can anyone please tell me how can u convert the view

matrixs LOOK AT vector to degrees?

like when ( its pointing at ( 0.5, 0.5, 0.5 ) u knw its

45 degrees but how do u mathematically convert the 0.5 to

45 degrees

thanks a lot

can anyone please tell me how can u convert the view

matrixs LOOK AT vector to degrees?

like when ( its pointing at ( 0.5, 0.5, 0.5 ) u knw its

45 degrees but how do u mathematically convert the 0.5 to

45 degrees

thanks a lot

[Please do not mail me a copy of your followup]

Are you familiar with trigonometry?Quote:>like when ( its pointing at ( 0.5, 0.5, 0.5 ) u knw its

>45 degrees but how do u mathematically convert the 0.5 to

>45 degrees

--

Ask me about my upcoming book on Direct3D from Addison-Wesley!

Direct3D Book http://www.xmission.com/~legalize/book/

izfree: Open source tools for Windows Installer

http://izfree.sourceforge.net

yes i am

just let me know the concept or whatever that can lead

somewhere

i used to know how u could do it but now i forgot ....

>-----Original Message-----

>[Please do not mail me a copy of your followup]

>[Please do not mail me a copy of your followup]

>>like when ( its pointing at ( 0.5, 0.5, 0.5 ) u knw its

>>45 degrees but how do u mathematically convert the 0.5

to

>>45 degrees

>Are you familiar with trigonometry?

>--

>Ask me about my upcoming book on Direct3D from Addison-

Wesley!

> Direct3D Book http://www.xmission.com/~legalize/book/

> izfree: Open source tools for Windows Installer

> http://izfree.sourceforge.net

>.

[Please do not mail me a copy of your followup]

Great, that will make things easier. The identities you need are forQuote:>>Are you familiar with trigonometry?

>yes i am

converting from polar coordinates to rectangular coordinates.

If you have a 2D point with cartesian coordinates <x,y>, then you

can convert the point to polar coordinates with a magnitude r and

angle theta with respect to the x axis:

r = sqrt(x*x + y*y)

theta = atan2(y, x)

^ y

| + (x,y)

| /

| r /

| /

| /

|/ theta

+--------------> x

(You can also use the function atan(y/x), but atan2() handles the

signs properly to give the proper quadrant.)

To go from polar to rectangular coordinates:

x = r*cos(theta)

y = r*sin(theta)

(With a little algebraic manipulation and the identities:

tan(theta) = sin(theta)/cos(theta)

cos(theta)*cos(theta) + sin(theta)*sin(theta) = 1

you can solve backwards for r and theta given x and y.)

Things are similar for 3D points, except that now you have a

magnitude r and two angles theta and phi, where theta and phi are

angles with respect to two of the principal axes, usually the angle

of the point when projected into the xy and xz planes.

This discussion computes the angle of a point with respect to the

origin. To find the relative angle between two points you can

apply a translation to make one of the points coincident to the

origin and use the same logic.

With vectors, we have dot and cross product operations that can also

be useful for computing angles.

For two vectors a and b,

a dot b = length(a)*length(b)*cos(theta)

where theta is the smallest angle between the two vectors. This

works for 2-d and 3-d vectors.

length(a cross b) = length(a)*length(b)*sin(theta)

and again, theta is the smallest angle between the two vectors.

D3DX provides 2D and 3D vector classes with functions for dot and

cross products.

--

Ask me about my upcoming book on Direct3D from Addison-Wesley!

Direct3D Book http://www.xmission.com/~legalize/book/

izfree: Open source tools for Windows Installer

http://izfree.sourceforge.net

Rich,

Forgive me for my last post, I was totally wrong about you!!! SORRY!

T.

> [Please do not mail me a copy of your followup]

> >>Are you familiar with trigonometry?

> >yes i am

> Great, that will make things easier. The identities you need are for

> converting from polar coordinates to rectangular coordinates.

> If you have a 2D point with cartesian coordinates <x,y>, then you

> can convert the point to polar coordinates with a magnitude r and

> angle theta with respect to the x axis:

> r = sqrt(x*x + y*y)

> theta = atan2(y, x)

> ^ y

> | + (x,y)

> | /

> | r /

> | /

> | /

> |/ theta

> +--------------> x

> (You can also use the function atan(y/x), but atan2() handles the

> signs properly to give the proper quadrant.)

