■ロータリーエンジンはこれから何処へ向かうのか?(その38)

x=ecos(n-1)t+Rcost

y=esin(n-1)t+Rsint

のくびれ点を常に正(n-1)角形が通るようにできるだろうか?

x'=-(n-1)esin(n-1)t-Rsint

y'=(n-1)ecos(n-1)t+Rcost

x''=-(n-1)^2ecos(n-1)t-Rcost

y''=-(n-1)^2esin(n-1)t-Rsint

x'=-(n-1)esin(n-1)t-Rsint

y''=-(n-1)^2esin(n-1)t-Rsint

x''=-(n-1)^2ecos(n-1)t-Rcost

y'=(n-1)ecos(n-1)t+Rcost

x'y''-x''y'=n(n-1)eRcos(n-2)t+(n-1)^3e^2+R^2

t=mπ/(n-2)、mは1,2,・・・,n-1

t=π/(n-2)

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t0:(x0,y0)

t1=t0+2π/(n-1):(x1,y1)

t2=t0-2π/(n-1):(x2,y2)も同じ曲線上にある

正n-1角形の1辺の長さを2Lとすると

Rsinπ/(n-1)=L

しかし、これだけでは一意に求めることはできないと思われる

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x0=ecos(n-1)t0+Rcost0

y0=esin(n-1)t0+Rsint0

x1=ecos(n-1)t0+Rcos(t0+2π/(n-1))

y1=esin(n-1)t0+Rsin(t0+2π/(n-1))

(Rcos(t0+2π/(n-1))-Rcost0)^2+(Rsin(t0+2π/(n-1))-Rsin0)^2=4L^2

2R^2-2R^2cos(t0+2π/(n-1))cost0-2R^2sin(t0+2π/(n-1))sint0

=2R^2-2R^2cos(2π/(n-1))

=2R^2・2sin^2π/(n-1)=4L^2・・・恒等式

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x=e+R,y=0を通って

傾きがπ-(n-3)/2(n-1)π=(n+1)π/2(n-1)の直線

y=tan(n+1)π/2(n-1){x-e-R}が

x=ecos(n-1)π/(n-2)π+Rcosπ/(n-2)

y=esin(n-1)π/(n-2)+Rsinπ/(n-2)

と通るとしたら・・・

R/e={-tan(n+1)π/2(n-1){cos(n-1)π/(n-2)-1}+sin(n-1)π/(n-2)}/{tan(n+1)π/2(n-1){cosπ/(n-2)-1}-sinπ/(n-2)}

R/e={((1+cos2π/(n-1))/sin2π/(n-1){-cosπ/(n-2)-1}-sinπ/(n-2)}/{-(1+cos2π/(n-1))/sin2π/(n-1){cosπ/(n-2)-1}-sinπ/(n-2)}

R/e={(1+cos2π/(n-1)){cosπ/(n-2)+1}-sinπ/(n-2)sin2π/(n-1)}/{(1+cos2π/(n-1)){cosπ/(n-2)-1}-sinπ/(n-2)sin2π/(n-1)}

n=4のとき、R/e=3.73206

n=5のとき、R/e=6.46411

n=6のとき、R/e=10.0559

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