Uses of a Tanto.

[res1cue]

For this image (assuming dimensions not shown are equivalent) the profile on the right is the long/slow taper and will be easier (i.e. require less force) to penetrate through whatever substance to a given depth due to the reduced surface area compared to the profile on the left which better approximates the western-tanto shape.
But what I was typing about previously regarded spine-width/taper, though the point is the same: force divided-by surface area, the pith of edge-angle "sharpness". The profile on the right will penetrate the car-hood more easily, i.e. it requires less strength with the tapered or japanese-tanto blade than the western-tanto.
Now, if the steel/HT require the extra support-metal of a western-tanto to accomplish this task without significant edge/blade damage, I would consider the extra required force a worthy sacrifice!

chiral.grolim, I think this is where we disagree. Youre reasoning is logically correct, but there are other factors. So lets say in that diagram, instead of something uniform like a stack of cardboard, it is just a car hood with nothing below it. Let's say the two blades are both made of sufficiently thick identical steel stock, and are heat treated right, etc, but one has a 1" taper, and one has a 2" taper

Yes, the 2" taper would require less force to penetrate further into the steel door. But if I am not sufficiently strong, and I just stab with all my force into the car hood, I may only get 1.25" penetration on the americanized tanto, which is enough to continue to slide in easily.

For the 2" taper, I would penetrate further, say 1.75". But then, I am stuck with .25" of widening blade, that I will be unable to shove into the solid steel car hood, since I don't have the benefit of momentum and gravity.

The 1" taper capitilizes on that initial contact, so I quickly punch a hole big enough to accomodate the full width of the blade, and then it will slide in easy.

Now given a strong enough person, it may still be possible, but a short taper is still ideal for this kind of "armor piercing" effect. Now if I have a FULL tapered dagger style blade, I may be able to get a whole 2.5" of penetration initially, but I will be unable to push the remaining length of the blade in.

This might be kind of confusing, and I apologize that I am not explaining this really well.

:thumbup::thumbup: I would insert the phrase "more easily" at the end of the bolded section (since there are examples of this on non-tanto styles), but THAT is imho one of the biggest advantages of the design. You can add extra support to that "armor-piercing" tip while keeping a thin main edge...
It could also be ground the other way around if user prefers (as seen on some WC Davis knives, though his aren't tanto).

:thumbup: Absolutely, and yeah I forgot, there are some knives to do feature the double grind, and I wish more people did!

 

[crocop5247]

Tantos arent good for outdoors, I find they are better for self defence.

 

[stabman]

Tantos arent good for outdoors

I disagree.
My Cold Steel Master Tanto is awesome for woods work.
Not as great as the Junglas (I like big knives), but every bit as useful as any other knife in it's blade size.:thumbup:

 

[TwinStick]

I guess all of this depends on the person. MOST of us are not experts in martial arts, edged weapons techniques, or even fighting for that matter. For me, and only speaking for me, i think a tanto edge is perfect. It fits the bill as an EDC, will cut cardboard, open pkgs, cut food & is still a formidible defensive knife, when used how most of us would use it, in a "snap cut" or quick flick, a stab or slash. In my experience of 23 yrs on the job, most criminals are not experts either, rather they are cowards, & would likely pick an easier target when presented with a citizen with the means to defend themself. Not always, but most of the time. I like my tanto a lot. Yet, it is not my only blade type. Thats why we are all here, LOL, because we like edged instruments.

 

[Czechmate]

my gunsite shears through cardboard pretty easily; so for me it's used as a utilitarian blade. It's also shredded pop cans, windshield washer jugs and coffee cans without too much damage. The serrations have chipped though - wish they weren't there.

My recon tanto (aus8 version) has seen some really hard use and held up amazingly well in my opinion.

 

[chiral.grolim]

...Youre reasoning is logically correct, but there are other factors...

Yes, the 2" taper would require less force to penetrate further into the steel door.

…I may only get 1.25" penetration on the americanized tanto, which is enough to continue to slide in easily.

For the 2" taper, I would penetrate further, say 1.75”.

The 1" taper capitilizes on that initial contact, so I quickly punch a hole big enough to accomodate the full width of the blade, and then it will slide in easy.

