How to rotate a line around one of its vertexes

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I am making my first raycasting engine, and would like to rotate a line over an angle θ

How does one do this? Would it be possible to show me some basic C++ code or some pseudocode?

This image describes my problem:
enter image description here

Optional question

I decided to make all of this in graphics.h, because it is the simplest graphics header for C/C++.

edited Jan 3 at 12:08

kvantour

10.8k41734

asked Jan 3 at 10:40

Sir.Rosticciano

106

Also interested, because cross product works only in 3 dimensions

– ishidex2
Jan 3 at 10:43

Link doesn't work, I get the image cannot be displayed.

– SPlatten
Jan 3 at 10:43

It works for me :/

– ishidex2
Jan 3 at 10:45

If you are doing linear algebra, pick a linear algebra library and express it using values of appropriate types. Then you can write code that looks like the maths B = P + M * (A - P); can be valid C++

– Caleth
Jan 3 at 12:08

This website can be of help to you: geomalgorithms.com

– kvantour
Jan 3 at 12:08

add a comment |

I am making my first raycasting engine, and would like to rotate a line over an angle θ

How does one do this? Would it be possible to show me some basic C++ code or some pseudocode?

This image describes my problem:
enter image description here

Optional question

I decided to make all of this in graphics.h, because it is the simplest graphics header for C/C++.

edited Jan 3 at 12:08

kvantour

10.8k41734

asked Jan 3 at 10:40

Sir.Rosticciano

106

Also interested, because cross product works only in 3 dimensions

– ishidex2
Jan 3 at 10:43

Link doesn't work, I get the image cannot be displayed.

– SPlatten
Jan 3 at 10:43

It works for me :/

– ishidex2
Jan 3 at 10:45

If you are doing linear algebra, pick a linear algebra library and express it using values of appropriate types. Then you can write code that looks like the maths B = P + M * (A - P); can be valid C++

– Caleth
Jan 3 at 12:08

This website can be of help to you: geomalgorithms.com

– kvantour
Jan 3 at 12:08

add a comment |

I am making my first raycasting engine, and would like to rotate a line over an angle θ

How does one do this? Would it be possible to show me some basic C++ code or some pseudocode?

This image describes my problem:
enter image description here

Optional question

I decided to make all of this in graphics.h, because it is the simplest graphics header for C/C++.

edited Jan 3 at 12:08

kvantour

10.8k41734

asked Jan 3 at 10:40

Sir.Rosticciano

106

I am making my first raycasting engine, and would like to rotate a line over an angle θ

How does one do this? Would it be possible to show me some basic C++ code or some pseudocode?

This image describes my problem:
enter image description here

Optional question

I decided to make all of this in graphics.h, because it is the simplest graphics header for C/C++.

c++ algorithm geometry raycasting cartesian-coordinates

edited Jan 3 at 12:08

kvantour

10.8k41734

asked Jan 3 at 10:40

Sir.Rosticciano

106

edited Jan 3 at 12:08

kvantour

10.8k41734

asked Jan 3 at 10:40

Sir.Rosticciano

106

edited Jan 3 at 12:08

kvantour

10.8k41734

edited Jan 3 at 12:08

kvantour

10.8k41734

edited Jan 3 at 12:08

kvantour

10.8k41734

asked Jan 3 at 10:40

Sir.Rosticciano

106

asked Jan 3 at 10:40

Sir.Rosticciano

106

asked Jan 3 at 10:40

Sir.Rosticciano

106

Also interested, because cross product works only in 3 dimensions

– ishidex2
Jan 3 at 10:43

Link doesn't work, I get the image cannot be displayed.

– SPlatten
Jan 3 at 10:43

It works for me :/

– ishidex2
Jan 3 at 10:45

If you are doing linear algebra, pick a linear algebra library and express it using values of appropriate types. Then you can write code that looks like the maths B = P + M * (A - P); can be valid C++

– Caleth
Jan 3 at 12:08

This website can be of help to you: geomalgorithms.com

– kvantour
Jan 3 at 12:08

add a comment |

Also interested, because cross product works only in 3 dimensions

– ishidex2
Jan 3 at 10:43

Link doesn't work, I get the image cannot be displayed.

– SPlatten
Jan 3 at 10:43

It works for me :/

– ishidex2
Jan 3 at 10:45

If you are doing linear algebra, pick a linear algebra library and express it using values of appropriate types. Then you can write code that looks like the maths B = P + M * (A - P); can be valid C++

– Caleth
Jan 3 at 12:08

This website can be of help to you: geomalgorithms.com

– kvantour
Jan 3 at 12:08

Also interested, because cross product works only in 3 dimensions

– ishidex2
Jan 3 at 10:43

Link doesn't work, I get the image cannot be displayed.

