. Similarly if the B side is 4 inches long, B squared is 16.Now add these two squared amounts together so A squared plus B squared would be equal to 9+16 or 25.Now lets talk about the square root. That is a number that multiplied by itself equals the original number. So if the number is 25, the square root of 25 is 5, since 5 times 5 is 25. We will call this operation SQRT'So knowing that a triangle has two sides of 3 and 4 inches long the hypotenuse is 5 inches long.So A squared plus B squared equals C squared.The SQRT of any number squared is the original numberconfused smileyQRT(number*number) = numberSo SQRT(A squared + B squared) = COK let me know if this is obvious or confusing or else the rest of this discussion will be hopeless.This promises to be an interesting discussion ...Let's also state how to compute surface area. Knowing the length and width of a rectangular surface, we multiply these two numbers to get the area. If the length and width are the same it is a square and the area is l squared or w squared - the same amount.So if something is 4x4 the area is 16.Bear with me laying out the basics here ...I'll go on with this. Just drop in if something is unclear.Let's make Pythagoras relevant.Let's make a theoretical 'scratch' on a surface. A simplified scratch. We will say that the scratch will have the shape of a v cross-section running parallel to the sides of a rectangular surface area from the front to the back of this surface. Think of a freshly plowed field with rows of furrows or a piece of corrugated aluminum.So let's define an area of 4x4. This could be 4 inches by 4 inches or 4 millimeters by 4 millimeters or 4 whatever by 4 whatever. Let's call it 4 units by 4 units. So the total surface area is 16 square units.Once you get the basics out of the way it gets a bit more interesting. Am I at least sounding coherent and understandable so far? You're probably finding it boring so far.OK so if we have a perfectly flat 4x4 surface it is 16 square units. Now let's say that we have a scratch whose width is one unit wide AND one unit deep. And these scratches are running continuously over the 4 unit length.We can look at this v shaped scratch as having an identical left side and right side. What is the length of the scratch in cross section? Well if it is 1 unit deep and one unit long, then the length of the two sides of the scratch can be computed using the Pythagorean theorem. For the moment let's make it 2 units wide and one unit deep for simplicity. So it would take two scratches to cover the entire surface.So now the two sides of the triangle are each 1 unit long so the left side of the scratch is Sqrt( 1 squared + 1 squared) or sqrt of 2 or 1.414214... or roughly 1.41. So now the length of the scratch is 1.41*2 or 2.83 and since there are 2 scratches the total length of the X side has become 5.66 units, whereas originally as a flat surface it was 4 units long. Now the length is still 4 so the surface area is 22.63 units vs the original 16 units.It becomes obvious that the length of the furrows (y axis of the area) is always going to be 4 so we can simply look at the length change increase or decrease of the corrugated up and down edge (referred to as the x axis of the area).Are you with me so far?Now let's make the scratches half as deep and half as long.sqrt(1/2 squared + 1/2 squared) X2 is the per scratch length but now there are twice as many scratches so 4 scratches.so Sqrt (1/4+1/4) X2X4 is sqrt(1/2)*8 or 0.707*8 or 5.66 (actually 5.656854)Odd coincidence? we get the same result for a scratch pattern twice as fine!Now let's go 10 times finer.sqrt (0.1 squared+ 0.1 squared)*2*20sqrt (0.01*2) *40sqrt(0.02) *400.1414 * 405.66 AGAIN!So without belaboring the point, if we go even finer or coarser we keep getting the same answer. If you wish to go much finer be careful to have enough decimal points so you don't get round off errors.Finer or coarser scratches DON'T change the Surface area! This is a false concept that the fineness or coarseness of the scratch pattern is related to a change in surface area.But there's more. Now that we have got this basic and to me quite interesting concept laid out, let's play some more with the numbers.....We will change the RATIO of the height and width of the scratchessmile emoticonLets make the scratch depth 1 and the scratch width also 1 Now each side of the scratch is sqrt(1 squared + 1/2 squared) or sqrt(1.25) or 1.12 so the length is 1.12*2*4 or 8.94Let's make the scratch depth and width 10x finer:sqrt(0.1 squared + 0.05 squared)*2*40sqrt(0.01+0.0025)*80sqrt(0.0125)*800.112*808.94 AGAIN!So now SO LONG AS THE ASPECT RATIO is constant the surface area remains the same!It is the ASPECT RATIO that is critical - not the scratch size.I'll leave it as an exercise to see that:1. As the scratches get deeper relative to the width the surface area goes up. So a scratch 10 or even 100 times deeper than it is wide has a MUCH greater surface area. If you think about this it (eventually) becomes intuitively obvious.2. As the width of the scratch goes up for the same depth, the change in surface area gets smaller and smaller, ultimately never exceeding the surface area of a perfectly flat surface - ie zero depth. When you plug a zero depth, then you see that sqrt(0 squared + width squared) = equals the width, so the math holds up without requiring exceptions.Given that especially most synthetic abrasives are spheres, cubes etc with relatively the same width and lengths of the particles we now know that these axially symmetric particles don't change the surface area as the particles get finer or coarser.Intuitively you can think of it as each scratch is smaller but there are more of them and these two factors cancel out.For natural stone particles, they tend to be less symmetric, sometimes resembling corn flakes so the scratch widths tend to be greater than the scratch depths, so you are even less likely to increase surface area with these natural stone particles AND that as the particles get smaller with similar aspect ratios that this would not account for any increased surface areas. A great deal of anecdotal evidence suggests that natural stone finishes hold up longer or have a more gradual breakdown, so having even less increases in surface area than a symmetric synthetic particle would produce is hardly the explanation for the observed behavior of a natural stone finish.I find this whole thing interesting to contemplate and a good starting point to further speculate the nature of edge behavior as a function of grit size and that it brings out a characteristic of the aspect ratio of scratches as having some unknown or at least not discussed effects if they can be controlled more precisely.---Ken

