Question

弧长5.3米,高0.3米,半径该多少

Answer 1

弧长C=5.3米,高H=0.3米,半径R该多少?

Rn+1=(1-(Rn*COS(C/(2*Rn))-Rn+H)/((C/2)*SIN(C/(2*Rn))-H))*Rn

R0=11

R1=11.5848

R2=11.653

R3=11.6538

R4=11.6538

R=11.6538