Now since p'p" = p""'p"""' it is evident that for a pole within the ovals p' and p' must belong to different branches while for a pole outside of the ovals they must belong to the same branch. Since p'p" is constant the inverse of the curve described by p' is similar to that described by p'. Thus the inverse of a Cartesian with respect to either focus is a similar curve, the inner inverting into an outer branch if the focus is the first or second, but into an inner branch if it is the third focus. Thus the inverse of the acnodal limagon with respect to the double focus is an ellipse similar to that which vanishes at that point, but its inverse with respect to the single focus is similar to itself. The inverse of the crunodal limaçon with respect to the double point which is at once on each branch consists of the two branches of an hyperbola with asymptotes parallel to the tangents at the double point. From (21) we also see that for the third focus the product of a secant and its external segment is constant, and the four tangents to the curve from this point are equal. From (22) we have p'+p" - p'"' — p'''' a constant. 1 1 (23) Now when A1 is the first focus is the least of the three coefficients and by (21) p' and p' have the same sign, while the pole being within the ovals pr and p'''' are of the opposite sign. Let p"" belong to the same branch as p', then p'-p" and p"-p"""' will denote the parts of a straight line passing through A, intercepted by the ovals and will have the same sign. Therefore (23) expresses that the sum of the intercepts of the ovals upon a linepassing through the first focus is constant. If A, is the second focus, 7 is the middle coefficient and p' and p" have opposite signs. Therefore (23) expresses that the difference of the intercepts upon a line passing through the second focus is constant. If A, is the third focus p'-p" and p"-p"!!! will be the intercepts between the branches, and will be of opposite signs since p' and p" now belong to the same branch; therefore (21) will expressthat for a line passing through the third focus the difference of the intercepts between the two branches is constant. 1 The bipolar equation of the Cartesian takes a simple form also when referred to a focus and the point C as poles. To transform to A1 and any other point of the axis, combine the equation with the general relation (4) so as to eliminate p2. Thus, squaring, we have 2 m2p12± 2mna ̧1 + n2az2 = l2p22; multiplying by a, and substituting from (4), 1 2 2 2 0. (m2α2 + l2α1)p12± 2mna2a3p1 + l2аzPz2 + n2α2αz2 + ľ2ajα2αz This is referred to the focus A, and any other point As, giving to a, and the values in (20) Ag coincides with C. Since a, is here the distance between the poles, we substitute from (20) а1 a relation involving the square of one coordinate with the first power of the other. If in this equation we suppose n = 0 or m = 0 we obtain the equation of a circle as before. If n = 1, (24) becomes this is therefore the bipolar equation of the limaçon referred to its node and the centre of the primitive circle. If m<l we have the crunodal limaçon; if m>l, the lower sign gives the acnodal limagon, the upper sign giving an equation satisfied only by A1, the point (0, a2). If in (25) we make m='l we have (26) 0 2 the bipolar equation of the intermediate case, the Cardioid, which is therefore a Cartesian whose three foci coincide at the cusp. Since in (26) P3 is the distance of a point from the centre of the primitive circle, P3 2 is the square of the tangent upon this circle; and (26) expresses that the square of the tangent from any point of the Cardioid to the primitive circle is a mean proportional to the diameter and the distance of the point from the cusp. We may now find the polar equation of the Cartesian referred to C by transformation from (24), making, see (1), p=p and p1 = p2-2a,p cos +ag. For the sake of abridgement we put for the present Uz2. mn ে 2 0. (22) then becomes p2± 2pa,1 + (p2 — s)α22 Transforming by the above = 0. [o2 + (p3 — 8)a,3]2 = 4p3u‚2(p2 — 2a, cos 0 + a23) expanding p2(8 + p2)a22p2 + 8p2a23p cos 0 + [(p2 — s)3 — 4p2]a‚1 : By adding and subtracting (s+p3)3at we may write this equation in the form [p2 - (s + p3)a,'] + 8p3a, cos — 4p3(s + 1)a,* = 0. In rectangular coordinates 4 [x2 + y2 — (8 + p3)a,3]3 + 8p3a23[x — }(8 + 1)a2] = 0. 2 In this equation a2 = lc the value of the distance CA1; substituting and restoring the values of p and 8, [x2 + y2 — (l3m2 + m3n2 + n3l3)c2]2 + 8l3m3n2c3[x — } (m2+n3+lc)]=0... (27) an equation of the form 82+k3L=0 where S=0 is the equation of a circle whose centre is at C, and L=0 is the equation of a straight line perpendicular to the axis. If we denote the distances CA1, CA2, CA3, by a1, a, and ag we may substitute in (27) lc = a1 m2c = a2 n3c = a, which gives [x2+y2—(a1a2+ɑ2αz+ɑ ̧a1)]2+8a ̧a ̧az[x—†(α1+a2+ag)]=0........