{"id":187,"date":"2021-02-17T23:24:44","date_gmt":"2021-02-17T15:24:44","guid":{"rendered":"https:\/\/blog.cruciferslab.net\/?p=187"},"modified":"2024-03-08T10:33:22","modified_gmt":"2024-03-08T02:33:22","slug":"%e5%86%8d%e6%8e%a2%e9%be%8d%e5%8d%9a%e5%a3%ab%e5%9c%93%e7%9b%a4%e6%8b%bc%e7%89%87-2","status":"publish","type":"post","link":"https:\/\/blog.cruciferslab.net\/?p=187","title":{"rendered":"\u518d\u63a2\u9f8d\u535a\u58eb\u5713\u76e4\u62fc\u7247 (2)"},"content":{"rendered":"\n

\u9019\u7bc7\u662f\u7cfb\u5217\u6587\u7684\u7b2c\u4e8c\u7bc7\uff0c\u4e3b\u8981\u4f86\u63a2\u8a0e 2-5 \u7d44\u5408\u7684\u8a73\u7d30\u5c3a\u5bf8\u3002\u5176\u4ed6\u7cfb\u5217\u6587\u53ef\u7531\u6a19\u7c64<\/a>\u9023\u7d50\u3002<\/p>\n\n\n\n\n\n\n\n

4 \u55ae\u4f4d\u7684\u8449\u5b50<\/h2>\n\n\n\n

\u63a2\u8a0e 2-5 \u9019\u500b\u7279\u6b8a\u7d44\u5408\u6700\u597d\u7684\u65b9\u6cd5\u9084\u662f\u8ddf\u8457\u7b2c\u4e09\u7bc7\u958b\u982d\u63d0\u5230\u7684\uff0c\u5f9e 2-5-1-7 \u9019\u500b 4 \u55ae\u4f4d\u7684\u8449\u5b50\u958b\u59cb\uff1b\u540c\u6a23\u4e0b\u5716\u6211\u628a\u9019\u500b\u7d44\u5408\u7576\u4e2d\u7684\u62fc\u7247\u57fa\u5e95\u4e09\u89d2\u5f62\u7d66\u756b\u51fa\u4f86\u4e86\uff1a<\/p>\n\n\n

\n
\"\"<\/figure>\n<\/div>\n\n\n

\u9996\u5148\u53ef\u4ee5\u770b\u5230\u865b\u7dda\u90e8\u4efd\u6709\u4e00\u689d\u5f88\u76f4\u7684\u76f4\u7dda\uff0c\u9019\u689d\u5176\u5be6\u662f\u9019\u500b\u8449\u5b50\u7684\u4e2d\u7dda\u30021 \u548c 7 \u865f\u62fc\u7247\u7684 \\(30\\degree\\) \u5728\u6b64\u767c\u63ee\u4f5c\u7528\uff1a\u7531\u65bc\u5169\u62fc\u7247\u9760\u5916\u7684\u5f27\u90fd\u662f 2 \u55ae\u4f4d (\\(60\\degree\\))\uff0c\u5c0d\u61c9\u9019\u5f27\u7684\u5713\u5468\u89d2\u5373\u662f \\(30\\degree\\)\uff0c\u7531\u6b64\u5373\u8b49\u660e\u9019\u5169\u500b\u5713\u5468\u89d2\u5728\u6b64\u5171\u7dda\uff0c\u9019\u7dda\u5c31\u662f\u8449\u5b50\u4e2d\u7dda\u3002\u9019\u4e00\u9ede\u9084\u80fd\u4ee5\u5169\u908a\u4ea4\u754c\u7684\u9019\u4e00\u89d2\u9032\u884c\u9a57\u8b49\uff1a7 \u865f\u62fc\u7247\u7684\u9019\u89d2\u662f \\(105\\degree\\)\uff0c1 \u865f\u62fc\u7247\u7684\u5e95\u89d2\u662f \\(75\\degree\\)\uff0c\u6e4a\u8d77\u4f86\u6b63\u597d\u662f\u4e00\u500b\u5e73\u89d2\u3002<\/p>\n\n\n\n

