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[原创] 均分田地,田埂最短问题

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发表于 2019-7-5 21:47:22 | 显示全部楼层
对于135# N=8 构型III,即128# 第三图的计算过程及结果:

设\(AD=BC=2c,AA1=BB1=2d,CC1=DD1=2e,AE=BE=2b,CD=CF=2h,HE=2a,HJ=HI=2k,FF1=2m\)

\(\widehat{AD},\widehat{AA1},\widehat{DD1},\widehat{EA},\widehat{DF},\widehat{HI}\)圆弧段的半弧度,半径分别为\(t_1,t_2,t_3,t_4,t_5,t_6,R_1,R_2,R_3,R_4,R_5,R_6\)

\(\angle HOI=\alpha_4,\angle IOA1=\alpha_3,\angle A1OA=\alpha_1,\angle AOD1=\alpha_2,\angle DAD1=\theta,\angle FDF1=\beta\)

\(OA=2z,AD1=2u,DF1=2v,EO=2x\)

对曲三边形HIJ有:

\((\frac{k}{\sin(t_6)})^2(2t_6-\sin(2t_6))+2k^2\sin(\frac{2\pi}{3}-2t_6)+\frac{1}{2}(\frac{2\pi}{3}-4t_6-\sin(\frac{2\pi}{3}-4t_6))=\frac{\pi}{8}\)

在三角形HIE中:

\(\frac{2a+2x}{\sin(t_6)}=\frac{2k}{\sin(\alpha_4)}=\frac{1}{\sin(\frac{2\pi}{3}+t_6)}\)

\(\alpha_4=\frac{\pi}{3}-2t_6\)

对曲五边形 IHEAA1:

\(\frac{3}{2}\frac{\pi}{8}+2bx\sin(\frac{\pi}{3}-t_4)-\frac{1}{2}(\frac{k}{\sin(t_6)})^2(2t_6-\sin(2t_6))-\frac{1}{2}(\frac{d}{\sin(t_2)})^2(2t_2-\sin(2t_2))-\frac{1}{2}(\frac{b}{\sin(t_4)})^2(2t_4-\sin(2t_4))=\frac{1}{2}(\alpha_3+\alpha_4)+2d\sin(t_2)\)

在三角形AEO中:

\(\cos(\alpha_1+\alpha_3+\alpha_4)=\frac{x^2-b^2+z^2}{2xz}\)

在三角形OAA1中:

\(\frac{d}{\sin(\alpha_1)}=\frac{z}{\sin(t_2)}\)

在三角形ODD1中:

\(OD^2=1+4e^2-4e\cos(t_3)=(2h\sin(\frac{\pi}{3}+t_5))^2+(2b\cos(\frac{\pi}{3}-t_4)+2c\cos(2t_5-t_1)-2x)^2\)

对曲六边形AEBCFD:

\((\frac{b}{\sin(t_4)})^2(2t_4-\sin(2t_4))+(\frac{c}{\sin(t_1)})^2(2t_1-\sin(2t_1))-(\frac{h}{\sin(t_5)})^2(2t_5-\sin(2t_5))+2b^2\sin(\frac{2\pi}{3}-2t_4)+4(b\sin(\frac{\pi}{3}-t_4)+h\sin(\frac{\pi}{3}+t_5))c\cos(2t_5-t_1)+2h^2\sin(\frac{2\pi}{3}+2t_5)=\frac{\pi}{8}\)

在三角形ADD1中:

\(u^2=c^2+e^2-2ce\cos(\frac{2\pi}{3}+t_3+t_1)\)

\(\frac{e}{\sin(\theta)}=\frac{u}{\sin(\frac{2\pi}{3}+t_3+t_1)}\)

对曲四边形ADD1A1:

\(\frac{1}{2}(\frac{d}{\sin(t_2)})^2(2t_2-\sin(2t_2))-\frac{1}{2}(\frac{c}{\sin(t_1)})^2(2t_1-\sin(2t_1))-\frac{1}{2}(\frac{e}{\sin(t_3)})^2(2t_3-\sin(2t_3))+2ce\sin(\frac{2\pi}{3}+t_1+t_3)+2cd\sin(\frac{2\pi}{3}-t_2+t_1-\theta)+\frac{1}{2}(\alpha_1+\alpha_2-\sin(\alpha_1+\alpha_2))=\frac{\pi}{8}\)

对三角形A1OD1边A1D1边上的高线:

\((\sin(\frac{\alpha_1+\alpha_2}{2})^2=u^2+d^2-2ud\cos(\frac{2\pi}{3}-t_2+t_1-\theta)\)

\(\frac{\pi}{3}-2t_2+2t_1+2t_3=\alpha_1+\alpha_2\)

对三角形D1OF1边D1F1上高线:

\(\sin(\frac{\pi}{6}-t_3-t_5)^2=e^2+v^2-2ev\cos(\frac{2\pi}{3}-t_3-t_5-\beta)\)

\(\pi-\alpha_1-\alpha_2-\alpha_3-\alpha_4=\frac{\pi}{3}-2t_3-2t_5\)

对三角形ADF中:

\(AF^2=4c^2+4h^2-8ch\cos(\frac{2\pi}{3}-t_1+t_5)=(2h\cos(\frac{\pi}{3}+t_5)+2c\cos(2t_5-t_1))^2+(2b\sin(\frac{\pi}{3}-t_4))^2\)

在三角形DFF1中:

\(v^2=m^2+h^2-2mh\cos(\frac{2\pi}{3}-t_5)\)

\(\frac{m}{\sin(\beta)}=\frac{v}{\sin(\frac{2\pi}{3}-t_5)}\)

对直线EF1计算:

\(2b\cos(\frac{\pi}{3}-t_4)+2c\cos(2t_5-t_1)+2h\cos(\frac{\pi}{3}+t_5)+2m=1+2x\)

对曲四边形DFF1D1:

\(\frac{1}{2}(\frac{h}{\sin(t_5)})^2(2t_5-\sin(2t_5))+(\frac{1}{2}(\frac{e}{\sin(t_3)})^2(2t_3-\sin(2t_3))+2mh\sin(\frac{2\pi}{3}-t_5)+2ev\sin(\frac{2\pi}{3}-t_3-t_5-\beta)+\frac{1}{2}(\frac{\pi}{3}-2t_3-2t_5-\sin(\frac{\pi}{3}-2t_5-2t_3))=\frac{\pi}{8}\)

对三角形OIJ边IJ上的高线:

\(\sin(\alpha_4)=2x+2a+2k\cos(\frac{\pi}{3}-t_6)\)

对A点三段圆弧倒数和为0有:

\(-\frac{\sin(t_4)}{b}+\frac{\sin(t_2)}{d}+\frac{\sin(t_1)}{c}=0\)

对D点三段圆弧倒数和为0有:

\(-\frac{\sin(t_3)}{e}+\frac{\sin(t_1)}{c}+\frac{\sin(t_5)}{h}=0\)

求解以上方程组可以得到:

\(a = 0.02840056761679086764020987, b =0 .2182797276912673582001118, c =0 .1686269439984292607342213, d = 0.3145185737042352252996932, e =0 .2820335668015998276599048, h =0 .1978111062683137000494497, k = 0.3566176463472298808105121, m =0 .2745741673631369728856403, t_1 = -0.01217620413037148299980614, t_2 = 0.06597845755042942615413943, t_3 = 0.02014347510207000591370788, t_4 = 0.03, t_5 = 0.02841416378334742503373155, t_6 = 0.2582619959693443811907420, x =0 .1515594488990822511554991, z =0 .1924756306598008399453179\)

\(u =0 .3952103420482784349235128, v =0 .4076020364562293526008298, \theta =0 .6625869040892034057693143, \beta =0 .6344279097489557805807071, \alpha_1 =0 .1079446937887457577534671, \alpha_2 =0 .823230484250390181920271, \alpha_3 =0 .7185153201909963309737759, \alpha_4 =0 .5306735592579089837727300\)

\(R_1 = -13.84923487116798059726764, R_2 = 4.770449216032525622843149, R_3 = 14.00218379629254590107556, R_4 = 7.277082436288434658427021, R_5 = 6.962644683183725020715182, R_6 = 1.396307140844249755301803\)

\(L = 4R_1t_1+4R_2t_2+4R_3t_3+4R_4t_4+4R_5t_5+4R_6t_6+2a+2m=6.774725075060182050341294\)

画图:

N=8 MATHE.gif

点评

同样,BB1没有垂直外圆周  发表于 2019-7-9 20:12
结果不是最优,只会偏大而不会偏小,而计算误差不会有这么大  发表于 2019-7-6 15:42
是的,你的结果并不是完全对称的方程计算,可能会有偏差~  发表于 2019-7-6 14:03
计算结果好像比我128#结果偏大  发表于 2019-7-6 13:45
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-7-6 16:16:28 | 显示全部楼层
128#贴出的结果的确精度稍微有点偏低,但是偏离不大,现在给个精度达到14位左右并且将图片旋转以后如下
(16:14) gp > A*M  (对应F点)
%92 = [1.5221940048656770969668838045971264686 E-14, -0.45199296262539374991788743807025850018]
(16:14) gp > B*M  (对应D点)
%93 = [0.34792727669992322412439040518738488792, -0.26317013002611812014442235552468224459]
(16:14) gp > C*M  (对应A点)
%94 = [0.36891130201476497247239041246685665175, 0.073367794368926618454937162567301213095]
(16:14) gp > D*M  (对应E点)
%95 = [9.6266224032440448824658132300685953087 E-16, 0.30439934704387116974972434987494265639]
(16:14) gp > E*M  (对应B点)
%96 = [-0.36891130201476073218511302554708704401, 0.073367794368937373771202039117446019282]
(16:14) gp > F*M  (对应C点)
%97 = [-0.34792727669992471405096251519433141347, -0.26317013002610144706503836404694688901]
(16:14) gp > G*M  (对应H点)
%98 = [-7.8024697399450363776688078985712668202 E-16, 0.35992003303179014220651761131323385136]
(16:14) gp > H*M  (对应F1点)
%99 = [2.5713938924237539237 E-39, -0.99999999999999533767255619528913135140]
(16:14) gp > II*M  (对应D1点)
%101 = [0.81410606598645220762711319872499335987, -0.58071620721662866360043846867110779235]
(16:14) gp > J*M  (对应A1点)
%102 = [0.95152505575446494854129337474955352273, 0.30757124096942762140252139573424375903]
(16:14) gp > K*M  (对应I点)
%103 = [0.50611437817529398439811662312914894913, 0.86246636815834434674715915972627758409]
(16:14) gp > L*M  (对应J点)
%104 = [-0.50611437817529200886685415671993837552, 0.86246636815834500310243560072933772730]
(16:14) gp > N*M  (对应C1点)
%105 = [-0.81410606598646085335243238536530256230, -0.58071620721660398952140683415606915653]
        Theta(B=>A)=0.02637144815747
        Theta(B=>C)=-0.00952911763614
        Theta(C=>D)=0.03590056579361
        Theta(D=>E)=0.03590056579361
        Theta(E=>F)=-0.00952911763615
        Theta(F=>A)=-0.02637144815747
        Theta(G=>D)=-0.00000000000000
        Theta(A=>H)=-0.00000000000001
        Theta(N=>F)=-0.02163324491014
        Theta(M=>E)=-0.06957907954494
        Theta(L=>G)=0.25826199596937
        Theta(K=>G)=-0.25826199596937
        Theta(J=>C)=0.06957907954494
        Theta(I=>B)=0.02163324491014
总长度6.76800596131092
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-7-9 18:35:53 | 显示全部楼层
关于123# N=9 构型II 及129# 的计算过程及结果:

设\(AJ=2a,BC=IH=2b,AB=AI=2c,IP=KB=2d,CD=HG=2e,CH=2h,DE=GF=2m,DL=GQ=2n,EF=2p,EM=FN=2u\)

圆弧段\(\widehat{BC},\widehat{IP},\widehat{CD},\widehat{CH},\widehat{DE},\widehat{DL},\widehat{EF},\widehat{EM}\)的半弧度及半径分别为\(t_0,t_1,t_2,t_3,t_4,t_5,t_6,t_7,t_8,R_0,R_1,R_2,R_3,R_4,R_5,R_6,R_7,R_8\)

\(BJ=2y,IG=2z_1,IO=2z,GN=2w\),单位圆圆心\(O_1\)到CH中点的距离为\(2x\),\(\angle ABJ=\alpha_1,\angle HGI=\beta_1,\angle HIG=\beta_2,\angle GIO=\beta_3,\angle FGN=\theta\)

对曲四边形\(ABKJ\)有:

\(-\frac{1}{2}(\frac{c}{\sin(t_1)})^2(2t_1-\sin(2t_1))+\frac{1}{2}(\frac{d}{\sin(t_2)})^2(2t_2-\sin(2t_2))+2ac\sin(\frac{2\pi}{3}+t_1)+2yd\sin(\frac{2\pi}{3}+t_1-t_2-\alpha_1)+\frac{1}{2}(\frac{\pi}{3}+2t_1-2t_2-\sin(\frac{\pi}{3}+2t_1-2t_2))=\frac{\pi}{9}\)

对三角形\(O_1PI\)中计算\(OI^2\):

\((2b\sin(\frac{\pi}{6}-t_0-t_4)+h)^2+(2b\cos(\frac{\pi}{6}-t_0-t_4)+2x)^2=1+4d^2-4d\cos(t_2)\)

对单位圆半径\(O_1J\):

\(2a+2c\cos(\frac{\pi}{3}-t_1)+2b\cos(\frac{\pi}{6}-t_0-t_4)+2x=1\)

对曲五边形\(ABCHI\)有:

\((\frac{c}{\sin(t_1)})^2(2t_1-\sin(2t_1))+(\frac{b}{\sin(t_0)})^2(2t_0-\sin(2t_0))+\frac{1}{2}(\frac{h}{\sin(t_4)})^2(2t_4-\sin(2t_4))+2c^2\sin(\frac{2\pi}{3}-2t_1)+(2h+4c\sin(\frac{\pi}{3}-t_1))b\cos(\frac{\pi}{6}-t_0-t_4)=\frac{\pi}{9}\)

对于曲五边形\(PIHGO\)有:

\(-\frac{1}{2}(\frac{d}{\sin(t_2)})^2(2t_2-\sin(2t_2))-\frac{1}{2}(\frac{b}{\sin(t_0)})^2(2t_0-\sin(2t_0))-\frac{1}{2}(\frac{e}{\sin(t_3)})^2(2t_3-\sin(2t_3))-\frac{1}{2}(\frac{n}{\sin(t_6)})^2(2t_6-\sin(2t_6))+2be\sin(\frac{2\pi}{3}+t_0+t_3)+2z_1n\sin(\frac{2\pi}{3}+t_3+t_6-\beta_1)+2dz\sin(\frac{2\pi}{3}+t_2+t_0-\beta_2-\beta_3)+\frac{1}{2}(2(t_0+t_2+t_3+t_6)-\sin(2(t_0+t_2+t_3+t_6)))=\frac{\pi}{9}\)

对三角形\(O_1GO\)有:

\((2e\sin(\frac{\pi}{3}-t_4+t_3)-2x)^2+(2e\cos(\frac{\pi}{3}-t_4+t_3)+h)^2=1+4n^2-4n\cos(t_6)\)

对曲六边形\(HCDEFG\)有:

\(-\frac{1}{2}(\frac{h}{\sin(t_4)})^2(2t_4-\sin(2t_4))+\frac{1}{2}(\frac{e}{\sin(t_3)})^2(2t_3-\sin(2t_3))+\frac{1}{2}(\frac{m}{\sin(t_5)})^2(2t_5-\sin(2t_5))+\frac{1}{2}(\frac{p}{\sin(t_7)})^2(2t_7-\sin(2t_7))+(4h+4e\cos(\frac{\pi}{3}-t_4+t_3))e\sin(\frac{\pi}{3}-t_4+t_3)+(4p+4m\cos(\frac{\pi}{3}+t_5+t_7))m\sin(\frac{\pi}{3}+t_5+t_7)=\frac{\pi}{9}\)

对\(DG\)计算:

\(2h+4e\cos(\frac{\pi}{3}-t_4+t_3)=2p+4m\cos(\frac{\pi}{3}+t_5+t_7)\)

对曲四边形\(OGFN\)有:

\(-\frac{1}{2}(\frac{m}{\sin(t_5)})^2(2t_5-\sin(2t_5))+\frac{1}{2}(\frac{n}{\sin(t_6)})^2(2t_6-\sin(2t_6))-\frac{1}{2}(\frac{u}{\sin(t_8)})^2(2t_8-\sin(2t_8))+2mu\sin(\frac{2\pi}{3}+t_5+t_8)+2nw\sin(\frac{2\pi}{3}-t_6+t_5-\theta)+\frac{1}{2}(\frac{\pi}{3}+2(t_5-t_6+t_8)-\sin(\frac{\pi}{3}+2(t_5-t_6+t_8))=\frac{\pi}{9}\)

对曲四边形\(FEMN\)有:

\(-\frac{1}{2}(\frac{p}{\sin(t_7)})^2(2t_7-\sin(2t_7))+\frac{1}{2}(\frac{u}{\sin(t_8)})^2(2t_8-\sin(2t_8))+(4p+4u\cos(\frac{\pi}{3}-t_7+t_8))u\sin(\frac{\pi}{3}-t_7+t_8)+\frac{1}{2}(\frac{\pi}{3}+2t_7-4t_8-\sin(\frac{\pi}{3}+2t_7-4t_8))=\frac{\pi}{9}\)

对三角形\(O_1FN\)计算\(O_1F^2\):

\((2e\sin(\frac{\pi}{3}-t_4+t_3)+2m\sin(\frac{\pi}{3}+t_5+t_7)-2x)^2+p^2=1+4u^2-4u\cos(t_8)\)

计算三角形\(O_1MN\)底边上的高线:

\(\cos(\frac{\pi}{6}+t_7-2t_8)=2e\sin(\frac{\pi}{3}-t_4+t3)+2m\sin(\frac{\pi}{3}+t_5+t_7)-2x+2u\sin(\frac{\pi}{3}-t_7+t_8)\)

计算三角形\(IPQ\)中底边PQ半长:

\(\sin(t_0+t_2+t_3+t_6)^2=d^2+z^2-2dz\cos(\frac{2\pi}{3}+t_2+t_0-\beta_2-\beta_3)\)

计算三角形\(KBJ\)底边KJ半长:

\(\sin(\frac{\pi}{6}+t_1-t_2)^2=y^2+d^2-2yd\cos(\frac{2\pi}{3}-t_2+t_1-\alpha_1)\)

对I点三圆弧倒数和为0:

\(\frac{\sin(t_0)}{b}=\frac{\sin(t_1)}{c}+\frac{\sin(t_2)}{d}\)

对H点三圆弧倒数和为0:

\(\frac{\sin(t_0)}{b}=\frac{\sin(t_4)}{h}+\frac{\sin(t_3)}{e}\)

对G点三圆弧倒数和为0:

\(\frac{\sin(t_3)}{e}=\frac{\sin(t_5)}{m}+\frac{\sin(t_6)}{n}\)

对F点三圆弧倒数和为0:

\(\frac{\sin(t_5)}{m}=\frac{\sin(t_7)}{p}+\frac{\sin(t_8)}{u}\)

对\(\angle AIH\)有:

\(2(t_0+t_1)+t_4=\frac{\pi}{6}\)

对三角形\(ABJ\)中:

\(a^2+c^2-2ac\cos(\frac{2\pi}{3}+t_1)=y^2\)

\(a\sin(\frac{2\pi}{3}+t_1)=y\sin(\alpha_1)\)

在三角形\(HGJ\)中:

\(e\sin(\frac{2\pi}{3}+t_3+t_0)=z_1\sin(\beta_2)\)

\(e^2+b^2-2eb\cos(\frac{2\pi}{3}+t_3+t_0)=z_1^2\)

在三角形\(FNG\)中:

\(w^2=m^2+u^2-2mu\cos(\frac{2\pi}{3}+t_5+t_8)\)

\(u\sin(\frac{2\pi}{3}+t_5+t_8)=w\sin(\theta)\)

计算\(BI\)长度:

\(4c\sin(\frac{\pi}{3}-t_1)=2h+4b\cos(\frac{\pi}{3}+t_4+t_0)\)

计算\(\angle HGF\):

\(2(t_5+t_3)+t_7=t_4\)

计算\(O_1MN\)底边MN的半长:

\(\sin(\frac{\pi}{6}+t_7-2t_8)=p+2u\cos(\frac{\pi}{3}-t_7+t_8)\)


求解以上方程组得到:

\(a = 0.1953662988355465590166726, \alpha_1 = 0.4193484969967193186331015, b = 0.1548223482206260003862638,\beta_1 = 0.5380404025457994667394824, \beta_2 = 0.3414907559923121161887345, \beta_3 = 0.7804392654386446442070010, c = 0.2481471255507734971805962, d = 0.2419565373319284707189872, e = 0.1011773988931060019917954, h = 0.3184815112068733435603822, m = 0.1982431166126087069250304, n = 0.2770433577131363813859600, p = 0.2346931985914641379958828, t_0 = 0.1225718959021155608545727, t_1 = 0.08431179687878592662763358, t_2 = 0.1091819075434388507795040, t_3 = 0.04509449675637060237142475, t_4 = 0.1098313900364958981126947, t_5 = 0.01338489929972743817899745, t_6 = 0.1049233060225625283346378, t_7 = -0.007127402075700182988149704, t_8 = 0.02264056041368733926708800, \theta = 0.5481381479511567156016647, u = 0.2312793361147615243312529, w = 0.3761257088948647501967653, x = 0.01459903867378488498267554, y = 0.3938572935956257668232283, z = 0.3830039465673149558613384, z_1 = 0.2327648622426534307665223\)

\(R_0 = 1.266282939155316529541030, R_1 = 2.946697498890872153499132, R_2 = 2.220495381975475846705067, R_3 = 0.8275240333125474870819006, R_4 = 2.905569318228195006887039, R_5 = 14.81139541573680061609157, R_6 = 2.645287740606731998162278, R_7 = -32.92857385821097761979464, R_8 = 10.21613828638155047443456\)

\(L=4R_0t_0 + 4R_1t_1 + 4R_2t_2 + 4R_3t_3 + 2R_4t_4 + 4R_5t_5 + 4R_6t_6 + 2R_7t_7 + 4R_8t_8 + 2a=7.060397324113186962020917\)

画图得到:

N=9 mathe III.gif

点评

我是先假定一个方向,然后计算方程,若为负值则应反向;我作图时因为没有检查方向因此会有几个方向是方的,这个我可以调整过来  发表于 2019-7-10 09:02
猜测是你的面积计算中部分圆弧部分的面积弄错了符号  发表于 2019-7-10 08:00
这个方向应该是错的,可以调整,不影响上面结果~我是根据上面的方程求长度,然后根据长度求各点坐标,  发表于 2019-7-9 20:42
看的你弧线KB不垂直外圆周,所以肯定有地方没有弄对  发表于 2019-7-9 20:03
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-7-9 19:39:38 | 显示全部楼层
我们的结果还是不一致,你能把坐标算出来比较一下吗?
另外发现我们图中几条弧线方向是反的
Result:
        A(-0.14034883297077,-0.51956269364913)
                A=>K(-0.02264572630075)
                A=>B(0.00683521819668)
                A=>C(0.01365160332952)
        B(-0.49509092786390,-0.21101696778947)
                B=>A(-0.00683521819668)
                B=>J(0.02264572630075)
                B=>H(-0.01365160332952)
        C(0.23324810294360,-0.38818437788779)
                C=>L(-0.10370943688338)
                C=>A(-0.01365160332952)
                C=>F(0.04488378327096)
        D(0.52099796897131,-0.02132162455011)
                D=>M(0.10911694634698)
                D=>E(0.08407578100476)
                D=>F(-0.12260582929225)
        E(0.39963476483586,0.45946925135651)
                E=>N(-0.00000000000001)
                E=>D(-0.08407578100476)
                E=>G(0.08407578100477)
        F(0.25889655965727,-0.18717476347492)
                F=>D(0.12260582929225)
                F=>C(-0.04488378327096)
                F=>I(-0.11023555500427)
        G(-0.09333789642328,0.51301221470541)
                G=>O(-0.10911694634698)
                G=>E(-0.08407578100477)
                G=>I(0.12260582929224)
        H(-0.41677003655140,0.17718500330407)
                H=>B(0.01365160332952)
                H=>I(-0.04488378327096)
                H=>P(0.10370943688338)
        I(-0.22125657276086,0.23045031936272)
                I=>F(0.11023555500427)
                I=>H(0.04488378327096)
                I=>G(-0.12260582929224)
        J(-0.92737520205603,-0.37413264307134)
                J=>B(-0.02264572630075)
        K(-0.24196527304681,-0.97028491003384)
                K=>A(0.02264572630075)
        L(0.61940339218229,-0.78507288690482)
                L=>C(0.10370943688338)
        M(0.98990918452933,-0.14170323350043)
                M=>D(-0.10911694634698)
        N(0.65626874680239,0.75452722414133)
                 N=>E(0.00000000000001)
        O(-0.27755897549098,0.96070860052588)
                O=>G(0.10911694634698)
        P(-0.86335656971023,0.50459432570945)
                P=>H(-0.10370943688338)
Total arc len 7.31494334143164
c9.png
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-7-9 19:55:19 | 显示全部楼层
以下是我计算出来的16个点的坐标(第三个数据为x值,第四个数据为y值)

[1, A, 0, 0.6081982336],
[2, B, -0.4064358484410108365557833, 0.3258725705758326961948072],
[3, I, 0.4064358484410108365557833, 0.3258725705758326961948072],
[4, C, -0.3174685274742648501340152, 0.02743896664],
[5, H, 0.3174685274742648501340152, 0.02743896664],
[6, J, 0, 1],
[7, K, -0.8394695934734930843059842, 0.5434066632214296442896714],
[8, P, 0.8394695934734930843059842, 0.5434066632214296442896714],
[9, D, -0.4293360431522275720909470, -0.1423422737136997505712269],
[10, G, 0.4293360431522275720909470, -0.1423422737136997505712269],
[11, L, -0.9786682800448140617287873, -0.2054468243466552066387008],
[12, O, 0.9786682800448140617287873, -0.2054468243466552066387008],
[13, E, -0.2344692550410086808263137, -0.4858354615518537109200934],
[14, F, 0.2344692550410086808263137, -0.4858354615518537109200934],
[15, M, -0.4530467255765997151816732, -0.8914867718840931570525416],
[16, N, 0.4530467255765997151816732, -0.8914867718840931570525416]


你应该是设的各个点的坐标值然后利用数值计算得到的结果吧.我觉得你在设各个点的坐标值时,利用对称原则设坐标

例如:利用对称性可以如下设参数,然后求解,得到的结果应该是准确的;并且减少了很多未知数

[1, A, 0, 1 - 2*a],
[2, B, -x1, y1],
[3, I, x1, y1],
[4, C, -x2, 2*x],
[5, H, x2, 2*x],
[6, J, 0, 1],
[7, K, -x3, y3],
[8, P, x3, y3],
[9, D, -x4, y4],
[10, G, x4, y4],
[11, L, -x5, y5],
[12, O, x5, y5],
[13, E, -x6, y6],
[14, F, x6, y6],
[15, M, -x7, y7],
[16, N, x7, y7]]

点评

我选两个面积方程检验一下我们的结果哈,主要是精度比对~  发表于 2019-7-10 09:05
我给的初始值精度已经很高了,自然可以收敛。  发表于 2019-7-10 08:00
我上面列的方程,只要给一个大概的初始值,就可以算出想要的精确位数~我就是用你给的值作为初始值计算的哈  发表于 2019-7-9 21:09
是的,数值解,对初始值很敏感,稍微偏离,就发散了  发表于 2019-7-9 20:19
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-7-10 21:53:42 | 显示全部楼层
对楼上144# mathe 关于N=9 的结果的验算(可能需要另一个人核算是否正确?)