> To go from polar to rectangular coordinates:

> x = r*cos(theta)

> y = r*sin(theta)

> (With a little algebraic manipulation and the identities:

> tan(theta) = sin(theta)/cos(theta)

> cos(theta)*cos(theta) + sin(theta)*sin(theta) = 1

> you can solve backwards for r and theta given x and y.)

> Things are similar for 3D points, except that now you have a

> magnitude r and two angles theta and phi, where theta and phi are

> angles with respect to two of the principal axes, usually the angle

> of the point when projected into the xy and xz planes.

> This discussion computes the angle of a point with respect to the

> origin. To find the relative angle between two points you can

> apply a translation to make one of the points coincident to the

> origin and use the same logic.

> With vectors, we have dot and cross product operations that can also

> be useful for computing angles.

> For two vectors a and b,

> a dot b = length(a)*length(b)*cos(theta)

> where theta is the smallest angle between the two vectors. This

> works for 2-d and 3-d vectors.

> length(a cross b) = length(a)*length(b)*sin(theta)

> and again, theta is the smallest angle between the two vectors.

> D3DX provides 2D and 3D vector classes with functions for dot and

> cross products.

> --

> Ask me about my upcoming book on Direct3D from Addison-Wesley!

> Direct3D Book http://www.xmission.com/~legalize/book/

> izfree: Open source tools for Windows Installer

> http://izfree.sourceforge.net

John,

You can forget getting any answer from the bozo who replied earlier. He's on

some kind of self-appointed pedestal. If you look, you'll notice he responds

to almost every post, but doesn't come close to actually answering the

question much solving the problem. So hopefully if you ignore him, he'll

just go away.

Now, let's see if I can help you with your problem....I can show you how to

go the other way, given latitude and longitude values you can convert those

to Eye, Up vectors for "look at" type things:

LAT & LON are always computed first by converting degrees to radians...

m_fLat = (ANGLE1 * (PI / 180));

m_fLong = (ANGLE2 * (PI / 180));

Now, given those two values you can compute the following:

m_fEyeX = cos(m_fLat)*sin(m_fLong);

m_fEyeY = sin(m_fLat);

m_fEyeZ = cos(m_fLat)*cos(m_fLong);

m_fUpX = -sin(m_fLat)*sin(m_fLong);

m_fUpY = cos(m_fLat);

m_fUpZ = -sin(m_fLat)*cos(m_fLong);

Given that you can now use Eye and Up for camera orientations...

LookAt( m_fEyeX, m_fEyeY, m_fEyeZ,

0.0, 0.0, 0.0,

m_fUpX, m_fUpY, m_fUpZ );

This will point your camera at the origin ( 0, 0, 0).

Note: You may also have to do some translating depending on where you want

to position your camera.

Hope this helps.

T.

> yes i am

> just let me know the concept or whatever that can lead

> somewhere

> i used to know how u could do it but now i forgot ....

> >-----Original Message-----

> >[Please do not mail me a copy of your followup]

> >>like when ( its pointing at ( 0.5, 0.5, 0.5 ) u knw its

> >>45 degrees but how do u mathematically convert the 0.5

> to

> >>45 degrees

> >Are you familiar with trigonometry?

> >--

> >Ask me about my upcoming book on Direct3D from Addison-

> Wesley!

> > Direct3D Book http://www.xmission.com/~legalize/book/

> > izfree: Open source tools for Windows Installer

> > http://izfree.sourceforge.net

> >.

Considering Rich is probably the most useful person on this board...althoughQuote:> John,

> You can forget getting any answer from the bozo who replied earlier. He's

on

> some kind of self-appointed pedestal. If you look, you'll notice he

responds

> to almost every post, but doesn't come close to actually answering the

> question much solving the problem.

he won't hold your hand or think for you.

The original post is off topic anyways. Try a math newsgropu or

comp.graphics.algorithms.

Right on, couldn't have said it better myself!Quote:> Considering Rich is probably the most useful person on this

board...although

> he won't hold your hand or think for you.

--

Eric DeBrosse

http://www.blown.com/dx/

Microsoft Visual Basic DirectX MVP

The opinions expressed in this message are my own personal views and

do not reflect the official views of Microsoft Corporation. The MVP program

does not constitute employment or contractual obligation with Microsoft.

any of you guys able to successfully import a model from poser to lightwave

and then add bones to it??(yeah-i know....using poser.....im just too lazy)

any help would be appreciated

Ash

=============<^>=============

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