Blade Thinness?

[UnknownVT]

Was there ever a traditional ratio or formula to work out the blade thinness/thickness in relation to the length for flexibility or stiffness?

For most general purpose usage I have preference for thinner blades, as they usually offer less resistance when cutting through things. But obviously if the blade is too thin they would flex too much. For example those very thin razor blades and the majority of true fishing/filleting knives.

Out of curiosity I took a look at some of what I considered the more common thin bladed knives - not surprisingly (for me ) the SAK and the Opinel (#8)
Note: this is not a quantative "analysis" - just a simple "look-see" - as I don't have a micrometer to do accurate measurements -

Typical thinner bladed knives -

Blade thinness -

Details:

Not surprisingly the thinnest blade of this bunch was the ($4) Victorinox/Forschner 4" Paring knife - that's probably one of the reasons (other than cheapness) that this knife is so popular - because it is very sharp and cuts through things well - as a good paring knife should do. For a cheapo knife it actually has a good distal taper. Again not surpringly it has quite a lot of flex in the blade.

The next most flexible by feel was the Wenger SAK (85mm closed - their standard size) if one squints carefully at the photo one can just make out it is thinner than the smaller 84mm Victorinox SAK (look carefully at the tangs near the handle).

The Opinel #8 is somewhat flexible in the ballpark of the 84mm and 91mm Victorinox SAKs - perhaps the SAKs are a bit stiffer - but not enough for me to make any big deal over it - this is the degree of flex I find "tolerable"/acceptable for an EDC general purpose blade -
although before this I never really complained about the Vic Paring knife's flexibility - so maybe there is some "bias" here.

The 94mm Victorinx SAK - as in the Soldier and the "Pioneer" Alox series blades are noticably thicker and also noticably stiffer - where I would put the Opinel #8 and the 84mm and 91mm Vic SAKs at the borderline between flexible and stiff - the 94mm Alox SAKs would be classified by me as stiff bladed - as are the Kershaw Leek which is noticably thinner than the Vapor.

Comments please... especially if there is some ratio or formula for working out the relationship between blade length and thickness.

--
Vincent

http://UnknownVincent.cjb.net/
http://UnknownVT.cjb.net/

 

[Chiro75]

I don't think there is a formula, at least not one that I know of. I have several knives with blades under 2" long and they're cut from 1/4" thick stock! With some exceptions, I always like thinner knives better, too.

 

[Roadrunner]

This thread has made me determined to get an Opinel. About the formula/ration, I have no idea. But I do like thin flat ground blades.

 

[Dirk]

I have not heard of a formula for thickness to length. I have seen a calculator to determine edge angle for sharpening.

Thickness, length and width are genarally determined by function of the blade. In many cases it is only determined by a look the company or maker is going for. Function being a secondary issue.

In my view, small folders 4" or less should be concerned with cutting. So, I don't want them to be too thick. At the same time, I don;t want them to be as thin a s a razor.

 

[roshi]

There is no universal formula. The stiffness of course depends not only on the cross section and length of the blade but the material.

The form of the knife blade should be ditated by its function but we humans also tend to like to add personal taste to the mix.

I prefer thinner wider blades but with some degree of "pointiness" since I'm after a knife for slicing materials, removing splinters, and last but not least important self defense which requires thrusting efficiency.

Others prefer a heavy weight blade that can pry and chop.

Fortunately for all of us, there are more than enough knives out there to satisfy all needs and most wants.

Thanks for the great pics!

 

[Morgoth412]

Vincent,
As you know, I EDC a SAK, and agree that the thin flatground blades are the best for EDCing. Most well known slipjoint companies and SAK's tend to offer the perfect ratio between thinness and flexibility. It seems even with a course edge, the thin blades of slipjoints will slice better then most tacticals, and with a razor edge their isn't much of a comparison.

Thin flatground tacticals like the Calypso Jr. and Military/Paramilitary are a good option if you want a one-hander that comes closer to a slipjoint in cutting ability.
-Kevin

 

[Gossman Knives]

I've never felt that blade thickness really mattered when slicing and cutting. I do most of my cutting during deer season. I've used thin flat ground blades, hollow ground blades and thicker convex ground blades and didn't notice any difference even though I prefer the convex grind. I filleted the sinew off the backstrap of a deer with a 3/16" thick convex ground blade with the ease of a flat ground fillet blade. IMO, I feel the degree of the edge has more to do with slicing and cutting then the thickness of the blade. When you cut or slice, the material you are cutting pushes out of the way as the blade goes down. The thicker the blade the farther the material is pushed. Think of how a boat floats. A small bass boat will float the same as an ocean liner even though their weight and size are extremely different. One just displaces more water then the other.
Scott

 

[Dirk]

Scott - Very true, but also think of resistance. Passing something with a relatively broad frontal area through air, water or whetever will give more resistance than something of similar construction, only with a narrower frontal area. To displace more material will see an increase in resistance. However, when you start speaking about cutting something like meat, you start introducing other things into that equation that can make it a bit more confusing.

