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## Kain the Scion of Balance (legacy of Kain) [Discussion]

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### Re: Kain the Scion of Balance (legacy of Kain) [Discussion]

Not using Brinell's hardness test and considering two claws for one hand, using the speed for the calc for Kain, but on the block instead, I ended up with 6,280 psi, or 43.3 MPa.
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### Re: Kain the Scion of Balance (legacy of Kain) [Discussion]

I'll try this again. The momentum I'm working with is 49.8 kg m/s. If I assumed that it took 1/1,000 of a second for Raziel's thrust to come to a full stop, then Raziel would have exerted a force of 49,800 N. Raziel's claws are 2.68 mm. The area of this would be using a circle, as 2.68 mm is the diameter. So his claws would have an area of 5.64 mm^2, or 0.222 in. 49,800 N is equal to 11,196 lb. This would give us 50,430 psi, or 347.7 MPa. This assumes the thrust into the block stopped at 1 millisecond.

Edit: I'm going to take the 9.3 kg. I get various times. Here they are.

• 373 ms
• 342 ms
• 316 ms
• 323 ms
• 338 ms

So I will add these and then divide by 5 to find the average time. This is 1.692 s / 5 = 0.3384 s. I doubled the distance of Raziel's arm the last time I calculated him striking the cube. 68.58 cm. * 2 gives us 137.16 cm., or 1.3716 m. Distance over time gives us 4.0531914893617021 m/s. Since momentum is p = mv, then that means Raziel struck with a momentum of 37.69468085106382953 kg m/s. I cannot find out how long it takes for Raziel's arm to come to a complete stop, though. Coming into contact with another object will slow his strikes down.

Raziel's claws are 64 px., so taking his height of 720 px., I'll end up with 0.0888888888888889
193.04 cm. * 0.0888888888888889 = 17.159111111111113256 cm.

That's how long Raziel's claws are, excluding the thumbs. Well, if distance over speed still holds true here, then the time it would take to cover 17.159111111111113256 cm. over 4.0531914893617021 m/s, then Raziel would come to a complete stop in 0.04233481481481482 s, or 42 ms. Force is Δp / Δt. This would result in 890.394372006862 N. Raziel's claws are 0.268111111111113256 cm. in diameter, so finding the area of a circle (since the tip of his claws represent a circle) is A = π r^2, which gives us 0.05645722220808630093009 cm^2

890.394372006862 N / 0.05645722220808630093009 cm^2 = 157.711332081679 MPa (22,874.20235813744596 psi).

I realize that the object appears metal and sounds like metal being struck, but even with this in mind, I am not saying the cube is made of limestone, only that it has the hardness and density of limestone, as it is the low-end. Aluminium has a density of 2.7 g/cm^3, while limestone has 2.6 g/cm^3.
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### Re: Kain the Scion of Balance (legacy of Kain) [Discussion]

Problem with an area of 5.6 mm squared is that is twice the size of the head off a pin;

http://icons.iconarchive.com/icons/icon ... e-icon.png

Which is rather large for the tip, hence why trying to discern size from a pixel is nigh impossible for something so small since that would suggest Raziels very claw tip is larger in area than even my pudgy sausage finger tips. hence why micrometers made more sense to me. Either way though, this impulse calculation here uses far more assumptions than our previous brinell scale calculation. I think your getting somewhere though, its definatly progress from the 2 newton or 50 newton figure and is getting believable.

Raziel's claws are 0.268111111111113256 cm. in diameter, so finding the area of a circle (since the tip of his claws represent a circle) is A = π r^2, which gives us 0.05645722220808630093009 cm^2

But Raziels claws are more like cones arent they? wouldnt you want to find the surface area of the cone, not a circle?

I realize that the object appears metal and sounds like metal being struck, but even with this in mind, I am not saying the cube is made of limestone, only that it has the hardness and density of limestone, as it is the low-end. Aluminium has a density of 2.7 g/cm^3, while limestone has 2.6 g/cm^3.

So what would the calculation be if it was made of iron? Aluminium is looks quite distinctive itself and is a fairly advanced metal as far as a medieval civilization like those in LoK would use, iron imo seems more likely based on that, plus we know they use iron already. Same if we did the brinell scale again with iron instead of limestone.

Also to get more complicated, how would one calcluate how much force it takes to hit a solid and turn it into a liquid or gas. To cover the fact when Razil strikes the blocks they do not get perforated by piecies hes moving from within or deformed in any way.

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### Re: Kain the Scion of Balance (legacy of Kain) [Discussion]

I'm not using "far more assumptions". We have the diameter of the tip of the claw. My only error was with respect to using a surface area of a circle, rather than a cone. Good suggestion. I didn't think of that because I reasoned that the point (not head) of a pin would be circular if we shrunk down to a very small size. I'll need to write the calcs on paper first and then I'll get back to this. I'm on my phone and am busy.

Edit: you know, I ended up with a larger number for the surface area of a cone. My last calc only worked on the tip of the claws, not the widest part, which we know that since D = d, we'd work with the thickest part. The tip of the claws (thumb not included) would be 0.14 cm^2, I think. But the width of the widest part will be concerned here, since a good portion of his claws go through.
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### Re: Kain the Scion of Balance (legacy of Kain) [Discussion]

Corrected some of my errors concerning pressure in the "physical stats" section at the bottom of the first page.

