Thread: The Secret to The Secret to Cambering Bevel Up Plane Blades

Hi Paul

Brent posted this in a discussion on the UK forum a few years ago. His results were disputed.

I think that I would prefer to accept the maths of our own Bob Loss, who is professor of Applied Physics at Curtin University (Perth, Australia).

I asked Bob to analyse the data before I published the article. Bob wrote:

Hi Derek,

I have analysed the situation regarding the additional metal you need to remove for higher angle bevels when you do plane blade edge rounding.

Because primary bevels typically range between 20 and 45 degrees it turns out that the additional metal that needs to be removed can be closely approximated by the ratio of the tangent of the bevel angle across that range.

Here is the setup I have used for a 2" blade. By way of example I considered removing 1/16" at the sides so this can be approximated by a triangle along the sedge of 1" long by 1/16" high.
Now looking at the sides of the blade, the only variable here is the height of the small triangle shown at the tip - this is determined by the tangent of the angle x the 1/16" shown
The volume of metal removed is thus a long thin tetragon. OK - it's not a pure tetragon but rather a long slightly curved tetragon.
However, for such small angles, the ratio of the relative volumes of the curved tetragon for say a 25° bevel and a 30° bevel will be the same ratio as that for the ratio of the relative volumes of the straight tetragon.

Below I plot the "angle of primary bevel" used as a function of the mount of metal that needs to be removed relative to the 25 degree bevel. By way of example, a 30 degree bevel requires 25% (or 1.25 times) more metal to be removed relative to the 25 degree bevel. A 45 degree bevel requires 2.15 times more metal to be removed and a 15 degree bevel requires only about half the metal to be removed.

It does not matter if you want to remove 1/16" or 1/32 or 1/64th, its all relative. The curve applies to all 1/16" removals, or all the 1/32" removals etc.

If you want to compare a 1/16" round removal with a 1/32" removal you have to take these into account separately - but that is intuitive so there is no need to argue that . Its surprising though, for the same bevel angle a 1/32 removal requires 4 times more metal be removed than a 1/64" removal and a 1/16" requires 4 times more be removed than a 1/32. I think that would confuse people so maybe you don't need to go into that

I hope this is understandable and useful.

Bob

This extract is taken from my article.

Regards from Perth

Derek

hmmm. Why did he use a 'surface area model', instead of spherical geometry?

Originally Posted by Clinton1

hmmm. Why did he use a 'surface area model', instead of spherical geometry?

Great idea. I may be standing on my head, but this is what I am thinking ...

If you draw two (parallel) horizontal lines, and mark out a section of the bottom line AB, then for any point C on the top line the triangle ABC has the same area. They are all equivalent to an equilateral triangle built on AB which is then stretched off to the left or right, maintaining the same height.

The same applies to a rectangle which is 'pushed over' to make a parallelogram of the same height, because you can break the rectangle down into 2 triangles that keep the same area.

So I'm thinking the volume of a plane-iron type shape with a radiused edge inclined at 50 degrees (and a rear edge also inclined at 50 degrees) will be the same as the volume of the same profile running vertically at 90 degrees. You just need to knock off the extra rear-edge leaning bit that isn't part of the actual plane-iron.

I think the rest will be easier on paper.

Paul

Assuming people will draw a circular arc segment on the back of the blade, this gives a ellipse for the projection of the blade bellow the sole with the minor axis of the ellipse reduced buy the sin of the angle of the bedding angle.

Lets take the examples of
1. Bevel down
50mm at bedded at 45 degrees and a 200 mm arc.

2. Bevel up
50mm at bedded at 12 degrees and an unknown arc radius x.

3. let p1 be the projection of the blade past the sole of the plane where the edge of the arc coincident with the plane of the sole of 1.
4. let p2 be the projection of the blade past the sole of the plane where the edge of the arc coincident with the plane of the sole of 2.