… a short taper is still ideal for this kind of "armor piercing" effect. Now if I have a FULL tapered dagger style blade, I may be able to get a whole 2.5" of penetration initially, but I will be unable to push the remaining length of the blade in.

Well, I'm trying to reason by physics.

You seem to be concerned with the cutting-medium's cumulative resistance to penetration, call it "rP", a coefficient established by the physical properties of the medium whatever it may be (stacked cardboard, a block of wood, water, a thin sheet of metal, etc.), incorporating surface tension and drag/friction forces to oppose perpendicular penetration.
For the example presented, the cutting-medium is constant, the same material being cut by both blades, and so "rP" is identical in opposition to the force bestowed upon the blades via the energy of the stab, "F".
Now, let's assume that the force of the stab "F" is exactly enough to allow full penetration of the 1"-taper tanto blade - i.e. penetration to the end of the taper to full width/thickness in both dimensions, whereupon penetration-resistance (directly proportional to the angle and surface area still attempting penetration) drops to ~0 as the angle of penetration achieves 90-degrees (cos90), so the blade slides on through the punched hole.

The medium's resistance to penetration "rP" effects deceleration of initial stabbing force "F". F=(mass x velocity)/(T2 - T1), so if "F" is exactly enough to overcome "rP" at T1, F=~0 at T2 when full penetration is achieved (i.e. velocity = 0 ), and rP=F

Examine the cross-sectional surface area of each blade as it penetrates the medium over the course of time (T2 - T1). At T1 when the points make contact, surface-area is equivalent for each, so at constant force F/A=P is the same for each, and "P" here refers directly back to "rP" as it resists the stabbing force. The ability of the medium to resist penetration is inversely proportional to the surface area upon which the stabbing-force "F" is acting. At any given depth of penetration thereafter, you will note that the cross-section surface area of the 2"taper is less than that of the 1"-taper. At 1" penetration, the 2" taper's surface area is 1/4 that of the 1" taper… that means it delivers 4X the pressure to the surface of the medium being stabbed. This translates back such that F=PxA: it required 1/4 "F" to bring the 2"-taper 1" through the medium.

(Mathematicians, please jump on me wherever I err.)

We defined T2 as the time at which the 1"-taper achieves full 1" penetration and "F" is expended. Well, at 1" deep, the 2"-taper has only expended 25% of "F", and if "F" relates directly to the amount of time expended to achieve this depth of penetration, then 1"-depth was achieved in 1/4 the time it took the 1"-taper to achieve that depth – a much quicker punch. And we still possess 75% of our energy to force the 2"-taper the rest of the way through the medium. How much further do you think you can penetrate into the medium with 75% of your force-ability still remaining after 1” depth?

Now reverse the situation: what if we only possessed the strength to force the 2"-taper 1" into the medium? How far would we get with the 1"-taper?

So that's my support of the position. I am not a physicist or mathematician, so I'll gladly accept correction to the calculations. Why do you think you won't be able to force the 2"-taper through the medium as easily as the 1" taper?


I contend that the “armor-piercing” attribute of the western-tanto is not about reduced speed or force of penetration, because they are not reduced. What IS reduced is damage suffered to the point upon impact against “armor”-like material due to the increased support-metal of a thick tip-grind. Glass/skull-crusher pommels function on the same principle of resilience.

 

[Tyronethepro]

I own 2 americanized tanto blades right now, both are CRKT and I love them both. I own the Ed Halligan Stiff Kiss and I wear it on my pack strap while hunting (Bears and Cougars are problem animals in the areas where I hunt and I like to have a knife I can draw if I set my firearm down). Also I own the CRKT SOFTB, Special Operations Tactical Fixed Blade, It batons through wood and I trust it for any outdoor use, it gets sharp and stays sharp, AUS-8 Steel and its part serrated. All-In-All I feel they're good for multiple roles.
Just my 2 cents on the subject.

 

[TwinStick]

my gunsite shears through cardboard pretty easily; so for me it's used as a utilitarian blade. it's also shredded pop cans, windshield washer jugs and coffee cans without too much damage. The serrations have chipped though - wish they weren't there.

my recon tanto (aus8 version) has seen some really hard use and held up amazingly well in my opinion.

x 2 !!!

