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.

Of Right-Angled Plain Triangles.

CASE I. In a Right-Angled Plain Triangle, The Base and the Angle at the Base being given, To find the Perpendicular.

SUppose that the Line CA (in the following Figure) in the Right-Angled Trian∣gle, were a Tree, Tower, or Steeple, and that you would know the Height thereof; you must observe with your Instrument the Angle CBA, and measure the Distance BA.

So have you in the Right-Angled Triangle ABC, the Base 405 Foot (Miles or Leagues the denomination might have been as well) and the Angle at the Base 32 deg. and it is required to find the Perpendicular AC.

Now because the Angle CBA is given, the Angle BCA is also given, it being the Complement of the other to 90 deg. and therefore the Angle BCA is 58 Degrees: Then to find the Perpendicular CA, the Proportion is,

As the Sine of the Angle BCA 58 deg. (which is)	9928420
Is to the Logarithm of the Side BA 405 Foot	2607455
So is the Sine of the Angle CBA 32 deg. (which is)	9724210
The Sum of the Second and Third added	12331665
The first Number substracted from the Sum	9928420
To the Logarithm of the Side CA	2403245

The nearest Absolute Number answering to this Logarithm 2403245, is 253 fere; and that is the Length of the Side CA in Miles or Leagues, or the Height of the Tree, Tower, or Steeple, which was required.

A GENERAL RULE.

IN all Proportions wrought by Sines and Logarithms, you must observe this for a General Rule, (viz.) To add the second and third Numbers together, and from the Sum of them to substract the first Number; so shall the Remainder answer your Question demanded, As by the former Work you may perceive, where the Loga∣rithm of the Side BA 2607455 (which is the second Term) is added to the Sine of the Angle CBA 9724210 (which is the third Term) and from the Sum of them, namely from 12331665, is substracted 9928420, the Sine of the Angle BCA,

which is the first Number, and there remaineth 2, 403245, which is the Logarithm of 253 almost, and that is the Length of the Side required.

[illustration] geometrical diagram

To resolve the same Work by the Line of Sines and Numbers.

YOu may work these Proportions more easily by help of the Line of Sines,* 1.1 Tan∣gents, and Numbers, on your Scale, the Proportion being as before.

Therefore if you set one Foot of your Compasses at 58 deg. in the Line of Sines, and extend the other Foot to 405 in the Line of Numbers, the same will reach from the Sine of 32 deg. to 253 in the Line of Numbers, which is the Length of the Side AC, which was required.

Or otherwise, Extend the Compasses from the Sine of 32 deg. to the Sine of 58 deg. in the Line of Sines; the same Extent will reach from 405 in the Line of Num∣bers, to 253, as before, the Work is much abbreviated, there being no need of Pen, Ink, nor Paper, or Tables; but only of your Compasses.

CASE II. The Base and the Angle at the Base being given, To find the Hypothenusa.

IN the same Triangle ABC, Let there be given (as before) the Base AB 405 Foot, Miles, Leagues, or Perches, and the Angle ABC 32 deg. and let it be required to find the Hypothenusa BC. Now because the Angle CBA is given, the other Angle BCA is also given; and the Proportion is,

As the Sine of the Angle BCA 58 deg.	9, 928420
To the Logarithm of the Side 405 Foot	2, 607455
So is the Sine of the Angle CAB 90 deg.	10, 000000
The Sum of the second and third Number	12, 667455 added.
To the Logarithm of the Side BC, which is	2, 679035

The Absolute Number answering to this Logarithm is 478; and so many Feet, Miles, Leagues, Perches is the Hypothenusa, according to the denomination of the Question; that is, whether it be Feet, Perches, Miles, or Leagues. By either of these the Work is the same way.

By the Line of Numbers and Sines.

AS was said before, the Work is altogether the same with the Tables; For the Proportion being,

As the Sine of the Angle BCA 58 degrees

Is to the Length of the Side BA 405 Foot:

So is the Sine of the Angle CAB 90 Degrees,

To the Length of the Side CB 478.

Extend the Compasses from the Sine of 58 deg. to 405 in the Line of Numbers; the same Extent will reach from the Sine of 90 deg. to 478 in the Line of Numbers, and that is the Length of the Side BC. Or you may extend the Compasses from the Sine of 58 deg. to 90 deg. the same Extent will reach 405 to 478, as before.

CASE III. The Hypothenusal and Angle at the Base being given, To find the Perpendicular.

IN the same Triangle let there be given the Hypothenusal BC 478 Feet, Poles, Miles, Leagues, and the Angle at the Base CBA 32 deg. To find the Perpen∣dicular CA.

The Angle CAB is a Right Angle, or 90 Degrees; Therefore the Proportion is,

As the Sine of the Angle CAB 90 deg	10, 000000
Is to the Logarithm of the Side BC 478	2, 679428
So is the Sine of the Angle CBA 32 deg.	9, 724210
To the Logarithm of the Side AC 253	12, 403638

The Number answering to this Logarithm is 253 fere; and that is the Length of the Side CA in Feet, Poles, Miles, or Leagues.

Here the Work is something abbreviated; for the Angle CAB being a Right Angle, and being the first Term, when the second and third Terms are added together, the first is easily substracted from it, by cancelling the Figure next your left hand, as you see in the Example; and so the rest of that Number is the Logarithm of the Num∣ber sought.

By the Line of Sines and Numbers.

EXtend the Compasses from the Sine of 90 Degrees, to 478; the same Extent will reach from the Sine of 32 Degrees, to 253.

Or, Extend the Compasses from the Sine of 90 Degrees, to the Sine of 32 Degrees; the same Extent will reach from 478, to 253; and that is the S••le CA.

CASE IV. The Hypothenusal and Angle at the Base being given, To find the Base.

LEt there be given in the Triangle the Hypothenusal BC, and the Angle at the Base CBA; and by consequence the Angle BCA the Complement of the other to 90 degrees: Then to find BA, the Proportion is,

[illustration]

As the Sine of 90 deg. CAB	10, 000000
To the Hypothenusal CB 478	2, 679428
So is the Sine of the Angle ACB 58	9, 928420
To the Logarithm of the Base AB	12, 607848

The nearest Number answering to 2, 607848, is the Logarithm of 405: And so many Foot or Poles, or if the Question be Miles or Leagues, is the Base or Parallel of Longitude AB.

Now you see the former Figure is turned, and therefore very fitly may have other Denominations (or Names) So that in the Art of Navigation, it will not be unfit to call one of these Sides the Parallel-Side, as AB, or Side of Longitude, or Meridian Distance; the other the Perpendicular-Side, or the Side of Latitude, as CA; and the Hypothenusal, the Side of Distance CB, and the Arches to lay down from the Chords, as before-directed.

By the Line of Sines and Numbers.

THe Angle given, as before, Extend the Compasses from the Sine of 90 deg. unto 478. the same Extent will reach from the Sine of 58 deg. to 405 in the Line of Numbers.

Or, Extend the Compasses from the Sine of 90 deg. to the Sine of 58 deg. the same Extent will reach from 478 to 405, which is the Length of the Base turned up, or Parallel-Line of Longitude, as before said, AB.

CASE V. Let the Perpendicular be the Difference of Latitude 253 Leagues, and the Angle at C, S. W. b. W. 1 deg. 45 min. Westerly, or 58 deg. Let it be given to find the Hypothenusal or Distance upon the Rhomb.

IF the Perpendicular or Difference of Latitude 253 Leagues AC be given, and the Angle at ACB, S. W. b. W. 1 deg. 45 Westerly, or 58 deg. Then by consequence the Angle ABC, or Complement of the Rhomb is also given; taking the first out of 90 deg. then the Hypothenusal may be found thus.

As the Complement Sine of the Rhomb 32 deg. at B	9, 724210
Is to the Logarithm of the Difference of Latitude 253	12, 403121
So is the Sine of the Angle or Radius 90 deg.	10, 000000
To the Logarithm of the Hypothenusal, or Distance upon the Rhomb or Course sailed 478	2, 678911

Here because the Angle CAB is a Right Angle, or 90 Degrees the Radius, and comes in the third place, I therefore only put an Ʋnity before the second Term, and so substract the first Term, and the Remainder is 2, 678911; the Absolute Number answering thereunto is 478, the Side required.

By the Line of Numbers.

EXtend the Compasses from the Sine of 32 deg. to 253 deg. the same Distance will reach from the Sine of 90 deg. to 478, the Side required.

Or, The Distance between the Sine of 32 deg. and 90 deg. will be equal to the Distance between 253 and 478, and giveth the Side required.

CASE VI. The Hypothenusal or Distance Sailed, and the Perpendicular or Diffe∣rence of Latitude given, To find the Rhomb or Angle ABC.

IN the foregoing Triangle, there is given the Hypothenusal or Distance sailed, CB 478 Leagu••s, and the Perpendicular or 253 Leagues difference of Latitude, and it is required to find the Angle ABC, and by it the Rhomb.

As the Logarithm of the Hypothenusal CB 478 Leagues	2, 679428
Is to the Right Angle or Radius 90 deg. CAB	10, 000000
So is the Logarithm of the Perpendicular 253 CA	2, 403121
To the Complement Sine of the Rhomb, or Sine of the An∣gle ABC 32 deg.	9, 723693

The nearest Number answering to 9, 723693, is the Sine of 32 deg. which de∣ducted from 90 deg. there remains the Angle of the Rhomb 58 deg. or S. W. b. W. 1 deg. 45 VVesterly.

By the Line of Numbers.

EXtend the Compasses from 478, to the Sine of 90; the same Distance will reach from 253, to 32 deg.

Or, Extend the Compasses from 478, to 253; the same Extent will reach from the Sine of 90, to the Sine of 32 deg, which is the inquired Angle ABC, and the Com∣plement of the Rhomb.

CASE VII. The Hypothenusal, and the Parallel of Longitude, and the Radius given, To find the Rhomb or Course Sailed.

IN the foregoing Triangle there is given the Hypothenusal or Distance Sailed, CB 478 Leagues, and the Right Angle CAB 90 deg. the Radius, and the Parallel of Longitude or Base 405 Leagues, to find the Course or Rhomb sailed, or the Angle ACB.

As the Hypothenusal or Distance sailed 478 CB	2, 679428
To the Right Angle CAB Radius or Sine of 90 deg.	10, 000000
So is the Parallel of Longitude, or Base AB 405 Leagues	12, 607455
To the Sine of the Angle of the Rhomb or Course sailed 58 or S. W. b. W. 1 deg. 45 Westerly	9, 928027

By the Lines of Sines and Numbers.

EXtend the Compasses from 478 in the Line of Numbers, to the Sine of 90 deg. the same Extent will reach from 405, to the Sine of 32 deg.

Or, Extend the Compasses from 478 to 405; the same Extent will reach from the Sine of 90, to the Sine of 32 deg. ACB, the Angle of the Rhomb or Course sailed, which was required.

Notes

• * 1.1

  Place here the Line of Numbers, Sines, and Tangents.