树莓派控制（TANK） [英] Raspberry pi controlled (TANK)

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本文介绍了树莓派控制（TANK）的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

大家好，我创建了一个Raspberry pi控制坦克，为此我需要一个如下代码：

（4轮驱动，通过L298N电机驱动控制）



（这应该通过电脑控制）

当按下以下键时，应发生以下情况



*按向上箭头 - 坦克前进（所有4个车轮向前）

*按下箭头 - 坦克向后（所有4个车轮向后）

*按左箭头 - 坦克向左（左2个车轮向后，右2个车轮向前）

*按右箭头 - 坦克向右（右2个车轮向后，左2个车轮向前）



和额外的：

一个鼠标应该用来控制两个舵机，用于炮塔系统。（就像一个fpv游戏）



我尝试了什么：



通过OriginalGriff回答解决方案＃1：



啊好的，谢谢。无论如何我没有想到这么快的回复。

好​​吧至少可以纠正这个：



//开始



hello guys , im creating an Raspberry pi controlled Tank , and for that i need an code which are as follows:
(4 wheel drive , controlled through L298N Motor Driver)

(this should be controlled through computer)
when pressed the following keys the following should happen

*press up arrow- the tank goes forward(all 4 wheels go forward)
*press down arrow - the tank goes backwards (all 4 wheels go backwards)
*press left arrow- tank goes left (left 2 wheels go backward, right 2 wheels go forward)
*press right arrow- tank goes right(right 2 wheels go backwards, left 2 wheels go forward)

and additional:
an mouse should be abbled to control an two servos ,for turret system.(like an fpv games)

What I have tried:

Answering to Solution #1 by OriginalGriff:

ah okay thanks for that. reply, anyhow i didnt thought of this much quick resppnse.
okay atleast can u correct this :

//start

# import curses and GPIO
import curses
import RPi.GPIO as GPIO

GPIO.setwarnings(False)

#set GPIO numbering mode and define output pins
GPIO.setmode(GPIO.BOARD)

motor1a = 7
motor1b = 11
motor1e = 22

motor2a = 13
motor2b = 16
motor2e = 15

GPIO.setup(motor1a,GPIO.OUT)
GPIO.setup(motor1b,GPIO.OUT)
GPIO.setup(motor1e,GPIO.OUT)
GPIO.setup(motor2a,GPIO.OUT)
GPIO.setup(motor2b,GPIO.OUT)
GPIO.setup(motor2e,GPIO.OUT)

# Get the curses window, turn off echoing of keyboard to screen, turn on
# instant (no waiting) key response, and use special values for cursor keys
screen = curses.initscr()
curses.noecho()  
curses.cbreak()
curses.halfdelay(3)
screen.keypad(True)

try:
        while True:   
            char = screen.getch()
            if char == ord('q'):
                break
            elif char == curses.KEY_UP:
                GPIO.output(motor1a,GPIO.HIGH)
                GPIO.output(motor1b,GPIO.LOW)
                GPIO.output(motor1e,GPIO.HIGH)
                GPIO.output(motor2a,GPIO.HIGH)
  GPIO.output(motor2b,GPIO.LOW)
  GPIO.output(motor2e,GPIO.HIGH)
            elif char == curses.KEY_DOWN:
                GPIO.output(motor1a,GPIO.LOW)
                GPIO.output(motor1b,GPIO.HIGH)
                GPIO.output(motor1e,GPIO.HIGH)
                GPIO.output(motor2a,GPIO.LOW)
  GPIO.output(motor2b,GPIO.HIGH)
  GPIO.output(motor2e,GPIO.HIGH)
            elif char == curses.KEY_RIGHT:
                GPIO.output(motor1a,GPIO.HIGH)
                GPIO.output(motor1b,GPIO.LOW)
                GPIO.output(motor1e,GPIO.HIGH)
                GPIO.output(motor2a,GPIO.LOW)
  GPIO.output(motor2b,GPIO.HIGH)
  GPIO.output(motor2e,GPIO.HIGH)
            elif char == curses.KEY_LEFT:
                GPIO.output(motor1a,GPIO.LOW)
                GPIO.output(motor1b,GPIO.HIGH)
                GPIO.output(motor1e,GPIO.HIGH)
                GPIO.output(motor2a,GPIO.HIGH)
  GPIO.output(motor2b,GPIO.LOW)
  GPIO.output(motor2e,GPIO.HIGH)
            elif char == 10:
                GPIO.output(motor1a,GPIO.LOW)
                GPIO.output(motor1b,GPIO.LOW)
                GPIO.output(motor1e,GPIO.LOW)
                GPIO.output(motor2a,GPIO.LOW)
  GPIO.output(motor2b,GPIO.LOW)
  GPIO.output(motor2e,GPIO.LOW)
             
finally:
    #Close down curses properly, inc turn echo back on!
    curses.nocbreak(); screen.keypad(0); curses.echo()
    curses.endwin()
    GPIO.cleanup()





