De proiectione geographica de Lisliana in mappa generali imperii Russici usitata

which is reduced to this: $α (a - b) ω = \cos. a - \cos. b$; whence we immediately conclude $ω = \frac{\cos. a - \cos. b}{α (b - a)}$, which value will be expressed in parts of one degree.

§. 15. After we have therefore rendered the errors of the projection equal at both termini A and B, let us furthermore equate them to the maximum error that can occur anywhere within the interval AB, which error, since it falls at the midpoint X, whose latitude is $= \frac{a+b}{2}$, the error at this location will be $α (90^\circ - \frac{a+b}{2} + z) ω - \cos. \frac{a+b}{2}$, which, since it tends in the opposite direction, must be set as

let this error, therefore, be set equal to the errors found for $a$ and $b$, and these two equations will arise:

§. 16. But indeed, the equality of the errors at termini A and B has already supplied us with this equation: $ω = \frac{\cos. a - \cos. b}{α (b - a)}$, which value, substituted into either of the preceding equations, will supply this equation:

which is reduced to this form:

from which equation it will be easy to define the distance $z$.

§. 17. Let us now apply these to the case of the map of the Russian Empire, where $a = 40^\circ$ and $b = 70^\circ$, and hence $\frac{a+b}{2} = 55^\circ$. Therefore, from this