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利用日期、经纬度求日出日落时间 C语言程序代码(zz)
阅读量:4975 次
发布时间:2019-06-12

本文共 5948 字,大约阅读时间需要 19 分钟。

先贴在这了,后面应该用得着
 
 
  1. #define PI 3.1415926
  2. #include<math.h>
  3. #include<iostream>
  4. using namespace std;
  5. int days_of_month_1[]={
    31,28,31,30,31,30,31,31,30,31,30,31};
  6. int days_of_month_2[]={
    31,29,31,30,31,30,31,31,30,31,30,31};
  7. long double h=-0.833;
  8. //定义全局变量
  9. void input_date(int c[]){
  10. int i;
  11. cout<<"Enter the date (form: 2009 03 10):"<<endl;
  12. for(i=0;i<3;i++){
  13. cin>>c[i];
  14. }
  15. }
  16. //输入日期
  17. void input_glat(int c[]){
  18. int i;
  19. cout<<"Enter the degree of latitude(range: 0°- 60°,form: 40 40 40 (means 40°40′40″)):"<<endl;
  20. for(i=0;i<3;i++){
  21. cin>>c[i];
  22. }
  23. }
  24. //输入纬度
  25. void input_glong(int c[]){
  26. int i;
  27. cout<<"Enter the degree of longitude(west is negativ,form: 40 40 40 (means 40°40′40″)):"<<endl;
  28. for(i=0;i<3;i++){
  29. cin>>c[i];
  30. }
  31. }
  32. //输入经度
  33. int leap_year(int year){
  34. if(((year%400==0) || (year%100!=0) && (year%4==0))) return 1;
  35. else return 0;
  36. }
  37. //判断是否为闰年:若为闰年,返回1;若非闰年,返回0
  38. int days(int year, int month, int date){
  39. int i,a=0;
  40. for(i=2000;i<year;i++){
  41. if(leap_year(i)) a=a+366;
  42. else a=a+365;
  43. }
  44. if(leap_year(year)){
  45. for(i=0;i<month-1;i++){
  46. a=a+days_of_month_2[i];
  47. }
  48. }
  49. else {
  50. for(i=0;i<month-1;i++){
  51. a=a+days_of_month_1[i];
  52. }
  53. }
  54. a=a+date;
  55. return a;
  56. }
  57. //求从格林威治时间公元2000年1月1日到计算日天数days
  58. long double t_century(int days, long double UTo){
  59. return ((long double)days+UTo/360)/36525;
  60. }
  61. //求格林威治时间公元2000年1月1日到计算日的世纪数t
  62. long double L_sun(long double t_century){
  63. return (280.460+36000.770*t_century);
  64. }
  65. //求太阳的平黄径
  66. long double G_sun(long double t_century){
  67. return (357.528+35999.050*t_century);
  68. }
  69. //求太阳的平近点角
  70. long double ecliptic_longitude(long double L_sun,long double G_sun){
  71. return (L_sun+1.915*sin(G_sun*PI/180)+0.02*sin(2*G_sun*PI/180));
  72. }
  73. //求黄道经度
  74. long double earth_tilt(long double t_century){
  75. return (23.4393-0.0130*t_century);
  76. }
  77. //求地球倾角
  78. long double sun_deviation(long double earth_tilt, long double ecliptic_longitude){
  79. return (180/PI*asin(sin(PI/180*earth_tilt)*sin(PI/180*ecliptic_longitude)));
  80. }
  81. //求太阳偏差
  82. long double GHA(long double UTo, long double G_sun, long double ecliptic_longitude){
  83. return (UTo-180-1.915*sin(G_sun*PI/180)-0.02*sin(2*G_sun*PI/180)+2.466*sin(2*ecliptic_longitude*PI/180)-0.053*sin(4*ecliptic_longitude*PI/180));
  84. }
  85. //求格林威治时间的太阳时间角GHA
  86. long double e(long double h, long double glat, long double sun_deviation){
  87. return 180/PI*acos((sin(h*PI/180)-sin(glat*PI/180)*sin(sun_deviation*PI/180))/(cos(glat*PI/180)*cos(sun_deviation*PI/180)));
  88. }
  89. //求修正值e
  90. long double UT_rise(long double UTo, long double GHA, long double glong, long double e){
  91. return (UTo-(GHA+glong+e));
  92. }
  93. //求日出时间
  94. long double UT_set(long double UTo, long double GHA, long double glong, long double e){
  95. return (UTo-(GHA+glong-e));
  96. }
  97. //求日落时间
  98. long double result_rise(long double UT, long double UTo, long double glong, long double glat, int year, int month, int date){
  99. long double d;
  100. if(UT>=UTo) d=UT-UTo;
  101. else d=UTo-UT;
  102. if(d>=0.1) {
  103. UTo=UT;
