i have this... | ^ | | | | | | | || I want this... | | ^ | | | | | | | | ||make sense? I know the normals etc. I just need a formula. I've searched the forums, but there are so many fague and unsure answers. Thanks. [edited by - code_evo on January 27, 2004 9:52:33 PM] [edited by - code_evo on January 27, 2004 9:53:03 PM]

# calculating reflection vector (i have no math book here)

Started by code_evo, Jan 27 2004 02:51 PM

4 replies to this topic

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#1
Members - Reputation: **146**

Posted 27 January 2004 - 02:51 PM

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#2
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Posted 27 January 2004 - 04:29 PM

If N is the normal and v is the initial vector. The reflected vector should be u = - 2 * |N.v| * N + v or something like that. Oh yeah, I assumed that v is pointing towards the surface.

[edited by - keeping on January 27, 2004 11:31:17 PM]

[edited by - keeping on January 27, 2004 11:32:40 PM]

[edited by - keeping on January 27, 2004 11:31:17 PM]

[edited by - keeping on January 27, 2004 11:32:40 PM]

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#5
Members - Reputation: **98**

Posted 05 April 2011 - 11:53 PM

So...7 years after the original post in this thread, I'm looking into this problem on my own. I'm referred to this thread.

Between the time I was referred to this thread, and the time I actually

So...For all my 'dummies-in-arms' out there. Here is how to reflect a vector in DirectX:

Given two vectors, a Normal and any arbitrary vector, the reflection is such (VisualBasic):

Dim vNormal as Vector3 = New Vector3(0, 1, 0) ' <--- Can be any normalized vector (Vector3.Normalize()). (0, 1, 0) is already Normalized.

Dim vArbitrary as Vector3 = New Vector3(-4, 2.1, 0) ' <--- Can be any vector, such as your Camera Position, or a Light Position, etc.

Dim vCross as Vector3 = Nothing

' There are two steps to reflecting a vector. The fist is to Cross the two vectors:

vCross = Vector3.Cross(vNormal, vArbitrary)

' Crossing the vectors gives you a vector perpendicular to your Normal and your Arbitrary.

' The next step is getting the cross of your new vector (vCross) and the Arbitrary. This will give you a vector that is perpendicular to the perpendicular of your Normal and your Arbitrary. I know, sounds complicated. It ain't:

vCross = Vector3.Cross(vArbitrary, vCross)

' There it is. After the two lines are executed, you now have a reflected vector in vCross.

' Us mouth-breathers need to stick together or we'll be overwhelmed by the brains, I hope this was useful.

' Keelah Sel'ai.

Between the time I was referred to this thread, and the time I actually

*looked*at it, I came up with a solution, and I would like to point out that my notation is for DirectX users. I'm no mathematician and flunked most advanced courses in high school, so the previous responses; (u = - 2 * |N.v| * N + v) are gibberish to me. For those of us that want to use the damn tool without know how the damn thing works, I've come up with some quickie vector reflection code...So...For all my 'dummies-in-arms' out there. Here is how to reflect a vector in DirectX:

Given two vectors, a Normal and any arbitrary vector, the reflection is such (VisualBasic):

Dim vNormal as Vector3 = New Vector3(0, 1, 0) ' <--- Can be any normalized vector (Vector3.Normalize()). (0, 1, 0) is already Normalized.

Dim vArbitrary as Vector3 = New Vector3(-4, 2.1, 0) ' <--- Can be any vector, such as your Camera Position, or a Light Position, etc.

Dim vCross as Vector3 = Nothing

' There are two steps to reflecting a vector. The fist is to Cross the two vectors:

vCross = Vector3.Cross(vNormal, vArbitrary)

' Crossing the vectors gives you a vector perpendicular to your Normal and your Arbitrary.

' The next step is getting the cross of your new vector (vCross) and the Arbitrary. This will give you a vector that is perpendicular to the perpendicular of your Normal and your Arbitrary. I know, sounds complicated. It ain't:

vCross = Vector3.Cross(vArbitrary, vCross)

' There it is. After the two lines are executed, you now have a reflected vector in vCross.

' Us mouth-breathers need to stick together or we'll be overwhelmed by the brains, I hope this was useful.

' Keelah Sel'ai.