MAKER: theSTEMpedia／译：趣无尽 Cherry（转载请注明出处）
人眼在观察景物时，光信号传入大脑神经，需经过一段短暂的时间，光的作用结束后，视觉形象并不立即消失，这种残留的视觉称“后像”，也被称为是“视觉暂留”。
人眼每秒处理的图像不超过 10-12 个。当一系列图像很快的连续性的滚动时，就会让我们产生幻觉，感觉画面在动。这也是制作动画电影的基本原则。
本期给大家带来一款别出心裁的 LED 显示器，用 Arduino、电机和一排 LED 制作的视觉暂留显示器。
是不是很酷呢，接下来我们先来看一段视频，看看是如何制作的吧。
材料清单
Arduino Nano×1
红外线传感器×1
洞洞板×1
LED 灯×若干
电阻×若干
直流电机×1
电池×1
电线×若干
热熔胶×1
烙铁×1
白纸×1
Arduino IDE（软件）
3D打印
3D 打印部分主要分为旋转支架和固定底座两个部分。
所需 3D 打印的 STL 文件请在项目文件库中下载。
安装 LED
3D 部件打印完毕后，就可以进入操作环节了。
1、剪取一块洞洞板，大小与旋转支架部分相匹配。
2、安装 LED 灯，我使用八个红色的 LED 灯。关于灯的颜色，你可以自由选择。
3、将所有 LED 灯的负极端口都焊接到洞洞板上。
4、添加电阻，防止 LED 灯因电压（5V）过高而烧坏。我使用 220ohm 的电阻，将每个电阻焊接到每个 LED 灯的正极。
5、将跳线电缆连接到电阻的另一个脚上。
连接 LED 与 Arduino Nano
连接接口的方式如下：
LED0：Nano 的 D2 。
LED1： Nano 的 D3 。
LED2： Nano 的 D4 。
LED3： Nano 的 D5 。
LED4： Nano 的 D6 。
LED5： Nano 的 D7 。
LED6： Nano 的 D8 。
LED7： Nano 的 D9 。
负极为：Nano 的 Ground 。
安装直流电机
将直流电机安装到固定底座上并通电。
添加红外线传感器
我们需要一个点作为文本读写开始的起点。在固定底座下面粘贴一块白纸块作为起点，红外线传感器能检测到这张白纸块。
如果改变白纸的位置，那么文本显示的位置也将随之发生改变。
红外线传感器与 Nano 连接方式如下 :
GND：Nano 的 GND。
Vin： Nano 的 5V。
Out：Nano 的 D10。
组装部件
1、连接电源，给 Arduino Nano 通电。
2、将 Arduino Nano 安装到旋转的部件中。
3、将旋转部件安装到固定底座上。
Arduino 编程
/*
* This Code demonstrates
* LED POV DISPLAY
*
* Components used
* 1) Arduino Nano
* 2) DC motor
* 3) LEDs
* 4) IR sensor
* 5) Battery
* 6) Wire
*
* Connect leds to 2 to 9 digital pin of Nano
* Connect IR sensor out pin to 10 digital pin of Nano
*
* code written by Palak Mehta on March 29,2019
*/
////////////////////////design a pattern of display the number and alphabates////////////////////////////////
int NUMBER9[]={1,1,1,1,0,0,0,1, 1,0,0,1,0,0,0,1, 1,0,0,1,0,0,0,1, 1,0,0,1,0,0,0,1, 1,1,1,1,1,1,1,1};
int NUMBER8[]={0,1,1,0,1,1,1,0, 1,0,0,1,0,0,0,1, 1,0,0,1,0,0,0,1, 1,0,0,1,0,0,0,1, 0,1,1,0,1,1,1,0};
int NUMBER7[]={1,0,0,0,0,0,0,0, 1,0,0,0,1,0,0,0, 1,0,0,0,1,0,0,0, 1,0,0,1,1,1,1,1, 1,1,1,0,1,0,0,0};
int NUMBER6[]={1,1,1,1,1,1,1,1, 1,0,0,0,1,0,0,1, 1,0,0,0,1,0,0,1, 1,0,0,0,1,0,0,1, 1,0,0,0,1,1,1,1};
int NUMBER5[]={1,1,1,1,1,0,0,1, 1,0,0,0,1,0,0,1, 1,0,0,0,1,0,0,1, 1,0,0,0,1,0,0,1, 1,0,0,0,1,1,1,1};
int NUMBER2[]= {1,0,0,0,0,0,1,1, 1,0,0,0,0,1,0,1, 1,0,0,0,1,0,0,1, 1,0,0,1,0,0,0,1, 0,1,1,0,0,0,0,1};
int NUMBER1[]= {0,0,1,0,0,0,0,0, 0,1,0,0,0,0,0,0, 1,1,1,1,1,1,1,1, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0};
