Imágenes de páginas
PDF
[ocr errors]

through o, because foe is a right angle. In like manner, it will pass through A, because cs is perpendicular to the middle of A. . . . Ao. Therefore ro = rs = rot; con- ... seq. Sor = . OSr, and CrO = . OST. cor = 90° — cro = 90° – 2 osr = D 90° — (Do — BE) = 90°-90°,+ BE = BE. That is, cor = inclination of the plane of the O great circle to the plane of projection. Hence, or is manifestly = sec. inclination, and cr = tan inclination, to rad co. Thus, with the radius = sec D, and at the distance from the centre = tan D, it will be easy to describe the circle. Again, take A1 = 2BE = 2AP = 2n, and draw oi, the point of intersection r with Ds will be the centre, and ro the radius. Also, since meas. of cor = BE, meas. of ocr = or = 90°– BE; we have con + ocR = 90°, and conseq. . cRo = 90°. , Hence, by drawing or r perpendicular to FE, we find the centre r and the radius ro: also cr = sin BE = sin D, or = cos D. 40. or, therefore, will always make with oc an equal equal to the inclination or distance of the two poles. Let there be, then, a second circle whose inclination shall be cor'; the radius of its projection will be ro, and the radii ro, r'o, will make at their points of intersection an angle ror' which will be their difference of inclination, or the mutual inclination of the two circles. This is a particular case of the general theorem. 41. Suppose now that AP = 90°, P will coincide with *, and the pole of the circle will be upon the limit of

[graphic]
[ocr errors][ocr errors][merged small]

the projection. Let BH = BE = A polar distance of the circle to be projected, the chord HE will be perpendicular to DB. Draw on and of s, Gs will be the diameter sought. Bisect Gs in n, n will be the centre, n E = m H = ns = ng, will be the radius of the projected circle. Meas. of Eng = 2nse = op — BE = 90°– BE = 90°–BCE Cne + no e = 90°... cen = 90°... En = tan EE = tan A, and cn = see BE = sec A. These values serve for all circles which have their o on the circumference of the circle of projection. f these are great circles, then A = 90°, and tan A, sec 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. o 42. Let obe the point of observation, or place of the eye, A the pole of the pro- A. jection, BDECB the plane of projection, PD an arc of a o great circle which has its origin at any point whatever M" of the circle oppf, Pt the B tangent of PD; ct will be the secant, and s the projection of P. Draw st: then from the rectilineal triangle sct we shall have, St? = Cs” + Ct” - 2cs. Ct. cos Sct = tan? AP + seco PD – 2 tan AAP sec PD cos pe. = tan” #AP + tan” PD + 1–2 tan AP sec PI) cos DE = seco #AP + tano PD – 2 tan #APsec 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 coST) E; - 1

[ocr errors]

Substituting this in the last value of st”, we have

[graphic]
[graphic]
[graphic]
[ocr errors][ocr errors][ocr errors][ocr errors]

= seco AP + tano PD – seco AP = tan? pp. Therefore st = tan PD = Pt. - * e Consequently, the tangent pt of an arc of a great circle terminated at the i. of projection, is projected into a right line equal to it. , Let It, and ot', 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', #st', are (from the above) equal in all respects: therefore the angles opposite to the common side tt' will be the same in both: conseguently, the tangents of any two arcs terminated at the plane of projection, are projected into lines which are reectively equal to them, and which form an equal-angle. #. 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

[ocr errors]
[graphic]
[ocr errors]

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. ratan? (66°32')=tan&3°16' tropicofcapricornr=tan? (113°28')=tană6°44' for the antarctic circle r = tan? (156°32') = tan'78°16' for any latitude L.... r = tan (90°–1)=tanț5°–$1. or, if the lat.be south r=tan (45°-F #L). As for the meridians, whose planes all pass through the 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). This kind of projection, the easiest 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,

[ocr errors]
[ocr errors]
[ocr errors]
[graphic]
« AnteriorContinuar »