5 \u865f\u62fc\u7247\u7684\u5713\u5468\u89d2\u548c 2-5 \u4e4b\u9593\u7684\u908a<\/h2>\n\n\n\n

\u5c0d\u65bc\u6211\u5011\u6709\u8208\u8da3\u7684 2-5 \uff0c\u9019\u500b\u5e73\u89d2\u7684\u8ca2\u737b\u5c31\u5728\u65bc\u5b83\u63d0\u4f9b\u4e86\u4e00\u500b\u8a08\u7b97\u5207\u958b 2-5 \u4e2d\u9593\u7684\u9019\u500b\u5f27\u7684\u65b9\u6cd5\u30025 \u865f\u62fc\u7247\u9760\u5916\u7684\u5f27\u662f\u300c1.5\u300d\u55ae\u4f4d\u5f27\uff0c\u4e5f\u5c31\u662f\u5b83\u7684\u5f26\u9577\u662f \\(\\sqrt3-1\\)\uff1b1 \u865f\u548c 5 \u865f\u62fc\u5728\u4e00\u8d77\u7684\u9019\u689d\u5f27\u662f 2 \u55ae\u4f4d\u5f27\uff0c\u5373\u5f26\u9577\u70ba 1\u3002\u95dc\u9375\u9ede\u5728\u65bc\u9019\u500b\u593e\u89d2\uff0c\u7531\u65bc\u5e73\u89d2\u7684\u95dc\u4fc2\uff0c\u5b83\u6b63\u662f\u5c0d\u61c9 2 \u865f\u62fc\u7247\u9760\u5916\u7684\u5f27\u7684\u5713\u5468\u89d2\uff1b\u4f46\u9019\u5f27\u537b\u662f 4 \u55ae\u4f4d\u6e1b\u53bb\u300c1.5\u300d\u55ae\u4f4d\uff0c\u5373 \\(\\frac{2\\pi}3-\\theta_s\\)\u3002\u9019\u500b\u89d2 (\u5f27\u7684\u4e00\u534a) \u7684\u9918\u5f26\u503c\u70ba\uff1a $$\\begin{alignat}{1}
\\cos\\theta_2&=\\cos\\left(\\frac\\pi3-\\frac{\\theta_s}2\\right)
\\\\&=\\cos\\frac\\pi3\\cdot\\cos\\frac{\\theta_s}2+\\sin\\frac\\pi3\\cdot\\sin\\frac{\\theta_s}2
\\\\&=\\frac12\\cdot\\sqrt\\frac{1+\\cos\\theta_s}{2}+\\frac{\\sqrt3}2\\cdot\\sqrt\\frac{1-\\cos\\theta_s}{2}
\\\\&=\\frac12\\sqrt\\frac{\\sqrt3}2+\\frac{\\sqrt3}2\\sqrt\\frac{2-\\sqrt3}2
\\\\&=\\frac{\\sqrt{\\sqrt3}}{2\\sqrt2}+\\frac{\\sqrt3}2\\cdot\\frac{\\sqrt3-1}2
\\\\&=\\frac14(3-\\sqrt3+\\sqrt{2\\sqrt3})
\\end{alignat}$$ \u7136\u5f8c\u7531\u9918\u5f26\u5b9a\u7406\uff0c\u9019\u689d 2-5 \u4e4b\u9593\u7684\u5f26\u9577\u53ef\u4ee5\u5982\u6b64\u8a08\u7b97\uff1a$$\\begin{alignat}{1}
c^2&=(\\sqrt3-1)^2+1^2-2\\cdot(\\sqrt3-1)\\cdot1\\cdot\\cos\\theta_2
\\\\&=8-4\\sqrt3-\\frac12(\\sqrt3-1)\\sqrt{2\\sqrt3}=8-4\\sqrt3-K_s
\\\\c&=\\sqrt{8-4\\sqrt3-K_s}\\approx0.624937
\\end{alignat}$$ \u5176\u5c0d\u61c9\u5f27\u89d2\u70ba \\(\\phi_{25}=2\\arcsin\\frac c2\\)\uff0c\u4e0d\u904e\u7531\u65bc \\(c\\) \u7684\u6700\u5916\u9762\u662f\u6839\u865f\uff0c\u9019\u88e1\u53ef\u4ee5\u7528\u500d\u89d2\u516c\u5f0f\u5316\u7c21\uff1a$$\\begin{alignat}{1}
\\sin^2\\frac{\\phi_{25}}2&=\\left(\\frac c2\\right)^2=2-\\sqrt3-\\frac14K_s
\\\\\\cos\\phi_{25}&=1-2\\sin^2\\frac{\\phi_{25}}2=1-(4-2\\sqrt3-\\frac12K_s)=-3+2\\sqrt3+\\frac12K_s
\\\\\\phi_{25}&=\\arccos\\left(2\\sqrt3-3+\\frac12K_s\\right)
\\end{alignat}$$ \u8fd1\u4f3c\u503c\u70ba \\(\\phi_{25}\\approx1.21387\\times\\frac\\pi6\\)\u3002\u9019\u5169\u500b\u6578\u503c\u5373\u662f\u4e4b\u524d\u7b2c\u4e09\u7bc7\u4e2d\u63d0\u5230\u7684\u8a08\u7b97\u7d50\u679c\u3002<\/p>\n\n\n\n