//结束<上面代码中的
，电机一直运行直到按下下一个键，所以如何使其工作如按键运行，并在键被释放时停止



// end
in the above code , the motor keeps running until next key is pressed , so how to make it to work like "press key to run , and stop when the key is relesed"

推荐答案

引用：

因为我需要一个如下代码：

for that i need an code which are as follows:



它不是那样的工作。

我们不为你工作。

如果你想让别人写你的代码，你必须支付 - 我建议你去Freelancer.com并在那里问。



但要注意：你得到你付出的代价。支付花生，买猴子。



发展的概念就像这句话所暗示的那样：系统地运用科学和技术知识来满足特定的目标或要求。 BusinessDictionary.com [ ^ ]

这与有一个不一样快速谷歌并放弃，如果我找不到完全正确的代码。

所以要么付钱给别人去做，要么学会如何自己写。我们不是来为你做这件事。


It doesn't quite work like that.
We do not do your work for you.
If you want someone to write your code, you have to pay - I suggest you go to Freelancer.com and ask there.

But be aware: you get what you pay for. Pay peanuts, get monkeys.

The idea of "development" is as the word suggests: "The systematic use of scientific and technical knowledge to meet specific objectives or requirements." BusinessDictionary.com[^]
That's not the same thing as "have a quick google and give up if I can't find exactly the right code".
So either pay someone to do it, or learn how to write it yourself. We aren't here to do it for you.


是的，谢谢你。无论如何我没有想到这么快的回复。

好​​吧至少可以纠正这个：



//开始



yeah okay thanks for that. reply, anyhow i didnt thought of this much quick resppnse.
okay atleast can u correct this :

//start

# import curses and GPIO
import curses
import RPi.GPIO as GPIO

GPIO.setwarnings(False)

#set GPIO numbering mode and define output pins
GPIO.setmode(GPIO.BOARD)

motor1a = 7
motor1b = 11
motor1e = 22

motor2a = 13
motor2b = 16
motor2e = 15

GPIO.setup(motor1a,GPIO.OUT)
GPIO.setup(motor1b,GPIO.OUT)
GPIO.setup(motor1e,GPIO.OUT)
GPIO.setup(motor2a,GPIO.OUT)
GPIO.setup(motor2b,GPIO.OUT)
GPIO.setup(motor2e,GPIO.OUT)

# Get the curses window, turn off echoing of keyboard to screen, turn on
# instant (no waiting) key response, and use special values for cursor keys
screen = curses.initscr()
curses.noecho()  
curses.cbreak()
curses.halfdelay(3)
screen.keypad(True)

try:
        while True:   
            char = screen.getch()
            if char == ord('q'):
                break
            elif char == curses.KEY_UP:
                GPIO.output(motor1a,GPIO.HIGH)
                GPIO.output(motor1b,GPIO.LOW)
                GPIO.output(motor1e,GPIO.HIGH)
                GPIO.output(motor2a,GPIO.HIGH)
  GPIO.output(motor2b,GPIO.LOW)
  GPIO.output(motor2e,GPIO.HIGH)
            elif char == curses.KEY_DOWN:
                GPIO.output(motor1a,GPIO.LOW)
                GPIO.output(motor1b,GPIO.HIGH)
                GPIO.output(motor1e,GPIO.HIGH)
                GPIO.output(motor2a,GPIO.LOW)
  GPIO.output(motor2b,GPIO.HIGH)
  GPIO.output(motor2e,GPIO.HIGH)
            elif char == curses.KEY_RIGHT:
                GPIO.output(motor1a,GPIO.HIGH)
                GPIO.output(motor1b,GPIO.LOW)
                GPIO.output(motor1e,GPIO.HIGH)
                GPIO.output(motor2a,GPIO.LOW)
  GPIO.output(motor2b,GPIO.HIGH)
  GPIO.output(motor2e,GPIO.HIGH)
            elif char == curses.KEY_LEFT:
                GPIO.output(motor1a,GPIO.LOW)
                GPIO.output(motor1b,GPIO.HIGH)
                GPIO.output(motor1e,GPIO.HIGH)
                GPIO.output(motor2a,GPIO.HIGH)
  GPIO.output(motor2b,GPIO.LOW)
  GPIO.output(motor2e,GPIO.HIGH)
            elif char == 10:
                GPIO.output(motor1a,GPIO.LOW)
                GPIO.output(motor1b,GPIO.LOW)
                GPIO.output(motor1e,GPIO.LOW)
                GPIO.output(motor2a,GPIO.LOW)
  GPIO.output(motor2b,GPIO.LOW)
  GPIO.output(motor2e,GPIO.LOW)
             
finally:
    #Close down curses properly, inc turn echo back on!
    curses.nocbreak(); screen.keypad(0); curses.echo()
    curses.endwin()
    GPIO.cleanup()





//结束<在上面的代码中，
，电机一直运行直到按下下一个键，所以如何使其工作如按键运行，并在键被释放时停止



// end
in the above code , the motor keeps running until next key is pressed , so how to make it to work like "press key to run , and stop when the key is relesed"


这篇关于树莓派控制（TANK）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！

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