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 Third Scituation. One Place having North Latitude, and the other Place having South La∣titude, of different Quantities, and the Difference of their Longitudes less than 90. As I omit one Place having North Latitude, as Lundy, 51 deg. 22 min. the other South Latitude, as the Rio de la Plata, 35 deg. 00 min. Difference of Longitude betwixt them 45 deg. 55 min. I demand the Distance, the Angle of Position, and the greatest La∣titude or Obliquity of the Great Circle that passeth over these two Places.

AFter you have described the outward Meridian NESA, take from the Line of Chords the Latitude 51 deg. 22 min. and lay it from E to P, and draw the Line PCO and HCM at Right Angles to P, take off the half Tangent Line the Dif∣ference

[illustration] geometrical diagram

of Longitude 45 deg. 55 min. and lay it from E to F, and draw the Me∣ridian-Circle NFS, whereon lay the Latitude of Rio dela Plata 35 deg. from F to R, by taking 35 out of the Line of Chords, and laying it from E to 35, and the ½ Tangent of FE from C to Pole, and draw the prick'd Line Pole 35, which cuts the Circle NFS in R, the Rio dela Plata: and through R draw the Circle PRO, and measure MN on the half Tangents, you will find the Angle of Position to be RPE 36 deg. 2 min. Then take the half Tangent of 36 deg. 2 min. and lay from C to K, and draw the prick'd Line from K through R, and it will cut the Line at T; therefore measure TP on the Line of Chords, and that is the measure of RP 95 deg. 18 min. for the Distance, or 1906 Leagues, or 5718 Miles: The greatest Latitude or Obliquity is from AE to L; and VLW is the Parallel of 68 deg. 21 min. the greatest Obliquity required,

PROBL. V.

Then by the Tables,

As Radius, To the Co-sine of Differ. of Longitude 45 d. 55 m.	984242
So is the Co-tangent of the greater Latitude 51 deg. 22 min.	990267
To the Tangent of the first Arch 29 deg. 5 min.	974509

The less Latitude 35 deg. and 90 deg. makes 125 deg. Take the first Arch 29 deg. 5 min. therefrom, and there remains 95 deg. 55 min. Take this out of 180 deg. and there remains 84 deg. 5 min. the second Arch: Then

As Co-sine of the first Arch 29 deg. 5 min.	994005
Is to the Co-sine of the second Arch 84 deg. 15 min.	901318
So is the Sine of the greater Latitude 51 deg. 22 min.	989273
Out of	180d	00′
Take	84	42	To the Co-sine of 84 d. 42 m.	890591 Sum
And there remains	95	18	The true Distance 95 d. 18 m.	896586

which was required.

Now to find the Obliquity, Take both their Latitudes as if they were North, or both South, and the Complement of the Difference of Longitude to 180 deg. which here is 134 deg. 05 min. half that is 67 deg. 2 min. 30″: both the Latitudes added to∣gether make 86 deg. 22 min. half that is 43 deg. 11 min. it being too little, I added about 1 deg. 20 min. to the half, to find the mean and true Latitude 44 deg. 31 min. by which I find the Obliquity, as I proved by this Operation.

As Radius, To Co-tangent of the Latitude 44 deg. 31 min.	1000732
So is Co-sine of half the Difference of Longitude 67 deg. 2 min.	959158
To the Co-tangent of the Obliquity 68 deg. 21 min.	959890

Now to find whether 68 deg. 21 min. be indeed the true Obliquity, make these Proofs of it.

As Radius, To Co-tangent of Obliquity 68 deg. 21 min.	959890
Take from it the Tangent of the Latitude 51 deg. 22 min.	990267
There remains the Co-sine of Differ. of Longitude 60 deg. 14 min.	969623

Again,

As Radius, To Co-tangent of Obliquity 68 deg. 21 min.	1959890
Take out the Co tangent of the other Latitude 35 deg. 00 min.	1015477
There remains Co-sine of Differ. of Longitude 73 deg. 51 min.	944413

Now both the Longitudes found,* 1.1 73 deg, 51 min. and 60 deg. 14 min. added

together, makes just 134 deg. 05 min. the Difference of Longitude at first propounded betwixt those two Places; which proves, That the greatest Obliquity of the Great Circle that passeth directly over these two Places, the Island Lundy and Rio dela Pla∣ta, so scituate, is 68 deg. 21 min.

Now if it so happen that both the Latitudes be of the same Quantity, as one Place North Latitude 11 deg. 30 min. and the other Place South Latitude 11 deg. 30 m. and the Difference of Longitude betwixt the two Longitudes 55 deg. 48 min. To find the true Great Circles Distance betwixt such Places, first divide the Difference of Lon∣gitude into two equal parts, and then take one Latitude and half the Difference of Longitude, and find the Distance belonging to one Latitude, which doubled, yields the whole Distance betwixt the Places propounded: As Longitude 55 deg. 48 min. halfed is 27 deg. 54 min. and Latitude 11 deg. 54 min. Then work thus.

As Radius, To Co-sine of Differ. of Longitude 27 d. 54 m.	994633
So is the Co-sine of the Latitude 11 deg. 30 min.	999119
To the Co-sine of 30 deg. half the Distance	993752

which doubled is 60 deg. the whole Distance betwixt one Place South Latitude 11 deg. 30 min. and and another North Latitude 11 deg. 30 min. having 55 deg. 48 min. Difference of Longitude. And so work for any two Places so scituate.

Geographical Questions. Two Ships being at Sea, their Difference of Longitude was 53 deg. Now upon a day they observed the Sun being between them; the North Ship found the Sun's Meridian Height 33 deg. and the South Ship 77 deg. the middle Latitude between them was 15 deg. North of the Aequi∣noctial Line: I demand the Angle of Position, and Distance from the North Ship to the South?

* 1.2FIrst add the two Meridian Altitudes Complement together, 33 deg. and 77 deg. Complement 57 and 13, the Sum is 70, the half Sum is 35 deg. the middle La∣titude 15; add the middle Latitude and half Sum together, it makes 50 deg. the

[illustration] geometrical diagram

North Ships Latitude; and substract the middle Latitude from the half Sum, and the Remain is 120 deg. the Latitude of the South Ship. The North Ships Latitude is laid from Q to N 50 deg. the Difference of Longitude QF 53 deg. Through F de∣scribe the Great Circle Meridian PFB, on which lay down the South Ships Latitude 20 deg. as FC, and so draw the Great Circle NCD through the Intersection of the prick'd Line IH, with the Meridian at C; for that is the Latitude of the second Ship: So the Angle of Position is NCQ, whose measure is CG on the half Tangents 48 deg. 58 min. from the South Westwards; and the Distance is NH 48 deg. 22 min. that is, 1683 ⅔ Leagues, or 5051 Miles, the nearest Distance of the two Ships, which was required. How to do it by the Tables, you have been shewed in the last Example.

Notes

• * 1.1

  60	14
  73	51
  134	05
• * 1.2

  Meridian Altitude.

  33 d. Compl. 57 deg.

  77 d. Compl. 13 deg.

  Sum 70

  ½ Sum 35

  Mid. Lat. 15

  Nor. Ship 50 lat.

  South Ship 20 lat.