 

[Gossman Knives]

Scott - Very true, but also think of resistance. Passing something with a relatively broad frontal area through air, water or whetever will give more resistance than something of similar construction, only with a narrower frontal area. To displace more material will see an increase in resistance. However, when you start speaking about cutting something like meat, you start introducing other things into that equation that can make it a bit more confusing.

Dirk,
Very good point. the stiffer the material the more resistance you will encounter cutting down. Cardboard is a good example. My regular job, I cut alot of cardboard. While my main tool is a case cutter, I have taken various knives to work to "test" them against cardboard. I do find that a thinner blade will cut "easier" not necessarily better then a thicker blade. Again I stress edge degree as an important factor.
Scott

 

[UnknownVT]

Again I stress edge degree as an important factor.

Edge angle is very important as is blade thickness/thinness - lots of things contribute to a great blade - that's the difference between one that's great and one that's just so-so.

Just to give a silly - absurdum example - if one took a 1" thick piece of (premimum) steel and ground in for a final edge of pretty acute 20deg per side that would give a 1.37" wide blade with a zero bevel (ie: no discontinuities since the edge bevel is the blade face) - but a spine of 1 inch!

There would be no doubt that this "blade" could score paper well - it probably would also cut hanging paper well too - since the final edge is very "sharp" at 20deg per side.

But it probably would be pretty terrible at cutting through things - mainly because it is just way too thick.

--
Vincent

http://UnknownVincent.cjb.net/
http://UnknownVT.cjb.net/

 

[brownshoe]

The best paper cutter I ever saw was a power cutter in a print shop. It used a chisel ground blade of about 1/4" thickness. It had an angle of about 15 degrees. Of course it push-cut, but it could go throgh about 2 or 3" of paper at a time.

 

[Cliff Stamp]

Vincent the bend induced in a blade can be calculated from :

(1/R) = M_B / (E I_A)

R = radius of bend

M_B = torque moment (total torque)

E = Youngs Modulus (property of the material)

I_A = Integral (x^2 dA) where x is in the direction of the torque and dA is perpendicular

For a rectangle I_A = (1/12) W H^3 where H is in the direction of the applied torque

Related is the deflection at a point in a bar when a force is applied at the end :

y(L) = - F L^3 / (3 * E * I_A)

So for example if you substitute in I_A for a bar you get :

y ~ (L/H)^3

or

y ~ L^3 / W

which would allow you to determine the dimensions needed for a desired flex over a specific length.

As for blade geometry not making a difference when cutting flesh, the force on a knife is made up of three parts :

1) force against the edge (influenced by sharpness)
2) wedging (influenced by cross section)
3) drag (influenced by cross section and surface texture)

Unless a material is capable of exerting a wedging action, and flesh isn't, pretty much all you will note is the first one and thus just sharpness matters.

So for example cutting light paper requires a very sharp blade only, cutting a thick turnip however requires a slim cross section as the turnip will wedge the blade heavily.

-Cliff

 

[UnknownVT]

the bend induced in a blade can be calculated from :
(1/R) = M_B / (E I_A)
<snip>which would allow you to determine the dimensions needed for a desired flex over a specific length.

Wonderful! - Many thanks Cliff.

Is that formula for calculating the bend/flex in a uniform thickness of material -
how does the blade grind affect the flex calculation?

Wouldn't the blade cross-section also contribute fairly significantly to the flex -
eg: a hollow grind?
or as another example the classic fuller (commonly known as the "blood groove")?

As for blade geometry not making a difference when cutting flesh, the force on a knife is made up of three parts :
1) force against the edge (influenced by sharpness)
2) wedging (influenced by cross section)
3) drag (influenced by cross section and surface texture)
Unless a material is capable of exerting a wedging action, and flesh isn't, pretty much all you will note is the first one and thus just sharpness matters.

While this does seem right - experience tells me that cutting through flesh - raw and/or cooked - thinness does make a significant difference - so perhaps another contributing factor is the drag? eg: slicers are never thick or wedge shaped.

So for example cutting light paper requires a very sharp blade only, cutting a thick turnip however requires a slim cross section as the turnip will wedge the blade heavily.