 

[res1cue]

Well, I'm trying to reason by physics.

Reasoning with physics is always tempting, but it is very tricky to translate reality into physics terms. There are a couple key things about your physics model that I don't agree with, which may be why we are arriving at different conclusions

You seem to be concerned with the cutting-medium's cumulative resistance to penetration, call it "rP", a coefficient established by the physical properties of the medium whatever it may be (stacked cardboard, a block of wood, water, a thin sheet of metal, etc.), incorporating surface tension and drag/friction forces to oppose perpendicular penetration.

This is fine, but remember that my original post was saying that a thin but highly resilient material being penetrated by a stabbing motion will behave differently than something soft and uniform.

Before we get TOOO technical, I am just going to reiterate what we have been talking about:

What I have been saying is that a 1" taper will be easier to stab through something like a car hood, but a 2" taper will be easier to stab deeper into a stack of cardboard.

As I understand it you have been saying that a 2" taper will penetrate better regardless of the medium.

For the example presented, the cutting-medium is constant, the same material being cut by both blades, and so "rP" is identical in opposition to the force bestowed upon the blades via the energy of the stab, "F".
Now, let's assume that the force of the stab "F" is exactly enough to allow full penetration of the 1"-taper tanto blade - i.e. penetration to the end of the taper to full width/thickness in both dimensions, whereupon penetration-resistance (directly proportional to the angle and surface area still attempting penetration) drops to ~0 as the angle of penetration achieves 90-degrees (cos90), so the blade slides on through the punched hole.

The medium's resistance to penetration "rP" effects deceleration of initial stabbing force "F". F=(mass x velocity)/(T2 - T1), so if "F" is exactly enough to overcome "rP" at T1, F=~0 at T2 when full penetration is achieved (i.e. velocity = 0 ), and rP=F

Examine the cross-sectional surface area of each blade as it penetrates the medium over the course of time (T2 - T1). At T1 when the points make contact, surface-area is equivalent for each, so at constant force F/A=P is the same for each, and "P" here refers directly back to "rP" as it resists the stabbing force. The ability of the medium to resist penetration is inversely proportional to the surface area upon which the stabbing-force "F" is acting. At any given depth of penetration thereafter, you will note that the cross-section surface area of the 2"taper is less than that of the 1"-taper. At 1" penetration, the 2" taper's surface area is 1/4 that of the 1" taper…

actually I think it is 1/2, assuming twice as long of a taper, and the taper being a straight line, but minor details

that means it delivers 4X the pressure to the surface of the medium being stabbed. This translates back such that F=PxA: it required 1/4 "F" to bring the 2"-taper 1" through the medium.

What you have underlined is ONLY true if we are talking about a continuous medium like the stack of cardboard.

Actually if we are attempting to penetrate through a car hood and that's all, youre rP resistance coefficient only exists at ONE point at any point of time: the point where the EDGE contacts the car hood steel. rP is basically ONLY the steel's resistance to shear stress (steel on steel friction is negligible in proportion to this). The only thing that affects rP assuming forces are equal is the angle, and the 2" taper does have advantage in this.

The other thing that your model does not account for is that the force F not constant. It's not even linear, it is going to decrease exponentially, because you are incapable of adding significant amount of force after the knife has made contact with the car hood.

We defined T2 as the time at which the 1"-taper achieves full 1" penetration and "F" is expended. Well, at 1" deep, the 2"-taper has only expended 25% of "F", and if "F" relates directly to the amount of time expended to achieve this depth of penetration, then 1"-depth was achieved in 1/4 the time it took the 1"-taper to achieve that depth – a much quicker punch. And we still possess 75% of our energy to force the 2"-taper the rest of the way through the medium. How much further do you think you can penetrate into the medium with 75% of your force-ability still remaining after 1” depth?

Again, in this situation, at 1" penetration the 1" taper will have cleared the width completely, but the 2" taper will have cleared only 1/2 the width of the blade, will have a whole another inch to shear steel, and with dwindling force.

Now reverse the situation: what if we only possessed the strength to force the 2"-taper 1" into the medium? How far would we get with the 1"-taper?

