Hi Everyone,

Very often, I encounter people who mistakenly think that a guided-rod set-up cannot sharpen a perfect V-edge. This subject of constant knife angles and pivoting-rod sharpeners always seems to come up, and there are tons of misconceptions. So I wrote this explanation for another forum, but I thought it would be of interest here. Eventhough, I'm sure that at some level or other, all WEPS fans/users already understand the material here, either intuitively, or mathematically.

Below, we will consider two cases: a knife with a circular belly, and a knife with a perfect straight edge. We will show that for these two cases, a guided-rod sharpener that pivots at a point (ie: a spherical rod-end or ball-and-socket joint), will sharpen a perfect V-edge. I won't discuss the circular case for very long; most misconceptions are for when the knife edge is a straight line. However, for edges which are not circular nor straight lines, a pivoting-rod sharpener may not sharpen a perfect V-edge.

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There are two cases where a pivoting-rod sharpener can make an exact and uniform angle along the knife edge.

(1) The first is case of a cone. For the cone, the knife belly is circular, and it's center is lined up through the pivot of the sharpener. In other words, if you take the pivot and project it perpendicularly onto the plane of the knife, it would land on top of the circle's center. Then, in this case, the sharpener will grind a bevel that is a right-circular cone (as opposed to an oblique cone). The angle of the bevel is the angle between the slope of the cone and it's base, and the apex of the cone is at the pivot of the sharpener.
https://en.wikipedia.org/wiki/Cone_(geometry)

(2) However, there is a second case where the pivoting-rod sharpener will also make an exact and uniform angle. If the knife edge is a straight line (has zero belly), then the pivoting-rod sharpener will make a perfect V-edge. This is a very commonly misunderstood fact about dihedral angles, and has caused endless argument. I'd rather not get into endless arguments, so I will present a mathematical proof and also an example illustration. If any of you still disagree after that, that's fine, but I'm unlikely to continue with the discussion for very long after that. I will respond if you find a valid error in the mathematical proof, or if you have a very carefully and well done demonstration. To challenge the proof below, you will need 3-dimensional geometry at the high-school level (ie: theorems, axioms, about planes and lines, etc.), or if you have had college level linear-algebra and/or vector-calculus, that is more than enough.

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First I'll present a quick illustration. For now, please pretend that the guide-rod is a line (ie: infinitely thin) and that the sharpening stone is also infinitely thin.

In the illustration, I tried to use line-thickness to do some perspective hinting. You can think of the thick black triangle as being the closest object to the viewer. Notice the two green angles; those are the dihedral angle of the knife bevel. Notice that the guide rod positions (red) are all contained inside the plane of the knife bevel. Therefore there is no problem in sharpening a perfect V-edge.



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Second, here is the technical proof.

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Axiom1: Three non-colinear points determine a plane.
Comment: Given three points, no two of which are identical, and which are non-colinear (ie: do not lie in a line), then there is exists a unique plane through those three points.

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Axiom2: Two distinct points determine a line.
Comment: Given two points which are not identical, then there exists a unique line through those two points.

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Axiom3: Two distinct points in a plane determine a line contained in the plane.
Comment: If I pick two different points in the plane, those two points determine a line. That line lies inside the plane.

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Theorem1: A line and a point, which is not on the line, determine a plane. This plane contains the line and the point.
Proof:
Let L be a line.
Let C be a point that does not lie on L.

(Part1): Prove there is a plane P that contains the line L and the point C.
Pick any two distinct points A and B on the line L. The three points A, B, and C are distinct (P cannot be A or B because P is not on the line L). The three points A,B,C determine a unique plane P (Axiom1). The plane P contains the points A,B,C, and the line L (Axiom 1, Axiom3).


(Part2): Prove that the plane P is unique.
Proof by contradiction: Suppose there is a second plane Q, not equal to P, that contains the line L and the point C.
If we can show that Q contains three non-colinear points which also lie in P, then Q and P would have to be the same plane.
Let A and B be distinct points that lie on L, as described above, so that the points A,B,C determine P.