  104. UT=UT_rise(UTo,GHA(UTo,G_sun(t_century(days(year,month,date),UTo)),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo)),G_sun(t_century(days(year,month,date),UTo)))),glong,e(h,glat,sun_deviation(earth_tilt(t_century(days(year,month,date),UTo)),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo)),G_sun(t_century(days(year,month,date),UTo))))));
  105. result_rise(UT,UTo,glong,glat,year,month,date);
  106. }
  107. return UT;
  108. }
  109. //判断并返回结果(日出)
  110. long double result_set(long double UT, long double UTo, long double glong, long double glat, int year, int month, int date){
  111. long double d;
  112. if(UT>=UTo) d=UT-UTo;
  113. else d=UTo-UT;
  114. if(d>=0.1){
  115. UTo=UT;
  116. UT=UT_set(UTo,GHA(UTo,G_sun(t_century(days(year,month,date),UTo)),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo)),G_sun(t_century(days(year,month,date),UTo)))),glong,e(h,glat,sun_deviation(earth_tilt(t_century(days(year,month,date),UTo)),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo)),G_sun(t_century(days(year,month,date),UTo))))));
  117. result_set(UT,UTo,glong,glat,year,month,date);
  118. }
  119. return UT;
  120. }
  121. //判断并返回结果(日落)
  122. int Zone(long double glong){
  123. if(glong>=0) return (int)((int)(glong/15.0)+1);
  124. else return (int)((int)(glong/15.0)-1);
  125. }
  126. //求时区
  127. void output(long double rise, long double set, long double glong){
  128. if((int)(60*(rise/15+Zone(glong)-(int)(rise/15+Zone(glong))))<10)
  129. cout<<"The time at which the sun rises is "<<(int)(rise/15+Zone(glong))<<":0"<<(int)(60*(rise/15+Zone(glong)-(int)(rise/15+Zone(glong))))<<" .\n";
  130. else cout<<"The time at which the sun rises is "<<(int)(rise/15+Zone(glong))<<":"<<(int)(60*(rise/15+Zone(glong)-(int)(rise/15+Zone(glong))))<<" .\n";
  131. if((int)(60*(set/15+Zone(glong)-(int)(set/15+Zone(glong))))<10)
  132. cout<<"The time at which the sun sets is "<<(int)(set/15+Zone(glong))<<": "<<(int)(60*(set/15+Zone(glong)-(int)(set/15+Zone(glong))))<<" .\n";
  133. else cout<<"The time at which the sun sets is "<<(int)(set/15+Zone(glong))<<":"<<(int)(60*(set/15+Zone(glong)-(int)(set/15+Zone(glong))))<<" .\n";
  134. }
  135. //打印结果
  136. int main(){
  137. long double UTo=180.0;
  138. int year,month,date;
  139. long double glat,glong;
  140. int c[3];
  141. input_date(c);
  142. year=c[0];
  143. month=c[1];
  144. date=c[2];
  145. input_glat(c);
  146. glat=c[0]+c[1]/60+c[2]/3600;
  147. input_glong(c);
  148. glong=c[0]+c[1]/60+c[2]/3600;
  149. long double rise,set;
  150. rise=result_rise(UT_rise(UTo,GHA(UTo,G_sun(t_century(days(year,month,date),UTo)),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo)),G_sun(t_century(days(year,month,date),UTo)))),glong,e(h,glat,sun_deviation(earth_tilt(t_century(days(year,month,date),UTo)),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo)),G_sun(t_century(days(year,month,date),UTo)))))),UTo,glong,glat,year,month,date);
  151. set=result_set(UT_set(UTo,GHA(UTo,G_sun(t_century(days(year,month,date),UTo)),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo)),G_sun(t_century(days(year,month,date),UTo)))),glong,e(h,glat,sun_deviation(earth_tilt(t_century(days(year,month,date),UTo)),ecliptic_longitude(L_sun(t_century(days(year,month,date),UTo)),G_sun(t_century(days(year,month,date),UTo)))))),UTo,glong,glat,year,month,date);
  152. output(rise,set,glong);
  153. system("pause");
  154. return 0;
  155. }

转载于:https://www.cnblogs.com/strinkbug/p/5942419.html

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