Fig.

1. By the Globe. XXVI.

Apply the Quadrant of Altitude, or Brass Plate, to the Two Places, and the Number of Degrees thereof contained between the Two Places, is their Distance, which will be found to be 50 Deg: 32 Min.

II. By Trigonometrical Calculation.
The Quadrant of Altitude (or Brass Plate) applied to the Two
Places, is represented by the Arch H B, and the Arches of the
Two Meridians, which pass through the Two Places, are PBN
and PHM; and P B and PH, are equal to the Complement of
of the Latitude of both the Places, viz.. 58 Deg. 20 Min. So
that now you have constituted upon the Globe an Oblique Spherical
Triangle PBH; in which you have given, (1.) The Two Sides
P B and PH, both equal to 58 Deg. 20 Min the Complement
of the Latitude. _(2.) The Angle BPH 60 Deg.. 15. Min. the
Difference of the Longitude of the Two given Places. To find
the Side B H, their Distance.. For which this is

The Canon for Calculation. By CASE I. of R. A. S. T.
As the Radius, Sine 90 Deg..

.:
Is to the Co-fine of the Common Latitude (PH or PB) 58 D. 20M..
So is the Sine of half the Difference of Longitude, (half BPH)

30 Deg. 07 Min..
To tó Sine of half the Distance (half B.H) 25 Deg. 16 Min.
The Double whereof, 5o Deg. 32 Min. is the Distance BH,
which in Miles is 3032 Miles.

PRO B. V.
Two Places, which differ both in Longitude and Latitude ; to find

their Distance.
N this there are Three various Positions.

1. If one of the Places lye under the Aquinoctial, and so have no Latitude ; and the other under fome Parallel of Latitude be: tween the Aquino&tial, and one of the Poles: As London, in 51 Deg. 30 Min. of North Latitude at L; and St. Thomas lisand under the Æquino&tial at T, but differ in Longitude 18. Deg. For finding the Distance of these. Two Places.

In , .

I. Upon

Fig.

a

1. Upon the Terrestrial Globe. Bring London to the Brass Meridian, and over it, fix the Qua. XXVI. drant of Altitude : The Globe being in this position, bring the Quadrant of Altitude to lye just over St. Thomas Island, and you will find it cut the Quadrant of Altitude, in 54 Deg. 45 Min. for the Distance of the Two Places.

II. By Trigonometrical Calculation. The Globe resting in the former Position, you will find constituted upon it a Right-angled Spherical Triangle L ÆT, composed of, (1.) L Æ, an Arch of the Brass, Meridian. (2.) Æ T, an Arch of the Aquino&tial. And, (3.) LT, an Arch of a Great Circle (made by the Quadrant of Altitude) paffing through both the Places: And in this Triangle, you have given, (besides the Right Angle at Æ) (1.) The Perpendicular LÆ, the Latitude of London, 51 Deg. 30 Min. (2.) The Angle at L, the Difference of Longitude 18 Deg. 10 Min. To find the Hypotenuse LT, the Distance.

The Canon for Calculation. By CAS E XIV. of R. A. S. T.
As Tang. of the Latitude L Æ, gi Deg. 30 Min.

Is to the Radius;
So is the Co-fine of ÆLT, 18 Deg. 10 Min.

To the Co-tangent of LT, 52 Deg. 55 Min.
Which reduced into Miles, makes 3175 Miles, for Distance be.

tween London, and St. Thomas Island.
2. If both the Places proposed shall be without the Æquino&tial,
but both of them, either on the North or South Side thereof: As
London in 31 Deg. 30 Min. at L, and Hierufalem in 31 Deg.
40 Min. at H, both on the North Side of the Æquino&tial; and
their Difference of Longitude 46 Deg: To find their Distance

Upon the Terrestrial Globe. Bring one of the Places, as London, to the Brass Meridian, and over it screw the Quadrant of Altitude, and keep the Globe there fixed, then move the Quadrant of Altitude, till it lye over HieruSalem, and you shall find it to lye under 39 Deg. 32 Min. of the Quadrant : And that is the Distance of the Two Places.

59

20

Fig.

By Trigonometrical Calculation. XXVI.

The Globe being in the former Position, you will find upon it an Oblique-angled Spherical Triangle, composed of P L, an Arch of the Brass Meridian: OfP H, an Arch of the Meridian that paf. seth over Hierufalem ; and of L H, an Arch of the Quadrant of Altitude. In which Triangle you have given, (1.) PL, the Complement of the Latitude of London, 38 Deg. 30 Min.' (2.) The Side P H (the Complement of the Latitude of Hierusalém 59 Deg. 20 Min. (3.) The Angle L PH, the Difference of Longi. tude of the Two Places, 46 Deg. To find the Two Angles of Poo sition at L and H: And, (2.) The Side L H, the Distance of the Two Places.