5 \u865f\u62fc\u7247\u7684\u9762\u7a4d<\/h2>\n\n\n\n

\u90a3\u9ebc\uff0c\u9019\u6a23\u5373\u53ef\u6c42\u51fa 5 \u865f\u62fc\u7247\u7684\u9762\u7a4d\u4e86\uff1b\u9019\u500b\u7279\u6b8a\u593e\u89d2\u7684\u6b63\u5f26\u503c\u4ee5\u985e\u4f3c\u4e0a\u6cd5\u6c42\u5f97\u70ba\uff1a$$\\begin{alignat}{1}
\\sin\\theta_2&=\\sin\\left(\\frac\\pi3-\\frac{\\theta_s}2\\right)
\\\\&=\\frac{\\sqrt3}2\\sqrt\\frac{\\sqrt3}2-\\frac12\\sqrt\\frac{2-\\sqrt3}2
\\\\&=\\frac{\\sqrt{3\\sqrt3}}{2\\sqrt2}-\\frac12\\cdot\\frac{\\sqrt3-1}2
\\\\&=\\frac14\\left(1-\\sqrt3+\\sqrt{6\\sqrt3}\\right)
\\end{alignat}$$ \u90a3\u9ebc\u4e09\u89d2\u5f62\u9762\u7a4d\u5373\u70ba $$\\begin{alignat}{1}
\\Delta_5&=\\frac12\\cdot(\\sqrt3-1)\\cdot1\\cdot\\frac14(1-\\sqrt3+\\sqrt{6\\sqrt3})
\\\\&=\\frac18(2\\sqrt3-4)+\\frac18(\\sqrt3-1)\\sqrt{6\\sqrt3}
\\\\&=\\frac{\\sqrt3}4-\\frac12+\\frac14\\sqrt3\\cdot\\frac12(\\sqrt3-1)\\sqrt{2\\sqrt3}
\\\\&=\\frac{\\sqrt3}4-\\frac12+\\frac{\\sqrt{3}}4K_s
\\end{alignat}$$ \u5f13\u5f62\u9762\u7a4d\u65b9\u9762\uff0c2 \u55ae\u4f4d\u5f13\u5f62\u9762\u7a4d \\(\\frac\\pi6-\\frac{\\sqrt3}4\\) \u548c\u300c1.5\u300d\u55ae\u4f4d\u5f13\u5f62\u9762\u7a4d \\(\\frac12(\\theta_s-K_s)\\)\u4e26\u4e0d\u56f0\u96e3\uff0c\u4f46\u9019\u689d\u7279\u6b8a\u908a\u7684\u5f13\u5f62\u9762\u7a4d\u5c31\u8981\u7b97\u4e00\u4e0b\u4e86\uff1a$$\\begin{alignat}{1}
\\sin\\phi_{25}&=\\sin\\arccos\\left(2\\sqrt3-3+\\frac12K_s\\right)
\\\\&=\\sqrt{1-\\left(2\\sqrt3-3+\\frac12K_s\\right)^2}
\\\\&=\\sqrt{1-\\left(21-12\\sqrt3+(2\\sqrt3-3)K_s+\\frac12(2\\sqrt3-3)\\right)}
\\\\&=\\sqrt{-\\frac{77}4+\\frac{23\\sqrt3}2-(2\\sqrt3-3)K_s}
\\\\&=\\frac12\\sqrt{-77+46\\sqrt3-(8\\sqrt3-12)K_s}
\\\\\\textrm{seg}(\\phi_{25})&=\\frac12(\\phi_{25}-\\sin\\phi_{25})
\\\\&=\\frac12\\phi_{25}-\\frac14\\sqrt{-77+46\\sqrt3-(8\\sqrt3-12)K_s}
\\end{alignat}$$ \u6211\u5011\u8981\u6c42\u7684 5 \u865f\u9762\u7a4d\u5c31\u662f $$\\begin{alignat}{1}
A_5&=\\left(\\frac{\\sqrt3}4-\\frac12+\\frac{\\sqrt{3}}4K_s\\right)+\\left(\\frac\\pi6-\\frac{\\sqrt3}4\\right)+\\frac12(\\theta_s-K_s)
\\\\&-\\left(\\frac12\\phi_{25}-\\frac14\\sqrt{-77+46\\sqrt3-(8\\sqrt3-12)K_s}\\right)
\\end{alignat}$$ \u9019\u7576\u4e2d\u53ea\u6709 \\(\\frac{\\sqrt3}4\\)\u53ef\u4ee5\u6d88\u6389\uff0c\u5176\u4ed6\u9805\u9577\u5f97\u5b8c\u5168\u4e0d\u50cf\u6240\u4ee5\u90fd\u5f97\u7559\u8457\u3002\u5176\u8fd1\u4f3c\u503c\u70ba\\(A_5\\approx0.33173\\)\u3002<\/p>\n\n\n\n

2 \u865f\u62fc\u7247\u7684\u9762\u7a4d<\/h2>\n\n\n\n

2 \u865f\u62fc\u7247\u73fe\u5728\u6709\u5169\u7a2e\u7b97\u6cd5\uff1a\u4e00\u662f\u7c21\u55ae\u7684\uff0c\u7531\u4e0a\u9762\u7684\u8449\u5b50\u9762\u7a4d\u6263\u53bb\u5df2\u77e5\u7684\u4e09\u7247\uff1b\u53e6\u4e00\u662f\u8f03\u7e41\u7684\uff0c\u6c42\u51fa 2 \u865f\u62fc\u7247\u7684\u5916\u908a\u548c\u5713\u5468\u89d2\u518d\u5faa\u4e0a\u6cd5\u6c42\u5f97\u9762\u7a4d\u3002\u9019\u88e1\u6253\u7b97\u5169\u8005\u90fd\u505a\uff0c\u5148\u4ee5\u8f03\u7e41\u7684\u505a\u6cd5\u6c42\u5f97\u7d50\u679c\uff0c\u518d\u4ee5\u5176\u6c42\u51fa\u8449\u5b50\u9762\u7a4d\u64da\u4ee5\u9a57\u7b97\u3002<\/p>\n\n\n\n

\u5713\u5468\u89d2\u548c 5 \u865f\u985e\u4f3c\uff0c\u53ea\u662f\u9019\u6b21\u5c0d\u61c9\u7684\u5f27\u5c31\u662f\u300c1.5\u300d\u55ae\u4f4d\u5f27\u672c\u8eab\uff0c\u6240\u4ee5\u5c31\u662f \\(\\frac{\\theta_s}2\\)\uff0c\u5176\u6b63\u5f26\u503c\u4e0a\u9762\u6c42\u904e\u662f \\(\\frac{\\sqrt3-1}2\\)\uff1b\u5916\u908a\u5247\u70ba\u4e0a\u9762\u6c42\u904e\u7684\u6b64\u5f27\u534a\u89d2 (\u5713\u5468\u89d2) \u7684\u6b63\u5f26\u503c \\(\\sin\\theta_2=\\frac14(1-\\sqrt3+\\sqrt{6\\sqrt3})\\) \u4e4b\u5169\u500d\u3002\u65bc\u662f\u4e09\u89d2\u5f62\u9762\u7a4d\u70ba $$\\begin{alignat}{1}
\\Delta_2&=\\frac12\\cdot(\\sqrt3-1)\\cdot\\frac12(1-\\sqrt3+\\sqrt{6\\sqrt3})\\cdot\\frac12(\\sqrt3-1)
\\\\&=\\frac18(10-6\\sqrt3)+\\frac18(\\sqrt3-1)^2\\sqrt{6\\sqrt3}
\\\\&=\\frac14(5-3\\sqrt3)+\\frac14(\\sqrt3-1)\\sqrt3\\cdot\\frac12(\\sqrt3-1)\\sqrt{2\\sqrt3}
\\\\&=\\frac14(5-3\\sqrt3)+\\frac14(3-\\sqrt3)K_s
\\end{alignat}$$ \u5916\u908a\u7684\u5f13\u5f62\u9762\u7a4d\u70ba $$\\begin{alignat}{1}
\\seg(2\\theta_2)&=\\theta_2-\\frac12\\sin2\\theta_2=\\theta_2-\\sin\\theta_2\\cos\\theta_2
\\\\&=\\frac\\pi3-\\frac{\\theta_s}2-\\left(\\frac14(1-\\sqrt3+\\sqrt{6\\sqrt3})\\right)\\left(\\frac14(3-\\sqrt3+\\sqrt{2\\sqrt3})\\right)
\\\\&=\\frac\\pi3-\\frac{\\theta_s}2-\\frac1{16}(1-\\sqrt3+\\sqrt{6\\sqrt3})(3-\\sqrt3+\\sqrt{2\\sqrt3})
\\\\&=\\frac\\pi3-\\frac{\\theta_s}2-\\frac1{16}((1-\\sqrt3)(3-\\sqrt3)+((1-\\sqrt3)+\\sqrt3(3-\\sqrt3))\\sqrt{2\\sqrt3}+\\sqrt{6\\sqrt3\\cdot2\\sqrt3})
\\\\&=\\frac\\pi3-\\frac{\\theta_s}2-\\frac1{16}(6-4\\sqrt3+(-2+2\\sqrt3)\\sqrt{2\\sqrt3}+6)
\\\\&=\\frac\\pi3-\\frac{\\theta_s}2-\\frac1{16}(12-4\\sqrt3+4\\cdot\\frac12(\\sqrt3-1)\\sqrt{2\\sqrt3})
\\\\&=\\frac\\pi3-\\frac{\\theta_s}2-\\frac14(3-\\sqrt3+K_s)
\\end{alignat}$$ \u65bc\u662f\u6700\u7d42 2 \u865f\u62fc\u7247\u7684\u9762\u7a4d\u70ba $$\\begin{alignat}{1}
A_2&=\\left(\\frac14(5-3\\sqrt3)+\\frac14(3-\\sqrt3)K_s\\right)+\\left(\\frac\\pi3-\\frac{\\theta_s}2-\\frac14(3-\\sqrt3+K_s)\\right)-\\frac12(\\theta_s-K_s)
\\\\&+\\left(\\frac12\\phi_{25}-\\frac14\\sqrt{-77+46\\sqrt3-(8\\sqrt3-12)K_s}\\right)
\\\\&=\\frac12-\\frac{\\sqrt3}2+\\left(1-\\frac{\\sqrt3}4\\right)K_s+\\frac\\pi3-\\theta_s+\\frac12\\phi_{25}-\\frac14\\sqrt{-77+46\\sqrt3-(8\\sqrt3-12)K_s}
\\end{alignat}$$ \u5176\u8fd1\u4f3c\u503c\u662f\\(A_2\\approx0.338931\\)\u3002<\/p>\n\n\n\n

