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Recreation of this calc made by Lina Shields, that due to some issues explained here. Calc will still use the scaling from the previous one, credit to Lina Shields.

The values obtained were:

• Cone length (H) = 0.558 m
• Cone radious = 1.15/2 = 0.575 m
• Frustum radious = 0.668/2 = 0.334 m
• Frustum length (h) = 0.227 m

Two radious, two areas, and those are:

• A1 = pi*0.575^2 = 1.039 m^2
• A2 = pi*0.334^2 = 0.35 m^2

Since the cone isn't circular at all, is made of several triangles, I cleaned an equation to compensate that: k = n/(2*pi)*sin[360/n], where n is the number of division/sides of the diamond (equation were cleaned from the Area of a Semi-circunference and Area of a Triangle). The diamond in the seems to have six sides (correct me if I miscounted), so k = 6/(2*pi)*sin[360/6] = 0.827, so real areas would be:

• k*A1 = 0.86 m^2 = Ac
• k*A2 = 0.29 m^2 = Af

Volume of the diamond:

• Cone: A1*H/3 = 0.86*0.558/3 = 0.16 m^3
• Frustum (truncated cone): 1/3*[Ac + Af + (Ac*Af)^(0.5)]*h = 1/3*[0.86 + 0.29 + (0.86*0.29)^(0.5)]*0.227 = 0.125 m^3
• Total: v = 0.16 + 0.125 = 0.285 m^3

Diamond density is 3510 kg/m^3, so the mass is 0.285*3510 = 1000.35 kg, that was the grafite mass that were converted to diamond, that the volume of grafite converted were 1000.35/2260 = ~0.442 m^3 = 442632.74 cc.

Recently, I made a blog about converting carbone to diamond, but since hasn't been accepted yet, I'll going to use the value in the Lina calc and going with two results.

• 442632.74*156545 = 69.292G J = 16.56 tnt tons
• 442632.74*172252.5 = 76.244G J = 18.22 tnt tons (Brute Force constant, not accepted yet)

Both results are 8-B (City Block) After some discussion, Kkapoios made a another blog where constant could be more accurate, being 106 656.6 J/cc, so the yield would be 442632.74*106656.6 = 47.209G J = 11.283 tnt tons, barely 8-B (City Block)

NOTE: Cerebella also destroyed the rest of the grafite, but since the 72.4% of the volume weren't converted into diamond, even if the rest were reduced to dust, the amount of energy is negligible compared with the ones obtained from above; so, consider those result pretty similar to the real ones, with an error of less that 4%

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