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For, since AP : PM :: CB : BF, and, by the similar triangles API, ABF, Al’: PI::AB : BF; Therefore PM ; PI:: CB : AB. But CB is the half of AB; therefore, also, PI is the half of PM. 1073. THEoREM IX. If two lines be drawn from the foci of an ellipse to any point in the curve, these two lines will make equal angles with a tangent passing through that point. Let TM (fig. 418.) be a tangent touching the curve w at the point M, and let F., f be the two foci, join \ G --~/ FM, fM, then will the angle FMT be equal to the - M / angle fMR. For draw the ordinate PM, and draw fR parallel to FM, then will the triangles TFM and T__
By the similar triangles TFM, Tf R, TF : # : FM : fR. It therefore appears that fM is equal to fB, therefore the angle fMR is equal to the angle f RM : but because FM and f R are parallel lines, the angle FMIT is equal to the angle f RM; therefore he angle FMT is equal to the angle fMR. Coroll. 1. Hence a line drawn perpendicular to a tangent through the point of contact will bisect the angle FM f, or the opposite angle DMG. For let MN be perpendicular to the tangent TR. Then, because the angle NMT and NMR are right angles, they are equal to one another; and since the angles FMT and fMR are also equal to one another, the remaining angles NMF and NM fare equal to one another. Again, because the opposite angles FMN and IMG are equal to one another, and the opposite angle fMN and IMD are equal to one another; therefore the straight line MI, which is the line MN produced, will also bisect the angle DMG. Coroll. 2. The tangent will bisect the angle formed by one of the radius vectors, and the prolongation of the other. For prolong FM to G. Then, because the angles RMN and RMI are right angles, they are equal to one another; and because the angles NM f and IMD are equal to one another, the remaining angles RMG and RM fare equal to one another. Scholium. Hence we have an easy method of drawing a tangent to any given point M in the curve, or of drawing a perpendicular through a given point in the curve, which is the usual mode of drawing the joints for masonic arches. Thus, in order to draw the line IM perpendicular to the curve: produce FM to G, and fM to D, and draw MI bisecting the angle DMG; then IM will be perpendicular to the tangent TR, and conse Tuently to the curve. As in optics the angle of incidence is always found equal to the angle of reflection, it appears that the foundation of that law follows from this theorem; for rays of light issuing from one focus, and meeting the curve in any point, will be reflected into lines drawn from these points to the other focus: thus the ray fM is reflected into MF; and this is the reason why the points Ffare called foci, or burning points. In like manner, a sound in one focus is reflected in the other focus. 1074. THEoREM X. Every parallelogram which has its sides parallel to two conjugate diameters and circumscribes an ellipsis is equal to the rectangle of the two axes. Let CM and CI (fig. 419.) be two semi conjugate diame- R. E. ters. Complete the parallelogram CIDM. Produce CA M > and MD to meet in T, and let AT meet DI in t. Draw IH and PM ordinates to the axis, and draw half the minor Taxis CE. Produce DM to K, and draw CK perpendicular to DK; then will the parallelogram CIDM be equal to the rectangle, whose sides are CA and CE; or four times the Fig. 419. rectangle CIDM will be equal to the rectangle made of the two axes AB and GE.
The ellipsis is of so frequent occurrence in architectural works, that an acquaintance with all the properties of the curve, and the modes of describing it, is of great importance to the architect. Excepting the circle, which may be called an ellipsis in which the two foci coincide, it is the most generally employed curve in architecture.
1075. ProBLEM I. To describe an ellipsis.
Let two pins at E and F (fig. 420.) be fixed in a plane within a string whose ends are
made fast at C. If the point C be drawn P D o equally tight while it is moved forward t ... -in the plane till it returns to the place b : from which it commenced, it will describe >~~ £2. \
an ellipsis. &- lf
P - - -1076. PRob. II. The two diameters --> -: w: * * AB and ED of an ellipse being given in > → N \ Ż position and magnitude, to describe the curve \" through points. Let the two diameters cut each other at C (fig. 421.). Draw AF and BG parallel to ED. Divide AC and AF each into the same number of equal parts, and draw lines, as in the figure, through the points of division; viz. those from the line A F to the point D, and the lines through AC to the point E; then through the points of intersection of the corresponding lines draw the curve AD, and in the same manner find the curve BD; then ADB will be the semi-ellipsis. It is evident that the same method also extends to a circle by making CD equal to CA: (fig. 422.); and it appears that the two lines forming any point of the curve to be drawn will make a right angle with each other. For these lines terminate at the extremities of the diameter ED, and the point of concourse being in the curve, the angle made by them must be a right angle; that is, the angle EAD, or El D, or Ei D. or Ek D, is a right angle: and from this property we have the following method of drawing the segment of a circle through points found in the curve. Fig. 122. Thus, let AB be the chord, and CD be the versed sine of an arc of a circle, to describe the arc. Through D draw Hl (fig. 423.) parallel to AB; join AD and DB, draw AH perpendicular to AD, and BI perpendicular to BD; divide
AC and HD each into the same number of equal parts, H_* F_2_s_D G | and join the corresponding points; divide A F into the \ * same number of equal parts, and through the points of di- * | vision draw lines to D, and through the corresponding A 1 e s C B points where these lines meet the former draw a curve Fig. 425.
