The mariners magazine, or, Sturmy's mathematical and practical arts containing the description and use of the scale of scales, it being a mathematical ruler, that resolves most mathematical conclusions, and likewise the making and use of the crostaff, quadrant, and the quadrat, nocturnals, and other most useful instruments for all artists and navigators : the art of navigation, resolved geometrically, instrumentally, and by calculation, and by that late excellent invention of logarithms, in the three principal kinds of sailing : with new tables of the longitude and latitude of the most eminent places ... : together with a discourse of the practick part of navigation ..., a new way of surveying land ..., the art of gauging all sorts of vessels ..., the art of dialling by a gnomical scale ... : whereunto is annexed, an abridgment of the penalties and forfeitures, by acts of parliaments appointed, relating to the customs and navigation : also a compendium of fortification, both geometrically and instrumentally / by Capt. Samuel Sturmy.

The second Scituation.

SEcondly, There may be two Places scituated in divers Parallels of Latitude, be∣twixt the Artick and Antartick Poles, that may have one Degree and Minute of Latitude, yet may have several Degrees of Longitude.

[illustration] geometrical diagram

PROBL. III.

Admit there be two Places both in the Latitude of 51 deg. 22 min. and their Difference of Longitude be 52 deg. 55 min.

1. To find the nearest Distance of those two Places.

2. The Direct Position of the one Place from the other.

How to do these Questions Geographically.

TAke off the Line of Chords the Latitude of the Place 51 deg. 22 min. and lay from AE to X, and from Q to C; and take of the ½ Tangents the same Latitude, and lay from K to O; and through these three Points draw the Parallel of Latitude XOC; the Difference of Longitude laid from Q to L, draw the Meridian Circle NLS, the second Place is at R, and first at C the Meridian-circle cuts at R: Therefore draw the Circle from C through R to B, and measure HI on the ½ Tan∣gents, and you will find it 68 deg. 46 min. for the Angle of Direct Position HCI. Now from the ½ Tangents take 68 deg. 46 min. and lay it from the Center K to E, and from E draw through the Point of Intersection at R the prickt Line ERF; and because it cuts the Line in F, therefore measure CF on the Line of Chords, and you will find it 32 deg. 18 min. for the true Great Circles Distance, which is 646 Leagues, or 1938 Miles.

In the Seventh Problem of sailing by Mercator's Chart, yo may see there was re∣quired the Distance of these two Places measured in the Parallel, and found to be 660 5/10: but here is required the nearest Distance in the Arch of a Great Circle: Work thus by the Tables.

For the Distance,

As the Radius, Is to the Sine Comple of the Lat. 38 d. 38 m. RN	979541
So is the Sine of the Differ of Longitude 26 d. 27 m. RFN	964876
To the Sine of half the Distance 16 deg. 09 min. RF	944417

Which doubled is 32 deg. 18 min. and this converted into Leagues and Miles, as be∣fore, is 646 Leagues, and 1938 Miles, the nearest Distance, and less than the Di∣stance measured in a Parallel by Miles 42.

To find the Direct Position,

As Radius 90, Is to the Sine-Compl. of the Lat. 58 d. 38 m. RN	989273
So is the Tangent of ½ the Differ. of Longitude 260 d. 27 m. RNF	969678
To the Co-tangent of the Angle of Position 68 d. 46 m. NRF	958951

Which sheweth, that if you will go the nearest way from C to R, you must not go West, though both be under one Parallel; but must first shape your Course from C from North 68 deg. 46 min Westerly, that is almost W. N. W. and so by little and little inclining to W. b. N. and then W. and W. b. S. and almost W. S. W. as before.

How to find the Obliquity.

TWo Places having Latitude both the same, as 51 deg. 22 min. and towards the same Pole, whether North or South, and Difference of Longitude 52 deg. 55 min. or any number of Degrees under 90: If above 90, take it out of 180, and work with the Remainer the same manner of way.

PROBL. IV.

As Radius 90, To Co-tangent of the Latitudes 51 d. 22 m. RN	990267
So is the Co-sine of ½ the Differ of Longitude 26 d. 27 m. RNF	995197
To the Compl. Tan. of the greatest Obliquity 54 d. 25 m. NF	985464

So that the greatest Obliquity is 54 deg. 25 min. And the same Proportion will hold for any Question of this nature.

We might proceed to frame many Questions touching those two Places; but these being the most material, I leave the rest to your own Practice, to use as much brevity as I may. I might have shewn you the Side and Angles; but in regard they are Sphe∣rical I omit it, and shall demonstrate them at last in Plano: But you may follow these Rules, if you cannot apprehend the Diagram; but some may desire the Tri∣angle, therefore I lay it down.

[illustration] geometrical diagram

In this Triangle CRN, let the two Places be C and R, and let N be the North Pole; then CN or RN either of them are 38 deg. 38 min. the Complement of the Latitude and the Angle CNR is the Difference of Lon∣gitude: There is required CFR the nearest Distance, and the Direct Position of the one to the other, NCR or NRC; for in this Case those two Angles are equal: And seeing NC and NR are equal, therefore let fall the Perpendicular NF, the Triangle NCR is divided into two Right-angled Triangles CNF and RNF, which are every way equal.