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perspective. But, as Lagrange has remarked, one may regard geographical maps from a more general point of view as representations of the surface of the globe, for which purpose we have but to draw meridians and parallels according to any given law; then any place we have to fix must take that position with reference to these lines that it has on the sphere with reference to the circles of latitude and longitude. Let the law which connects latitude and longitude, and with the rectangular coordinates x and y in the representation be such that dx mdo+ndw, and dy = m'do + n'dw. In fig. 6 let the lines intersecting in the parallelogram PQRS be the representations of the meridians rp, sq and parallels rs, pq intersecting in the indefinitely small rectangle pqrs on the surface of the sphere. The coordinates of P being x and y, while those of p are and w the coordinates of the other points will stand thus

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w,

=

w+dw

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Fig. 6.

Let Ppq, Prs (fig. 7) be two contiguous meridians crossed by parallels rp, sq, and Op'd, Or's the straight lines representing these angle at O. Let the co-latitude meridians. If the angle at P is du, this also is the value of the

Pp-u, Pq-u+ du; Op' = p, Oq=p+dp, the circular arcs p'r', q's representing the parallels pr, qs. If the radius of the sphere be unity,

p'q'dp; p'r'= pdμ,

S

pq = du;

or -sinudu.

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x + mdp+ndw Thus we easily see that PR (m2+m22)1dp; and PQ =(x2+n'2)ldw; also the area of the parallelogram PQRS is equal to (m'n - mn')dødw. If 90° are the angles of the parallelogram, then

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If the lines of latitude and of longitude intersect at right angles, theu mn + m'n=0. Since the length of pr is do, its representation PR is too great in the proportion of (m2+m22); 1; and pq being in length cos pdw, its representation PQ is too great in the ratio of (n2+n2): cosp. Hence the condition that the rectangle PQRS is similar to the rectangle pqrs is (m2 + m22) cos2= n2+n2, together with mn+m'n'=0; or, which is the same, the condition of similarity is expressed by

- n'm cos ; n=m' cos p.

P

then p'qopq and 'r'-o'pr. That is to say, σ, o' may be regarded as the relative Fig. 7. scales, at co-latitude u, of the representation, & applying to meridional measurements, o' to measurements perpendicular to the meridian. A small square situated in colatitude u, having one side in the direction of the meridian-the length of its side being i-is represented by a rectangle whose sides are io and is; its area consequently is oo'.

If it were possible to make a perfect representation, then we should have σ = = 1, σ = 1 throughout. This, however, is impossible. We may make σ = I throughout by taking pu. This is known as the Equidistant Projection, a very simple and effective method of representation.

sin

Or we may make σ'= 1 throughout. This gives p = a perspective projection, namely, the Orthographic. Or we may require that areas be strictly represented in the development. This will be effected by making oo' = 1, or pdp sin udu, the integral of which is p= 2 sinu, which is the Equivalent Projection of Lambert, sometimes referred to as Lorgna's Projection. In this system there is misrepresentation of form, but no misrepresentation of areas. Or we may require a projection in which all small parts are to be represented in their true forms. For instance, a small square on the spherical surface is to be represented as a small square in the development. This condition will be αρ du attained by making σ =σ', or the integral of P sin u which is, c being an arbitrary constant, p= ctanu. This, again, is a perspective projection, namely, the Stereographic. In this, though all small parts of the surface are represented in their correct shapes, yet, the scale varying from one part of the map to another, the whole is not a

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Since the area of the rectangle pqrs is cos dodo, the exaggeration of area in the representation will be expressed by m'n - mn': cos . Thus when the nature of the lines representing the circles of latitude and longitude is defined These two last projections are, as it were, at the extremes we can at once calculate the error or exaggeration of scale of the scale; each, perfect in its own way, is in other reat any part of the map, whether measured in the direction spects very objectionable. We may avoid both extremes of a meridian or of a parallel; and also the misrepresenta-1 and σ'=1, so as to have a perfect picture of the by the following considerations. Although we cannot make tion of angles.

The lines representing in a map the meridians and parallels on the sphere are constructed either on the principles of true perspective or by artificial systems of developments. The perspective drawings are indeed included as a particular case of development in which, with reference to a certain point selected as the centre of the portion of spherical surface to be represented, all the other points are represented in their true azimuths, the rectilinear distances from the centre of the drawing being a certain function of the corresponding true distances on the spherical surface. For simplicity we shall first apply this method to the projection or development of parallels and meridians when the pole is the centre. According to what has been said above, the meridians are now straight lines diverging from the pole, dividing the 360° into equal angles; and the parallels are represented by circles having the pole as centre, the radius of the parallel whose co-latitude is u being p, a certain function of u. The particular function selected determines the nature of the development.

local errors of the representation, we may make (σ - 1)2+ spherical surface, yet considering σ- -1 and σ'-1 as the (-1) a minimum over the whole surface to be represented. To effect this we must multiply this expression by the element of surface to which it applies, viz, sinududp, and then integrate from the centre to the (circular) limits to be represented, then the total misrepresentation is to be of the map. Let ẞ be the spherical radius of the segment

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1+2 tan a

case the plane of the drawing may be supposed to pass through the centre of the sphere. Let the circle (fig. 8) represent the plane of the equator on which we propose to make an orthographic representation of meridians and parallels. The centre of this circle is clearly the projection of the pole, and the parallels are projected into circles having the pole for a common centre. The diameters aa', bb' being at right angles, let the semicircle bab' be divided into the required number of equal parts; the diameters drawn through these points are the projections of meridians. The distances of

Fig. 8.

C, of d, and of e from the diameter aa' are the radii of the successive circles representing the parallels. It is clear that, when the points of division are very close, the parallels will be very much crowded towards the outside of the map; so much so, that this projection is not much used. For an orthographic projection of the globe on a meridian plane, let qnrs (fig. 9) be the meridian, ns the axis of rotation, then qr is the projection of the equator. The parallels will be represented by straight lines passing through the g points of equal division; these lines are, like the equator, perpendicular The meridians will in this case be ellipses described on us as

to ns.

a common major axis, the distances

Fig. 9.

of c, of d, and of e from ns being the minor semiaxes. Let us next construct an orthographic projection of the sphere on the horizon of any place. Set off the angle aop

Put = cot2, then ‹ is a maximum when a =, and the (fig. 10) from the radius oa, equal to the latitude. Drop the corresponding value of e is

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2

For simplicity of explanation we have supposed this method of development so applied as to have the pole in the centre. There is, however, no necessity for this, and any point on the surface of the sphere may be taken as the centre. All that is necessary is to calculate by spherical trigonometry the azimuth and distance, with reference to the assumed centre, of all the points of intersection of meridians and parallels within the space which is to be represented in a plane. Then the azimuth is represented unaltered, and any spherical distance u is represented by p. Thus we get all the points of intersection transferred to the representation, and it remains merely to draw continuous lines through these points, which lines will be the meridians and parallels in the representation.

m

perpendicular pP on ea, then P is the projection of the pole On ao produced take ob=pP, then ob is the minor semi axis of the ellipse representing the equator, its major axis being gr at right angles to ao. The points in which the meridians meet this elliptic equator are determined by lines q drawn parallel to aob through the points of equal subdivision cdefgh. Take two points, as d and gr which projections on the equator; then i are 90° apart, and let ik be their Fig. 10. is the pole of the meridian which passes through k. This meridian is of course an ellipse, and is described with reference to i exactly as the equator was described with reference to P. Produce io to l, and make lo equal to half the shortest chord that

can be drawn through i; then lo is the semiaxis of the elliptic the meridian, and

The exaggeration in such systems, it is important to remember, whether of linear scale, area, or angle, is the same for a given distance from the centre, whatever be the azimuth; that is, the exaggeration is a function of the dis-major axis is the dia

tance from the centre only.

We shall now examine and exemplify some of the most important systems of projection and development, commencing with

Perspective Projections.

In perspective drawings of the sphere, the plane on which the representation is actually made may generally be any plane perpendicular to the line joining the centre of the sphere and the point of vision. If V be the point of vision, P any point on the spherical surface, then p, the point in which the straight line VP intersects the plane of the representation, is the projection of P.

In the orthographic projection, the point of vision is at an infinite distance and the rays consequently parallel; in this

meter perpendicular to iol.

For the parallels : let it be required to describe the parallel whose co-latitude is u; take pm=pn=u, and let m'n' be the projections of m and n on oPa; then m'n' is FIG. 11.-Orthographic Projection. the minor axis of the ellipse representing the parallel. Its centre is of course midway between m' and n', and the greater axis is equal to mu. Thus the construction is obvious. When pm is less than pa, the whole of the ellipse

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Stereographic Projection.-In this case the point of vision is on the surface, and the projection is made on the plane of the great circle whose pole is V. Let kp/V (fig. 12) be a great circle through the point of vision, and ors the trace of the plane of projection. Let c be the centre of a small circle whose radius is cpcl; the straight line pl repre sents this small circle in orthographic projection.

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Fig. 12.

We have first to show that the stereographic projection of the small circle pl is itself a circle; that is to say, a straight line through V, moving along the circumference of pl, traces a circle on the plane of projection ors. This line generates an oblique cone standing on a circular base, its axis being cV (since the angle pVc-angle cV); this cone is divided symmetrically by the plane of the great circle kpl, and also by the plane which passes through the axis Ve, perpendicular to the plane kpl. Now Vr.Vp, being Vo sec kVp.Vk cos kVpVo Vk, is equal to Vs-Vi; therefore the triangles Vrs, Vlp are similar, and it follows that the section of the cone by the plane rs is similar to the section by the plane pl. But the latter is a circle, hence also the projection is a circle; and since the representation of every infinitely small circle on the surface is itself a circle, it follows that in this projection the representation of small parts is (as we have before shown) strictly similar. Another inference is that the angle in which two lines on the sphere intersect is represented by the same angle in the projection. This may otherwise be proved by means of fig. 13, where Vok is the diameter of the sphere passing through the point of vision, fgh the plane of projection, kt a great circle, passing of

k

Fig. 13.

e

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d

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k

m

Pop', and vP, P' cutting Ir in pp': these are the projections
of the poles, through which
all the circles representing
meridians have to pass. All
their centres then will be in a
line smn which crosses pp' at
right angles through its middle
point m. Now to describe the
meridian whose west longitude
is w, draw pn making the angle
opn=90°-w, then is the centre
of the required circle, whose
direction as it passes through p
will make an angle opg=w with
pp'. The lengths of the several lines are

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Fig. 14.

optanu; op'-cotžu; om = cotu; MN cosec u cotw. Again, for the parallels, take Pb = Pc equal to the co-latitude, say c, of the parallel to be projected; join vb, vc cutting Ir in e, d. Then ed is the diameter of the circle which is the required projection; its centre is of course the middle point of ed, and the lengths of the lines are

od-tan (u-c); oe-tan (u+c,.

The line sn itself is the projection of a parallel, namely, that
180°-u, a parallel which passes
of which the co-latitude c=
through the point of vision.

cos,

A very interesting connexion, noted by Professor Cayley, exists between the stereographic projection of the sphere on a meridian plane (i.e., when a point on the equator occupies the centre of the drawing) and the projection on The very same circles the horizon of any place whatever. that represent parallels and meridians in the one case represent them in the other case also. In fig. 15, abs being a projection in which an equatorial point is in the centre, draw any chord ab perpendicular to the centre meridian and on ab as diameter describe a circle, when the property referred to will be observed. This smaller circle is now the stereographic projection of the sphere on the horizon of some place whose co-latitude we may call u. The radius of the first circle being unity, let ac = sinx, then by what has been proved above co= cotu cosx; therefore u=x, and ac sinu. the meridian circles dividing the 360° at the pole into equal angles, must be actually the same in both systems, yet a parallel circle whose co-latitude is c in the direct projection abs belongs in the oblique system to some other co-latitude To determine the connexion between c and c', consider the point t (not marked), in which one of the parallel In the direct system, p being circles crosses the line soc. the pole,

as c'.

course through V, and ouv
the line of iutersection of
these two planes. A tangent
plane to the surface at t
cuts the plane of projection
in the line res perpendicular
to o; tv is a tangent to the
circle kt at t, tr and ts are
any two tangents to the surface at t. Now the angle vtu (u
being the projection of t) is 90° - otV = 90° - oVt ouV = tuv,
therefore to is equal to uv; and since tvs and uvs are right
angles, it follows that the angles vts and vus are equal.
Hence the angle rts also is equal to its projection rus;
that is, any angle formed by two intersecting lines on the
surface is truly represented in the stereographic projection.
We have seen that the projection of any circle of the
sphere is itself a circle. But in the case in which the circle
to be projected passes through V, the projection becomes,
for a great circle, a line through the centre of the sphere; and in the oblique,
otherwise, a line anywhere. It follows that meridians and
parallels are represented in a projection on the horizon of
any place by two systems of orthogonally cutting circles, which, replacing ac by its value sin
one system passing through two fixed points, namely, the
poles; and the projected meridians as they pass through
the poles show the proper differences of longitude.

To construct a stereographic projection of the sphere on the horizon of a given place. Draw the circle vlkr (fig. 14) with the diameters kv, Ir at right angles; the latter is to represent the central méridian. Take kol equal to the co-latitude of the given place, say u; draw the diameter

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fos (u - c') 1+cot u cot c' thererore tanc tan c' tanu is the required relation.

Notwithstanding the facility of construction, the stereographic projection is not much used in map-making. But it may be made very useful as a means of graphical interpolation for drawing other projections in which points are represented in their true azimuths, but with an arbitrary

[MATHEMATICAL

law of distance, as p=f(u). We may thus avoid the calcu- | This vanishes when = 1, that is, if the projection be stereolation of all the distances and azimuths (with reference to graphic; or for u=0, that is, at the centre of the map. the selected centre point) of the intersections of meridians At a distance of 90° from the centre, the greatest alteration and parallels. Construct a stereographic projection of the is 90°-2 cot-1/h (See Philosoph. Mag., April 1862.) globe on the horizon of the given place; then on this pro- The constants h and k can be determined, so that the jection draw concentric circles (according to the stereo total misrepresentation, viz., graphic law) representing the loci of points whose distances from the centre are consecutively 5°, 10°, 15°, 20°, &c., up to the required limit, and a system of radial lines at inter- shall be a minimnm, ß being the greatest value of u, or the vals of 5. Then to construct any other projection,-com-spherical radius of the map. mence by drawing concentric circles, of which the radii are sions for σ and o' the integration is effected without diffiOn substituting the expres culty. Put

M = {(σ - 1)2 + (o' - 1)} sin udu,

λ=

1-cos
; v = (h-1)x,
h+ cos B
H=v-(h+1) log. (x+1),

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azimuth of P at G, then the spherical triangle whose sides | for a map of Africa, which is included between latitudes 40° are 90° -7, 90°-4, and u gives these relations

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north and 40° south, and 40° of longitude east and west of a central meridian.

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