Then neither would go all the way through b/c you would be a weakling

I contend that the “armor-piercing” attribute of the western-tanto is not about reduced speed or force of penetration, because they are not reduced. What IS reduced is damage suffered to the point upon impact against “armor”-like material due to the increased support-metal of a thick tip-grind. Glass/skull-crusher pommels function on the same principle of resilience.

We are in agreement here :thumbup:

 

[6gunsally]

I carried a Cold Steel Tanto (SanMai III) that could slice through a full coke can in one pass. Could'nt find anyone strong enough to penetrate body armor. Big and bulky, I do see them carried by some of the riders on TV's "Sons of Anarchy". But, stopped carrying due to sharpening issues. Some knives need more than a quick diamond rod.

 

[chiral.grolim]

...There are a couple key things about your physics model that I don't agree with, which may be why we are arriving at different conclusions...

...Before we get TOOO technical, I am just going to reiterate what we have been talking about:

What I have been saying is that a 1" taper will be easier to stab through something like a car hood, but a 2" taper will be easier to stab deeper into a stack of cardboard.

As I understand it you have been saying that a 2" taper will penetrate better regardless of the medium.

Good reiteration.:thumbup:

The 2"-taper has a lower edge/tip angle throughout penetration, allowing it to concentrate the same force over a smaller area compared to the 1"-taper: Pressure. I'll explain how I calculated the "4x" value below.

Oh, I did stipulate much earlier that, if it was the nature of the medium to fracture apart on contact, e.g. wood along the grain, then deeper "penetration" could be achieved with the wider edge/tip-profile given sufficient force in the blow to transfer penetration ahead of the blade (i.e. a 12" log splits fully through despite that the axe-blade only proceeded 3" into the wood).

actually I think it is 1/2, assuming twice as long of a taper, and the taper being a straight line...

Perhaps pressure is the wrong way to go entirely, but i got the 1/4 or 4x value by modeling the cross-section of each blade at 1" penetration. I assumed that the taper-length is measured along the spine and that grinds are full-flat, so at 1" penetration the 1"-taper blade's cross-section includes the full spine thickness (call it "x") and also the full width from spine to belly-edge (call it "y"). The surface area (A) of the triangular cross-section is calculated: A= (xy)*(1/2).
At 1" penetration, the 2"-taper blade's cross-section has only achieved 1/2 the full spine thickness (x) and also only 1/2 the full spine-to-belly width (y), such that it's cross-sectional area is calculated: B= (x/2)*(y/2)*(1/2) = [(xy)/4]*(1/2) = (xy)*(1/2)*(1/4) = A*(1/4). The surface area of the 2"-taper blade is 1/4 that of the 1"-taper blade at 1" penetration.

What you have underlined is ONLY true if we are talking about a continuous medium like the stack of cardboard.

Actually if we are attempting to penetrate through a car hood and that's all, youre rP resistance coefficient only exists at ONE point at any point of time: the point where the EDGE contacts the car hood steel. rP is basically ONLY the steel's resistance to shear stress (steel on steel friction is negligible in proportion to this). The only thing that affects rP assuming forces are equal is the angle, and the 2" taper does have advantage in this.

The other thing that your model does not account for is that the force F not constant. It's not even linear, it is going to decrease exponentially, because you are incapable of adding significant amount of force after the knife has made contact with the car hood.

This is where I was hoping a physicist would jump in and present the proper calculation.
In a continuous medium, like fluid, drag/friction resistances are different due to continued application along the entire length, and especially the point, of the penetrating object as it proceeds. Stacked cardboard is actually about the same as a single sheet of metal because each sheet of cardboard acts independently of the others in the stack and can be examined individually. We can look at a sheet's thickness as "continuous" medium for as thick as it is, but if we reduce the thickness to "0" for modeling purposes, you are right that the edge/tip-angle is an important resistance factor, and indeed the 2"taper again has the advantage. You can use the cosine-function to map the medium's ability to resist penetration of a blade with a given bevel-angle (non-linear relationship). Note that for a 30 degree bevel-grind (60-degree inclusive), there is a 60-degree angle between the edge and the medium's surface (90-30), and cos60 = 0.5. Now if the the 1"-taper has the 30-degree bevel, we know that the 2"-taper bevel is ~16.2 degrees. Calculate 90-16.2=73.8, cos73.8= ~0.28. At any given depth of penetration up until full-width/thickness is achieved, resistance to penetration of the 2"-taper is 0.56 that of the 1"-taper, ~1/2.