将mathe结果的各点画出(没有将圆弧段画出,只画出了弦长,已将图形旋转90度)

N=9 mathe.gif

对曲四边形\(ABJI\)及\(AHNI\)有:

\(R_1 = 2.952434852351175847858680, R_2 = 2.222749259771117083160569, R_9 = 9.776253372*10^{12}, a = 0.1955250674490900000000000, \alpha_1 = 0.4198756913147222859145249, c = 0.2479359260338913984946207, d = 0.2420585981996870372683824, t_1 = 0.08407578100478, t_2 = 0.10911694634697, t_9 = 2.000000000*10^{-14}, y = 0.3937764655711861469369061\)

\(R_1 = 2.952434852351725373992478, R_2 = 2.222749259770709140044597, R_9 = 9.776253372*10^{12}, a = 0.1955250674490900000000000, \alpha_1 = 0.4198756913147786336845914, c = 0.2479359260338492857411436, d = 0.2420585981996647073400482, t_1 = 0.08407578100475, t_2 = 0.10911694634698, t_9 = 2.000000000*10^{-14}, y = 0.3937764655711446600034546\)

分别代入面积方程:(理论结果应为0)

\(-\frac{1}{2}(\frac{c}{\sin(t_1)})^2(2t_1-\sin(2t_1))+\frac{1}{2}(\frac{d}{\sin(t_2)})^2(2t_2-\sin(2t_2))+2ac\sin(\frac{2\pi}{3}+t_1)+2yd\sin(\frac{2\pi}{3}+t_1-t_2-\alpha_1)+\frac{1}{2}(\frac{\pi}{3}+2t_1-2t_2-\sin(\frac{\pi}{3}+2t_1-2t_2))-\frac{\pi}{9}\)

得到 :\(3.26161982481*10^{-14}\)和\(-3.16349730782*10^{-14}\)

对曲五边形\(BCDKJ\)及\(HGFMN\)有:

\(R_0 = 1.268075152575563057060586, R_2 = 2.222749259770709140044597, R_3 = 2.258136399836979329989031, R_6 = 2.674498619926179370559759, b = 0.1550841807339082490886190, \beta_1 = 0.5382145846702955929800172, \beta_2 = 0.3414933539631021531741976, \beta_3 = 0.7932658356241924491572141, d = 0.2420585981996647073400482, e = 0.1013196777755968963630534, n = 0.2768737961584571036419171, t_0 = 0.12260582929225, t_2 = 0.10911694634698, t_3 = 0.04488378327095, t_6 = 0.10370943688337, z = 0.3850323421693768036283513, z_1 = 0.2331245724851209616701347\)

\(R_0 = 1.268075152575466121747851, R_2 = 2.222749259771117083160569, R_3 = 2.258136399836599982715141, R_6 = 2.674498619925946506569440, b = 0.1550841807338838084465590, \beta_1 = 0.5382145846702483060951915, \beta_2 = 0.3414933539631494400590222, \beta_3 = 0.7932658356241914533069152, d = 0.2420585981996870372683824, e = 0.1013196777756024341605942, n = 0.2768737961584595980024110, t_0 = 0.12260582929224, t_2 = 0.10911694634697, t_3 = 0.04488378327096, t_6 = 0.10370943688338, z = 0.3850323421693771578837145, z_1 = 0.2331245724851026872311345\)

代入面积方程(理论结果应为0)

\(-\frac{1}{2}(\frac{d}{\sin(t_2)})^2(2t_2-\sin(2t_2))-\frac{1}{2}(\frac{b}{\sin(t_0)})^2(2t_0-\sin(2t_0))-\frac{1}{2}(\frac{e}{\sin(t_3)})^2(2t_3-\sin(2t_3))-\frac{1}{2}(\frac{n}{\sin(t_6)})^2(2t_6-\sin(2t_6))+2be\sin(\frac{2\pi}{3}+t_0+t_3)+2z_1n\sin(\frac{2\pi}{3}+t_3+t_6-\beta_1)+2dz\sin(\frac{2\pi}{3}+t_2+t_0-\beta_2-\beta_3)+\frac{1}{2}(2(t_0+t_2+t_3+t_6)-\sin(2(t_0+t_2+t_3+t_6)))-\frac{\pi}{9}\)

得到:\(-1.2708699747*10^{-15}\)和\( -2.5691767489*10^{-15}\)

对曲四边形\(DE1L1K\)及\(FE2L2M\)有:

\(R_5 = 14.50511692360809172141812, R_6 = 2.674498619926179370559759, R_8 = 10.20224675749035855802393, m = 0.1980119518974119046381624, n = 0.2768737961584571036419171, t_5 = 0.01365160332952, t_6 = 0.10370943688337, t_8 = 0.02264572630074, \theta = 0.5479979594814039071946316, u = 0.2310175411286317402650472, w = 0.3757222319230400163881143\)

\(R_5 = 14.50511692360799774744916, R_6 = 2.674498619925946506569440, R_8 = 10.20224675748571936951393, m = 0.1980119518974106217826626, n = 0.2768737961584595980024110, t_5 = 0.01365160332952, t_6 = 0.10370943688338, t_8 = 0.02264572630075, \theta = 0.5479979594813965565557471, u = 0.2310175411286286877596648, w = 0.3757222319230372217230239\)

代入面积方程(理论结果应为0)

得到:\(-1.9316795620*10^{-15}\)和\(-3.4033833418*10^{-15}\)

对曲四边形\(E1E2L2L1\)有:

\(R_7 = -34.39210954263664531611740, R_8 = 10.20224675748571936951393, p = 0.2350757424934750000000000, t_7 = -0.00683521819668, t_8 = 0.02264572630074, u = 0.2310175411286317402650472\)

代入面积方程(理论结果应为0)

\(-\frac{1}{2}(\frac{p}{\sin(t_7)})^2(2t_7-\sin(2t_7))+\frac{1}{2}(\frac{u}{\sin(t_8)})^2(2t_8-\sin(2t_8))+(4p+4u\cos(\frac{\pi}{3}-t_7+t_8))u\sin(\frac{\pi}{3}-t_7+t_8)+\frac{1}{2}(\frac{\pi}{3}+2t_7-4t_8-\sin(\frac{\pi}{3}+2t_7-4t_8))-\frac{\pi}{9}\)

得到 \(8.0528226597*10^{-15}\)

