14068An ESSAY
this Appearance paſſes thro’ that of the Point
L, it will be a Part of C X. But becauſe that
Line, drawn from the Point L, paſſes thro’ the
propos’d Point; the Repreſentation of the ſaid
Point is alſo in C X; and ſo in X, the common
Interſection of C X, and T X.
L, it will be a Part of C X. But becauſe that
Line, drawn from the Point L, paſſes thro’ the
propos’d Point; the Repreſentation of the ſaid
Point is alſo in C X; and ſo in X, the common
Interſection of C X, and T X.
Remark.
83.
If the Point T ſhould be at too great a
Diſtance; or if T B X, or C X, ſhould too ob-
liquely cut each other; the perſpective Plane
muſt then be ſuppos’d to be reduc’d to a 1181. pendicular, or upright one; and the Repreſen-
tation of a Point, above the Geometrical Plane,
(whoſe Seat is L, and Height M E) muſt be
found .
2250.Diſtance; or if T B X, or C X, ſhould too ob-
liquely cut each other; the perſpective Plane
muſt then be ſuppos’d to be reduc’d to a 1181. pendicular, or upright one; and the Repreſen-
tation of a Point, above the Geometrical Plane,
(whoſe Seat is L, and Height M E) muſt be
found .
Problem III.
84.
To find the Repreſentation of a Line, perpendi-
cular to the Geometrical Plane`.
cular to the Geometrical Plane`.
The Appearance of the Extremity of the Per-
33Fig. 45. pendicular muſt be found , in conſidering 4482. ſaid Extremity as a Point above the Geometri-
cal Plane, by the Height of the propos’d Perpendi-
cular: Then if a Line be drawn from the Point
D, to the Point of Sight; its Interſection 5516. T X, will give the Appearance a of the Seat of
the Perpendicular propos’d.
33Fig. 45. pendicular muſt be found , in conſidering 4482. ſaid Extremity as a Point above the Geometri-
cal Plane, by the Height of the propos’d Perpendi-
cular: Then if a Line be drawn from the Point
D, to the Point of Sight; its Interſection 5516. T X, will give the Appearance a of the Seat of
the Perpendicular propos’d.
Note, When there is a Neceſſity of having re-
courſe to the Remarks of the foregoing Problem,
in order to find the Point X; then the Point a
may be found, in drawing A S and D V, and
afterwards joining the Points B and X by a
Line. And when B X and D V cut each
courſe to the Remarks of the foregoing Problem,
in order to find the Point X; then the Point a
may be found, in drawing A S and D V, and
afterwards joining the Points B and X by a
Line. And when B X and D V cut each
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