標(biāo)題:
基于51單片機(jī)的智能紅外循跡小車(chē)程序
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作者:
cqupt.qwert
時(shí)間:
2019-6-14 19:17
標(biāo)題:
基于51單片機(jī)的智能紅外循跡小車(chē)程序
最近剛結(jié)束單片機(jī)講座的課,在寫(xiě)關(guān)于智能汽車(chē)的文章的同時(shí)自己做了一個(gè)簡(jiǎn)單的智能循跡小車(chē),這里采用的是光電循跡模塊,使用比較簡(jiǎn)單,
紅外對(duì)管輸出的電壓模擬信號(hào)通過(guò)比較器轉(zhuǎn)換為了數(shù)字電平,只需要根據(jù)輸出的電平高低控制電機(jī)的正反轉(zhuǎn),這邊附上部分單片機(jī)代碼
#include "reg52.h"
typedef unsigned int u16;
typedef unsigned char u8;
sbit left1 = P0^0;
sbit left2 = P0^1;
sbit right1 = P0^2;
sbit right2 = P0^3;
sbit IN1 = P1^4;
sbit IN2 = P1^3;
sbit IN3 = P1^2;
sbit IN4 = P1^1;
void xunji();
void main()
{
while(1)
{
xunji();
}
}
void TurnRight()
{
IN1 = 0;
IN2 = 1;
IN3 = 0;
IN4 = 1;
}
void TurnLeft()
{
IN1 = 1;
IN2 = 0;
IN3 = 1;
IN4 = 0;
}
void Forward()
{
IN1 = 1;
IN2 = 0;
IN3 = 0;
IN4 = 1;
}
void Stop()
{
IN1 = 0;
IN2 = 0;
IN3 = 0;
IN4 = 0;
}
void xunji()
{
unsigned char flag = 0;
if((left1 == 0)&&(left2 == 0)&&(right1 == 0)&&(right2 == 0)) //0 0 0 0
flag = 0;
if((left1 == 0)&&(left2 == 0)&&(right1 == 0)&&(right2 == 1)) //0 0 0 1
flag = 1;
if((left1 == 0)&&(left2 == 0)&&(right1 == 1)&&(right2 == 0)) //0 0 1 0
flag = 0;
if((left1 == 0)&&(left2 == 0)&&(right1 == 1)&&(right2 == 1)) //0 0 1 1
flag = 1;
if((left1 == 0)&&(left2 == 1)&&(right1 == 0)&&(right2 == 0)) //0 1 0 0
flag = 0;
if((left1 == 0)&&(left2 == 1)&&(right1 == 0)&&(right2 == 1)) //0 1 0 1
flag = 1;
if((left1 == 0)&&(left2 == 1)&&(right1 == 1)&&(right2 == 0)) //0 1 1 0
flag = 0;
if((left1 == 0)&&(left2 == 1)&&(right1 == 1)&&(right2 == 1)) //0 1 1 1
flag = 1;
if((left1 == 1)&&(left2 == 0)&&(right1 == 0)&&(right2 == 0)) //1 0 0 0
flag = 2;
if((left1 == 1)&&(left2 == 0)&&(right1 == 0)&&(right2 == 1)) //1 0 0 1
flag = 0;
if((left1 == 1)&&(left2 == 0)&&(right1 == 1)&&(right2 == 0)) //1 0 1 0
flag = 2;
if((left1 == 1)&&(left2 == 0)&&(right1 == 1)&&(right2 == 1)) //1 0 1 1
flag = 0;
if((left1 == 1)&&(left2 == 1)&&(right1 == 0)&&(right2 == 0)) //1 1 0 0
flag = 2;
if((left1 == 1)&&(left2 == 1)&&(right1 == 0)&&(right2 == 1)) //1 1 0 1
flag = 0;
if((left1 == 1)&&(left2 == 1)&&(right1 == 1)&&(right2 == 0)) //1 1 1 0
flag = 2;
if((left1 == 1)&&(left2 == 1)&&(right1 == 1)&&(right2 == 1)) //1 1 1 1
flag = 3;
switch(flag)
{
case(0):Forward();break;
case(1):TurnRight();break;
case(2):TurnLeft();break;
default:Stop();break;
}
}
作者:
zafu
時(shí)間:
2019-7-20 15:10
很不錯(cuò),請(qǐng)樓主分享一下全部代碼,
1539429523@qq.com
作者:
?小萌新?
時(shí)間:
2019-10-29 17:18
樓主可不可以給我發(fā)一份完整代碼,急用
1078786909@qq.com
歡迎光臨 (http://www.torrancerestoration.com/bbs/)
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