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the projection. Let BH = BE = A polar distance of the circle to be projected, the chord HE will be per.pendicular to DB. Draw on and ors, Gs will be the diameter sought. Bisect Gs in n, n will be the centre, nE = nil = ns = ng, will be the radius of the projected circle. Meas. of Eng = 2nse = od — BE = 90°– BE = 90°–BCE cne + no e = 90°... cen = 90°.". En = tan BE = tan A, and cn = sec BE = sec A. These values serve for all circles which have their pole on the circumference of the circle of projection, f these are great circles, then A = 90°, and tan A, see A are infinite: consequently, the centres of the projections falling at an infinite distance from c, the projections themselves will be right lines passing through c, and intersecting under the angles which such circles, taken two and two, form on the sphere. 42. Let obe the point of observation, or place of the eye, A the pole of the projection, BDEcb the plane of projection, PD an arc of a great circle which has its origin at any point whatever M" of the circle ob PE, Pt the E tangent of PD; ct will be the secant, and s the projection of P. Draw st: then from the rectilineal triangle sco we shall have, st” = Cs” + cł? – 2Cs. Ct. cos sct = tan” #AP + seco PD – 2 tan AP sec PD cos DE = tan” AP + tano PD + 1–2 tan AAP sec PD cos DE = seco AP + tano PD – 2 tan #AP sec PD cos D.E. But the spherical triangle PDE right angled in E gives, (chap. vi. equa. 6), cos PD = cos PE cos DE = sin AP cos DE; 1 1. coso, "" in Arco Do" Substituting this in the last value of st”, we have

and, therefore, sec PD =


st" = seco 3AP + tano PD – stan war coins Sin Ap cos de 2 tan #Ap 2sinor cosor = seco AP + tano PD – seco AP = tan? pp. Thereforest = tan PD = Pt. Consequently, the tangent Pt of an arc of a great circle terminated at the . of projection, is projected into a right line equal to it. Let Pi, and Pl', two such tangents, be connected by the right line it' which will be in the plane of projection. Let 'st, st', be the projections of those tangents; the triangles tet', tst', are (from the above) equal in all respects: therefore the angles opposite to the common side tt' will be the same in both: consequently, the tangents of any two arcs terminated at the plane of projection, are projected into lines which are respectively equal to them, and which form an equal angle. Hence, two circles which intersect in P on the sphere, form on the projection an angle equal to that which they make on the sphere; because, at the point of intersection the elements of the arcs coincide with those of their tangents. Therefore, all great circles intersect mutually on the plane #.projection under the same angle as on the sphere; so also do little circles which intersect at the same points, and have, by consequence, common tangents. 43. By way of showing the application of these principles, let us suppose that the eye is at the south pole of the equator. The plane of projection will then be the equator itself; the centre of the equator will represent the north pole; AP (fig. to art. 35) will be = 0; the projections of the parallels to the equator will all

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have for a common centre that of the projection; and the radii of those circles will be the tangents of the halves of the polar distances. Thus, for the polar circle .... r=tan; ( 23°28′)=tan 11°44' tropic of cancer. retan ; (66°32')=tan33°16 tropicofoapricornr= tank (113°28')=tană6°44' for the antarctic circle r = tan? (156°32') = tan'78°16' for any latitude L.... r =tan (90°–1)=tan 45°–31. or, if the south r=tan (45°-H L). As for the meri- B dians, whose planes allpass throughthe place of the eye, they all become diameters , which divide the equator in its several degrees, and form at the centre of the projection angles equal to the dif. ferences of longitude. For these circles d = co, and r = oc (art. 38). p This kind of pro- jection, the easiest - II of all to describe, serves very conveniently for eclipses of the sun. The meridians and the parallels are herein divided mutually into degrees: those of the parallels are equal; those of the meridians unequal; for the expression for one of their degrees is,

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44. On a planisphere of this kind, the stars are placed according to their right ascensions and declinations. Then the ecliptic is traced, as well as its poles, the circles of latitude all intersecting mutually at those two poles, and then the parallels to the ecliptic. , Thus, on the radius marked 270°, at a distance cr from the centre = tan 11° 44', we mark the north pole a of the ecliptic. From that point, with an opening of the compasses = cosec 23° 28′ we mark the point E, or, which amounts to the same, we make ce. = cot 23° 28′. Through the point E we draw the indefinite perpendicular vex, which is evidently the locus of the centres of all the circles of latitude intersecting mutually in II and w: E will be the centre of that circle of latitude which passes through 0° and 180°, or of the equinoctial colure, which will be the circle vaxII. a BII will be the solstitial colure. -

In general, make Erg’ = longitude: from the centre g" with the radius G'+ we describe a circle, it will be the circle of latitude which answers to the longitude supposed. Repeating the same operation on the other side of the line wr, we shall have the circles of latitude of the other hemisphere. For the circles parallel to the ecliptic we employ the formulae

tan #(p + A) + tan #(D - A) d = −; tan * (p + A) – tan * (p — A)

p being = 23° 27' 49", or nearly 23°28′, the distance between the poles of the ecliptic and the equator; and A the polar distance of the parallel. When A > 0, the sign of the second term will be changed. .

45. The principal defect of this projection is the little resemblance and proportion between the arcs of the sphere and their projections. Thus, the arcs wa and AIT represent arcs of 90°; a D represents an arc of 113° 28, and pz, though greater, only represents 66° 32'; the arcs au, wo, wh, &c, are of 90°; or represents 23° 28 = DAH = inclination of the ecliptic: A He to the equator ADQ. It is true, however, that the greatest inequalities are out of the circle ADQB which is properly the projection. . If we regard the circle rvilxr as a map of half the terrestrial globe, then rv, wy, we, we, will represent arcs of 90°, though rv will be the only one of those four which is actually a quadrant.

46. Another inconvenience of this projection, is the difficulty of finding the true distance of two points of which we have the projections. Yet, let M and N in the diagram to art. 42, be two such points: produce cM, cN, to m and n respectively; the arc mn will give us . McN which is the same as on the sphere. CM and cn are the tangents of the half distances from the pole of the projection. The spherical triangle will give (see chap. vi. equa. 2),

cos MN = cos cM cos cn + sin cM sincN cosc
(1 – tan” #cM) (1 – tan” #cN) +4tan #cM tan #cN cose
- (1 + tan” &cm) (1 + tan” cN)
_ (1 – cu”) (? – co") + 4cm. on cos c
(1 + coo) (1 + coo)

Take cM' = cM and draw om'M”, AM" will be equal to the arc represented by cm. Proceed similarly for cN. Then cos MN may be computed from the above.

The third member of the equation is obtained from the second by substituting for cos CM, sin cM, their values in tan #CM, &c. deduced by means of equa. R, chap. iv.

#. The projections here treated serving for the usual purposes of astronomy, we need not enter upon the explication of the other kinds of projection devised by geometers for different purposes. The principal of these is the gnomonic projection, in which the eye is supposed at #: centre of the sphere, and the plane of projection a tangent plane to the sphere at any assumed point. All the o within adequate limits have their projections at the extremities of the tangents of their


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