Texture coordinates

OpenGL OpenGL: Advanced Coding
1

I am creating a terrain and I have a texture that I would like to map over the entire surface. When specifying each vertex within this terrain I would like to attach it to a certaing part of the texture using glTexCoord2f(), but it doesn’t want to work…the values I pass to glTexCoord2f can be as small as 1/2048,1/2048 …is this a problem…??

If this is a problem can you please tell a way that it can be done .thanks.

Steve

2

Exactly what problem do you have? You only want the texture to be at certain polys, not at all? Then try to use GL_CLAMP instead of GL_REPEAT in the texture settings.

3

1/2048 gives you a result of 0.00048828125… or there abouts.

so using floats glTexCoord2f()you are loosing quite some precision.

try
glTexCoord2d()

4

If you have a landscape with a grid of 2049x2049 vertices (!), values like 1/2048, 2/2048, etc seem “normal” to me (! again).

Tell us more.

5

It could be because of integer roundoff. Try using 1.0f/2048.0f.

j

6

I have tried 1.0/(double)2048 etc but the same problem persists…

Thanks,
Steve

7

IIRC certain cards read ‘voodoo’ had problems with large numbers ‘100+’ as texture coords perhaps you’re seeing the opposite.

though i believe the problem lies elsewhere

are u using
for (x=0;x<2048;x++)
glTexCoord2f( ( 1.0/2048.0 ), ( 1.0/2048.0 ))

instead of
for (x=0;x<2048;x++)
glTexCoord2f( x*( 1.0/2048.0 ), y*( 1.0/2048.0 ))

8

Thanks zed (and everyone else!) what you suggested fixed it (:-))

Steve

9

Originally posted by dans:
1/2048 gives you a result of 0.00048828125… or there abouts.

so using floats glTexCoord2f()you are loosing quite some precision.

Wrong.
You see - usual floating point formats are binary based, not decimal.
1/2048 = 2^-11, it is exact number in binary representation - “float” has more then enough precision for it.

10

Serge, that is something I was wondering when dans posted his message yesterday: perhaps you will be able to tell me:

Taken from MSDN:

float 4 none 3.4E +/- 38 (7 digits)
double 8 none 1.7E +/- 308 (15 digits)
long double 10 none 1.2E +/- 4932 (19 digits)

What does the (x digits) means ??? Does it mean that a float has 7 digits precision ?

Thanks for any explanation…

Regards.

Eric

P.S.: phew, OpenGL is far, far away from this question…

11

yep the x digits gives you the precision of the type (acutally i think float has only 6 digits precision but i am not sure)

e.g with float 12.456 is different from 12.456 but
12.4561 ist the same as 12.4567

hmmm hope that helps
Chris

12

Originally posted by Eric:
float 4 none 3.4E +/- 38 (7 digits)
double 8 none 1.7E +/- 308 (15 digits)
long double 10 none 1.2E +/- 4932 (19 digits)

What does the (x digits) means ??? Does it mean that a float has 7 digits precision ?

It means that float has enough precision to hold 7 significant decimal digits for sure.

float(9.1) = 9.100000381… // first 7 decimal digits are correct

// binary form: float = (+/-) 1.xxx * 2^n
// exponent : -127<n<128
// mantissa : 24bit, 1(integer)+23(fraction)
// another point of view : float = (+/-)24bit_integer * 2^(n-23)

13

Thanks, both of you !

Regards.

Eric