But this shear resistance is around the entire perimeter of the penetrating blade. The 2"-taper not only has angle advantage almost double that of the 1"-taper, but at 1" penetration the perimeter of the 2"-taper is only 1/2 that of the 1"-taper, meaning it has double the angle-advantage over the 1"-taper at that point. That's again ~4x the advantage or 4x less force required to continue penetrating.

Now, from the above I hypothesize that the non-linear deceleration/loss of force imparted to the blades (F=ma) during the stab due to the nature of "rP" is ~4x lower for the 2"-taper than the 1" taper. Now, if full deceleration is achieved precisely when the 1"-taper accomplishes "full" penetration, the 2"-taper will require 1/4 the force to achieve the same 1" depth and 2/4 the force to achieve "full" penetration at 2", leaving 50% of the employed force to continue penetration! Put another way, the 2" taper required half the force or half the time to accomplish twice the penetration of the 1" taper, and all that remains is to slide the full-thickness through the open hole, which we have plenty of force left to accomplish.

Ug, math :barf:. Does anyone have a good graphing program to demonstrate to all of us which theory is correct, i.e. deceleration of pyramidal-objects penetrating a resistant surface with low deflection properties? I am by no means convinced that my calculations are accurate, they are simply what I have

 

[TwinStick]

Too much science. Real world use is worth more IMHO. LOL

 

[EZ Bake]

They're great for cutting pills (tablets) in half

 

[cziv]

I carried a Cold Steel Tanto (SanMai III) that could slice through a full coke can in one pass. Could'nt find anyone strong enough to penetrate body armor. Big and bulky, I do see them carried by some of the riders on TV's "Sons of Anarchy". But, stopped carrying due to sharpening issues. Some knives need more than a quick diamond rod.

What's up with sharpening issues on the CS Tanto 6" blade model? Is it something to do with the San Mai 3, which I still don't understand what the h*ll that means except layered steels I think?

I've wanted one forever but I always balk, thinking "what do you need this for?" I find it hard to undertand that a knife would slice through a coke can but it's even sharper tip lacked penetration.

 

[res1cue]

I am not really sure, my CS SM3 Master tanto is very easy to sharpen. VG1 is quite easy to work with

 

[res1cue]

Chiral.grolim, I didn't even know you responded!

Good reiteration.:thumbup:

The 2"-taper has a lower edge/tip angle throughout penetration, allowing it to concentrate the same force over a smaller area compared to the 1"-taper: Pressure. I'll explain how I calculated the "4x" value below.

Oh, I did stipulate much earlier that, if it was the nature of the medium to fracture apart on contact, e.g. wood along the grain, then deeper "penetration" could be achieved with the wider edge/tip-profile given sufficient force in the blow to transfer penetration ahead of the blade (i.e. a 12" log splits fully through despite that the axe-blade only proceeded 3" into the wood).



Perhaps pressure is the wrong way to go entirely, but i got the 1/4 or 4x value by modeling the cross-section of each blade at 1" penetration. I assumed that the taper-length is measured along the spine and that grinds are full-flat, so at 1" penetration the 1"-taper blade's cross-section includes the full spine thickness (call it "x") and also the full width from spine to belly-edge (call it "y"). The surface area (A) of the triangular cross-section is calculated: A= (xy)*(1/2).
At 1" penetration, the 2"-taper blade's cross-section has only achieved 1/2 the full spine thickness (x) and also only 1/2 the full spine-to-belly width (y), such that it's cross-sectional area is calculated: B= (x/2)*(y/2)*(1/2) = [(xy)/4]*(1/2) = (xy)*(1/2)*(1/4) = A*(1/4). The surface area of the 2"-taper blade is 1/4 that of the 1"-taper blade at 1" penetration.