The Canons for Calculation.
By CASE JII. and CASE IX. of A.O. S. T..

4.O.S.

D. M.
The Side {P #}is { 38 39
Their. Sum is

96 50
Their Difference is

50 The Half Sum is

48 25 The Half Difference is

55
Being thus prepared, I say,
(1.) As the Sine of half the Sum of the Sides PL and PH, 48

Deg. 25 Min.
Is to the Sine of half their Difference, 9 Dég. 55 Min.
So is the Co-tangent of half the Difference of Longitude (i. e)

half the Angle L PH, 23 Deg.
To the Tangent of 28 Deg. 29 Min.

Which is the half Difference of the Angles P L H and PH L. (2.) As the Co-sine of half the Sum of the Sides P L and P H,

41 Deg. 35 Min.
Is to the Co-line of half their Difference, 80 Deg. 2 Min.
So is the Co-tangent of half the Difference of Longitude, i. e.

half the Angle L PH 23 Deg.
To the Tangent of 74 Deg. 2 Min.
Which is the Tangent of half the Sum of the Two Angles
PLH and PHL..

The

19

9

Fig. XXVI.

:

102

D. M.
The Half Sum is

74 02
The Half Difference is 28
Their Sum

31 equal to C PLH.
Their Difference

45 32 equal to <PHL.
(3.) For the Side L H, which is the Distance,
As the Sine of PHL, 45 Deg. 33 Min.
Is to the Sine of PL, 38 Deg. 30 Min.

So is the Sine of L PH, the Differ. of Longitude, 46 Dug.
To the Sine of L H, 30 Deg. 51 Min. .
Which 38 Deg. 31 Min. reduced in Miles, makes 2331 Miles

for the Distance of the Two Places.
3. If the Two Places proposed should be so situate, that one have
North, and the other South, Latitude, and under different Meri-
dians. As suppose Constantinople, lying in North Latitude 47

D.
at C. and the Cape of Good Hope, lying in 35 Deg. of South La-
titude, at V; and differing in Longitude 59 Dég. To find the Di.
stance of these Two Places

Upon the Terrestrial Globe.
Bring one of the Places (as Constantinople) to the Brass Meridi-
an, and there keep the Globe; then apply the Quadrant of Altis
tude (or rather a thin Plate of Brass divided as that is) to the Two
Places, and you shall find 97 Deg. 42 Min. of the Quadrant (or
Brass Plate) to be contained between them, and that is their
Distance.

By Trigonometrical Calculation.
The Globe resting in its former Position, you will discover up- .
on it an Oblique-angled Spherical Triangle, composed of P C, an
Arch of the Brass Meridian; P., V, an Arch of a Meridian, passing
through the Place in South Latitude: And of C V, the Quadrant
(or Brass Plate) representing the Arch of a Great Circle passing
through both the Places : - In which Triangle you have given,
(1.) The Side PC, the Complement of the Latitude of Constanti

.
nople, 43 Deg. (2.) The Side PV, the South Latitude of the
Cape of Good Hope, with 90 Deg. added, which make 125 Deg.
And, (3.) The Angle V PC, the Difference of Longitude, 59 Deg.
To find, (1.) The Angles of Position PVC and PCV: And;
(2.) The third Side VC, the Distance of the Two Places

The

Fig. The Canons for Calculations, as in the last, by CASE III and XXYI. IX. of A. O.S. T.

(1.) As the Sine of the Half Sum of the Sides PV and PC, 84 D.

Ís to the Sine of half their Difference, 41 Deg.
So is the Co-tangent of half the Difference of Longitude, i.e.

(half the Angle V PC) 29 Deg. 30 Min.
To the Tangent of 49 Deg. 23 Min.

Which is the Half Difference of the Angles PVC and PCV. . (2.) As the Co-fine of half the sum of the Sides, P. V and P.C,

6 Deg.
Is to the Co-fine of half their Difference, 49 Deg.
So is the Co-tangent of half the Difference of Longitude, 61 Deg.
To the Tangent of 85 Deg. 32 Min.
Which is half the Sum of the Two Angles, PC.V and PVC.

D. M.
The half Sum is

85 32
The half Difference is

49 23 Their Sum

134 55 equal to <P CV. Their Difference

36 09 equal to PVC.
The Two Angles of Position.
(3.) For the Side V C, which is the Distance of the Two Places.

As the Sine of PVC, 36 Deg. 9 Min.
Is to the Sine of PC, 43 Deg.

So is the Sine of V'PĆ, 59 Deg.
To the Sine of 82 Deg. 18 Min.
Whose Complement to 180 Deg. is 97 Deg. 42 Min. for the

Side V C, which is the Distance of the Two Places. Which

in Miles is 5862. And these are all the Varieties of Positions that any Two

Places upon the Globe can be fituate.

ANCILLA