对曲五边形\(AHGCB\)有:

\(R_0 = 1.268075152575563057060586, R_1 = 2.952434852351725373992478, R_4 = 2.892231836848885934114376, b = 0.1550841807339082490886190, c = 0.2479359260338492857411436, h = 0.3181814499566550000000000, t_0 = 0.12260582929225, t_1 = 0.08407578100475, t_4 = 0.11023555500427\)

代入面积方程(理论结果应为0)

得到\(-1.25611227620*10^{-14}\)

对曲六边形\(GFE2E1DC\)有:

\(R_3 = 2.258136399836979329989031, R_4 = 2.892231836848885934114376, R_5 = 14.50511692360809172141812, R_7 = -34.39210954263664531611740, e = 0.1013196777755968963630534, h = 0.3181814499566550000000000, m = 0.1980119518974119046381624, p =0.2350757424934750000000000, t_3 = 0.04488378327095, t_4 = 0.11023555500427, t_5 = 0.01365160332952, t_7 = -0.00683521819668\)

代入面积方程(理论结果应为0)

\(-\frac{1}{2}(\frac{h}{\sin(t_4)})^2(2t_4-\sin(2t_4))+\frac{1}{2}(\frac{e}{\sin(t_3)})^2(2t_3-\sin(2t_3))+\frac{1}{2}(\frac{m}{\sin(t_5)})^2(2t_5-\sin(2t_5))+\frac{1}{2}(\frac{p}{\sin(t_7)})^2(2t_7-\sin(2t_7))+(4h+4e\cos(\frac{\pi}{3}-t_4+t_3))e\sin(\frac{\pi}{3}-t_4+t_3)+(4p+4m\cos(\frac{\pi}{3}+t_5+t_7))m\sin(\frac{\pi}{3}+t_5+t_7)-\frac{\pi}{9}\)

得到:\(-1.6588417051*10^{-15}\)

点评

面积公式没错,是我算你参数p取成负数,若除掉符号就对了,结果为8.05282265 97*10^(-15),第六个已一样,结果为-1.6588417051*10^(–15)  发表于 2019-7-11 09:15
找到你计算对曲四边形E1E2L2L1面积部分的错误了。 公式$(4p+4u cos(pi/3-t_7+t_i))u\sin(Pi/3-t_7+t_8)$计算梯形面积,但是p是负数,所以错了。把p改为-p代入,得出梯形面积为0.2803759074824,同我的相同  发表于 2019-7-11 07:04
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-7-10 22:23:40 | 显示全部楼层
(22:18) gp > M=[-0.45419827613860,-0.89090062630617]
%1 = [-0.45419827613860000000000000000000000000, -0.89090062630617000000000000000000000000]
(22:19) gp > N=[0.45419827613851,-0.89090062630621]
%2 = [0.45419827613851000000000000000000000000, -0.89090062630621000000000000000000000000]
(22:19) gp > F=[0.23507574249345,-0.48413074973539]
%3 = [0.23507574249345000000000000000000000000, -0.48413074973539000000000000000000000000]
(22:19) gp > E=[-0.23507574249349,-0.48413074973537]
%4 = [-0.23507574249349000000000000000000000000, -0.48413074973537000000000000000000000000]
(22:19) gp > atan(M[2]/M[1])
%5 = 1.0993242219947246517039813472958398982
(22:19) gp > Pi/2-%5
%6 = 0.47147210480017196752734034434391154390
(22:20) gp > area(u,v)=(u[1]*v[2]-u[2]*v[1])/2
%7 = (u,v)->(u[1]*v[2]-u[2]*v[1])/2
(22:20) gp > area(M,N)+area(N,F)+area(F,E)+area(E,M)
%8 = 0.28037590748241500145424619100000000000
(22:21) gp > (2*%6-sin(2*%6))/2
%9 = 0.066826576121111248819354036965516325113
(22:21) gp > %8+%9
%10 = 0.34720248360352625027360022796551632512
(22:21) gp > Pi/9
%11 = 0.34906585039886591538473815369772254269
验算区域E1E2L2L1的面积(我上面标注为E,F,M,N等了),忽略三条较直的曲边(看成直线段),可以看出%10(近似面积)很接近Pi/9
使用坐标计算多边形面积的好处是公式形式统一,便于计算机编程计算
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-7-11 21:16:41 | 显示全部楼层
很奇怪:

对于楼上146#坐标转换成143#的弦长后(与143#符号统一后)即:

[1, A, 0, 1 - 2*a]
[2, B, -x1, y1],
[3, I, x1, y1],
[4, C, -h, 2*x],
[5, H, h, 2*x],
[6, J, 0, 1],
[7, K, -x3, y3],
[8, P, x3, y3],
[9, D, -x4, y4],
[10, G, x4, y4],
[11, L, -x5, y5],
[12, O, x5, y5],
[13, E, -p, y6],
[14, F, p, y6],
[15, M, -x7, y7],
[16, N, x7, y7]

根据144# 数据(旋转90度后,布局和143#一致)
x1 = 0.40709986713448,
y1 = 0.32582693799745,
x3 = 0.83990883262206,
y3 = 0.54272751255436,
x4 = 0.43074531885496,
y4 = -0.13982224088985,
x5 = 0.98257552176460,
y5 = -0.18586377815225,
y6 = -0.48413074973536,
x7 = 0.45419827613861,
y7 = -0.89090062630616

