Explain the Donuts as Old as Ears Part 3

Now what's left is what's happening in the nested for loop You may have seen r ₁ sin θ and r ₁ cos θ

These are used to make circles in 2d diagrams

and r ₂ are used to keep the distance between the circles so that they do not overlap

So, r ₂ > r ₁ because r ₂ starts from the origin to the center of the circle

Now, for overwhelming matrix multiplication, we will create a single row

C language

 singlerow circle = {2 cos(theta), sin(theta), 0};

java language

 singlerow circle = new singlerow(2 math.cos(theta), math.sin(theta), 0);

Now make 3 matrices ry, rx, rz, which will help us rotate the circle and donuts

java language

 // rotation on y-axis
matrix ry = new matrix(
  new singlerow(math.cos(phi), 0, math.sin(phi)),
  new singlerow(0, 1, 0),
  new singlerow(-math.sin(phi), 0, math.cos(phi))
);
// rotation on x-axis
matrix rx = new matrix(
  new singlerow(1, 0, 0),
  new singlerow(0, math.cos(a), math.sin(a)),
  new singlerow(0, -math.sin(a), math.cos(a))
);
// rotation on z-axis
matrix rz = new matrix(
  new singlerow(math.cos(b), math.sin(b), 0),
  new singlerow(-math.sin(b), math.cos(b), 0),
  new singlerow(0, 0, 1)
);

C language

 // rotation on y-axis
matrix ry = {{cos(phi), 0, sin(phi)}, {0, 1, 0}, {-sin(phi), 0, cos(phi)}};
// rotation on x-axis
matrix rx = {{1, 0, 0}, {0, cos(a), sin(a)}, {0, -sin(a), cos(a)}};
// rotation on z-axis
matrix rz = {{cos(b), sin(b), 0}, {-sin(b), cos(b), 0}, {0, 0, 1}};

Using the multiplication function we created earlier, we will get the rotated donut coordinates

C language

 singlerow donut = multiple(circle, ry);
singlerow rotatex = multiply(donut, rx);

// we will consider it as [nx, ny, nz]
singlerow spinningdonut = multiply(rotatex, rz);

java language

 singlerow donut = matrix.multiply(circle, ry);
singlerow rotatex = matrix.multiply(donut, rx);

// we will consider it as [nx, ny, nz]
singlerow spinningdonut = matrix.multiply(rotatex, rz);

We will make recinz, which will be the countdown of nz 5 (distance from camera)

 float recinz = 1 / (spinningdonut.a3 5);

int x = 40 30 * spinningdonut.a1 * recinz;
int y = 12 15 * spinningdonut.a2 * recinz;

// o is index of current buffer
int o = x screen_width * y;

screen_height / 2 should be 11, but we now choose 12
What are 30 and 15? I don't know and multiply by recinz, why? I don't know there are too many unsolved mysteries in the donut code

Now to make it 3d we need some parts to shine

To do this, we need to find
n = ny - nz
- 2 sinb cosphhi cosθ
- 2 sinb cosψ
2 cosb sina sinphi
2 cosa sinphi

n between 0 and √2 Now multiply n by 8, with a maximum of 11

 int l = n * 8

To print it with brightness, we will create an array of characters from the lowest to the highest brightness

 char charopts[] = ".,-~:;=!*#$@";

or

 char[] charopts = {'.', ',', '-', '~', ':', ';', '=', '!', '*', '#', '$', '@'};

Now is the last part to check whether:
x y recinz > zbuffer[0]
If so, then

 if (zBuffer[o]  0 ? L : 0];
  zBuffer[o] = reciNz;
}

If l is negative, use charopts[0]/period(.)

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