(28) 3 = If one of the constants as a1 O, the equation reduces to S2 O, the equation of a pair of circles coincident with S = 0. The form of the equation S2+ L = 0 shows that the Cartesian touches two points where this line cuts the circle S0. double tangent of the outer branch. the line L 0 in the Hence L= 0 is the If we now transform the equation to any other rectangular axes it will take the form S3L' = 0 in which S' = 0 is the transformed equation of S=0 and L'= 0, that of L= 0; therefore the general equation of the Cartesian in rectangular coordinates is of the form S2+k3L=0, the centre of the circle S determining the point C, while the axis is perpendicular to the line L = 0. Suppose the equation when transformed to the origin C and the axis as axis of x, to be (x2 + y2 · a2)2 + k3 (x − p) a3 then comparing eq. (28) we see that a, a, and a, are the roots of the cubic 2pa2 + a2a- 1/3 == 0. (30) If in (29), a<p, the line x = p actually touches the outer branch which has a reentrant portion like Nos. 1, 2 and 3 in the diagram. which is the case in No. 4, (in which a1 1, a2 = 4 and ag has contact of the third order with the curve. If a <p, as in No. 5, the double tangent does not really touch the curve. If a = p = Ρ 1), It is noticeable that with given values of a and p in (29) the Cartesian cannot be made to approach as near as we choose to the coincident circles S2=0 when these are cut by L = 0, for the whole Cartesian is always on S2 one side of L = 0; yet k = 0 reduces (29) to S2=0. In fact in this case, since pa, the cubic (30) will be found to have but one real root when k=0. Nevertheless taking a = 0, real values of a1 and a, can be found. For comparing (28) and (29) we see we are no longer bound to satisfy the condition p = (α1 + α2+ αg) when k = 0, hence we have only to fulfil the condition a1ɑ2 + ɑ2αz + azα1 = 0 which reduces to 1a1⁄2 =0. Thus the foci A1 and A, are real and subject only to the condition that a the radius shall be a mean proportional between them. 2 a1 Since in (29) the expression x2 + y2-a2 is the constant product of the segments of a chord or secant to the circle through (x, y) and x — p is the distance to the line xp, eq. (29) expresses that the distance from any point of the Cartesian to the straight line L=0 is proportional to the square of the product of the segments of a chord or to the fourth power of the tangent to the circle 80. ON SUPPLYING OMISSIONS IN LAND SURVEYING. BY P. H. PHILBRICK, PROF. OF CIVIL ENGINEERING, IOWA STATE UNIV. In our standard works on Land Surveying we are told that: "Any two omissions in a closed survey whether of the length, or of the direction, or of both, of one or more of the sides bounding the area surveyed, can always be supplied by a suitable application of the principle of Latitudes and Departures." Let us proceed to examine the truthfulness of the statement. Case I. When the length of one side and the bearing of another are wanting. (a). Let the deficient sides adjoin each other. Let ABCDEF (Fig. 1) represent the case; in which the length of BC and the bearing of DC are unknown. Solution. With D as a center and the known length of DC as a radius describe an arc. This arc will in general intersect the indefinite line BC in A two points (C & C") either one of which may be taken, consistently with the data, as the missing corner, thus giving the two figures ABCDEF and ABC'DEF either one of which may be considered as the field surveyed. (b). Let the deficient sides be separated from each other. Suppose the length of BC and the bearing of ED unknown. Solution. Draw DH parallel to BC. With E as a center and the known length of ED as radius describe an arc which arc will generally intersect DH in two points D and D' from which we draw DC and D'C'' giving the two figures ABCDEF and ABC''D'EF, the latter of which complies with all the data of the problem equally well with the first which by hypothesis represents the tract surveyed. In case the length of the side whose bearing is unknown is equal to the length of the perpendicular upon the side whose length is unknown; then whether the deficient sides are separated or not the two solutions unite in Thus if DC is equal to the perpendicular DC" upon BC the points Cand C coincide in C", and the two figures ABCDEF and ABCDEF become identical being represented by ABC""DEF. one. If more than one known side intervene between the deficient sides a similar construction will nevertheless suffice. Case II. When the lengths of two sides are wanting. (a). The deficient sides adjoin. In Fig. 1 let the lengths of BC and DC be wanting. From B and D with the given bearings of BC and DC we must run the lines BC and DC to their intersection C, thus there is but one solution in this case. (b). The deficient sides are separated. Let AB and DC (Fig. 2) running from A and D in known directions, be the sides whose lengths are unknown, B and C representing the undetermined corners. Draw GB From D draw DG with the length and bearing of the side which lies between AB and DC. parallel to DC to intersect AB in B, and BC parallel to GD to intersect DC in C. ABCDEF represents the tract surveyed. If however AB and DC are parallel G is on AB or its prolongation; for if not, GB and AB, which according to the present hypothesis are parallel, would not intersect and consequently the survey would not close. Since G then, lies upon AB, and GB and AB are parallel they must coincide and therefore any point upon AB may be taken as their intersection from which a line drawn parallel to GD to intersect DC will close the survey. Thus there are an infinite number of solutions in this case as is illustrated in fig. 3. If more than one side lie between the deficient sides a similar construction will still suffice. ing. Case III. When the bearing of two sides is want (a). Let the deficient sides adjoin each other.(fig. 4) Let C, D, E, F and A represent the known corners. The distances of the corner B from A and C being known but not its bearings. 1 Solution. With A and C as centers and radii equal to AB and CB respectively describe arcs. These arcs will always intersect in two points B and B' giving the two figures ABCDEF and AB' CDEF, either of which answers to all the conditions of the problem and therefore may be taken to represent the tract surveyed. The arcs described with the centers A and C must intersect since the survey must close, and they must intersect in two points for otherwise AB and BC would represent the same line. |