■周期的三重らせん構造(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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