Geforce FX Bilinear Anisotropic Filtering Question ??

[Xmas]

I agree, point sampling should be point sampling no matter what you set in the driver panel.
An application might prefer bilinear over trilinear for performance reasons. But systems get faster so sacrificing quality for speed might be pointless. However, if an application selects point sampling, it does so because filtering produces undesired visual artifacts. There's no other reason to take point sampling over bilinear filtering.

 

[OpenGL guy]

I meant that you should be able to exchange x and y without any change in the output, in other words giving no preference over any derivative.

I still don't understand. Take the hyperbola y^2 = x^2 - 1. This is far different from x^2 = y^2 - 1.

 

[KimB]

I still don't understand. Take the hyperbola y^2 = x^2 - 1. This is far different from x^2 = y^2 - 1.

Right, the x and y axes are transposed. A normal hyperbola will always show a difference between the x and y axes. This is where the absolute value comes in in the equation:
sqrt(|x^2 - y^2|)

If you'll note, switching the x and y in that equation will not change the result.

 

[OpenGL guy]

I still don't understand. Take the hyperbola y^2 = x^2 - 1. This is far different from x^2 = y^2 - 1.

Right, the x and y axes are transposed. A normal hyperbola will always show a difference between the x and y axes. This is where the absolute value comes in in the equation:
sqrt(|x^2 - y^2|)

If you'll note, switching the x and y in that equation will not change the result.

Sure and what you are describing is not a hyperbola either...

Edit: Typed parabola instead of hyperbola.

 

[KimB]

Yeah it is. Just set that equal to a constant. Anyway, I don't think it would work entirely properly with that exact equation, but a proper variation should work just fine.

Update:
Well, actually, that equation will generate two different hyperbolas, one symmetrix about the x axis, the other about the y axis.

 

[OpenGL guy]

Yeah it is. Just set that equal to a constant. Anyway, I don't think it would work entirely properly with that exact equation, but a proper variation should work just fine.

Update:
Well, actually, that equation will generate two different hyperbolas, one symmetrix about the x axis, the other about the y axis.

Exactly. This is not a hyperbola.

Edit: And both hyperbolas are symmetric about both the x and y axes.

 

[Joe DeFuria]

I like circles.

 

[Ante P]

I like circles.

Crop circles are actually the physical manifestations of discussions like this one.
At least that's my theory.

 

[Ailuros]

I like circles.

Hmmm I can only say that I'm glad that I'm not hypertrophic *ahem*

You may excuse the interruption

 

[Dio]

"I wish I was in Greenall Whitley Land...."

(You will need to be both from the UK and the right age to get THAT joke)

 

[Randell]

yesterday afternnon I was my last day at my current job.

 

[Mulciber]

what a sec, so what did we conclude from this thread???

 

[K.I.L.E.R]

Can we PLEASE have a summary?
Thanks