  [Geometric diagram] Let AB, Fig.6., be the given right line, joining the given points A and B. Let the direct, lateral, and vertical distances of the point A be TD, DC, and CA. Let the direct, lateral, and vertical distances of the point B be TD', DC', and C'B. Then, by ProblemI., the position of the point A on the plane of the picture is a. And similarly, the position of the point B on the plane of the picture is b. Join ab. Then ab is the line required. [p14] If the line AB is in a plane parallel to that of the picture, one end of the line AB must be at the same direct distance from the eye of the observer as the other. Therefore, in that case, DT is equal to D'T. Then the construction will be as in Fig.7.; and the student will find experimentally that ab is now parallel to AB. [Geometric diagram] And that ab is to AB as TS is to TD. Therefore, to draw any line in a plane parallel to that of the picture, we have only to fix the position of one of its extremities, a or b, and then to draw from a or b a line parallel to the given line, bearing the proportion to it that TS bears to TD. [p15] If the line AB is in a horizontal plane, the vertical distance of one of its extremities must be the same as that of the other. Therefore, in that case, AC equals BC' (Fig.6.). And the construction is as in Fig.8. [Geometric diagram] In Fig.8. produce ab to the sight-line, cutting the sight-line in V; the point V, thus determined, is called the Vanishing-Point of the line AB. Join TV. Then the student will find experimentally that TV is parallel to AB. [p16] If the line AB produced would pass through some point beneath or above the station-point, CD is to DT as C'D' is to D'T; in which case the point c coincides with the point c', and the line ab is vertical. Therefore every vertical line in a picture is, or may be, the perspective representation of a horizontal one which, produced, would pass beneath the feet or above the head of the spectator. |
|
