De proiectione geographica de Lisliana in mappa generali imperii Russici usitata (17uu) - Page 8

first, for the angle $ω$, we will obtain this equation:

$ω = \frac{\text{cos. } 40^\circ \text{ — cos. } 70^\circ}{30 α} = \frac{0,4240243}{0,5235987}$

whence $ω = 48'. 44''$ is found. Having found this value, therefore, the prior equation, with the values for $a$ and $b$ substituted, becomes

$α (85^\circ + 2 z) ω = \text{cos. } 40^\circ + \text{cos. } 55^\circ = 1,33962$

however, $α ω = \frac{0,42402}{30} = 0,0141$ was the case, and thus we will have

$85^\circ + 2 z = \frac{1,33962}{0,01415} = 95^\circ. 0'$ and therefore $z = 5^\circ$.

§. 18. We assumed here that the maximum error falls near the middle of the interval A B; but since it may differ from this location, let us seek this very point X where the error becomes the maximum. Let $x$, therefore, denote the latitude of this location, and because the error there will be $α (90^\circ - x + z) ω - \text{cos. } x$, let us set its differential equal to zero. Here, however, one must be careful not to write $- d x \text{ sin. } x$ for $d. \text{cos. } x$ in the customary manner, because here $x$ is assumed to be expressed in degrees, while the differential of the arc itself, which is $α x$, must be multiplied by $\text{sin. } x$. Since, therefore, it is

$d. \text{cos. } x = - α d x \text{ sin. } x$,

the differential of our formula will give

$- α ω d x + α d x \text{ sin. } x = 0$, whence it follows that $\text{sin. } x = ω$

where $ω$ is the fraction found above $= \frac{\text{cos. } a - \text{cos. } b}{α (b - a)}$, the value of which in our case is $\frac{0,4240243}{0,5235987} = \text{sin. } x$, whence $x = 54^\circ . 4'$, which location therefore hardly differs from the midpoint of the interval A B.

§. 19. Now that this value for $x$ has been found, the error at that location will be $α (90^\circ - x + z) ω - \text{cos. } x$, the negative of which, set equal to the error at termini A and B, will give this equation:

$α (180 - a - x + 2 z) ω = \text{cos. } a + \text{cos. } x$

from which the value of $z$ itself must be defined, specifically: since $x = 54 \frac{1}{15}^\circ$, the equation will be $85 \frac{14}{15} + 2 z = \frac{\text{cos. } a + \text{cos. } x}{α ω} = 95^\circ 56'$, and therefore $2 z = 10^\circ$ and $z = 5^\circ$ with $ω = 0,8098270$ in degrees, or $ω = 48'. 44''$.