\u9019\u88e1\u7d66\u7684 \\(A_2\\) \u548c \\(A_5\\) \u7684\u9762\u7a4d\u8fd1\u4f3c\u503c\u4e5f\u548c\u6211\u5728\u7b2c\u4e09\u7bc7\u6587\u7ae0\u88e1\u7d66\u51fa\u7684\u7531 Mathematica \u7684\u516c\u5f0f\u6240\u6c42\u51fa\u4f86\u7684\u9762\u7a4d\u76f8\u540c\u3002<\/p>\n\n\n\n

\u9a57\u7b97<\/h2>\n\n\n\n

\u90a3\u9ebc\u5c31\u4f86\u9032\u884c\u4e0a\u9762\u63d0\u5230\u7684\u9a57\u7b97\u4e86\u3002\u9996\u5148\u5148\u628a \\(A_2+A_5\\) \u7576\u4e2d\u53ef\u4ee5\u6d88\u7684\u6d88\u6389\u3002$$\\begin{alignat}{1}
A_2+A_5&=\\left(-\\frac12+\\frac{\\sqrt3}4K_s+\\frac\\pi6+\\frac12(\\theta_s-K_s)-\\seg(\\phi_{25})\\right)
\\\\&+\\left(\\frac12-\\frac{\\sqrt3}2+(1-\\frac{\\sqrt3}4)K_s+\\frac\\pi3-\\theta_s+\\seg(\\phi_{25})\\right)
\\\\&=\\frac\\pi2-\\frac{\\sqrt3}2-\\frac12(\\theta_s-K_s)
\\end{alignat}$$\u53e6\u4e00\u65b9\u9762\uff0c$$
A_1+A_7=\\left(\\frac12 – \\frac\\pi{12}\\right)+\\left(\\frac\\pi4-\\frac12+\\frac12(\\theta_s-K_s)\\right)
=\\frac\\pi6+\\frac12(\\theta_s-K_s)
$$ \u65bc\u662f\\(A_2+A_5+A_1+A_7=-\\frac{\\sqrt3}2+\\frac{2\\pi}3\\)\u3002
\u800c 4 \u55ae\u4f4d\u5f27\u7684\u5f13\u5f62\u9762\u7a4d\u662f \\(\\seg(\\frac{2\\pi}3)=\\frac12(\\frac{2\\pi}3-\\frac{\\sqrt3}2)\\)\uff0c\u5169\u500d\u6b63\u597d\u7b49\u65bc\u4e0a\u884c\u7684\u9762\u7a4d\u548c\u3002\u9a57\u7b97\u7121\u8aa4\u3002<\/p>\n\n\n\n

\u4e0b\u4e00\u7bc7\u8981\u4f86\u6311\u6230\u5927\u9b54\u738b\uff0c\u9019\u7d44 11-12-13 \u7684\u4e09\u7247\u62fc\u7247\u4e86\u3002\u4e4b\u524d\u7684\u6587\u7ae0\u88e1\u6211\u53ea\u6709\u7d66\u51fa\u8fd1\u4f3c\u6e2c\u91cf\uff0c\u4f46\u9019\u6b21\u5c31\u8981\u8a8d\u771f\u4f86\u7d30\u7b97\u5b83\u5011\u7684\u5c3a\u5bf8\u4e86\u3002<\/p>\n","protected":false},"excerpt":{"rendered":"

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