Let Q be a plane different from P, which is also determined by the line L and the point C.
Since Q is determined by L and C, it must contain the point C as we proved in (Part1).
Since Q is determined by L and C, it must contain the line L as we proved in (Part1).
But L contains points A and B.
So Q contains L, and L contains A and B.
Therefore, Q contains A and B.
Therefore, the plane Q contains the points A, B, and C. So Q must be the same plane as P (Axiom1).

Therefore, there is a unique plane P that contains a line L and a point C not on the line.

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Now we will apply Theorem1 to knife sharpening. Suppose our knife has a single edge which is a straight line. For example, maybe we are sharpening a straight razor, or one edge of a wood-chisel. I'll consider two cases, because the first case is much easier to understand, and then I'll look at a second case which is a modification of the first case.

Let L be the line representing the straight edge of the knife.
Let P be the plane which contains one bevel of the knife. Since the knife has a V-edge and has a straight edge, one of it's bevels is perfectly flat (ie: planar).

Simple Case:
Let's set up our sharpener so that the pivot C is in the plane P.
The plane P contains the line L and the point C (Theorem1).

For now, we will pretend our guide-rod is a line (ie: infinitely thin) and that our sharpening stone is also infinitely thin. We'll fix this later, after we understand this case. Let S be the plane which represents the surface of the sharpening stone. This means that S contains the surface of the stone as well as the guide rod itself.

Setup the guide-rod such that:
(1) The guide rod goes through the point C.
(2) The sharpening stone lies on the knife edge.

Because we placed the stone onto the knife, the stone surface, S, contains the knife edge L.

However, the plane S also contains the guide rod, and therefore, plane S contains the point C.

Therefore, the plane S contains the knife edge L, and the pivot points C.
Recall that earlier, that our knife bevel is a plane P which is determined also by the line L and the point C.

By Theorem1, the plane S and the plane P are the same plane (uniqueness proven in Part2 of Theorem1).

Notice that it did not matter where we put the sharpening stone onto the knife edge: the plane of the sharpening stone will be the same as the plane of the knife bevel.

Therefore, the sharpening stone will remain in perfect contact with the bevel of the knife and sharpen a perfect V-edge.

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In the above proof, we assumed that the guide-rod and the sharpening-stone were infinitely thin. The above proof can be modified to account for the thickness of the stone and the radius of the guide-rod. However, it is rather laborious to go through that proof. So instead of a full-blown proof, I will informally describe what happens. Basically, you can put the guide-rod in a plane which is parallel to the plane P, but offset by the thickness of the stone and the radius of the guide rod. Once you do this, the stone can always contact the knife bevel perfectly. If that makes sense to you intuitively, then no point in going through he informal argument below. But if it does not, the below is a brief sketch of what happens.

Basically, we still have the knife-edge as line L and the knife-bevel as plane P.
Let t be the thickness of the sharpening stone.
Let r be the radius of the guide-rod.

Then the distance between the central-axis of the guide rod and the stone surface is d=t+r.

Next, consider a plane Q which is parallel to P, but is a distance d away from the knife edge.

We set-up our sharpener so that the center-axis of the guide rod is always in plane Q.
Claim: We can rotate the sharpening stone around the rod axis until the stone's surface lies in the plane P.
Informal Proof: We have two parallel planes P and Q. In Q we have the center-axis of the guide rod. Let's call the center-axis of the guide rod line J.

Now let me use some 3d geometry from high school: The plane Q contains line J. Let U be a plane that is perpendicular to plane Q and which contains line J. (We can do this, because two distinct planes intersect at a line, and we can adjust the dihedral angle between the two planes until they are perpendicular.) Now the plane U will also intersect plane P at a line K. Since U is perpendicular to plane Q, it is also perpendicular to plane P (because P and Q are parallel). So the line K is parallel to line J and exactly a distance d from it. Therefore, we can rotate the surface of the sharpening stone until it contains the line K (because the stone surface is also d away from line J, the center axis of the guide rod).



