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1. The Distance between them. Suppose Cittadella at A, and Peckin at B; draw proper great Circles through these two Places, and there will be formed the Isosceles Triangles A P B, in which each Side AP or P B is equal to the Complentent of the Latitude, viz. 50° oo', and the Angle at P = 104° 30', the Difference of Longitude. But the Perpendicular PC divides it into two Right Angled Triangles APC and C P B equal to each other ; by either of which this Problem is satisfied.

Therefore in the Right-angled Triangle A PC, there is given AP = 50° oo', and the Angle APC = 52° 15'; to find AC = Half the Distance requi red, thus ;

10.0000000

As Radius
To the Sine of AP = 50° 00'

9.8842540 So is the Sine of APC = 52° 15'

9.8980060 To the Sine of AC = 37° 17' The Double whereof is 74° 34' = the Distance required in the Arch of a great Circle, viz. AB= 5182.2 English Miles nearly; and 4674 Italian ones.

9.7822600

2. To find how they bear from each other ; say,

As Radius

10.0000000 To the Co-line of 50° po' = AP 9.8080675 So is the Tangent of 529 15' = A PC 10.1111004 To the Co-tangent of 50° 18' =CAP 9.9191579 That is, Peckin bears N. E. and 5° 18'eafterly of Cittadella ; consequently Cittadella bears S.W. and 50 181 more westerly of Peckin. Whereas according to the Rhumbs, they, lie directly East and West of each other. Such a Difference there is between the Position

of

of a Rhumb and a great Circle on the Surface of the Globe.

PROBLEM IV.

To find the Distance and Bearing of two Places, the one lying under the Equinoctial, the other in any given Latitude ; their Difference of Longitude being

given also

London

Jumata

Practice. Example. The

P Middle of the İne of Sumatra lyeth under the Equator, and is juft an

1000 more East than the Longitude of London, whose Latitude is 512 32'; I demand the Distance of that Ife, and on

S what Point it bears from London?

First draw the proper great Circles (one of which LIO, passeth through London and the said INand, ) and you will have formed the Right-angled Triangle 102; in which there is given the Side Q = ÆL

= 51° 32' the Latitude of London ; and the Side 12= 80', the Complement of the Difference of Longitude Æ I to 180°, or a Semicircle.

I. To find the Side 10, the Complement of the Distance from London to 180°, or a Semicircle. Thus

;

VOL. II.

Q_q

As

As Radius

10.0000000 Is to the Co-line of 21 = 80° 00' 9.2396702 So is Co-fine of 20 = 51° 32'

' 9.7938317 To the Co-fine of O I= 83° 481 9.0335019 Which fubftract from 180° oo' There remains LI= 96° 12' the Arch of Distance between the two Places ; which reduced is 66857% English Miles ; and so far is Sumatra from London the nearest Way that even a Bird could fly to it.

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2. To find the Angle at 0 = P LI, the Point on which Sumatra bears from the N towards the E of London. Thus,

As the Sine of O I = 83° 48'

9.9974523 Is to Radius

10.0000000 So is the Sine of 1 2 = 80° ool

9.9933515 To the Sine of the Angle at 0 = 829 08' 9.9958992 Thus Sumatra lies E by N, and 3° 23' to the East of London.

3. To find the Angle at I, the Point on which London bears from the W to the N of Sumatra. Thus,

10.0000000

As the Sine of O I = 83° 48' 9.9974523 Is to Radius So is the Sine of 20= 519 321 9.8937452 To the Sine of the Angle at I = 51° 57' 9.8962929 Thus Sumatra hath London bearing from it NW, and 6° 57' to the North,

PRO

PROBLEM Y.

Given the Difference of Latitude and Longitude of any two Places ( on one side the Equator, ) to find their Distance and Position from each other.

Practice.

Londos

L

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Admit

Р the Latitude of London be 51° 32', the Latitude of

1

Ye Terusalem Jerusalem 32° 44' ;

Æ and their Difference of Longitude to be 37° 24' ; I demand their Dir

S tance from each other?

Draw the Circles and prepare the Scheme as here you see it, and you will have formed the Oblique Triangle LP I, in which there is given the Side L P = 38° 28', the Side I P = 57° 16'; and the Angle at P = 37° 24' the Difference of Longitude ; co find the Side I L. In order to this, let fall the Perpendicular A I, then in the Right-angled Triangle API, find AP, thus ;

VOL. II

Qq 2

As

As Radius
Is to the Co-fine of P

37° 24'
So is the Tangent of IP = 57° 16'
To the Tangent of AP =

512017

10.0000000

9.9000472 10.1919171 10.0919643

Then AP-LP = AL = 12° 33'. Now say;

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As the Co-sine of A P = 51 oi C. Ar.0.2012842 Is to the Co-fine of AL= 12 33 9.9894973 So is the Co-sine of IP = 57 16

9.7329803 To 'the Co-line of IL = 32 58 9.9237618

This 32° 58' reduced, make 2291 English Miles ; and so far is Jerusalem distant from London, according to the latest Observations.

Note, The Difference of Longitude by many Authors and Maps is made 50° 24', and then the Arch IL would be found 40° 49' = 2837 Miles ; 546 more than it is now known to be ; for this Reafon I did not follow the old Account of the Longitude of Jerusalem ; though in most other places I was obliged to do it for want of better.

2. To find the Position of Jerusalem from London ; or the Angle A LI. Thus ;

As

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