July 1, 1887.] 



♦ KNOWLEDGE 



201 



great circle passes through any poiut, it passes also through 

 the antipodes of that point. 



(2) If a great circle touches a small circle on the sphere, 

 it touches also the small circle antipodal to the former : for 

 instance, if a great circle touches latitude-parallel 30° north, 

 it touches also latitude-parallel 30° south. This needs no 

 demonstration, being really a corollary of 1 ; for the point in 

 which the great cii'cle touches one small circle has for its 

 antipode a corresponding point on the antipodal small circle, 

 and also, by 1, on the great circle : and there can be no other 

 point in which the great circle meets the antipodal small 

 circle ; for if there were, then, by 1, there would be corre- 

 sponding points of contact or of intersection with the 

 original small circle, which by our hypothesis is not the 

 case. 



To use the charts, however, the sejiman need not concern 

 himself either with the method of constructing them or with 

 the principles on which their use in gi'eat circle sailing de- 

 pends. All he need care for is rightly to apply the con- 

 structions which result from these principles. 



I propose to indicate only what are the processes neces- 

 sary for the five following problems : 



I. To find the great circle course between any two points, 

 as A and B, Fig. 3. 



II. To find the vertex, Y, or highest latitude reached, on 

 that course. 



III. To find the bearing at any point, as Q, on the 

 course. 



IT. To find the composite course, from port to port, touch- 

 ing any given limiting latitude. 



V. To find the di'^tance, A V B, to be traversed. 



Tlie constructions for these five problems are all included 

 in the following simple statements (P, Fig. 3, is the pole. 



e'EE' the equator), the italicised parts indicating the actual 

 constructions, the rest giving the reasons and demonstra- 

 tions : 



I. Find on the chart a and h the antipodes of A and B 

 (which of course is easy, as we have only to takeoff 180° on 

 the meridians A P nr, B P 6, not shown in the figure to 

 avoid crowding) ; then a circle throiigh any three of the points 

 A, a, B, b, is a <jreat circk of the sphere hy 1, and must pass 

 through the fourth point. Describe such a circle A B a b ; 

 A V B is the gnat circle course required. [It will be best to 

 take the three points A, B, and b, A being the point of de- 

 parture. Pencil the bisecting-perpendiculars to 6B, AB (not 

 shown in Fig. 3), inter.secting in C, around which point as 

 a centre describe in pencil the circle AB6. Note whether 

 it passes through a, for this serves as a test of the accuracy 

 of the result. You may also note whether the points e' 

 and E', in which it cuts the equator, are, as they should be, 

 on the extremities of a diameter through P ; for two great 

 circles on a sphere necessarily intersect on a diameter of the 

 sphere, and therefore, as P Ls the projection of the pole of 



the equator, two such points of intersection must lie on a 

 straight line through P.] 



II. A straight line vPOV, through P and C, cuts the great 

 circle course in V, the vertex required, Y l)eing the highest 

 latitude on one side of the equator, v that on the other. [V 

 does not necessarily fall on the actual great circle coui-se 

 between two points. For instance, QB is the gi-eat circle 

 course from Q to B, but Y lies outside QB.] 



III. Draw QT, tan^/ent to the coiirse, at Q. Then the 

 angle PQT gives the bearing of the course at Q from the due 

 northerly direction QP. By obser^■ing that QT is at right 

 angles to CQ, we can get the bearing without actually draw- 

 ing QT. Thus in the case illustrated in the figure, the 

 dii'ection QT is north of due east by an angle equal to CQP, 

 e;isily measured with a protractor. 



lY. Suppose A YB, Fig. 4, the great circle course, to have 

 its vertex Yin inconveniently or dangerously high latitudes. 

 Let L on PY be the highest latitude vhich the ship must 

 reach. Tale I, antipodal to L; then if we bisect I L, in G, 

 and describe round P as centre the circle c G c', it is obvious 

 that any circle having its centre on c G c', and radius 

 equal to GL or G I, will touch both the latitude-parallels 



HLK and klh ; and, by 2, will be a great circle. Therefore, 

 around A and B, as centres, with radiiis GL, describe circles 

 cutting c G c', in c and c' , and with c and c' a.s centres and 

 the same raditis describe circular arcs, AH, BK, tou<:hing 

 the latitude-parallel HLK in H and K. Then AHKB is 

 the composite co'nrse required. The distances along AH and 

 BK can be determined by the method shown in the suc- 

 ceeding section, and the distance along the latitude-parallel 

 HLK is of course easily determined, being an arc of a known 

 number of degrees in a known latitude. 



Y. [\Ye have to determine how many degrees there are 

 in the arc AYB, not as it appears in the chart, but as it 

 really is on the sphere.] Take p, 90 d_egrees of l/ititude from 

 V and V, Fig. 3, along «PV. (This of course is done at 

 once on the chart, which shows the degrees of latitude from 

 the pole P.) Then p is the pole of AYB. Find D, the 

 centre of the great circle A p, and F, the centre of the great 

 circle B p. (This we do b}' I. ; but most of the work is 

 already done. We have the bLsecting-perpendicular to B 6 ; 

 the bisecting-perpendicular to A a pa.sses through C ; then 

 the bisecting-perpendiculars to A p, ;' B, give us — by their 

 intersections with those to A « and B b — D and F at 

 once.) What we want is to determine the angle A p B, 

 between the arcs A /?, /> B ; and it is obvious that this is the 

 supplement of the angle D p F, which is easily measured 

 W'ith a protractor. The number of degrees, multiplied by 60, 

 gives the number of geographical miles, or knots, in the 

 distance AYB. 



An example of these methods is given in the accompanying 



