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How to find a normal vector in 2D

1

$begingroup$

I hope this is not a stupid question. I am having a hard time finding a normal in 2D. I am working on a game project and I am trying to perform vector reflection during collision. I know the rest of the math, but I am having a hard time creating the normal vector that is required.

Is there a way to build a normal vector from the point of collision? What do I need to create the normal vector? My position vector does not necessarily point in the direction of the object. My position can be (100, 200) but the object is a vertical object for example. What else can I do for this?

vectors

share|cite|improve this question

edited Jul 19 '15 at 17:18

K. Rmth

1,454822

asked Jul 19 '15 at 1:39

EthosikEthosik

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The normal vector is perpendicular to the surface...
  $endgroup$
  – mniip
  Jul 19 '15 at 2:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you can upload your picture somewhere else and post a link
  $endgroup$
  – coldnumber
  Jul 19 '15 at 2:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In two dimensions, a vector normal to the vector $(a,b)$ is $(-b,a)$ (or any multiple thereof).
  $endgroup$
  – Greg Martin
  Jul 19 '15 at 2:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added an image
  $endgroup$
  – Ethosik
  Jul 19 '15 at 16:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How are you defining the collision surfaces?
  $endgroup$
  – K. Rmth
  Jul 19 '15 at 17:14
  

  
  
  
  

 | 
show 6 more comments

1

$begingroup$

I hope this is not a stupid question. I am having a hard time finding a normal in 2D. I am working on a game project and I am trying to perform vector reflection during collision. I know the rest of the math, but I am having a hard time creating the normal vector that is required.

Is there a way to build a normal vector from the point of collision? What do I need to create the normal vector? My position vector does not necessarily point in the direction of the object. My position can be (100, 200) but the object is a vertical object for example. What else can I do for this?

vectors

share|cite|improve this question

edited Jul 19 '15 at 17:18

K. Rmth

1,454822

asked Jul 19 '15 at 1:39

EthosikEthosik

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The normal vector is perpendicular to the surface...
  $endgroup$
  – mniip
  Jul 19 '15 at 2:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you can upload your picture somewhere else and post a link
  $endgroup$
  – coldnumber
  Jul 19 '15 at 2:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In two dimensions, a vector normal to the vector $(a,b)$ is $(-b,a)$ (or any multiple thereof).
  $endgroup$
  – Greg Martin
  Jul 19 '15 at 2:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added an image
  $endgroup$
  – Ethosik
  Jul 19 '15 at 16:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How are you defining the collision surfaces?
  $endgroup$
  – K. Rmth
  Jul 19 '15 at 17:14
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

I hope this is not a stupid question. I am having a hard time finding a normal in 2D. I am working on a game project and I am trying to perform vector reflection during collision. I know the rest of the math, but I am having a hard time creating the normal vector that is required.

Is there a way to build a normal vector from the point of collision? What do I need to create the normal vector? My position vector does not necessarily point in the direction of the object. My position can be (100, 200) but the object is a vertical object for example. What else can I do for this?

vectors

share|cite|improve this question

edited Jul 19 '15 at 17:18

K. Rmth

1,454822

asked Jul 19 '15 at 1:39

EthosikEthosik

62

$endgroup$

share|cite|improve this question

edited Jul 19 '15 at 17:18

K. Rmth

1,454822

asked Jul 19 '15 at 1:39

EthosikEthosik

62

share|cite|improve this question

edited Jul 19 '15 at 17:18

K. Rmth

1,454822

asked Jul 19 '15 at 1:39

EthosikEthosik

62

edited Jul 19 '15 at 17:18

K. Rmth

1,454822

edited Jul 19 '15 at 17:18

K. Rmth

1,454822

asked Jul 19 '15 at 1:39

EthosikEthosik

62

asked Jul 19 '15 at 1:39

EthosikEthosik

62

2 Answers
2

active

oldest

votes

0

$begingroup$

$vec V_reflected =vec V_incident -2vec n(vec V_incident cdot vec n)$

share|cite|improve this answer

answered Jul 19 '15 at 19:32

K. RmthK. Rmth

1,454822

$endgroup$

add a comment | 

-1

$begingroup$

I also had this problem. you can do $vecv times vecp$ and consider $x$ and $y$ coordinates only.