"I am still discussing issues of steels and performance at this stage." -- Cliff Stamp, May his memory be a blessing
"Life is GOOD", -- Stefan_Wolf, May His Memory Be A Blessing
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Edited 1 time(s). Last edit at 10/09/2015 04:46PM by CliffStamp.
Re: Surface Area of Coarse vs Fine Stone
October 09, 2015 05:08PM
Jason, please use the quote feature when you are bringing in outside sources to clearly show what words are yours and what you are quoting. As well, please provide a link if possible to the source.

--

For someone using "theoretical" constantly, it would be nice if they understood what it actually meant as Ken doesn't appear to. If I dropped a ball I would not ask you "in theory" what would happen as it isn't theoretical any more because we all have practical experience in gravity. In the same way you don't say "in theory" what happens when an abrasive scratches a surface as this is well known and can be looked at in the literature.

His argument is wrong on many levels, both in regards to the level of approximation he makes and the practical reality which is ignores (his approximation leaves out actual significant effects).

On a first level, this is what he appears to argue :

Lets imagine the edge of a knife is made out of a number of triangles, for simplicity lets say they are equilateral (all equal side lengths) then :

-triangle edge length (l) = length of the knife edge (L) / number of triangles (n)

l = L/n

Total contact length is going up and down each triangle so 2*n*l = 2*n*l = 2*n*L/n=2*L

And notice it doesn't depend on the size of the triangles at all. However this ignores one very large practical effect. As the triangles get very small the material doesn't actually go up/down the triangles any more, it starts making contact with the triangle tops only. In this case the contact length (for very large n) is :

Lim (n-> infinity) of # of triangles * l = Lim (n->infinity) n*L/n=L * Lim (n-> infinity) (n/n) = L

Back to the edge length L.

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In short, as a first approximation, as edges get very coarse and have large triangular teeth on the edge then the edge length will approach 2L. As the edges get very fine the edge length will approach L.

There are other issues which can't be ignored such as the fact that when the teeth get large enough they have teeth as well and thus the real end bounds are 2^s*L and L, where s is a scale factor which depends on the coarseness of the grit.

--

And just think, you never thought calculus had any practical applications. You now just applied it to sharpening knives.
Re: Surface Area of Coarse vs Fine Stone
October 09, 2015 07:37PM
Cliff-
Sorry, I copied it from Facebook on my phone, and usually, at least with my browser on my old phone, it would put the quote in for me automatically, I will preview it next time.

As far as what Ken was talking about, I didn't follow what he was trying to describe, and it didn't make much sense how he went about it, I was hoping someone understood it better than me, but I liked the question, it's interesting, but I am not sure how practical.

"I am still discussing issues of steels and performance at this stage." -- Cliff Stamp, May his memory be a blessing
"Life is GOOD", -- Stefan_Wolf, May His Memory Be A Blessing
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Re: Surface Area of Coarse vs Fine Stone
October 09, 2015 09:03PM
I couldn't make head nor tails of what he was saying. Glad Cliff figured out enough to explain it.
Re: Surface Area of Coarse vs Fine Stone
October 09, 2015 10:00PM
Cliff-
I think he is talking about the triangles as the grit, ploughing into what is being abraded away and that for surface area of a coarse or fine a grit is the same, but not only does the conclusion Ken comes to I am not sure of, but it is assuming a number of things and disregarding others.
I thought this might be an interesting topic, but it's a little complex, and I haven't practiced my calculus in a very long time, and when you have two different bodies of geometric shape interacting, you need to use math to work on it or even descibe it.

"I am still discussing issues of steels and performance at this stage." -- Cliff Stamp, May his memory be a blessing
"Life is GOOD", -- Stefan_Wolf, May His Memory Be A Blessing
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Re: Surface Area of Coarse vs Fine Stone
October 09, 2015 11:28PM
Jason,

In several places he mentions edges :

"A great deal of anecdotal evidence suggests that natural stone finishes hold up longer or have a more gradual breakdown, so having even less increases in surface area than a symmetric synthetic particle would produce is hardly the explanation for the observed behavior of a natural stone finish."

and

"I find this whole thing interesting to contemplate and a good starting point to further speculate the nature of edge behavior as a function of grit size ..."

It would help if he clarified by writing what he is asserting and what he is trying to conclude from it.

--

It reminds me of when Mike Stewart from Bark River has used physics in the past to "prove" batoning is abusive when the knife isn't perpendicular and other nonsensical claims. The problem with math and physics is that it only leads you to truth when you do it correctly.
Re: Surface Area of Coarse vs Fine Stone
October 11, 2015 04:43AM
Cliff,
This is a problem with reading text online, especially when the person who wrote it isn't interacting, and what they did write is long and unclear, because even the two quotes you used, I still hear that he is referring to the grit particle size and shape.

Normally, I would just ask the person what they are talking about and why, but there is a group of people who don't exactly like questions, and I don't want to be censored, I like participating with the knife community.

"I am still discussing issues of steels and performance at this stage." -- Cliff Stamp, May his memory be a blessing
"Life is GOOD", -- Stefan_Wolf, May His Memory Be A Blessing
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