View Full Version : View matrix for point light calculation should be inverse?

SasMaster

07-12-2012, 03:56 AM

Hi All. I have the following question.I am coding the point lights.I want to use camera space .I need to transform light position using camera (view) matrix.But I don't understand,

should be the same inverted camera matrix I use to move the camera? Or I should pass the camera (view) matrix without inverse ?

Thanks .

thokra

07-12-2012, 04:32 AM

Why would you use the inverse? The inverse view matrix is only necessary if you want to go from eye-space to world-space. So if you want to do your lighting calculations in eye-space simply transform every entity involved in the calculation to eye-space an give it a go.

SasMaster

07-12-2012, 04:54 AM

So if you want to do your lighting calculations in eye-space simply transform every entity involved in the calculation to eye-space an give it a go.

That is what I don't get. What you mean is to take the camera position (its model matrix) and use it to transform every entity ?

thokra

07-12-2012, 05:24 AM

What you mean is to take the camera position (its model matrix) and use it to transform every entity ?

Uhm, no. You want to perform calculations on entities which all reside in the same space - otherwise you'll get false results. This has nothing to do with the world-space camera position. The result of the transformation into eye-space is that the camera in this space is implicitly located at the origin - i.e. (0, 0, 0) - and all other objects are then defined relative to the camera's coordinate system.

Do you know how the transformation pipeline works and what the purpose of it is?

Edit: To explain it a little better, think of a point light at world-space position L, a camera at world-space position C and some world-space point P in space being lit by the light and looked-at by the camera. Let N be the normal at P.

In world-space the inverse light incidence I vector is simply

I_world = L_world - P_world

the inverse viewing direction is simply V

V_world = C_world - P_world

In a Phong shader you could now use I, V and N to determine the specular reflection at P. However, for this world-space calculation you actually need all world-space positions. In eye-space this is not the case anymore. If you transform P and L into eye-space using the view-matrix and N to eye-space using the inverse transpose of the view matrix you already know the camera position implicitly: it's simply at the origin. Now, the I vector is still calculated using

I_eye = L_eye - P_eye

V, however comes down to

V_eye = C_eye - P_eye = (0, 0, 0) - P_eye = -P_eye

See the difference? In any case, it doesn't matter in which space you calculate. Just be consistent. In some cases you might save some data, e.g. you don't need a cam position in eye-space, but on the other hand you have three additional transformations and have to think a little differently. All spaces have their purposes but mathematically it doesn't matter as long as the space for all entities in the calculation is the same.

HTH.

SasMaster

07-12-2012, 05:37 AM

I do understand.But it seems you don't get what I am asking. I just made a test.I passed to the light shader the camera matrix without inverting it.So now if I move the camera let's say to the right of the lighted

object which is in the middle of the scene I get the lights showing smaller and smaller and disappear abruptly fromthe object surface once the camera gets to some offset on X axis Which is this case just 200 units but the cam still looks at the surface ..

But if I pass the camera inverse matrix which I usually pass into vertex shader to calculate View perspective matrix then the lights positions stay all right. So what you just said doesn't look right unless we have

a misunderstanding here.

thokra

07-12-2012, 05:48 AM

camera inverse matrix

:doh: Ah, now I get it. You call camera inverse matrix what most people call the view-matrix. I thought you were asking about the inverse view-matrix which is definitely the wrong answer. Yeah, the view-matrix basically represents the opposite camera movement and rotation - so one could call it inverted. However, I've yet to see someone, besides you, to call the view-matrix the camera inverse matrix. :)

But I don't understand, should be the same inverted camera matrix I use to move the camera?

BTW, the view-matrix does actually the opposite of moving the camera, it moves everything else.

SasMaster

07-12-2012, 05:55 AM

BTW, the view-matrix does actually the opposite of moving the camera, it moves everything else.

I know it too :biggrin-new: . So your answer is that I am doing it the right way ?

thokra

07-12-2012, 06:09 AM

Looks like it. Are the results correct? If so, you're doing it right. I mean, verifying a trivial lighting setup isn't that hard. :)

Dark Photon

07-12-2012, 06:23 AM

...using camera (view) matrix.But I don't understand, should be the same inverted camera matrix I use to move the camera?

In more conventional terminology, the MODELING transform for the camera takes you from EYE-SPACE to WORLD-SPACE (EYE-SPACE being the OBJECT-SPACE of the camera).

If you want the VIEWING transform (WORLD-SPACE to EYE-SPACE transform), you intuitively just invert the camera's MODELING transform.

Powered by vBulletin® Version 4.2.2 Copyright © 2016 vBulletin Solutions, Inc. All rights reserved.