In fact for any $vecv=(x,y,0)$ and $vecp=(0,0,1)$ then $vecv times vecp=beginbmatrixveci & vecj & veck\x & y & z \ 0 & 0 & 1endbmatrix = yveci-xvecj+0veck=(y,-x)$

That implies that the normal vector $vecn=(vecVy, -vecVx)$

share|cite|improve this answer

edited Jul 29 '16 at 15:34

answered Jul 27 '16 at 20:12

Carlos AfonsoCarlos Afonso

25329

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The OP used $vec v$ for the velocity of the ball and $vec p$ for the position of the center of the rectangle. Those two things do not determine the direction of the normal vector at all, so it is really unclear what you think you are doing with the formulas in this answer.
  $endgroup$
  – David K
  Jul 29 '16 at 15:52
  

  
  
  
  

add a comment | 

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0

$begingroup$

$vec V_reflected =vec V_incident -2vec n(vec V_incident cdot vec n)$

share|cite|improve this answer

answered Jul 19 '15 at 19:32

K. RmthK. Rmth

1,454822

$endgroup$

add a comment | 

0

$begingroup$

$vec V_reflected =vec V_incident -2vec n(vec V_incident cdot vec n)$

share|cite|improve this answer

answered Jul 19 '15 at 19:32

K. RmthK. Rmth

1,454822

$endgroup$

add a comment | 

$begingroup$

$vec V_reflected =vec V_incident -2vec n(vec V_incident cdot vec n)$

share|cite|improve this answer

answered Jul 19 '15 at 19:32

K. RmthK. Rmth

1,454822

$endgroup$

share|cite|improve this answer

answered Jul 19 '15 at 19:32

K. RmthK. Rmth

1,454822

answered Jul 19 '15 at 19:32

K. RmthK. Rmth

1,454822

answered Jul 19 '15 at 19:32

K. RmthK. Rmth

1,454822

-1

$begingroup$

I also had this problem. you can do $vecv times vecp$ and consider $x$ and $y$ coordinates only.

In fact for any $vecv=(x,y,0)$ and $vecp=(0,0,1)$ then $vecv times vecp=beginbmatrixveci & vecj & veck\x & y & z \ 0 & 0 & 1endbmatrix = yveci-xvecj+0veck=(y,-x)$

That implies that the normal vector $vecn=(vecVy, -vecVx)$

share|cite|improve this answer

edited Jul 29 '16 at 15:34

answered Jul 27 '16 at 20:12

Carlos AfonsoCarlos Afonso

25329

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The OP used $vec v$ for the velocity of the ball and $vec p$ for the position of the center of the rectangle. Those two things do not determine the direction of the normal vector at all, so it is really unclear what you think you are doing with the formulas in this answer.
  $endgroup$
  – David K
  Jul 29 '16 at 15:52
  

  
  
  
  

add a comment | 

-1

$begingroup$

I also had this problem. you can do $vecv times vecp$ and consider $x$ and $y$ coordinates only.

In fact for any $vecv=(x,y,0)$ and $vecp=(0,0,1)$ then $vecv times vecp=beginbmatrixveci & vecj & veck\x & y & z \ 0 & 0 & 1endbmatrix = yveci-xvecj+0veck=(y,-x)$

That implies that the normal vector $vecn=(vecVy, -vecVx)$

share|cite|improve this answer

edited Jul 29 '16 at 15:34

answered Jul 27 '16 at 20:12

Carlos AfonsoCarlos Afonso

25329

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The OP used $vec v$ for the velocity of the ball and $vec p$ for the position of the center of the rectangle. Those two things do not determine the direction of the normal vector at all, so it is really unclear what you think you are doing with the formulas in this answer.
  $endgroup$
  – David K
  Jul 29 '16 at 15:52
  

  
  
  
  

add a comment | 

$begingroup$

I also had this problem. you can do $vecv times vecp$ and consider $x$ and $y$ coordinates only.

In fact for any $vecv=(x,y,0)$ and $vecp=(0,0,1)$ then $vecv times vecp=beginbmatrixveci & vecj & veck\x & y & z \ 0 & 0 & 1endbmatrix = yveci-xvecj+0veck=(y,-x)$

That implies that the normal vector $vecn=(vecVy, -vecVx)$

share|cite|improve this answer

edited Jul 29 '16 at 15:34

answered Jul 27 '16 at 20:12

Carlos AfonsoCarlos Afonso

25329

$endgroup$

share|cite|improve this answer

edited Jul 29 '16 at 15:34

answered Jul 27 '16 at 20:12

Carlos AfonsoCarlos Afonso

25329

answered Jul 27 '16 at 20:12

Carlos AfonsoCarlos Afonso

25329

answered Jul 27 '16 at 20:12

Carlos AfonsoCarlos Afonso

25329