知道弧长5750,拱高1070求半径 知道圆弧的弧长和拱高，怎么求半径

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\u5219\u6709L=2\u03c0r\u00d7\u03b8/360
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\u6839\u636e\u4f59\u5f26\u5b9a\u7406\uff0ccos(\u03b8/2)=(r-h)/r
\u7ed3\u5408\u4e24\u5f0f,\u03b8=180L/\uff08\u03c0r\uff09\uff0c\u5219\u03b8/2=90L/\uff08\u03c0r\uff09
cos\uff08\u03b8/2\uff09=1-h/r
cos[90L/(\u03c0r)]=1-h/r
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Rn+1=(1-(Rn*COS(C/(2*Rn))-Rn+H)/((C/2)*SIN(C/(2*Rn))-H))*Rn

R0=505
R1=479.915
R2=482.017
R3=482.034
R4=482.034

R=482.034\u6beb\u7c73

弦长一半的平方除以拱高加上拱高，再除以2就是半径

弧长C=5750，弧高H=1070。求半径R？
Rn+1=(1-(Rn*COS(C/(2*Rn))-Rn+H)/((C/2)*SIN(C/(2*Rn))-H))*Rn

R0=3862
R1=3549.126
R2=3668.595
R3=3668.787
R4=3668.787

R=3668.787

分我默默拖的

鐭ラ亾寮ч暱鍜屽崐寰勬庝箞绠鎷遍珮
绛旓細寮ч暱L锛屽崐寰凴 鎷遍珮锛漵in(L/2R)tan(L/4R)

宸茬煡寮ч暱涓7550姣背,鎷遍珮涓400姣背,鍦嗙殑鍗婂緞鏄灏?
绛旓細涓瑙ｅ喅涓嶄簡瓒呰秺鏂圭▼锛屽鏋滄槸鏈烘鎴栧伐绋嬮棶棰橈紝搴旇鎵句笓涓氳璁′汉鍛樿В鍐筹紝浣嗘槸杩樻槸缁欎綘鍒椾釜鏂圭▼鍦ㄤ笅闈細鍦嗗績瑙 伪 寮у害=R/ 7550=2*arccos(1-400/R)锛

鐭ラ亾寮ч暱鍜鎷遍珮鎬庝箞姹傚崐寰
绛旓細杩囧姬鐨勪袱涓鐐瑰仛鍦嗗姬鍒囩嚎鐨勫瀭绾匡紝涓庡姬楂樺欢闀跨嚎浜や簬鐐筄锛岀偣O鍗充负鍦嗗績銆備袱鏉″瀭绾挎墍澶硅鍗充负胃锛屽姬绔偣鑷砄鐨勮窛绂诲嵆涓簉銆傛牴鎹綑寮﹀畾鐞嗭紝cos(胃/2)=(r-h)/r 缁撳悎涓ゅ紡,胃=180L/锛埾r锛夛紝鍒櫸/2=90L/锛埾r锛塩os锛埼/2锛=1-h/r cos[90L/(蟺r)]=1-h/r 杩欐槸寮ч暱銆鎷遍珮鍜屽崐寰勭殑鍏崇郴...

D=6.9绫,鎷遍珮1.7绫,姹寮ч暱
绛旓細鍥

宸崇煡鐜勯暱1500mm,鎷遍珮750mm,姹寮ч暱
绛旓細璁惧崐寰勪负r r²-(r-750)²=(1500梅2)²1500r-562500=562500 r=750mm 鍦嗗績瑙=2Xarcsin(1500梅2梅750)=180掳 锛堝崐鍦嗭級寮ч暱=3.14X750X2X(180掳梅360掳) =2355姣背

鐜荤拑鐨寮ч暱涓1150mm,鎷遍珮涓70mm,姹傛洸鐜囦负澶氬皯,鍦ㄧ嚎绛夈
绛旓細鍗婂緞R=2350锛屽鸡闀 L=1138.56銆

寮ч暱6985鎷遍珮950姹傚崐寰
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鎷卞舰寮ч暱璁＄畻鍏紡,宸茬煡璺ㄥ害鍜鎷遍珮,姹傛嫳褰㈠姬闀 璺ㄥ害1200,鎷遍珮750...
绛旓細璺ㄥ害L=1200,鎷遍珮H=750,姹寮ч暱C?寮у崐寰勪负R,寮ф墍瀵圭殑鍦嗗績瑙掍负A.R^2=(R-H)^2+(L/2)^2R^2=R^2-2*R*H+H^2+L^2/4 2*R*H=H^2+L^2/4 R=H/2+L^2/(8*H)=750/2+1200^2/(8*750)=615A=2*ARC SIN((L/2)/R)=2*ARC SIN((1200/2)/615)...

宸茬煡寮ч暱鍜鎷遍珮濡備綍鐭ラ亾鍗婂緞鏄灏?
绛旓細鍥炵瓟锛氭暟瀛︿笂:宸茬煡寮ч暱L 銆鎷遍珮H,璁捐涓篴(寮у害)銆佸崐寰剅 鏈変袱涓柟绋嬪紡: L/2r= acosa=(r-H)/r 鍙槸杩欎釜鏂圭▼缁勮В璧锋潵姣旇緝楹荤儲

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