De proiectione geographica de Lisliana in mappa generali imperii Russici usitata

§. 10. Now, before all else, let us see how much this representation is going to depart from the truth at the extreme termini of the map, A and B. Let A $a$ be one degree of the parallel for terminus A, and B $b$ be such a degree for terminus B, which in reality ought to be $cos. a$ and $cos. b$. In order to investigate the quantity of these degrees on the map, let us first seek the angle P O $p$ corresponding to one degree, which will be

Let us, therefore, for the sake of brevity, set this angle = $ω$, so that $ω = \frac{δ (cos. p - cos. q)}{q - p}$ becomes true. Hence, therefore, if we take p = 50° and q = 60° as above, that angle P O $p$ will become $ω = 49'. 6''$. In this calculation, however, it must be properly observed that the interval $q - p$ should not be expressed in degrees, but in parts of the radius, where it should be noted that the quantity of one degree is 0.01745329. Hence, therefore, it is clear that the angles $ω$, which refer individual degrees of longitude to point O, are somewhat smaller than one degree.

§. 11. Here, however, looking at the matter in general, let us set that angle corresponding to one degree = $ω$, so that $ω = \frac{δ (cos. p - cos. q)}{q - p}$ becomes true, where it should be noted that because here the letters $p$ and $q$ are expressed in degrees, the interval $q - p$ must be multiplied by 0.01745329, for which, for the sake of brevity, let us write $α$, so that $ω = \frac{δ (cos. p - cos. q)}{α (q - p)}$ becomes true, where $1^\circ$ can be written in place of $δ$, if indeed we also desire the angle $ω$ in degrees. Furthermore, let us set the distance of point O beyond the pole = $z$ degrees. Since, therefore, the distance of place P from the pole is 90° - p, its distance from point O will be 90° - p + z, the value of which in parts of the radius will be $α (90° - p + z)$. This interval, however, was found before to be $PO = \frac{(q - p) cos. p}{cos. p - cos. q}$, which, because it is expressed in degrees, ought to be equated to the angle 90° - p + z, so that from this, $z = \frac{(q - p) cos. p}{cos. p - cos. q} - 90° + p$ becomes true.

§. 12. With these things set, because the distance of terminus A from the pole is 90° - a, the interval $A O = 90° - a + z$ will be, and in parts of the radius