int NUMBER0[]= {1,1,1,1,1,1,1,1, 1,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,1, 1,1,1,1,1,1,1,1};
int _[] = {0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0};
int A[] = {1,1,1,1,1,1,1,1, 1,0,0,1,0,0,0,0, 1,0,0,1,0,0,0,0, 1,0,0,1,0,0,0,0, 1,1,1,1,1,1,1,1};
int B[] = {1,1,1,1,1,1,1,1, 1,0,0,1,0,0,0,1, 1,0,0,1,0,0,0,1, 1,0,0,1,0,0,0,1, 0,1,1,0,1,1,1,0};
int C[] = {0,0,1,1,1,1,0,0, 0,1,0,0,0,0,1,0, 1,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,1};
int D[] = {1,1,1,1,1,1,1,1, 1,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,1, 0,1,0,0,0,0,1,0, 0,0,1,1,1,1,0,0};
int E[] = {1,1,1,1,1,1,1,1, 1,0,0,1,0,0,0,1, 1,0,0,1,0,0,0,1, 1,0,0,1,0,0,0,1, 1,0,0,1,0,0,0,1};
int F[] = {1,1,1,1,1,1,1,1, 1,0,0,1,0,0,0,0, 1,0,0,1,0,0,0,0, 1,0,0,1,0,0,0,0, 1,0,0,1,0,0,0,0};
int G[] = {0,1,1,1,1,1,1,1, 1,0,0,0,0,0,0,1, 1,0,0,0,1,0,0,1, 1,0,0,0,1,0,0,1, 1,0,0,0,1,1,1,0};
int H[] = {1,1,1,1,1,1,1,1, 0,0,0,0,1,0,0,0, 0,0,0,0,1,0,0,0, 0,0,0,0,1,0,0,0, 1,1,1,1,1,1,1,1};
int I[] = {1,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,1, 1,1,1,1,1,1,1,1, 1,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,1};
int J[] = {0,0,0,0,0,1,1,0, 0,0,0,0,1,0,0,1, 0,0,0,0,0,0,0,1, 0,0,0,0,0,0,0,1, 1,1,1,1,1,1,1,0};
int K[] = {1,1,1,1,1,1,1,1, 0,0,0,1,1,0,0,0, 0,0,1,0,0,1,0,0, 0,1,0,0,0,0,1,0, 1,0,0,0,0,0,0,1};
int L[] = {1,1,1,1,1,1,1,1, 0,0,0,0,0,0,0,1, 0,0,0,0,0,0,0,1, 0,0,0,0,0,0,0,1, 0,0,0,0,0,0,0,1};
int M[] = {1,1,1,1,1,1,1,1, 0,1,0,0,0,0,0,0, 0,0,1,0,0,0,0,0, 0,1,0,0,0,0,0,0, 1,1,1,1,1,1,1,1};
int N[] = {1,1,1,1,1,1,1,1, 0,0,1,0,0,0,0,0, 0,0,0,1,1,0,0,0, 0,0,0,0,0,1,0,0, 1,1,1,1,1,1,1,1};
int O[] = {0,1,1,1,1,1,1,0, 1,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,1, 0,1,1,1,1,1,1,0};
int P[] = {1,1,1,1,1,1,1,1, 1,0,0,1,0,0,0,0, 1,0,0,1,0,0,0,0, 1,0,0,1,0,0,0,0, 0,1,1,0,0,0,0,0};
int Q[] = {0,1,1,1,1,1,1,0, 1,0,0,0,0,0,0,1, 1,0,0,0,0,1,0,1, 0,1,1,1,1,1,1,0, 0,0,0,0,0,0,0,1};
int R[] = {1,1,1,1,1,1,1,1, 1,0,0,1,1,0,0,0, 1,0,0,1,0,1,0,0, 1,0,0,1,0,0,1,0, 0,1,1,0,0,0,0,1};
int S[] = {0,1,1,1,0,0,0,1, 1,0,0,0,1,0,0,1, 1,0,0,0,1,0,0,1, 1,0,0,0,1,0,0,1, 1,0,0,0,1,1,1,0};
