汽车行驶驱动力和行驶阻力平衡计算代码,C代码库,德仔网

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【C】汽车行驶驱动力和行驶阻力平衡计算
作者:Nick_Aaron / 发布于2015/4/8/ 522
汽车行驶驱动力和行驶阻力平衡计算，在ubuntu14.04的codeblocks13.12下测试运行成功。

	torque.c

	#include <stdio.h>
	#include <unistd.h>
	
	int main()
	{
	 double fun(double);
	 int Ua; double i0 = 5.83, ig1 = 6.09, ig2 = 3.09, ig3 = 1.71, ig4 = 1.00, CDA =2.77; /*题目给的各个参数*/
	 double Tq1, Tq2, Tq3, Tq4, Ft1, Ft2, Ft3, Ft4, Ff_plus_Fw, n1, n2, n3, n4, r = 0.367, eta = 0.85;
	
	 printf(" ╭┐┌╮&deg;．﹒\n");
	 printf(" ╭┘└┘└╮&there4;&deg;☆&deg;\n");
	 printf("└┐．．┌┘────╮&there4;&deg;\n");
	 printf("╭┴────┤HAPPY├╮ 此程序为计算汽车行驶阻力于驱动力\n");
	 printf("│ｏ ｏ│ 牛 │●&deg; 在不同档位下的平衡曲线\n");
	 printf("╰┬───╯　 　 │ &there4;&deg;﹒\n");
	 printf(" ☆ □|□/□□□□□/&there4;☆\n");
	 printf("\n 本程序遵循GPL2开源协议，\n \n\&ldquo; 版权没有，翻版不究\"\n");
	 printf("\n\n 三秒钟后开始计算\n");
	
	 sleep(3);
	
	 printf("&darr;-------------------------Ft1------------------------------&darr;\n");
	 for (Ua = 0; Ua < 101; Ua++)
	 {
	 n1 = Ua * ig1 / 0.0237; Tq1 = fun(n1);
	 Ft1 = Tq1 * ig1 * i0 * eta / r;
	 if (Ft1 > 0)
	 {
	 printf("\t\tFt1 = %fKN\n", Ft1);
	 printf("\t\tUa = %dKm/h\n", Ua);
	 }
	 }
	 printf("&uarr;-------------------------Ft1------------------------------&uarr;\n\n");
	
	 printf("&darr;-------------------------Ft2------------------------------&darr;\n");
	 for (Ua = 0; Ua < 101; Ua++)
	 {
	 n2 = Ua * ig2 / 0.0237;
	 Tq2 = fun(n2); Ft2 = Tq2 * ig2 * i0 * eta / r;
	 if (Ft2 > 0)
	 {
	 printf("\t\tFt2 = %fKN\n", Ft2);
	 printf("\t\tUa = %dKm/h\n", Ua);
	 }
	 }
	 printf("&uarr;-------------------------Ft2------------------------------&uarr;\n\n");
	
	 printf("&darr;-------------------------Ft3------------------------------&darr;\n");
	 for (Ua = 0; Ua < 101; Ua++)
	 {
	 n3 = Ua * ig3 / 0.0237;
	 Tq3 = fun(n3);
	 Ft3 = Tq3 * ig3 * i0 * eta / r;
	 if (Ft3 > 0)
	 {
	 printf("\t\tFt3 = %fKN\n", Ft3);
	 printf("\t\tUa = %dKm/h\n", Ua);
	 }
	 }
	 printf("&uarr;-------------------------Ft3------------------------------&uarr;\n\n");
	
	 printf("&darr;-------------------------Ft4------------------------------&darr;\n");
	 for (Ua = 0; Ua < 101; Ua++)
	 {
	 n4 = Ua * ig4 / 0.0237; Tq4 = fun(n4);
	 Ft4 = Tq4 * ig4 * i0 * eta / r;
	 if (Ft4 > 0)
	 {
	 printf("\t\tFt4 = %fKN\n", Ft4);
	 printf("\t\tUa = %dKm/h\n", Ua);
	 }
	 }
	 printf("&uarr;-------------------------Ft4------------------------------&uarr;\n\n");
	
	 printf("&darr;-------------------------Ff+Fw----------------------------&darr;\n");
	 for (Ua = 0; Ua < 101; Ua++)
	 {
	 Ff_plus_Fw = 494.312 + 0.131 * Ua * Ua;
	 if (Ff_plus_Fw > 0)
	 {
	 printf("\t\tFf+Fw = %fKN\n", Ff_plus_Fw);
	 printf("\t\tUa = %dKm/h\n", Ua);
	 }
	 }
	 printf("&uarr;-------------------------Ff+fw----------------------------&uarr;\n");
	
	 return 0;
	}
	
	double fun(double rec)
	{
	 double out;
	 out = -19.13 + 259.27 * (rec / 1000) - 165.44 * (rec / 1000) * (rec / 1000) /*这是一行写不下分几行的！*/
	 + 40.874 * (rec / 1000) * (rec / 1000) * (rec / 1000) - 3.8445
	 * (rec / 1000) * (rec / 1000) * (rec / 1000) * (rec / 1000); /*这是一行写不下分几行写的！*/
	 return out;
	}

