The description and use of the universall quadrat.: By which is performed, with great expedition, the whole doctrine of triangles, both plain and sphericall, two severall wayes with ease and exactness. Also the resolution of such propositions as are most usefull in astronomie, navigation, and dialling. By which is also performed the proportioning of lines and superficies: the measuring of all manner of land, board, glasse; timber, stone. &c. / By Thomas Stirrup, Philomathemat.

CHAP. XLVI. To finde how many minutes or miles, answer to one degree of longitude, in any latitude required.

IN sailing by the compasse, the course holds somtime upon a great circle, sometime upon a parallel to the E∣quator; but most commonly upon crooked lines wind∣ing towards one of the Poles, which lines are well known by the name of Rombs.

If the course hold upon a great circle, it is either North or South, under some one of the meridians, or else East or West under the Equator: and in these cases, every de∣gree requires an allowance of 20 leagues or 60 miles, eve∣ry 20 leagues making a degree difference in sailing upon those circles; so that here needs no further precept then this, divide your leagues sailed by 20, and you shall have the degrees of distance; and contrarily, multiply the de∣grees of distance by 20, so shall you have the number of leagues sailed.

But if the course hold East or West, upon any of the parallels to the Equator, to finde how many miles do an∣swer to one degree of the Equinoctial take this propor∣tion.

• As the radius,
• To the cosine of the latitude:
• So is 60 miles the measure of one degree at the equator,
• To the number of miles answering to one degree of lon∣gitude in the given latitude.

Thus in the latitude of 50 deg. if I place the threed up∣on the intersection of the contrary sine of 90 deg. with the right sine of 40 deg. (which is the complement to 50 deg.) it will cut the contrary parallel of 60, at the right parallel of 38 34/60, and so many miles do answer to one de∣gree of longitude, in the latitude of 50 deg.

Again, in the latitude of 52 deg. 30 min. I place the threed to the intersection of the contrary sine of 90 deg. with the right sine of 37 deg. 30 min. and it cutteth the contrary parallel of 60, at the right parallel of 36 31/60, and so many miles do answer to one degree of longitude, in the latitude of 52 deg 30 min. as was required.

Or if you place the bead to 60 on the side of the qua∣drat, and then open the threed to the angle of the latitude, as here to 50 deg. the bead will cut the contrary parallel of 38 34/60, as in the first example.

And if you open the threed to the angle of 52 deg 30 min. the bead will cut the contrary parallel of 36 31/60, as in the latter example.