AD. In the same manner the other half BD may be drawn. 1077. PRca. III. A diameter KH of an ellipsis leing given, and an ordinate DL, to find the limits of the other conjugate diameter. Bisect KH in I (fig. 424.), through I draw EA parallel to DL, and draw DC and KB perpendicular to EA; from the point L with the distance K describe M K an arc cutting EA at F, join LF, and produce LF to C; make IE in and IA each equal to L C; then will EA be a diameter conjugate V. to KH. 1078. Prob. IV. A diameter KH and an ordinate DL of an £f F->. ellipsis being given, to describe the curre. (fig. 424.) h Find the limits E and A of the other conjugate diameter by the preceding construction. Produce KB to q, and make Kq equal to I A or I E, and through the centre l of the curve and the point q, draw the straight line MN. Then, suppose the straight line KBq to be an inflexible rod, having the point B marked upon it. Move the rod round, so that the point q on the rod may be in the line MN, while the point B is in the line FA : then, at any instant of the motion, the place
of the point K on the plane whereon the figure is to be drawn may be marked; the points thus found will be in the curve. Instead of a rod, a slip of paper may be used, and in some cases a rod with adjustible points to slide in a cross groove, and a sliding head for a pencil is convenient; and such an instrument is called a trammel. When the diameters KH and EA (fig. 425.) are at right angles to each other, the straight line Kq coincides with the diameter KH, and consequently K the line MN, on which the point q of the inflexible line Kq moves, --~T =>!: will also fall upon the diameter KH. Therefore in this case no- / D thing more is required to find the limits of the other diameter, "Til 7 than to take the half diameters IK, KH of the given diameters, > * > and from the extremity L with that distance describe an arc H cutting the unlimited diameter in the point F, then drawing Fig. 425. LF, and producing it to q, and making IE and IA each equal to qL, EA will be the other diameter; and since the two diameters are at right angles to each other, they are the two axes given in position and magnitude, and thus the curve may be described as before. A method of describing the curve from any two conjugate diameters is occasionally of considerable use, and particularly so in perspective. For, in every representation of a circle in perspective, a diameter and a double ordinate may be determined by making one of the diameters of the original circle perpendicular to the plane of the picture and the other parallel to it; and then the representation of the diameter of the original circle, which is perpendicular to the intersecting line, will be a diameter of the ellipsis, which is the representation of that circle; and the representation of the diameter of the circle which is parallel to the intersecting line will become a double ordinate to the diameter of the ellipsis which is the perspective representation of the circle. 1079. PRoB. V. Through two given points A and B to describe an ellipsis, the centre C being given in position and the greater aris being given in magnitude only. About the centre C (fig. 426.) with a radius equal to half the greater axis describe a circle HEDG ; join AC and BC; draw AD perpendicular to AC, and BE perpendicular to BC, cutting the circumference in the points D and E; draw also 13 F parallel to AC, and find BF, which is a fourth propor- G tional to AD, AC, and BE ; through the point F and the centre C draw FG to cut the circle in H and G, and GH is the major axis of the ellipsis. By drawing an ordinate Bq, the curve may be described by the preceding problem, having the axis GH and Fig. 426. the ordinate Bq. 1080. ProB. VI. Through a given point in the major a is of a given ellipsis to describe another similar ellipsis wiich shall have the same centre and its major aris on the same straight line as that of the given ellipsis. Let A CBD (fig. 427.) be the given ellipsis, having AB for its major axis and CD for its minor axis, which are both given in position and magnitude. C k
On the axis CD make EI equal to EH, and on the axis AB make EF equal to EG. Then, having the major axis AB, and Fig. 127. the minor axis FG, the ellipsis FIGH may be described, and when drawn, it will be similar to the given ellipsis ADBC. 1081. Prob. VII. Through any giren point p, within the curve of a given ellipsis to describe another ellipsis which shall be similar and concentric to the given one. Let C (fig. 428.) be its centre. Draw the straight line Cp P, cutting the curve of the given ellipsis in P. In such curve take any other number of points Q, R, S, &c., and join QC, RC, SC, &c.; join PQ and draw pg parallel thereto cutting qC at q : join PR and draw pr parallel to PR, cutting RC at r, join PS and draw ps parallel to PS cutting SC in s. The whole being completed, and the curve p, s, t, u drawn through the points p, q, r, s, &c., the figure will be similar and concentric to the given ellipse P, S, T, U; or when Fig. 428 the points at the extremities for one half of the curve have been drawn, the other half may be found by producing the diameter to the opposite side, and making the part produced equal to the other part. 1082. ProB. VIII. About a given rectangle ABCD to describe an ellipsis which shall have its major and minor ares respectively parallel to the sides of the rectangle and its centre in the points of intersection of the two diagonals. Bisect the sides A I) and A B (fig. 429.) of the rectangle respectively at L and O
through L draw GH parallel to AB cutting the opposite side BC of the rectangle in M, and through the point O draw K1 parallel to AD or BC cutting the opposite side DC in N. In NK or NK produced, make NQ equal to NC, and join CQ ; draw QR parallel to G H cutting CB or CB produced in R; make EH and EG each equal to QC, as also EI and EK each equal to PC; then will GH be the major axis and KI the minor axis of the ellipsis required. The demonstration of this method, in which the line QK has Fig. 429. nothing to do with the construction, is as follows: — By the similar triangles CPM and CQR, we have CP : CM:: CQ : CR. But because MP is equal to MC= EN, and since CR is equal to RQ = EM, And, by construction, since PC is equal to EI or EK, and QC is equal to EG or EH, EI: EN:: EH : EM, or, alternately, EI : EH:: EN: EM. But EN is equal to MC, and EM equal to NC; Whence EI : E H :: MC : CN. But since the wholes are as the halves, we shall have K I : GH :: BC : CD. This problem is useful in its application to architecture about domes and pendentives, as well as in the construction of spheroidal ceilings and other details.