	

	

	#include <stdio.h>
	#include <unistd.h>
	
	int main()
	{
	 double fun(double);
	 int Ua; double i0 = 5.83, ig1 = 6.09, ig2 = 3.09, ig3 = 1.71, ig4 = 1.00, CDA =2.77; /*题目给的各个参数*/
	 double Tq1, Tq2, Tq3, Tq4, Ft1, Ft2, Ft3, Ft4, Ff_plus_Fw, n1, n2, n3, n4, r = 0.367, eta = 0.85;
	
	 printf(" ╭┐┌╮&deg;．﹒\n");
	 printf(" ╭┘└┘└╮&there4;&deg;☆&deg;\n");
	 printf("└┐．．┌┘────╮&there4;&deg;\n");
	 printf("╭┴────┤HAPPY├╮ 此程序为计算汽车行驶阻力于驱动力\n");
	 printf("│ｏ ｏ│ 牛 │●&deg; 在不同档位下的平衡曲线\n");
	 printf("╰┬───╯　 　 │ &there4;&deg;﹒\n");
	 printf(" ☆ □|□/□□□□□/&there4;☆\n");
	 printf("\n 本程序遵循GPL2开源协议，\n \n\&ldquo; 版权没有，翻版不究\"\n");
	 printf("\n\n 三秒钟后开始计算\n");
	
	 sleep(3);
	
	 printf("&darr;-------------------------Ft1------------------------------&darr;\n");
	 for (Ua = 0; Ua < 101; Ua++)
	 {
	 n1 = Ua * ig1 / 0.0237; Tq1 = fun(n1);
	 Ft1 = Tq1 * ig1 * i0 * eta / r;
	 if (Ft1 > 0)
	 {
	 printf("\t\tFt1 = %fKN\n", Ft1);
	 printf("\t\tUa = %dKm/h\n", Ua);
	 }
	 }
	 printf("&uarr;-------------------------Ft1------------------------------&uarr;\n\n");
	
	 printf("&darr;-------------------------Ft2------------------------------&darr;\n");
	 for (Ua = 0; Ua < 101; Ua++)
	 {
	 n2 = Ua * ig2 / 0.0237;
	 Tq2 = fun(n2); Ft2 = Tq2 * ig2 * i0 * eta / r;
	 if (Ft2 > 0)
	 {
	 printf("\t\tFt2 = %fKN\n", Ft2);
	 printf("\t\tUa = %dKm/h\n", Ua);
	 }
	 }
	 printf("&uarr;-------------------------Ft2------------------------------&uarr;\n\n");
	
	 printf("&darr;-------------------------Ft3------------------------------&darr;\n");
	 for (Ua = 0; Ua < 101; Ua++)
	 {
	 n3 = Ua * ig3 / 0.0237;
	 Tq3 = fun(n3);
	 Ft3 = Tq3 * ig3 * i0 * eta / r;
	 if (Ft3 > 0)
	 {
	 printf("\t\tFt3 = %fKN\n", Ft3);
	 printf("\t\tUa = %dKm/h\n", Ua);
	 }
	 }
	 printf("&uarr;-------------------------Ft3------------------------------&uarr;\n\n");
	
	 printf("&darr;-------------------------Ft4------------------------------&darr;\n");
	 for (Ua = 0; Ua < 101; Ua++)
	 {
	 n4 = Ua * ig4 / 0.0237; Tq4 = fun(n4);
	 Ft4 = Tq4 * ig4 * i0 * eta / r;
	 if (Ft4 > 0)
	 {
	 printf("\t\tFt4 = %fKN\n", Ft4);
	 printf("\t\tUa = %dKm/h\n", Ua);
	 }
	 }
	 printf("&uarr;-------------------------Ft4------------------------------&uarr;\n\n");
	
	 printf("&darr;-------------------------Ff+Fw----------------------------&darr;\n");
	 for (Ua = 0; Ua < 101; Ua++)
	 {
	 Ff_plus_Fw = 494.312 + 0.131 * Ua * Ua;
	 if (Ff_plus_Fw > 0)
	 {
	 printf("\t\tFf+Fw = %fKN\n", Ff_plus_Fw);
	 printf("\t\tUa = %dKm/h\n", Ua);
	 }
	 }
	 printf("&uarr;-------------------------Ff+fw----------------------------&uarr;\n");
	
	 return 0;
	}
	
	double fun(double rec)
	{
	 double out;
	 out = -19.13 + 259.27 * (rec / 1000) - 165.44 * (rec / 1000) * (rec / 1000) /*这是一行写不下分几行的！*/
	 + 40.874 * (rec / 1000) * (rec / 1000) * (rec / 1000) - 3.8445
	 * (rec / 1000) * (rec / 1000) * (rec / 1000) * (rec / 1000); /*这是一行写不下分几行写的！*/
	 return out;
	}
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