Yes, and it can be a lot more subtle than initially credited. The Kershaw Leek is actually quite thin bladed and it cut through a raw sweet potato (not a turnip - but pretty similar) by wedging - whereas the thin bladed Opinel #8 encountered a lot more drag. The wedging by the Kerhsaw Leek is probably the relatively significant discontinuity/shoulder between the edge bevel and the hollow ground face. So the end comparison was pretty similar in effort to cut but probably due to different mechanisms.

This was emphasized even more by cutting through Cheddar cheese - where the drag contributes a lot more - the Kershaw Leek probably wedged the cheese so avoiding the stick/drag - whereas the cheese definitely stuck to the convex continuous grind of the Opinel #8.

As always many thanks for the very informative input, Cliff.
Much appreciated.

--
Vincent

http://UnknownVincent.cjb.net/
http://UnknownVT.cjb.net/

 

[Cliff Stamp]

The effect of geometry is taken into account in the calculation of I_A. For wedge shaped blades I_A ~ W H^3. So the stiffness goes up linear with the width, and cubic with the thickness, which should seem sensible if you ever tried to bend blades you know that increasing width gains little, but even small changes in thickness make a huge difference. A 3/16" blade at 10" is easy to break, 1/4" is *very* difficult for this reason.

[fixed typo - had I_B instead of I_A]

-Cliff

 

[UnknownVT]

The effect of geometry is taken into account in the calculation of I_B which requires a knowledge of the geometry. For wedge shaped blades I_B ~ W H^3.

Is I_B a new variable to the formula you used previously?
Or is it somehow related to I_A as it is proportional to W H^3?

BUT I_A for a rectangular bar was also proportional to W H^3 -

For a rectangle I_A = (1/12) W H^3 where H is in the direction of the applied torque

so the flex of the rectangular shaped bar is affected proportionally the same as a wedge in terms of width and thickness?

Does I_B become very complex to calculate for non-uniform/regular shapes like hollow grinds and fullers?

Thanks,

--
Vincent

http://UnknownVincent.cjb.net/
http://UnknownVT.cjb.net/

 

[Cliff Stamp]

I_B should have been I_A, I hate equations in ascii.

Yes, wedges and rectangules have the same proportionality, for sabre grinds the partial height is a factor as well.

For hollows and convex it ends up depending on the radius of curvature of the wheel, and the way in which the grind is applied. I can work out a few examples if you are interested.

-Cliff

 

[TLM]

(1/R) = M_B / (E I_A)

Usually M_B is called bending moment, torque is for screwing...

For simple shapes I_A can be found tabulated in any structural engineering text, for more complex ones you either spend some time or use a CAD program giving you the values.

E is a material constant which is practically invariant in steels, so hardness does not affect stiffness.

TLM

 

[UnknownVT]

E is a material constant which is practically invariant in steels, so hardness does not affect stiffness.

Thanks TLM - so are you saying that for all steels the Young's modulous is a constant, so the flex is mainly due to the thickness and geometry?

Surely not, aren't some steels more flexible than others -
hence examples as "spring" steel?

Thanks,

--
Vincent

http://UnknownVincent.cjb.net/
http://UnknownVT.cjb.net/

 

[Cliff Stamp]

When you apply a force to a steel to bend it (stress), the amount that it bends (strain), is proportional to the force applied, this constant is youngs modulus, which as noted doesn't change a lot in steels.

However this does *NOT* mean the stiffness doesn't change. Anyone who has even pried with a knife, or bent any metal knows that this isn't true. For example try to bend a coathanger and a small nail and see if they are the same in stiffness. Bend a knife with an annealed spine compared to a full hard one and watch what happens.

The above equation only applies when the relationship between stress and strain is linear, this means the deformation is elastic, the metal springs back.

When metals get soft this relationship fails readily as they have little spring and thus move quickly out of the linear stress <-> strain relationship. As a steel gets harder, this relationship holds for higher and higher stress, making it "stiffer" or harder to bend at higher loads as when the relationship fails the amount of force required plateaus (in practical application it decreases rapidly because of the effect of a torque / leverage issue).

As for M_B it is just the integeral of the torque on the body, the bar must be under a torque to bend, as if there was no net torque it would simply be still (or accelerate), engineers usually call it a bending moment but in physics its just a net torque, that is to say a force applied in an opposing couple so as to induce a rotation, in this case its around the point of the blade being held in hand.

-Cliff

 

[Anarcus]

Fascinating.
I'm curious as to whether manufacturers use this formula when designing a blade . Do people specify any of these figures (other than implicitly) when ordering say a custom knife? Or indeed are they tabled specs for say K bars?
Tim