5. Let p1=p2

which give us
6. equation 6.gif
p1=sin(pi/4)*(200-((200^2-(50/2)^2)^(1/2)))
solving 6 give us a projection of ~1.11mm
7.equation 7.gif
p2=sin(pi/15)*(x-((x^2)-(50/2)^2)^(1/2)
NOTEs:
pi/4=45 degrees in radians
pi/15 = 12 degrees in radians
200 = radius on example 1
50 = blade width


combining 6 and 7 with 5 gives us

8.equation 8.gif
sin(pi/4)*(200-((200^2-(50/2)^2)^(1/2)))=sin(pi/15)*(x-((x^2)-(50/2)^2)^(1/2)
(copy and paste into Wolfram|Alpha: Computational Knowledge Engine and play with the values)
after a little algebra give us x=~61.263mm


notes: since this is a projection of a circle onto a plane the value of the eccentricity of the projected ellipse will be slightly different for each bedding angle for the same projection.

Funny how things can look clear as mud until ... later ... duhhhh

I'm going to be presumptuous and attempt to explain Josh's post ... and so I can put some questions ...

OK so assume two intersecting planes ... the sole of the plane and the face of the plane-iron.

(Please excuse the sketchup beginner)

-bevup1.JPG

The plane-iron is cambered, so the cutting edge is assumed to be an arc of a circle.
Where the arc meets the edges of the iron on each side {A and B} is exactly the point (ok line) where the theoretically zero-width surface passes through the sole.
The lowest point of the cutting edge - on the mid-line of the plane-iron - is at C.
The plane-iron rests on a bed, so crosses the sole at an angle ... call it theta θ.

-bevup1a.JPG

Ignoring the sole of the plane for the moment, the arc of the cutting edge is part of a circle centred at E.
Every point on the arc (cutting edge) is an equal distance R from E.
R is the length of the line CE.
The point F is the mid-point of AB, and the point where CE crosses the sole of the plane.
The length of CF is the maximum amount of the plane-iron protruding beneath the sole.

-bevup1c.JPG

Hopefully that is a correct illustration of the setup.

So this:

decodes as ...

200 = R

50 = AB and 50/2 = AF

Taking the triangle AEF, a right-hand triangle: R^2 = AF^2 + EF^2

so EF = sqrt( R^2 - AF^2 )

and so the length CF of protruding plane-iron = (R - EF) = as Josh has it above in the brackets.

So far so good.


The next bit I see what's happening but don't yet know why.

Consider projecting the curve of the cutting edge up onto the plane that the sole lies in.

-bevup1b.JPG

Vertically above C is D ... and the triangle CDF includes the bed angle ... 45 degrees (= pi/4 radians) in the first example

So the line CF projects up to the line DF in the plane that the sole lies in, and the length DF is given by Josh's first formula.

- -
OK ... having posted that and re-read it, I was wondering if Josh's "sin" shouldn't have been "cos" ... but now I realise that he was calculating the vertical depth CD ... which I hadn't clicked on to.

This is good because it ties into something I had a question about with Monsieur Beach's web-site.

He takes as an example a 0.010" (0.25mm) shaving. I wasn't clear whether that measurement should be CD, or a line projected from C at 90 degrees to the plane-iron up to the sole. A shaving comes in horizontally, but travels up the iron(*), curls, maybe lifts ahead of the cutting edge ... it's all a bit vague as a model.

He says "The following sketchup models are based on the camber needed to take a 0.01" thick shaving - 0.041" on a 2" blade used bevel up in a low angle (12 degree bedding angle) plane."

I assume the 41 thou refers to the 'depth' of the camber CF ...

but assuming CD = 10 thou and theta=12 degrees ... I get CF = 48 thou ...

or using the other concept of shaving thickness ... 47 thou.

I might try to find the previous discussions Derek said has taken place on a UK forum.

Cheers,
Paul

(*) Oooga. Oooga. Sloppy thinking.
In a bevel-down plane the shaving travels up the inclined plane in the model.
In a bevel-up plane the shaving travels up the bevel (so a significantly greater angle than theta).