I won't prove this next bit, but if you can visualize the above, then it is obvious that not only does the stone-surface contain the line K, but because of the way we set it up, the stone surface must also be parallel to plane P. This is because line K is actually the perpendicular projection of the guide-rod onto the plane P, and the guide rod, J, is parallel to P.

Sorry if the above bit is a bit complicated to prove. It is fairly easy to understand intuitively. Basically, if we keep the guide-rod in a plane Q, then the stone can sharpen a plane P which parallel to Q, and which is a distance d away.

The conclusion is, so long as the guide rod is in plane Q, we can get the sharpening stone to perfectly sharpen a bevel in plane P. So all we need to do, is to place the rod-pivot in plane Q, and then sharpen normally.

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Discussion:

So we now know that the following two cases can be handled perfectly by a pivoting-rod system that uses a spherical rod-end (ie: pivots about a single point):

(1) A circular knife belly where the center of the circle is lined-up with the pivot center. That is, if we were to do a perpendicular projection of the pivot onto the plane of the knife, it would land on the center of the circle.

(2) A knife with a straight edge that is a line.

In general, though, a knife belly is not perfectly circular, or the pivot is not lined up with the center, etc. So there will be some variation. But the straight part of the knife edge will be fine, and you can put the pivot at a point which is aligned with the center of a best-fit circle to the knife belly. So in practice, it is still pretty good.

Sincerely,
--Lagrangian

"What grit sharpens the mind?"--Zen Sharpening Koan

P.S. I'm wondering if there are other geometries that work, but I haven't figured out the mathematics yet.

This is great! Thank you for posting such an elegant explanation. I spent a year on week on another forum trying to explain the principles you've so clearly laid out. I even constructed a 3d model for people to download and manipulate in an eViewer. Still, people struggled with the concept. After that, I went over to the college here and offered a reward for the kind of proof you just presented! Thanks again for posting it.

Awesome, incredibly informative. Although the theory portion is a bit over my head Thanks!

Excellently! You mastered planar geometry perfectly. However, knifes with a direct edge is a rarity how to sustain a sharpening corner on an edge bend?

Hi Anthony,

This is marvellous! It is a proof from the ground up and deceptively simple. That is what we call elegance. I had never imagined it could be so simple.

I have a question about the proof of the case in which the stone is infinitely thin, but does have a surface.

The proof depends on being able to setup the system in a particular way:



This proof thus supposes that it is possible to setup the guide-rod system such that (1) and (2) are satisfied.

You wrote that your proof applies to any guided-rod sharpener that pivots at a point. My question is whether this is sufficient. Isn’t it also necessary that the stone is able to rotate on the rod? In other words: would it be possible to setup a guide-rod system (for any L and C) in such a way that (1) and (2) are satisfied if the stone were not able to rotate on the rod?

The reason for asking this is that I think this is the reason so many people intuitively have trouble in understanding that the angle of the edge along a straight line does not change.

wickededge wrote:


I'll admit, I was one of those. Had to figure it out for myself...





Great writeup!!!

Anthony, this reads like a clear and concise doctoral dissertation mate. What an excellent piece of work. You can work in my lab any day Anthony!! Super!

Leo

wow, thanks for the time you put into this! I will admit that it is slightly over my head... I understood it for the most part though I don't mean to be retarded, but what has been the argument for ages that you referred to in the beginning of your OP? "Very often, I encounter people who mistakenly think that a guided-rod set-up cannot sharpen a perfect V-edge. This subject of constant knife angles and pivoting-rod sharpeners always seems to come up, and there are tons of misconceptions." Why would people argue that guided systems cannot do this and what is their argument?

I did notice that you said in the end of your post that "In general, though, a knife belly is not perfectly circular, or the pivot is not lined up with the center, etc. So there will be some variation. But the straight part of the knife edge will be fine, and you can put the pivot at a point which is aligned with the center of a best-fit circle to the knife belly. So in practice, it is still pretty good."