其余参数:
R0 = 1.26807515257547, R1 = 2.95243485235118, R2 = 2.22274925977071, R3 = -2.25813639983660, R4 = 2.89223183684889, R5 = -14.5051169236080, R6 = 2.67449861992595, R7 = -34.3921095426366, R8 = 10.2022467574857, R9 = 9.776253372*10^12, a = 0.195525067449090, b = 0.155084180733908, c = 0.247935926033849, d = 0.242058598199665, h = 0.318181449956655, m = 0.198011951897411, n = 0.276873796158457, p = 0.235075742493475, u = 0.231017541128632, beta1 = 0.5382145846702483060951915, beta2 = 0.3414933539631494400590222, beta3 = 0.7932658356241914533069152, d = 0.2420585981996870372683824, e = 0.1013196777756024341605942, h = 0.3181814499566550000000000, n = 0.2768737961584595980024110, t0 = 0.12260582929224, t2 = 0.10911694634697, t3 = 0.04488378327096, t4 = 0.11023555500427, t6 = 0.10370943688338, x = 0.01433863302, z = 0.3850323421693771578837145, z1 = 0.2331245724851026872311345, alpha1 = 0.4198756913147786336845914, t5 = 0.01365160332952, t7 = -0.00683521819668, t8 = 0.02264572630074, theta = 0.5479979594814039071946316, u = 0.2310175411286317402650472, w = 0.3757222319230400163881143, y = 0.3937764655711446600034546, t1 = 0.08407578100475

代入143#的28个方程(将左项减去右项,理论为0)
得到
-2.74482602504*10^(-14), -2.4016594171251*10^(-12), -3.6893777913912*10^(-12), -2.0548953910*10^(-16), 0., -1.43084380173*10^(-14), -8.13987249352*10^(-14), -1.54929686033*10^(-14), -8.10162027856*10^(-15), 1.1*10^(-25), 1.062797521244*10^(-14), 1.031283963017*10^(-12), 5.6767702565*10^(-15), 1.5066627934*10^(-15), -5.8842214808*10^(-15), 4.0560778260*10^(-16), 2.200760991*10^(-16), 8.4718877556*10^(-15), 3.5748879151395*10^(-12), -3.6948220661324*10^(-12), 3.94421204790*10^(-15), 2.418974833212*10^(-14), 5.10074543328*10^(-14), 1.280928174052*10^(-13), -1.31312767095*10^(-14), -7.29434874565*10^(-15), -4.88730771072*10^(-14), 1.*10^(-14), 5.6705955218*10^(-15)


可以认为我143#的方程是正确的,并且mathe 给出的坐标也是正确(准确到12位)


但是:我直接求解143#得到

a = 0.1953662988355465590166726, alpha1 = 0.4193484969967193186331015, b = 0.1548223482206260003862638, beta1 = 0.5380404025457994667394824, beta2 = 0.3414907559923121161887345, beta3 = 0.7804392654386446442070010, c = 0.2481471255507734971805962, d = 0.2419565373319284707189872, e = 0.1011773988931060019917954, h = 0.3184815112068733435603822, m = 0.1982431166126087069250304, n = 0.2770433577131363813859600, p = 0.2346931985914641379958828, t0 = 0.1225718959021155608545727, t1 = 0.08431179687878592662763358, t2 = 0.1091819075434388507795040, t3 = 0.04509449675637060237142475, t4 = 0.1098313900364958981126947, t5 = 0.01338489929972743817899745, t6 = 0.1049233060225625283346378, t7 = -0.007127402075700182988149704, t8 = 0.02264056041368733926708800, theta = 0.5481381479511567156016647, u = 0.2312793361147615243312529, w = 0.3761257088948647501967653, x = 0.01459903867378488498267554, y = 0.3938572935956257668232283, z = 0.3830039465673149558613384, z1 = 0.2327648622426534307665223

并且反代回143#的28个方程(将左项减去右项,理论为0)
得到
-1.*10^(-25), 5.*10^(-25), 1.*10^(-25), -6.*10^(-26), 0., 0., -2.*10^(-25), 0., -4.*10^(-26), -1.*10^(-26), 1.*10^(-26), 0., -1.8*10^(-24), 2.*10^(-25), 0., 5.*10^(-26), 0., 1.6*10^(-24), -1.*10^(-25), 0., 9.*10^(-26), -1.2*10^(-25), 2.*10^(-25), 2.*10^(-25), -1.*10^(-25), -1.*10^(-26), 1.*10^(-25), 0., -1.*10^(-25)

精度明显比mathe 高出很多

我利用得到的精确解反过来算

x1 = 0.4073797579052223652840296, x3 = 0.8400944569950585337610475, y1 = 0.3258071465455285502905786, y3 = 0.5424401380117236012809872, x4 = 0.4312739987993303769629988, x5 = 0.9839351951253389529053724, y4 = -0.1388058480388596041110756, y5 = -0.1785259975288225657368824, y6 = -0.4831273868655789528677821, x7 = 0.4533673531691661799637804, y7 = -0.8907327589103144882491142

其中x7^2+y7^2-1=-0.0010531953(理论应为0),感觉这里的{x7,y7}似乎有点问题???
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-7-11 21:30:42 来自手机 | 显示全部楼层
你的表达式太复杂了,看不大明白。最大的可能是你用的方程数目不够。需要注意的是面积方程之间是不独立的。比如你的方案图是对称的,总共九个区域,那么给出四个区域的面积,对称性确定另外四个的面积,于是余下一个区域面积就是总面积减去其它八个面积之和,不是独立的方程。

点评

看来只有我的计算方案比较适合人工检查与修正哈  发表于 2019-7-11 22:05
表达式很长,没法列出来,只适合计算机内部表达  发表于 2019-7-11 22:00
哦,还要加9个弧度参数  发表于 2019-7-11 21:59
能否把你求解的方程列出来学习一下?  发表于 2019-7-11 21:56
我大概算了一下需要:5个面积方程,3个正交方程,7个交度约束,11个坐标方程,合计27个。若简化方程长度,需增加17个边长等式  发表于 2019-7-11 21:54
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 6 天前 | 显示全部楼层
c10.png
10个区域可以为7.83705523333973

点评

程序计算时间不重要,主要构造比较困难  发表于 6 天前
你的程序算出结果花了多少时间呢?  发表于 6 天前
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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