Again, I contend, that the drag (or shear resistance really) in a single carhood case, is ONLY significant at the point of the edge, surface area, I think is negligible. If you take a single suspended piece of cardboard, and poke a tanto through it slowly, you can see that along the flat sides of the blades, and along the spine, there is very little drag. The main resistance is the cardboard's resistance to be cut.

This is where I was hoping a physicist would jump in and present the proper calculation.

That's funny, I actually have a BA in physics, but its really really rusty.. been years since I touched that stuff :barf:

In a continuous medium, like fluid, drag/friction resistances are different due to continued application along the entire length, and especially the point, of the penetrating object as it proceeds. Stacked cardboard is actually about the same as a single sheet of metal because each sheet of cardboard acts independently of the others in the stack and can be examined individually.

I disagree with this part too, because stacked cardboard is like a continuous mass, and the two diffrent blades will behave differently as they penetrate. A stacked piece of cardboard will optimize the attributes of the 2" taper, because at any point after penetration, the tip will be piercing another piece of cardboard. All of your assumptions would be correct in this case, since the 1" taper would have to "punch" through each of these softer materials, where surface drag will be comparable cumulatively to the edge resistance.

In the car hood case, we are talking about edge resistance that is many many times greater than the friction of the steel on steel. Also, after the tip passes through the car hood, it does not have to do anything else

We can look at a sheet's thickness as "continuous" medium for as thick as it is, but if we reduce the thickness to "0" for modeling purposes, you are right that the edge/tip-angle is an important resistance factor, and indeed the 2"taper again has the advantage. You can use the cosine-function to map the medium's ability to resist penetration of a blade with a given bevel-angle (non-linear relationship). Note that for a 30 degree bevel-grind (60-degree inclusive), there is a 60-degree angle between the edge and the medium's surface (90-30), and cos60 = 0.5. Now if the the 1"-taper has the 30-degree bevel, we know that the 2"-taper bevel is ~16.2 degrees. Calculate 90-16.2=73.8, cos73.8= ~0.28. At any given depth of penetration up until full-width/thickness is achieved, resistance to penetration of the 2"-taper is 0.56 that of the 1"-taper, ~1/2.

But this shear resistance is around the entire perimeter of the penetrating blade. The 2"-taper not only has angle advantage almost double that of the 1"-taper, but at 1" penetration the perimeter of the 2"-taper is only 1/2 that of the 1"-taper, meaning it has double the angle-advantage over the 1"-taper at that point. That's again ~4x the advantage or 4x less force required to continue penetrating.

Now, from the above I hypothesize that the non-linear deceleration/loss of force imparted to the blades (F=ma) during the stab due to the nature of "rP" is ~4x lower for the 2"-taper than the 1" taper. Now, if full deceleration is achieved precisely when the 1"-taper accomplishes "full" penetration, the 2"-taper will require 1/4 the force to achieve the same 1" depth and 2/4 the force to achieve "full" penetration at 2", leaving 50% of the employed force to continue penetration! Put another way, the 2" taper required half the force or half the time to accomplish twice the penetration of the 1" taper, and all that remains is to slide the full-thickness through the open hole, which we have plenty of force left to accomplish.

My head hurts now, , I will reread this later and respond to you then

Ug, math :barf:. Does anyone have a good graphing program to demonstrate to all of us which theory is correct, i.e. deceleration of pyramidal-objects penetrating a resistant surface with low deflection properties? I am by no means convinced that my calculations are accurate, they are simply what I have

Exactly how I feel :thumbup:

 

[chiral.grolim]

Chiral.grolim, I didn't even know you responded!

No worries.

Again, I contend, that the drag (or shear resistance really) in a single carhood case, is ONLY significant at the point of the edge, surface area, I think is negligible. If you take a single suspended piece of cardboard, and poke a tanto through it slowly, you can see that along the flat sides of the blades, and along the spine, there is very little drag. The main resistance is the cardboard's resistance to be cut.

