Explaining donut like 5 years old Part-3

#java #c #cpp

Now all that remains is what happens inside nested for-loop
You might have seen that r₁sin θ and r₁cos θ

These are used for making a circle in 2D graph

And r₂ for keeping distance between circles so they don't overlap

So, r₂ > r₁ because r₂ starts from origin to centre of circle

Now, to do that overwhelming seen matrix multiply, we will create a singleRow

In C

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

In Java

singleRow circle = new singleRow(2 + Math.cos(theta), Math.sin(theta), 0);

Now make 3 matrix Ry, Rx, Rz that will help us in rotation of circle and donut

In Java

// 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)
);

In C

// 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}};

On using multiply function, we created previously, we will get spinning donut coordinates

In C

singleRow donut = multiply(circle, Ry);
singleRow rotateX = multiply(donut, Rx);

// We will consider it as [Nx, Ny, Nz]
singleRow spinningDonut = multiply(rotateX, Rz);

In Java

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 reciprocal 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 have been 11 but we will go with 12 for Now
what are 30 and 15? IDK
and multiply reciNz, why? IDK
Donut code has too many unsolved mysteries

Now to find make it 3D, we need to some part luminous

For that, we need to find
N = Ny - Nz
- 2 sinB cosϕ cosθ
- 2 sinB cosϕ
+ 2 cosB sinA sinϕ
+ 2 cosA sinϕ

N is between 0 to √2
Now multiply N by 8 which will be to max 11

int L = N * 8

To print it with luminacity, we will create an array of character from lowest to highest luminacity

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

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

Now the last part
check if:
x < screen width
y < screen height
reciNz > zBuffer[0]
If yes, then

if (zBuffer[o] < reciNz && y < screen_height && x < screen_width) {
  buffer[o] = charOpts[L > 0 ? L : 0];
  zBuffer[o] = reciNz;
}

If L is negative, use charOpts[0]/ period(.)

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Explaining donut like 5 years old Part-3

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