■周期的四面体らせん構造(その9)

 次に中心Oを求めなければならない.

  A(x,−bs,bc)

  B(x,bs,−bc)

  C(−x,c/2,s/2)

  D(−x,−c/2,−s/2)

  E(α,βc−γs,βs+γc)

  O(ξ,0,0)

  OA=(OB=OC)=OD

(x−ξ)^2+b^2s^2=(x+ξ)^2+c^2/4

−4xξ+b^2s^2−c^2/4=0

を満たすξが求まる.

  A(x−ξ,−bs,bc)

  D(x−ξ,bs,−bc)

  C(−x−ξ,c/2,s/2)

  B(−x−ξ,−c/2,−s/2)

  E(α−ξ,βc−γs,βs+γc)

  O(0,0,0)

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cos(∠COD)=(−x^2+ξ^2−bsc/2)/{((x−ξ)^2+b^2s^2)((x+ξ)^2+c^2/4)}^1/2

cos(∠AOB)=(−x^2+ξ^2−bsc/2)/{((x−ξ)^2+b^2s^2)((x+ξ)^2+c^2/4)}^1/2

∠COD=∠AOB

−4xξ+b^2s^2−c^2/4=0を代入すると,

(x−ξ)^2+b^2s^2)=x^2+ξ^2−2xξ+b^2s^2

=x^2+ξ^2−b^2s^2/2+c^2/8+b^2s^2

=x^2+ξ^2+b^2s^2/2+c^2/8

(x+ξ)^2+c^2/4=x^2+ξ^2+2xξ+c^2/4

=x^2+ξ^2+b^2s^2/2−c^2/8+c^2/4

=x^2+ξ^2+b^2s^2/2+c^2/8

どちらも等しい.

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