– SPlatten
Jan 3 at 10:43

It works for me :/

– ishidex2
Jan 3 at 10:45

If you are doing linear algebra, pick a linear algebra library and express it using values of appropriate types. Then you can write code that looks like the maths B = P + M * (A - P); can be valid C++

– Caleth
Jan 3 at 12:08

This website can be of help to you: geomalgorithms.com

– kvantour
Jan 3 at 12:08

add a comment |

3 Answers
3

active

oldest

votes

You want:

B = P + M * (A - P)

Where M is a 2D rotation matrix:

M = |  cos(ϴ)  -sin(ϴ) |

    |  sin(ϴ)   cos(ϴ) |

In C++ it could be written as:

float c = cos(theta), s = sin(theta);

float dx = ax - px, dy = ay - py;

float bx = px + c * dx - s * dy;

float by = py + s * dx + c * dy;

answered Jan 3 at 10:53

HolyBlackCat

17.2k33568

add a comment |

One simple algorithm:

Move the circle -P, so that P is at (0, 0).

Rotate A by the angle by multiplying it by the rotation matrix.

Move the circle P to restore its original position.

All these three steps can be done using one 3x3 matrix multiplication.

answered Jan 3 at 10:51

Maxim Egorushkin

90.1k11104192

add a comment |

The scalar product of two vectors have the following property:

vec(PA) . vec(PB) = rho cos theta

Taking the definition of our two vectors:

vec(PA) = (x_a-x_p, y_a-y_p)

vec(PB) = (x_b-x_p, y_b-y_p)

We can get:

(x_a-x_p)(x_b-x_p) + (y_a-y_p)(y_b-y_p) = rho cos theta (1)

Since PA=PB, we also have:

(x_a-x_p)^2 + (y_a-y_p)^2 = (x_b-x_p)^2 + (y_b-y_p)^2 (2)

From (1) and (2) you can derive x_band y_b with some arithmetic autopilot.

edited Jan 3 at 13:42

answered Jan 3 at 10:57

YSC

25.7k657112

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

You want:

B = P + M * (A - P)

Where M is a 2D rotation matrix:

M = |  cos(ϴ)  -sin(ϴ) |

    |  sin(ϴ)   cos(ϴ) |

In C++ it could be written as:

float c = cos(theta), s = sin(theta);

float dx = ax - px, dy = ay - py;

float bx = px + c * dx - s * dy;

float by = py + s * dx + c * dy;

answered Jan 3 at 10:53

HolyBlackCat

17.2k33568

add a comment |

You want:

B = P + M * (A - P)

Where M is a 2D rotation matrix:

M = |  cos(ϴ)  -sin(ϴ) |

    |  sin(ϴ)   cos(ϴ) |

In C++ it could be written as:

float c = cos(theta), s = sin(theta);

float dx = ax - px, dy = ay - py;

float bx = px + c * dx - s * dy;

float by = py + s * dx + c * dy;

answered Jan 3 at 10:53

HolyBlackCat

17.2k33568

add a comment |

You want:

B = P + M * (A - P)

Where M is a 2D rotation matrix:

M = |  cos(ϴ)  -sin(ϴ) |

    |  sin(ϴ)   cos(ϴ) |

In C++ it could be written as:

float c = cos(theta), s = sin(theta);

float dx = ax - px, dy = ay - py;

float bx = px + c * dx - s * dy;

float by = py + s * dx + c * dy;

answered Jan 3 at 10:53

HolyBlackCat

17.2k33568

You want:

B = P + M * (A - P)

Where M is a 2D rotation matrix:

M = |  cos(ϴ)  -sin(ϴ) |

    |  sin(ϴ)   cos(ϴ) |

In C++ it could be written as:

float c = cos(theta), s = sin(theta);

float dx = ax - px, dy = ay - py;

float bx = px + c * dx - s * dy;

float by = py + s * dx + c * dy;

answered Jan 3 at 10:53

HolyBlackCat

17.2k33568

answered Jan 3 at 10:53

HolyBlackCat

17.2k33568

answered Jan 3 at 10:53

HolyBlackCat

17.2k33568

answered Jan 3 at 10:53

HolyBlackCat

17.2k33568

add a comment |

One simple algorithm:

Move the circle -P, so that P is at (0, 0).

Rotate A by the angle by multiplying it by the rotation matrix.

Move the circle P to restore its original position.

