Um he stated the projection matrix is undocumented because its identity, ie, it is not ever messed with, and even if it was, you could glGet it.
On Sep 10, 2010, at 9:59 AM, George Toledo wrote: > > > On Fri, Sep 10, 2010 at 9:57 AM, vade <[email protected]> wrote: > Actually, with all due respect you did not pay attention. I believe Chris > answered it for you. Get the modelview and projection matrix and do some math. > > > I wasn't referring to Chris's reply, where he states that the projection > matrix is undocumented. > > -George > > > On Sep 10, 2010, at 8:47 AM, George Toledo wrote: > >> With all due respect, you're answering a question I'm not asking, and Dr. >> Wright was answering a question by guessing. >> >> On Sep 10, 2010 4:57 AM, "Louis Schultz" <[email protected]> wrote: >> > George, >> > >> > Dr Wright was sending you in the correct direction. Perspective assumes a >> > fixed distance from a fixed eye point to a fixed picture plane. The view >> > that eye has is perpendicular to the picture plane, and an object to be >> > represented on that picture plan is some distance behind that plane. >> > >> > The apparent effect of moving an object away is a function of its distance >> > from the eye AND the distance from the eye to the front of the picture >> > plane. Perhaps the attached image will help to make that more clear. >> > >> > >> > If we let DP = distance from eye to picture plane; DO = distance from eye >> > to the object, then a plane or line parallel to the picture plane would >> > appear to be (DE/DO) x (absolute size of the line or plane) >> > >> > You can also use that same formula and a bit of trigonometry to figure out >> > how to make a plane or line appear to tilt in relation to the picture >> > plane. In the attached image, imagine the hypotenuse of the blue 45 deg. >> > right triangle as an edge view of a 1 unit square (that square being >> > perpendicular to the diagram plane). If the edge at A were projected onto >> > a picture plane 2 units from the eye position, then it would measure 2/3 = >> > 0.667 units. If the line at B were project to the same plane, it would >> > measure 2/(3+sqrt0.5) = 0.54 units. >> > >> > To make it clear where those numbers come from, The hypotenuse is 1 unit, >> > and the square root of 0.5 is the hypotenuse length x cosine 45, which is >> > the angle of the plane in relation to the perpendicular view of the eye. >> > >> > In that case, two edges remain parallel to the picture plane, but with a >> > little work you can figure out how to make a plane appear at any distance >> > and at any angle to the picture plane. If it gets confusing, all you >> > should really need get it figured out are some quick sketches of top and >> > side views (picture plane appears as a line). >> > >> > One final note, perspective is a useful tool, but not a true depiction of >> > reality (whatever that is). The further an object in that system moves >> > from the line of the view, the more distortion creeps in. Consider a 1 >> > unit square 8 units behind the picture plane and parallel to it. The >> > square is centered on the line of view. The picture plane is 2 units from >> > the view point. That square would be 10 units from the eye. If it were >> > moved it 10 units away from the line of view in the same plane it would >> > actually then be 14 units from the eye point. It would project the same >> > size though based on our formula. That contradicts what we know to happen. >> > >> > We actually see in something more like spheres of vision rather than >> > planes. The distance between the eye and the picture plane has to be zero >> > for that to really work though, which is, to state the obvious, how you do >> > see the world. >> > >> > I mention that both as a warning to keep things somewhat centered if you >> > don't want them to look weird, but also as an encouragement to play with >> > the notion of curved picture "planes" if it strikes your fancy. The >> > artists Victor Vasarely and of course Escher might be inspirational for >> > that. >> > >> > >> > >> > >> > >> > >> > On Sep 9, 2010, at 8:38 PM, George Toledo wrote: >> > >> >> So, just so I know where this is at, the statement that there is no way >> >> to do this 100% accurately in QC is valid? >> >> >> >> -George Toledo >> >> >> >> On Thu, Sep 9, 2010 at 2:05 PM, Christopher Wright >> >> <[email protected]> wrote: >> >>> I basically want to know the exact number, from Apple (or really, from >> >>> anyone, but not a visual comparison/kinda close thing, like this). >> >> >> >> >> >> There isn't an exact number -- this sort of thing requires passing the >> >> coordinates through the projection matrix QC uses (undocumented, but it's >> >> been investigated on the list before in the past) as well as the model >> >> view matrix (which can be arbitrarily configured via Trackball and 3D >> >> Transformation). >> >> >> >> i'm going to assume this is for faking Z on a Billboard? If so, why not >> >> just use a sprite? If not, what other reason is there to fake Z >> >> positioning like this? (I'm not saying there isn't a reason, I'm just not >> >> able to think of one off the top of my head :). >> >> >> >> -- >> >> Christopher Wright >> >> [email protected] >> >> >> >> >> >> >> >> >> >> _______________________________________________ >> >> Do not post admin requests to the list. They will be ignored. >> >> Quartzcomposer-dev mailing list ([email protected]) >> >> Help/Unsubscribe/Update your Subscription: >> >> http://lists.apple.com/mailman/options/quartzcomposer-dev/lulu%40vt.edu >> >> >> >> This email sent to [email protected] >> > >> >> _______________________________________________ >> Do not post admin requests to the list. They will be ignored. >> Quartzcomposer-dev mailing list ([email protected]) >> Help/Unsubscribe/Update your Subscription: >> http://lists.apple.com/mailman/options/quartzcomposer-dev/doktorp%40mac.com >> >> This email sent to [email protected] > >
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