This is my fault, i should have been clearer. I agree with you regarding minimal steel-on-steel friction as the blade is passing perpendicularly (90-degrees) through the already-cut hole, especially if we ignore the width of the steel car-hood entirely.
The "surface area" I was talking about was that of the car-hood surface being cut and the cross-sectional area of the blade as it makes the cut (i.e. before the end of the taper), hence the 1" penetration depth-limit for the 1" taper-blade. Once the end of the taper is reached on either knife, the hole-cut is complete and the rest of the knife slides on through, as you stated, like a nail after the beveled point. When the hole is complete, there is zero pressure on the surface of the car-hood since it offers no perpendicular resistance then. But before the hole is complete, as the tip is still in the process of cutting, as the car-hood deflects and resists tip-penetration, pressure is being exerted on the car-hood surface over an area determined by the cross-section of the penetrating tip: P=F/A.
From >0" to <1" depths, the cross-sectional area of the penetrating 1"-taper is 4x that of the 2"-taper.

I disagree with this part too, because stacked cardboard is like a continuous mass, and the two diffrent blades will behave differently as they penetrate. A stacked piece of cardboard will optimize the attributes of the 2" taper, because at any point after penetration, the tip will be piercing another piece of cardboard. All of your assumptions would be correct in this case, since the 1" taper would have to "punch" through each of these softer materials, where surface drag will be comparable cumulatively to the edge resistance.

Each piece of cardboard in a stack has it's own surface-tension and deflection coefficient levels, just like a single sheet of metal (car-hood). It is only a "continuous mass" within its own individual thickness. For the stack of cardboard, or a stack of car-hoods for that matter, each sheet will independently respond to the blade-tips as they attempt penetration, i.e. "rP" does not change. Stacking 6 sheets of very small width, each with the same "rP" coefficient accumulates to 6*(rP) which is just another coefficient. So if instead of a single sheet of metal we are trying to penetrate a stack of sheets, we treat the entire stack as a single sheet with a single "rP" value equal to the cumulative "rP" values of each individual sheet.
Either way, to determine which blade will penetrate further with the same initial force, or with less force to the same depth, we look at a single "rP".

For this discussion (which is WAY off the OP, sorry, but it's still educational ), we want to know the precise level of force required for each blade, the 1"-taper and the 2"-taper, to cut fully through that single sheet (car-hood) such that, at the precise moment it completes the cut there is zero force/velocity remaining to push the blade any further.

IF "rP" relates to pressure-resistance, then it is directly proportional to the area over which the "normal force" of the stab is applied, which relates to cross-sectional area of each blade-tip.
If "rP" relates to angle of surface contact between the car-hood surface and the penetrating tip, then perhaps it is directly proportional to the cosine of that angle. To clarify, a perpendicular stab delivers force to the medium being cut along 2 plains, the vertical and the horizontal, related by the angle at which the blade-bevel contacts the surface being cut. For a 30-degree bevel, the angle of incidence is 90-30 = 60 degrees. The force-vector "F" during the stab is actually NOT perpendicular but is along the bevel as it intersects the medium at angle "G". The perpendicular component of this force is calculated (F)*sin(G). The horizontal component of this vector, which does not contribute to penetration depth but is lost to medium resistance "rP", is calculated (F)*cos(G).

- a stab made with no sharpened bevel is 100% deflected horizontally along the surface area = cos0.
- a stab made with a 30-degree bevel loses 50% of the force horizontally (cos60)
- a stab made with a 16.2-degree bevel loses ~28% of the force horizontally (cos73.8)
- a stab made with a hypothetical 90-degree bevel (zero cross-sectional area) exerts all force perfectly perpendicular to the surface such that 0% is lost (cos90).

I am not sure about the importance of the perimeter of the blade passing through the surface of the medium at a given bevel-angle, I am only certain that at a given depth >0 and <1" the perimeter of the 1"-taper is 2x that of the 2"-taper.

Hopefully that clarifies my math a bit... sorry for obfuscation.:barf:

 

[cziv]

Are you guys still at it with the Einstein math equations? lol

Tantos look cool, I only have one on a Recon-1 and I rounded it off for more versatility. Not sure if I succeeded but it push cuts a whole sheet of paper. Not sure on car hood yet. Maybe I'll try it this summer! :thumbup:

 

[Edwood7]

I love reading this tread, any one whana to add anyting new to it ?

 

[sh00ter01]

Snap Cuts! It's all about SNAP CUTS

-sh00ter