All these three steps can be done using one 3x3 matrix multiplication.

answered Jan 3 at 10:51

Maxim Egorushkin

90.1k11104192

add a comment |

One simple algorithm:

Move the circle -P, so that P is at (0, 0).

Rotate A by the angle by multiplying it by the rotation matrix.

Move the circle P to restore its original position.

All these three steps can be done using one 3x3 matrix multiplication.

answered Jan 3 at 10:51

Maxim Egorushkin

90.1k11104192

add a comment |

One simple algorithm:

Move the circle -P, so that P is at (0, 0).

Rotate A by the angle by multiplying it by the rotation matrix.

Move the circle P to restore its original position.

All these three steps can be done using one 3x3 matrix multiplication.

answered Jan 3 at 10:51

Maxim Egorushkin

90.1k11104192

One simple algorithm:

Move the circle -P, so that P is at (0, 0).

Rotate A by the angle by multiplying it by the rotation matrix.

Move the circle P to restore its original position.

All these three steps can be done using one 3x3 matrix multiplication.

answered Jan 3 at 10:51

Maxim Egorushkin

90.1k11104192

answered Jan 3 at 10:51

Maxim Egorushkin

90.1k11104192

answered Jan 3 at 10:51

Maxim Egorushkin

90.1k11104192

answered Jan 3 at 10:51

Maxim Egorushkin

90.1k11104192

add a comment |

The scalar product of two vectors have the following property:

vec(PA) . vec(PB) = rho cos theta

Taking the definition of our two vectors:

vec(PA) = (x_a-x_p, y_a-y_p)

vec(PB) = (x_b-x_p, y_b-y_p)

We can get:

(x_a-x_p)(x_b-x_p) + (y_a-y_p)(y_b-y_p) = rho cos theta (1)

Since PA=PB, we also have:

(x_a-x_p)^2 + (y_a-y_p)^2 = (x_b-x_p)^2 + (y_b-y_p)^2 (2)

From (1) and (2) you can derive x_band y_b with some arithmetic autopilot.

edited Jan 3 at 13:42

answered Jan 3 at 10:57

YSC

25.7k657112

add a comment |

The scalar product of two vectors have the following property:

vec(PA) . vec(PB) = rho cos theta

Taking the definition of our two vectors:

vec(PA) = (x_a-x_p, y_a-y_p)

vec(PB) = (x_b-x_p, y_b-y_p)

We can get:

(x_a-x_p)(x_b-x_p) + (y_a-y_p)(y_b-y_p) = rho cos theta (1)

Since PA=PB, we also have:

(x_a-x_p)^2 + (y_a-y_p)^2 = (x_b-x_p)^2 + (y_b-y_p)^2 (2)

From (1) and (2) you can derive x_band y_b with some arithmetic autopilot.

edited Jan 3 at 13:42

answered Jan 3 at 10:57

YSC

25.7k657112

add a comment |

The scalar product of two vectors have the following property:

vec(PA) . vec(PB) = rho cos theta

Taking the definition of our two vectors:

vec(PA) = (x_a-x_p, y_a-y_p)

vec(PB) = (x_b-x_p, y_b-y_p)

We can get:

(x_a-x_p)(x_b-x_p) + (y_a-y_p)(y_b-y_p) = rho cos theta (1)

Since PA=PB, we also have:

(x_a-x_p)^2 + (y_a-y_p)^2 = (x_b-x_p)^2 + (y_b-y_p)^2 (2)

From (1) and (2) you can derive x_band y_b with some arithmetic autopilot.

edited Jan 3 at 13:42

answered Jan 3 at 10:57

YSC

25.7k657112

The scalar product of two vectors have the following property:

vec(PA) . vec(PB) = rho cos theta

Taking the definition of our two vectors:

vec(PA) = (x_a-x_p, y_a-y_p)

vec(PB) = (x_b-x_p, y_b-y_p)

We can get:

(x_a-x_p)(x_b-x_p) + (y_a-y_p)(y_b-y_p) = rho cos theta (1)

Since PA=PB, we also have:

(x_a-x_p)^2 + (y_a-y_p)^2 = (x_b-x_p)^2 + (y_b-y_p)^2 (2)

From (1) and (2) you can derive x_band y_b with some arithmetic autopilot.

edited Jan 3 at 13:42

answered Jan 3 at 10:57

YSC

25.7k657112

edited Jan 3 at 13:42

answered Jan 3 at 10:57

YSC

25.7k657112

answered Jan 3 at 10:57

YSC

25.7k657112

answered Jan 3 at 10:57

YSC

25.7k657112

add a comment |

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