I do agree with this... there will be variation when you have a knife edge that is completely straight vs a knife edge that is in a perfect circle to the pivot point. the latter will have a perfectly consistant angle on the edge whereas a knife that is straight edged and long (let's say 2' long) the angle will change and become more narrow the further out you get (both directions) from the center, correct?

NickVas wrote:


Hi NickVas,

If the "bend" or belly of the knife is a circular arc, and if we align the sharpening pivot with the center of the circle, then the sharpening angle is constant along the knife edge.

If the knife edge has a circular belly (ie: bend), then the knife bevel is a section of a cone. Simply place the pivot at the apex of the cone, and then the guide rod is able to trace out the cone.

Hope that made sense; if not just ask me again.

Sincerely,
--Lagrangian

mark76 wrote:


Hi Mark76,

Ah, very good... Originally I didn't want to get too much into that part of the proof, because it's slightly complicated. But you are absolutely correct. So let me prove that you can always set-up the sharpening stone such that (1) and (2) are satisifed. To do so, I'll need some extra stuff, and I won't prove all of it, but it should be very convincing, I think.

If you have additional comments or questions, please let me know.
The extra part of the proof is below.

Sincerely,
--Lagrangian

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Axiom4: Given two lines, L1 and L2, exactly one of the following is true:
(1) L1 and L2 do not intersect
(2) L1 and L2 intersect at a single point
(3) L1 and L2 are the same line
Comment: Generally two lines intersect at a point, or don't intersect at all, or else they're the same line.

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Theorem2: Two distinct lines that intersect determine a unique plane
Proof:
Let the two distinct lines be L1 and L2.
Since L1 and L2 are distinct, they are not the same line. So their intersection must be a point (Axiom4).

We now choose three points that determine a unique plane that contains both lines.

Let A be the intersection of L1 and L2.
Let B be any point on the line L1 that is not the point A.
Let C be any point on the line L2 that is not the point A.
The points A,B,C are not colinear because the line L1 and L2 are distinct.
So A,B,C determines a unique plane P (Axiom1).
The plane P contains points A and B which determine L1. So plane P contains line L1. (Axiom3)
The plane P contains points A and C which determine L2. So plane P contains line L2. (Axiom3)

So the two lines L1 and L2 determine a plane P that contains both lines. Any plane that contains L1 and L2 will also contain the points A,B,C and therefore be the same as plane P (uniqueness by Axiom1).

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Theorem3: Suppose P1 and P2 are two planes that both contain a line G. Then by rotating plane P1 around the line G, we can rotate plane P1 into plane P2.
Proof:
I won't prove this here, because to so, I would need to add all sorts of theorems and axioms about rotations. That would be a lot of extra material. Instead, I will appeal to your intuition with a picture. Consider two planes P1 and P2 which both contain a line G (drawn below). We can imagine the line G as a "hinge" and we can swing P1 like a door until it rotates into P2. The curved arc with an arrow represents a rotation.


Please rest assured that this theorem is true and can be easily proven, although it would require some extra geometry and/or linear algebra.

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Next we will prove that we can setup the guide-rod such that:
(1) The guide rod goes through the point C.
(2) The sharpening stone lies on the knife edge.
Let our guide-rod be line G, which we will determine by the pivot-point C, and any point on the knife edge line L.
Line G intersects the knife edge, line L, at a single point.
Therefore line G and line L determine a plane Q that contains line G and line L (Theorem2).
Notice that the plane Q contains line G, and the stone surface S also contains line G.
Therefore we can always rotate the sharpening stone (plane S) around the guide-rod (line G) until the stone is in plane Q (Theorem3).

After doing this, the stone's plane S, will be the same as plane Q.
Because plane Q contains the knife edge line L, we know our stone's surface, plane S, contains the knife edge, line L. Therefore, the stone lies on the knife edge. So we have satisfied condition (1).

Our rotation around the guide-rod, line G, did not move the guide-rod, so the guide rod still goes through point C. So we have satisfied condition (2).

Thus we've shown it is possible to set-up the guide-rod such that (1) and (2) are satisfied.