Question concerning Shearing Force

[Captain_Torch]

From what I've heard, one can use shearing force to calculate slicing feats.

Calculations Introduction and References

Calculations may seem intimidating, but there is a very real and easy way to start getting into them. While being swamped with information may seem overwhelming, taking things piece-by-piece will go miles to easing your newfound endeavor. Though this is a hefty read, it is designed for an...

vsbattles.fandom.com

^This link uses 2 calcs as an example, however these calcs are quite outdated, and at least one of them is wrong according to the comments of the blog.

How should this method actually be used? From what I've understood it's:

E = Shear force * distance slash travelled
Shear force = shear strength * area sheared
Shear Strength = Ultimate Tensile Strength * 0.6

But I keep getting pretty high numbers when testing this method out, so I was curious if I am using it correctly, or if the method is even viable in the first place.

 

[KLOL506]

From what I've heard, one can use shearing force to calculate slicing feats.

Calculations Introduction and References

Calculations may seem intimidating, but there is a very real and easy way to start getting into them. While being swamped with information may seem overwhelming, taking things piece-by-piece will go miles to easing your newfound endeavor. Though this is a hefty read, it is designed for an...

vsbattles.fandom.com

^This link uses 2 calcs as an example, however these calcs are quite outdated, and at least one of them is wrong according to the comments of the blog.

How should this method actually be used? From what I've understood it's:

E = Shear force * distance slash travelled
Shear force = shear strength * area sheared
Shear Strength = Ultimate Tensile Strength * 0.6

But I keep getting pretty high numbers when testing this method out, so I was curious if I am using it correctly, or if the method is even viable in the first place.

I think we stopped using that method for cutting calcs altogether.

Now we just use: length of the cut area * width of the cut area * thickness of the blade edge * compressive strength/pulv. energy of the material

Coincidentally, it also works for breaking things in half instead of shattering the entire thing.

 

[ElajRuengies]

I think we stopped using that method for cutting calcs altogether.

Now we just use: length of the cut area * width of the cut area * thickness of the blade edge * compressive strength/pulv. energy of the material

Coincidentally, it also works for breaking things in half instead of shattering the entire thing.

What are the units? (Meters, MPA?)

 

[KLOL506]

What are the units? (Meters, MPA?)

MPa and cm for length, width and thickness, since it's basically the usual destruction value stuff.

Basically you're applying the compressive strength value to a volume of an object that is super thin. Volume is in cm^3.

 

[KLOL506]

Think of it like cutting a super thin cylinder or a super thin cuboid. Super thin because the thickness is the blade edge thickness.

For example. Think of a solid concrete block that with sides a b c, each side being 1.3 m, 0.65 m and 0.15 m respectively.

Now, you cut the block in half and the block was lying horizontally, so the destroyed area would be a * b. Carving knives have a blade edge thickness of 0.35 mm or 0.00035 m. Safe to assume sword blades can be potentially thicker, this link I found says bare minimum at least 2mm but not more than 4mm.

So the actual volume destroyed is: a * b * blade thickness. So 130 cm * 65 cm * 0.035 cm= 295.75 cm^3.

Pulv. energy of concrete as per our Calculations Page (Pulv. energy is just another name for compressive strength) is 40 MPa or 40 J/cc.

295.75*40= 11830 J (9-C+).

 

[Captain_Torch]

Think of it like cutting a super thin cylinder or a super thin cuboid. Super thin because the thickness is the blade edge thickness.

For example. Think of a solid concrete block that with sides a b c, each side being 1.3 m, 0.65 m and 0.15 m respectively.

Now, you cut the block in half and the block was lying horizontally, so the destroyed area would be a * b. Carving knives have a blade edge thickness of 0.35 mm or 0.00035 m.

So the actual volume destroyed is: a * b * blade thickness. So 130 cm * 65 cm * 0.035 cm= 295.75 cm^3.

Pulv. energy of concrete as per our Calculations Page (Pulv. energy is just another name for compressive strength) is 40 MPa or 40 J/cc.

295.75*40= 11830 J (9-C+).

Thanks! I assume that length * width varies per shape? So a circular area would be pi*r^2 instead.

 

[KLOL506]

Thanks! I assume that length * width varies per shape? So a circular area would be pi*r^2 instead.

Yeah, pretty much. Area affected depends on shape.