变换矩阵（翻转？）

Question

这是代码的一部分，可视化了我的 arduino 的旋转。不幸的是，滚动是 "inversed"...它会在每个滴答声中调用一些更新的值。

我所说的反转是指：如果我将 arduino 向右滚动，可视化将它向左滚动，反之亦然。

pushMatrix(); // begin object
translate(width/4, height/4); // set position
float c1 = cos(radians(roll));
float s1 = sin(radians(roll));
float c2 = cos(radians(pitch));
float s2 = sin(radians(pitch));
float c3 = cos(radians(yaw));
float s3 = sin(radians(yaw));
applyMatrix( c2*c3, s1*s3+c1*c3*s2, c3*s1*s2-c1*s3, 0,
             -s2  , c1*c2         , c2*s1         , 0,
             c2*s3, c1*s2*s3-c3*s1, c1*c3+s1*s2*s3, 0,
             0    , 0             , 0             , 1);

drawArduino();
popMatrix(); // end of object

有人看到我犯的错误了吗？

这是我计算传递给矩阵的值的方式：

int aix, aiy, aiz, tax, tay, taz;
int gix, giy, giz;
float ax, ay, az;
float gx, gy, gz;
float roll, pitch, heading;

CurieIMU.readMotionSensor(aix, aiy, aiz, gix, giy, giz);

// convert from raw data to gravity and degrees/second units
ax = convertRawAcceleration(aix);
ay = convertRawAcceleration(aiy);
az = convertRawAcceleration(aiz); //acceleration for 2g
gx = convertRawGyro(gix);
gy = convertRawGyro(giy);
gz = convertRawGyro(giz);

// update the filter, which computes orientation
filter.updateIMU(gx, gy, gz, ax, ay, az);


// print the heading, pitch and roll
roll = filter.getRoll();
pitch = filter.getPitch();
heading = filter.getYaw();

Answer 1

一个选项是反转滚动值：这可以通过从滚动值可以具有的最大值中减去滚动值来完成

inverted = max - current

我看到你在使用度数，所以这里的演示你可以运行如何在假设最大值为 360 的情况下反转值：

点击并拖动旋转
按一个键反转旋转

var roll, pitch, yaw;

var maxRoll = 360;
var invertRoll = false;


function setup(){
  createCanvas(300,300);
  rectMode(CENTER);
}
function draw(){
  var rollValue = roll;
  if(invertRoll){
    
    rollValue = maxRoll - roll;
    
    background(0);
  }else{
    background(255);
  }
  
  background(invertRoll ? color(0) : color(255));
  translate(width * 0.5, height * 0.5);
  rotate(radians(rollValue));
  rect(0,0,100,100);
}

function mouseDragged(){
  roll = map(mouseX,0,width,0,360);
}

function keyPressed(){
  invertRoll = !invertRoll;
}

<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/0.5.5/p5.min.js"></script>

如果有帮助，这里是基于您上面的代码片段的处理测试草图：

float roll, pitch, yaw;

float maxRoll = 360;
boolean invertRoll;

void setup() {
  size(400,400,P3D);
}

void draw() {
  background(255);

  pushMatrix(); // begin object
  translate(width/4, height/4); // set position

  float rollValue = roll;

  if(invertRoll) rollValue = maxRoll - roll;

  float c1 = cos(radians(rollValue));
  float s1 = sin(radians(rollValue));

  float c2 = cos(radians(pitch));
  float s2 = sin(radians(pitch));
  float c3 = cos(radians(yaw));
  float s3 = sin(radians(yaw));
  applyMatrix( c2*c3, s1*s3+c1*c3*s2, c3*s1*s2-c1*s3, 0, 
    -s2, c1*c2, c2*s1, 0, 
    c2*s3, c1*s2*s3-c3*s1, c1*c3+s1*s2*s3, 0, 
    0, 0, 0, 1);

  drawArduino();
  popMatrix(); // end of object
}

void drawArduino() {
  box(100);
}

void mouseDragged(){
  roll = map(mouseX,0,width,0,360);
}

void keyPressed(){
  invertRoll = !invertRoll;
}

另一种选择是将要反转的轴缩放 -1 然后平移偏移回到你想画的位置。

变换矩阵（翻转？）

Transformation Matrix (roll inversed?)

java

processing

transformation