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Make the Equinoctial Colure to contain, with the Brass Meridian, an Angle of 570 22', and there keep the Globe fast; then move it up or down, till, by the Quadrant of Altitude, you find the Equinoctial Colure and the Horizon contain an Angle of 57° 32'; the Globe being fixed in this position, you have the Triangle POS formed thereon ; in which the CoDeclination P S, will be found 66° 57' ; the Latitude PO, will be seen to be 50° 56' on the Meridian ; and the Amplitude of the Sun's Rising from the North So, will appear on the graduated Horizon to be 51° 48'.
Thus I have briefly shewn the Manner of solving Right-angled Spherical Triangles by the Globe or Sphere; I could indeed have been more large on this Head, but I abhor Tautology, and an unnecessary Prolixity ; for as it is an Affront to the Ingenious, ( a Word to the Wife being enough ) so it is a fruitless Labour to the Dull and Brainless; whose pitious Fate is, always to be ignorant and indocible by any Means whatever. I am well assured those who have but a tolerable Genius, and know how to apply particular Rules and Examples to general Uses, will, by what is here faid, understand what ever is unfaid on this Matter ; for Example, they know how the Angle PSO may be measured, as was the Angle S PO in any
Café ; viz. by inverting the Triangle. They also know that the last Cafe might have been resolved by changing the Angles into Sides, according to
Theorem 13, and its Corollary, with some other things of this kind, which I thought not needfull to infift upon.
C H A P. XVI. .
The eighth Method of solving Right-angled
Spherical Triangles, by the Stereographic Projection.
HIS Way is (generally speaking) more artful than useful ; not but that to a Person well
versed in Spherics, it is of peculiar Use and Service ; for this Method dispels all Ambiguity, and Errors, which attend the Solution by most other Methods ; and by a little Use, is very practicable and easy. So that if the Ingenuity, Certainty, East, and Expeditinusness, of any Method, be fufficient to reconmend it, this cannot fail of Acceptance with all those who have the least Genius and Taste for this Science. The Problems I shall all along refer to, are those of the Stereographic Projection in the Chapter beforegoing.
As in every Cafe, the Triangle may be made either at the Center or at the Circumference of the Primitive Circle, I shall accordingly present the Reader with (wo Schemes, the one having the Triangle at the Center, the other at the Circumference of the Primitive Circle ; with the Manner of constructing them, and of folving the Triangles in all the six Cafes, as follows.
BC = 44 52 and one of the Oblique Angles C =
1. To construct the
Scheme, so that the given Angle may be at the Center.
M Describe the Primi
tive -DEFG; cross it with two Dia
D meters DF, EG; km make the Angle at C
= 56° 57', and set the Half-Tan H
gent of - 44° 52', from the Center C to B; then thro' the Point and at Right Angles to draw the Oblique Circle 10 shall the Triangle
be made as required.
2. Let the Angle gi
ven be at the Primi
tive. Describe the Primi
tive FCED; cross it with two Dia. meters
FE, CD; bm make by ( Prob. 2,) the Angle at C =
56° 57', by drawing the Obli
R que Circle CGD; then ( by Prob. 7, )
draw the Parallel at 44° 52' distance from through the Point of Interfection draw the Diameter
so will the Triangle
be made as required.
1. To find the Leg A B.
In Fig 1, lay a Ruler from the Pole a of the Oblique Circle E AG, to the Point B, and it will cut the Primitive in b; measure Fb on the Line of Chords, and you will find it 36° 15'. The same you will find it in Fig. 2, if A B be measured on the Half-Tangents from go downwards.
2. To find the Leg A C. İn Fig. 1, this is found on the Half-Tangents to to bę 28° 30'; and in the Fig. 2, A C measured on the Chords will be found the same,
3. To find the Angle B.
In either Figure, lay a Ruler from the Angular Point B, and the Pole of the Oblique Circle a, and it will cut the Primitive in m ; then the Arch Mm measured on the Line of Chords will be found to be 42° 34'.
B C = 44 52 and one of the Legs, as A B = 36 15
1. Let the An
615 A then will the in
Half - Tan-
H draw the Paral
G lel CAK; Set the Half-Tangent of 56 15' from and draw the Oblique Circle then will the Triangle
be that which is required.
B to A, D A G ; A B C,
2. Let the Triangle be
at the Periphery of
the Primitive. Draw the Primitive
Circle CEFD; set the Half-Tangent
a of 459 08, from D
im and draw the Parallel
HPI. With the Half-Tan
gent of 53° 45', describe the Parallel
then through the Points draw the great Oblique Circle also through the Point draw the Right Circle thus you have the Triangle
made as required VOL. II.