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    文件類型: .py
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    發(fā)布日期: 2021-05-05
  • 語言: Python
  • 標(biāo)簽: 三邊定位??

資源簡介

自己開發(fā)python的三邊定位模塊,包含 solve_quadratic_equation(a, b, c):解二次方程 triangle_area(x1,y1,x2,y2,x3,y3):計算三角形面積 triangle_perimeter(x1,y1,x2,y2,x3,y3):計算三角形周長 circle_intersect(r1,r2,x1,y1,x2,y2):計算兩圓交點(diǎn) locate(X,Y,R):利用兩兩相交的三個圓的6個交點(diǎn),取合圍邊長最短的三個點(diǎn)的質(zhì)心作為目標(biāo)定位點(diǎn)

資源截圖

代碼片段和文件信息

import?math
combination_list?=?[(0?1?2)?(0?1?3)?(0?1?4)?(0?1?5)?(0?2?3)?(0?2?4)?(0?2?5)?(0?3?4)?(0?3?5)?(0?4?5)?(1?2?3)?(1?2?4)?(1?2?5)?(1?3?4)?(1?3?5)?(1?4?5)?(2?3?4)?(2?3?5)?(2?4?5)?(3?4?5)];
def?solve_quadratic_equation(a?b?c):
????if?a==0:
????????return?‘二次項不能為零!‘?;
????delta=b*b-4*a*c;
????if?delta>=0?and?a!=0:
????????x1?=?(-b+math.sqrt(delta))/(2*a);
????????x2?=?(-b-math.sqrt(delta))/(2*a);
????????return?[x1x2];
????else:
????????return?‘方程無實(shí)數(shù)解!‘?;

def?triangle_area(x1y1x2y2x3y3):
????return?abs(x1*y2-x2*y1+x3*y1-x1*y3+x2*y3-x3*y2)/2;

def?triangle_perimeter(x1y1x2y2x3y3):
????a?=?math.sqrt((x1-x2)**2+(y1-y2)**2);
????b?=?math.sqrt((x2-x3)**2+(y2-y3)**2);
????c?=?math.sqrt((x3-x1)**2+(y3-y1)**2);
????return?a+b+c;



‘‘‘
(x1-x)^2+(y1-y)^2=r1^2----①
(x2-x)^2+(y2-y)^2=r2^2----②
①-②==>y=-(x1-x2)/(y1-y2)x+(x1^2-x2^2+y1^2-y2^2+r2^2-r1^2)/2(y1-y2)----③
????????????或x=-(y1-y2)/(x1-x2)y+(y1^2-y2^2+x1^2-x2^2+r2^2-r1^2)/2(x1-x2)----④
當(dāng)x1=/=x2,④代入①,否則③代入①。
‘‘‘
def?circle_intersect(r1r2x1y1x2y2):
????d?=?math.sqrt((x1-x2)**2+(y1-y2)**2);
????if?d>r1+r2:
????????return?‘兩個圓無交點(diǎn)!‘;
????elif??x1!=x2:
????????k=(y1-y2)/(x1-x2);
????????D=(x1**2-x2**2+y1**2-y2**2+r2**2-r1**2)/(2*(x1-x2));
????????a=1+k**2;
????????b=(2*k*(x1-D)-2*y1);
????????c=y1**2+(x1-D)**2-r1**2;
???

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