De proiectione geographica de Lisliana in mappa generali imperii Russici usitata

the radius = a ( 90° — a + z ), which multiplied by ω will give degree A a, the quantity of which will therefore be [ δ ( 90° — a + z ) ( cos. p — cos. q ) ] / [ q — p ], and since this degree ought in reality to be = δ cos. a, the difference between these values will show the error of this projection at terminus A itself.
In the same way, for the other terminus B in this projection, the degree of the parallel will be [ δ ( 90° — b + z ) ( cos. p — cos. q ) ] / [ q — p ], which, since in reality it is = δ cos. b, the difference between these values will show the error of this projection at terminus B itself.

§. 13. It will therefore be convenient, first, to choose the two intermediate locations P and Q such that the errors at both termini A and B result in being equal to one another, whence is obtained this equation:

which is reduced to this form:

§. 14. Moreover, in order to make our investigation easier, let us introduce into the calculation, in place of the quantities p and q, the interval z expressed in degrees, by which point O is removed beyond the pole; and, furthermore, let us assume the angle ω, which corresponds to individual degrees of longitude around point O, or at which two adjacent meridians one degree apart are inclined to one another, and that this angle ω is given by degrees or the usual parts of a degree, by which means it will be permitted to write unity for the letter δ. In this way, therefore, one degree of the parallel at terminus A will be

Since, therefore, in these locations the quantity of these degrees is in reality cos. a and cos. b, the errors equated to one another will yield this equation: