Page images
PDF
EPUB
[ocr errors]

Let

[ocr errors]

Let Ppq, Prs (fig. 7) be two contiguous meridians crossed by parallels rp, sq, and Op'q', 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,

[ocr errors]

R

p'q'=dp; = du; pq

p'r' = pdμ, sinudu.

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

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. the law which connects latitude and longitude, and w, with the rectangular coordinates x and y in the representation be such that dx mdp + ndw, and dy = m'do + n'dw. In fig. 6 let the lines Fig. 6. 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

=

=

[merged small][ocr errors][ocr errors][merged small][merged small][merged small][subsumed][merged small][merged small][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][merged small]

Thus we easily see that PR = (m2+m22)}dp; and PQ (n2+n'2)idw; also the area of the parallelogram PQRS is equal to (m'n - mn')dødw. If 90° are the angles of the parallelogram, then

mn+m'n'

tan = m'n - mn'

If the lines of latitude and of longitude intersect at right angles, then mn+m'n'=0. Since the length of pris=do, its representation PR is too great in the proportion of (m2+m2): 1; and pq being in length cosødw, 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+n'2, 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.

Since the area of the rectangle pqrs is cos dodw, the exaggeration of area in the representation will be expressed by m'n - mn': cosp. Thus when the nature of the lines representing the circles of latitude and longitude is defined we can at once calculate the error or exaggeration of scale at any part of the map, whether measured in the direction of a meridian or of a parallel; and also the misrepresentation 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 sur

face 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.

σ =

dp

σ' dui

[merged small][ocr errors]

then p'qopq and r'r'-o'pr. That is to say, σ, '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 io'; its area consequently is 'oo'.

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

p=u.

Or we may make σ = 1 throughout. This gives p = sin u, a perspective projection, namely, the Orthographic. Or we may require that areas be strictly represented in the development. This will be effected by making σơ'=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 , the integral of attained by making σ =σ', or 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 similar representation of the original. The scale, σ= c secu, at any point, applies to all directions round that point.

[ocr errors][merged small]

These two last projections are, as it were, at the extremes of the scale; each, perfect in its own way, is in other respects very objectionable. We may avoid both extremes by the following considerations. Although we cannot make σ = 1 and σ'= 1, so as to have a perfect picture of the local errors of the representation, we may make (σ − 1)2+ spherical surface, yet considering σ-1 and '-1 as the sented. To effect this we must multiply this expression by (o' - 1)2 a minimum over the whole surface to be reprethe element of surface to which it applies, viz, sinudu du, the element of surface to which it applies, viz, sinu du du, 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

[merged small][merged small][subsumed][subsumed][ocr errors][subsumed][subsumed][merged small][merged small][merged small][merged small][merged small][merged small]
[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][ocr errors]

1+ tan2 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 projec-
tions of meridians. The distances of

C,

Fig. 8.

n

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 ns as

to ns.

a common major axis, the distances

Fig. 9.

e

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

π

[ocr errors]

method of development so applied as to have the pole in
For simplicity of explanation we have supposed this
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 meri-
dians and parallels within the space which is to be repre-
sented in a plane. Then the azimuth is represented unaltered,
and any spherical distance u is represented by p.
get all the points of intersection transferred to the repre-
sentation, and it remains merely to draw continuous lines
through these points, which lines will be the meridians and
parallels in the representation.

f

a

m

[ocr errors]

n

perpendicular pP on oa, then P is the projection of the pole. On ao produced take ob-pP, then ob is the minor semiaxis of the ellipse representing the which the meridians meet this ellipequator, its major axis being grat right angles to ao. The points in tic equator are determined by lines q drawn parallel to aob through the points of equal subdivision cdefgh Take two points, as d and g, which projections on the equator; then i are 90° apart, and let ik be their is the pole of the meridian which passes through . 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 7, and make lo equal to half the shortest chord that can be drawn through The exaggeration in such systems, it is important to rei; then lo is the semimember, whether of linear scale, area, or angle, is the same axis of the elliptic fór a given distance from the centre, whatever be the azimuth; that is, the exaggeration is a function of the dis-major axis is the diameridian, and the

tance from the centre only.

Thus we

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.

Fig. 10.

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 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 mn. Thus the construction is obvious. When pm is less than pa, the whole of the ellipse

FIG. 11.-Orthographic Projection.

[merged small][ocr errors]

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 cp=cl; the straight line pl represents this small circle in orthographic projection.

=

Fig. 12.

[ocr errors]
[ocr errors]

k

m

n

PoP', and vP, vP' 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 v
right angles through its middle
point m. Now to describe the S
meridian whose west longitude
is w, draw pn making the angle
opn=90°-w, then n 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
optanu; op=cotДu;

om=cotu; mn=cosec u cotw.

Fig. 14.

Again, for the parallels, take Pb = Pc equal to the co-latitude, say c, of the parallel to be projected; join vb, ve cutting r 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

[blocks in formation]

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

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 Vc, perpendicular to the plane kpl. Now Vr.Vp, being Vosec kVp.Vk cos kVp = Vo Vk, is equal to Vs Vl; 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 course through V, and ouv the line of intersection of these two planes. A tangent plane to the surface at t cuts the plane of projection in the line rvs perpendicular to ov; 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° - otV90° - oVt ouV=tuv, therefore tv 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 ris 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 the pole, 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; 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, 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.

Fig. 13.

=

[ocr errors]

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 meridian. Take koP equal to the co-latitude of the given place, say u; draw the diameter

a

A very interesting connexion, noted by Professor Cayley, exists between the stereographic projection of the sphere on a meridian plane (ie., when a point on the equator occupies the centre of the drawing) and the projection on the horizon of any place whatever. The very same circles that represent parallels and meri. dians 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 cos, 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 Fig. 15. of some place whose co-latitude we may call u. The radius of the first circle being unity, let acsinx, then by what has been proved above co= sinx cotucos; therefore u=x, and ac=sinu. Although 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 circles crosses the line soc. In the direct system, p being

as c'.

[blocks in formation]

law of distance, as p=f(u). We may thus avoid the calcu- | This vanishes when h=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 intervals of 50. Then to construct any other projection,-commence by drawing concentric circles, of which the radii are

M

= ƒ®3 {(o − 1)2 + (o' − 1)2 }; sin udu,

shall be a minimum, ß being the greatest value of u, or the
spherical radius of the map. On substituting the expres-
sions for σ and o' the integration is effected without diffi-
culty. Put
1-cos B
h+cos B

λ=

; v=(h−1)x,

H=v-(h+1) loge (λ+1),

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

H3
and must be determined so
H"

FIG. 16. Stereographic Projection.

previously calculated by the law p=f(u), for the successive values of u, 5°, 10°, 15°, 20°, &c., up to the limits as before, and a system of radial lines at intervals of 5°. This being completed, it remains to transfer the points of intersection from the stereographic to the new projection by graphic interpolation.

9

[ocr errors]

m

We now come to the general case in which the point of vision has any position outside the sphere. Let abcd (fig. 17) be the great circle section of the sphere by a plane passing through p c, the central point of the portion of surface to be represented, and V the point of vision. Let

perpendicular to Vc be the plane of representation, join mV cutting pj in f, then ƒ is the projection of any point m in the circle abc, and ef is the representation of cm.

a

e

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

Therefore M 4 sin2 1ẞ-
as to make H2: H'a maximum. In any particular case this
maximum can only be ascertained by trial, that is to say,
log H2-log H' must be calculated for certain equidistant
values of h, and then the particular value of h which cor-
responds to the required maximum can be obtained by
interpolation. Thus we find that if it be required to make
the best possible perspective representation of a hemisphere,
the values of h and I are h=147 and k=2.034; so that

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

Σ=

h+cos u 1+h cos u

[ocr errors]
[merged small][subsumed][merged small][merged small][ocr errors]

90

Fig. 18.

in this case is 1.61. Fig. 18 is a map of Asia having the meridians and parallels laid down on this system.

Figure 19 is a perspective representation of more than a | The co-ordinates originating at the centre, take the central hemisphere, the radius ß being 108°, and the distance h of the point of vision, 1.40.

The co-ordinates xy of any point in this perspective may be expressed in terms of the latitude and longitude of the corresponding point on the sphere in the following manner.

meridian for the axis of y and a line perpendicular to it for the axis of x. Let the latitude of the point G, which is to occupy the centre of the map, be y; if o, w be the latitude and longitude of any point P (the longitude being reckoned from the meridian of G), u the distance PG, and u the

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

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

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

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

[blocks in formation]
« EelmineJätka »