int T[] = {1,0,0,0,0,0,0,0, 1,0,0,0,0,0,0,0, 1,1,1,1,1,1,1,1, 1,0,0,0,0,0,0,0, 1,0,0,0,0,0,0,0};
int U[] = {1,1,1,1,1,1,1,0, 0,0,0,0,0,0,0,1, 0,0,0,0,0,0,0,1, 0,0,0,0,0,0,0,1, 1,1,1,1,1,1,1,0};
int V[] = {1,1,1,1,1,1,0,0, 0,0,0,0,0,0,1,0, 0,0,0,0,0,0,0,1, 0,0,0,0,0,0,1,0, 1,1,1,1,1,1,0,0};
int W[] = {1,1,1,1,1,1,1,1, 0,0,0,0,0,0,1,0, 0,0,0,0,0,1,0,0, 0,0,0,0,0,0,1,0, 1,1,1,1,1,1,1,1};
int X[] = {1,1,0,0,0,0,1,1, 0,0,1,0,0,1,0,0, 0,0,0,1,1,0,0,0, 0,0,1,0,0,1,0,0, 1,1,0,0,0,0,1,1};
int Y[] = {1,1,0,0,0,0,0,0, 0,0,1,0,0,0,0,0, 0,0,0,1,1,1,1,1, 0,0,1,0,0,0,0,0, 1,1,0,0,0,0,0,0};
int Z[] = {1,0,0,0,0,1,1,1, 1,0,0,0,1,0,0,1, 1,0,0,1,0,0,0,1, 1,0,1,0,0,0,0,1, 1,1,0,0,0,0,0,1};
int* alpha[]= {A,B,C,D,E,F,G,H,I,J,K,L,M,N};//,T,U,V,W,X,Y,Z};
int letterSpace;
int delayTime;
#define IR_pin 10
void setup()
{
Serial.begin(9600);
pinMode(IR_pin,INPUT);
for( int i = 2; i<10 ;i++ ) // setting the ports of the leds to OUTPUT
{
pinMode(i, OUTPUT);
}
letterSpace =4;// defining the space between the letters (ms)
delayTime =1;// defining the time dots appear (ms)
}
void printLetter(int letter[])
{
int y;
// printing the first y row of the letter
for (y=0; y<8; y++)
{
digitalWrite(y+2, letter[y]);
}
delay(delayTime);
// printing the second y row of the letter
for (y=0; y<8; y++)
{
digitalWrite(y+2, letter[y+8]);
}
delay(delayTime);
// printing the third y row of the letter
for (y=0; y<8; y++)
{
digitalWrite(y+2, letter[y+16]);
}
delay(delayTime);
for(y = 0; y<8; y++) {
digitalWrite(y+2, letter[y+24]);
}
delay(delayTime);
for(y = 0; y<8; y++) {
digitalWrite(y+2, letter[y+32]);
}
delay(delayTime);
// printing the space between the letters
for (y=0; y<8; y++)
{
digitalWrite(y+2, 0);
}
delay(letterSpace);
}
void loop()
{
if(digitalRead(IR_pin)==LOW)
{
printLetter (P);
printLetter (O);
printLetter (V);
printLetter (_);
printLetter (_);
printLetter (D);
printLetter (I);
printLetter (S);
printLetter (P);
printLetter (L);
printLetter (A);
printLetter (Y);
printLetter (_);
}
代码文件可以在项目文件库中下载。
完成
当红外线传感器检测到白纸时，LED 灯就开始发光并显示出视觉暂留的效果。由于视觉暂留原理，开关特定 LED 灯时，会显现出不同的图像相互重叠而形成的文字。
是不是简单又有趣？
译自：instructables.com/id/Persistence-of-Vision-PoV-Display-Using-Arduino/
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