1083. The direction of a plane cutting a cone, which produces the form called the hyperbola, has been already described; its most useful properties will form the subject of the following theorems, which we shall preface with a few definitions : — 1. The primary axis of an hyperbola is called the transverse axis. 2. A straight line drawn through the centre of an hyperbola and terminated at each extremity by the opposite curves is called a diameter. 3. The extremities of a diameter terminated by the two opposite curves are called the vertices of that diameter. 4. A straight line drawn from any point of a diameter to meet the curve parallel to a tangent at the extremity of that diameter is called an ordinate to the two abscissas. 5. A straight line which is bisected at right angles by the transverse axis in its centre, and which is a fourth proportional to the mean of the two abscissas, their ordinate, and the transverse axis, is called the conjugate axis. 6. A straight line which is a third proportional to the transverse and conjugate axis is called the latus rectum or parameter. * 7. The two points in the transverse axis cut by ordinates which are equal to the semi-parameter are caled the foci. 1084. ThroREM I. In the hyperbola the squares of the ordinates of the transverse axis are to each other as the rectangles of their abscissas. Let QVN (fig. 430.) be a section of the cone passing along the axis VD, the line of section of the directing plane, H B the line of axis of the cutting plane, the directing and cutting plane being perpendicular to the plane QVN. Let the cone be cut by two planes perpendicular to the axis passing through the two points P, H, meeting the plane of section in the lines PM, HI, which are ordinates to the circles and to the figure of the section, of the same time. **t By the similar triangles APL and AHN, AP : PL::AH : HN; Fig. 430 And by the similar triangles BPR and BHQ, BP: PK::BH : HQ. g. 430. Therefore, taking the rectangles of the corresponding terms, AP x BP: PL x PK::AH x
1085. THEoREM II. In the hyperbola, as the square of the transverse | aris is to the square of the conjugate aris, so is the rectangle of the abscissas to the square of their ordinate.
Let AB (fig. 431.) be the transverse axis, GE the conjugate axis, o—R —k C being the centre of the opposite curves; also let HI and PM be or- | dinates as before; then will
Coroll. Hence AB : GE*:: CP*-CA*: PM" (fig. 432.). For let the cutting plane of the opposite hyperbola intersect two circles parallel to the base in H1 and PM, and let the cone be cut by another plane parallel to the base, passing through the centre C of the transverse axis, and let mn be the diameter of the circle made by such plane.
Though Ct is not in the same plane, it is what is usually called the scui-conjugate axis, and it agrees with what has been demonstrated 9 in the first part of this proposition.
1086. THEoREM III. In the hyperbola, the square of the semi- Fig. 432 conjugate axis is to the square of the semi-transverse aris as the sum of the squares of the semi-conjugate aris and of the ordinate parallel to it is to the square of the abscissas.
Let AB (fig. 433.) be the transverse axis, GE the conjugate, C the cen- / tre of the figure, and PM an ordinate, then will \1/
1087. THEoREM IV. In the hyperbola, the square of the distance of the focus from the centre is equal to the sum of the squares of the semi-ares. Let AB (fig. 434.) be the transverse axis, CE the semi-conjugate. In
Therefore, extracting the root of each number, FM = C £r- CA.
And, subtracting the upper equation from the lower, fM-FM =2CA.
Coroll. 1. Hence is derived the common method of describing the hyperbolic curve mechanically. Thus : – In the transverse axis AB produced (fig. 435.), take the foci F, f, and any point I in the straight line AB so produced. Then, with the radii AI, BI, and the