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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