De proiectione geographica de Lisliana in mappa generali imperii Russici usitata

§. 23. So that this investigation may be conducted more easily, let our middle meridian AB be extended as much upward to O as downward to the equator at E, such that EA = 40°, AB = 30°, and BO = 25°; however, let the pole be at $Π$, with $Π$O = 5°, and let the circle described with center O through E represent the equator, even if we do not need it on our map, in which let an arc EF of 90 degrees be taken, as we have just defined, and the angle EOF = 90° . $ω$ = 72°, 53', with the interval OF being 95°. This point F will therefore be the common pole of all great circles that can be drawn normally to our meridian AG.

§. 24. But if, therefore, we take any point z within the interval, through which a great circle normal to meridian AB ought to be drawn, it will indeed be perpendicular to AB at z and will pass through point F. Its true shape, however, will be a highly transcendental curve; meanwhile, it will hardly differ noticeably from a circular arc, which will be drawn through points Z and F normally to the straight line AB, which let be arc ZF, to find the curvature of which, let a perpendicular be drawn from F to the straight line OE, and it will be

From this, therefore, it is clear that the straight line FG itself represents the quadrant of a great circle normal to meridian AB, which, since it contains nearly ninety degrees of the meridian, will hardly deviate from the truth. But indeed, if such a great circle normal to BA is drawn through terminus A, its arc AF will be somewhat larger than the straight line FG; meanwhile, however, the error can be easily tolerated. For the radius of such a circle would be 165°, 9477, which is already so great that its curvature on the map turns out to be hardly perceptible.