Kitten Lord
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### Re: Kain the Scion of Balance (legacy of Kain) [Discussion]

Kitten Lord
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### Re: Kain the Scion of Balance (legacy of Kain) [Discussion]

Since I am under the impression that Kain's skin is made up of iron, just as his original armor was when he was a human, I thought I'd go out of my way to find out just how much energy would be required to vaporize and atomize his skin right off of him. Does that make me a terrible person? Using Hamwi's formula, the weight of a male at 5 feet is 106 lbs (48 kg.). This is increased with every inch. Since Daniel Cabuco said Kain is 6'8", I thought I'd add an additional 20 inches. Hamwi's formula says that for every inch (2.54 cm.) added for a male, that's an additional 6 lbs. (2.7 kg.). This means that Kain weighs 226 lbs. (102.51 kg.) 16% of this will need to be removed, since that is the percentage that makes up a human body, so we're left with 36.16 lbs. (16.4 kg.).

I will need to convert this into the density of iron, since that's what I'm working with. Water has a density of 1,000 kg/m^3, whereas iron has a density of 7,874 kg/m^3. Taking the mass of Kain's skin and dividing it by the density of water will give me a volume of 0.0164 m^3. Next, to find the mass of the skin, I need to multiply this by 7,874 kg/m^3, which will give me 129.1 kg. While Kain's mass was affected by stripping him of his skin, giving us 66.35 kg., the addition of his skin as iron would mean Kain has a mass of 195.45 kg. (430.98 lbs.). Since Kain's skin would have a mass of 129.1 kg., we can continue. The heat of vaporization of iron is 340 kJ/mol, or 6.088 * 10^6 J/kg. Multiplying this by 129.1 kg. gives us 786 megajoules. 187.86 kilograms of TNT would be able to vaporize Kain's skin.

As for atomization, this will be higher. Atomization is the breaking of molecular bonds so that all molecules are broken to individual atoms. This is different from vaporization, which is colloquially used to mean atomization. Vaporization is simply turning a solid or liquid into vapor. They are still molecular. It takes 415 kJ/mol to atomize iron. This is equal to 7.431 * 10^6 J/kg. Once again, this will be multiplied by 129.1 kg., which gives us 959.3 megajoules. You'd need 229.29 tons of TNT to atomize Kain's skin right off of him. It should not be thought that because these are the energies required to vaporize or atomize Kain's skin, that it means that he will not be injured unless by this amount of energy.
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### Re: Kain the Scion of Balance (legacy of Kain) [Discussion]

This assumes though that his skin is made of iron. Clearly it is not actually made of iron, and we know its consistency is vastly stronger than iron since Raziel effortlessly moving blocks allows him to pierce iron, rock and brass with equal ease but even in a rage, could not scratch Kain.

Using this train of thought though, what is the closest material in reality that could resist at least one Petapascal of pressure? Maybe a neutron star material or some such? Carbon nanotubes? maybe that is the closest, what energy is required to vaporize that?

Kitten Lord
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### Re: Kain the Scion of Balance (legacy of Kain) [Discussion]

You're right, it does assume that Kain's skin is made of iron, although I don't think that's true. I think the hardness is in that range, and although Raziel can pierce through iron cubes, iron still has a range of hardness between 200 MPa to 1.18 GPa. So his skin could as hard as alpha ferrite (pure iron), which has a hardness of 270 MPa (80 HB). A W80 nuclear warhead will generate pressure of up to 6.5 PPa, which ranges anywhere from 5 to 150 kilotons of TNT.

I will be doing calculations all over again.
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### Re: Kain the Scion of Balance (legacy of Kain) [Discussion]

I'll be using gabbro, which has a density of 2,700 kg/m^3

Raziel's height: 81 pixels; 1.8034 m.
Crown: Assumed to be 81 pixels; 1.8034 m
Crown surface area: Assumed to be 81 pixels; 1.8034 m
Portion 1 height: 218 pixels; 4.8535950617283950172 m.
Portion 2 height: 218 pixels; 4.8535950617283950172 m.
Base height: 81 pixels; 1.8034 m.
Base width: 99 pixels; 2.20415555555555551548 m.

Volume of the crown
V = a^2 * h/3
V = 1.8034 m^2 * 1.8034/3
V = 1.9550368211013333333 m^3

Volume of frustum
V = 1/3 π h (R^2 + r^2 + sqrt(R^2 r^2))
V = 1/3 π 9.7071901234567900344 m. (1.10207777777777775774 m^2 + 0.9017 m^2 + sqrt(1.10207777777777775774 m^2 * 0.9017 m^2))
V = 30.7133910813887130144661572003514214582744623221020448514088195263 m^3

Volume of base
V = H * L * W
V = 1.8034 m. * 2.20415555555555551548 m. * 2.20415555555555551548 m.
V = 8.76146130938004906411745631435062018030146336 m^3

Total volume
V = 1.9550368211013333333 m^3 + 30.7133910813887130144661572003514214582744623221020448514088195263 m^3 + 8.76146130938004906411745631435062018030146336 m^3
V = 41.4298892118700954118836135147020416385759256821020448514088195263 m^3

Mass
2,700 kg/m^3 * 41.4298892118700954118836135147020416385759256821020448514088195263 m^3 = 111,860.70087204925761208575648969551242415499934167552109880381272101 kg., or 111.86 metric tons.
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