弧長C=6,拱高H=1,求弦長L?
弧半徑為R,弧所對的圓心角為A.
Rn+1=(1-(Rn*COS(C/(2*Rn))-Rn+H)/((C/2)*SIN(C/(2*Rn))-H))*Rn
R0=4
R1=4.28
R2=4.321
R3=4.322
R4=4.322
R=4.322
A=C/R=6/4.322=1.388246弧度=1.388246*180/π=79.54度
L=2*R*SIN(A/2)
=2*4.322*SIN(79.54/2)
=5.5297Rn+1=(1-(Rn*COS(C/(2*Rn))-Rn+H)/((C/2)*SIN(C/(2*Rn))-H))*Rn
當(dāng)R0=4時(shí)，R0+1=R1,即R1=[1-(4*COS0.75-4+1)/(3*SIN0.75-1)]*4=[1-0.9997/(-0.9607)]*4=8.162...不是R1=4.28啊，求助?。?！當(dāng)R0=4時(shí)，R0+1=R1,
即R1=[1-(4*COS0.75弧度-4+1)/(3*SIN0.75弧度-1)]*4
=[1-(4*COS42.972度-4+1)/(3*SIN42.972度-1)]*4
=[1-(-0.0732445245)/(1.04491628)]*4
=4.28