You diagrams are spot on.


if you put
equation10thou.gif
0.01=sin(pi/15)*(x-((x^2)-(2/2)^2)^(1/2)

into wolframalpha you get 10.4196 inches radius for x

and 0.0480975 inches for the camber.

near enough to the same as Brent Beach calculator

when he is talking about shaving thickness in practice I think he is taking into account that some proportion of the cutting edge is above the sole?

either that or its a typo since he has it as 48tho in his table:
"Typical cambers - for various bench planes."

Could you move point 'c' to the left and right of the centre line?

Originally Posted by Clinton1

Could you move point 'c' to the left and right of the centre line?

With my sharpening ... sure! Might get three 'C's ...

But seriously sure ... Derek and others get to their camber on diamond stones (or whatever) by making more strokes on the outer areas than the inner area ... symmetrically ... so logically you could also do otherwise.

Paul

I think I can tie where Josh got us to to my post regarding triangles and rectangles being mushed around.

-bevup2.JPG

Imagine in the upper block there are two diagrams superimposed. One a rectangular prism (P), and the other a rectangular prism with part of the front face swiped off it vertically (PC).

As I said regarding the two-dimensional analogy, if P were transformed back at an angle (phi), maintaining the same height, the volume is left unchanged. You can easily verify that if you cut off the new triangular-prism-shaped section that appears at the back and glue it onto the now-sloped front you get back the original prism P.

And the same applies to the (vertically) cambered prism PC.

So consequently, for any given Depth of Camber d, the volume removed to put that depth of camber on any angle of bevel phi is always the same.

But - and this is where it ties back to Josh - for different bed angles theta, you will need a different depth of camber if you want to maintain the same width of shaving.
Alternatively you can put on the same depth of camber at a steeper bed angle and accept a narrower shaving.

It took me quite a while to get the illustration ... I wanted to squash the top prism, but settled for making a second angled one. I put the above in bold for my benefit because I've just twigged that the bed angle is coming into play and it seems to me at the moment that the bevel angle is not ... but the brain is now basically numb, so I'm gunna let that sink in for a while.

I will point out that Josh's formula ... only involves the following terms ...

Bed angle, Radius of curvature of the camber, and Width of the blade.

Cheers,
Paul

I am starting to hate woodwork... I'm off to the bar.

:P

I mean shift point C to the left and right and create a tangent ogive to the left and right near the edges of the blade. (This should give a projected ellipse on the cutting edge, only on the extremities of the blade, which will be a small part of the horizontal axis of the ellipse... e.g. as per the drawn part of an ellipse when the most vertical section of a trammel is used). Once having the cutting edge drawn, re-work the model (along the long axis of the blade) for a constant bevel angle in order to re-work the geometry to the top of the blade. I think the devil is in the Gaussian curves that you will find. As it is, that blade will cut an elliptical scallop, won't it( )? rather than a flat with slightly scalloped edges. If that's as clear as mud, then I'm sorry, but think of it as .... as an ellipse is formed by plane projecting through a cone, the split the cone.

Originally Posted by Clinton1

I mean shift point C to the left and right and create a tangent ogive to the left and right near the edges of the blade.

OK .. I think you mean why not have a basically flat across blade that is just clipped off at the very corners. ?

People do do this. I can't remember if I made the point previously - I thought about it - that the scenario of superimposing a circular camber on to the blade is only one possible method for avoiding track marks ... one that can be analysed geometrically.

But some people will hold the plane-blade edge-vertical and use a stone to grind some material off from the corner of the leading edges.

Also maybe the camber scenario might be the only one to present the bevel to the wood at a constant angle across the whole width ... ? ... not sure about that.

Plus I guess that a cambered blade that can be lowered or raised is less an "all-or-nothing" proposition compared to a mostly flat cutting edge.

Might need some practical feedback from people on how they prepare their smoothing irons.

Cheers,
Paul