Viewing page 198 of 504

This transcription has been completed. Contact us with corrections.

7)    
[[Left margin]] See p. 144. for measuring the degrees in a spheroid corresponding to those in a sphere. [[/margin]]

A Question. By Mr. G. Witchell, [[underlined]] Teacher of the Mathematics [[/underlined]], at the Front House in [[underlined]] White Fryers gate, Fleet Street [[/underlined]]. (From Martin's or the General Mag. Jan. 7 1764. p. 34. and Solved by him March 1764. p. 139.)

Suppose a Circle to be described upon the transverse Axis of a given Ellipsis, (as a Diameter) and that a Right line be drawn, through two given Points in the Circumference of the Circle, to cut the Ellipsis; it is required to determine the lengths of those two Segments of the right -Line, which are intercepted between the Peripheries of the Circle and Ellipsis.

[[right margin]] Oil & Sulphur the Cause of all Vegetable colours. [[/margin]]
 
CONSTRUCTION.
Let BC (fig. 17.) be the trans. diam. of the given Ellipsis, and BEDC its circumscribing circle; thro' D,E, the two given Points, draw the line DE, producing it (if necessary) 'till it meets CB (produced also) in A; at any point (C) of the line AC, erect the perpendicular CH, meeting the right line, AH in H; in CH, produced, take Ch to CH, in the ratio of the greater axis of the given ellipsis, to the lesser, and joint h, A, then from g and f, the intersections of the right line Ah, with the Circle BEDC draw gI and fK, perpendicular to AC, intersecting AH, in G and F, then shall EG and FD be the required Segments.

[[right margin]] Luna Eclipses retard the moon in her periodical motion. [[/margin]]

DEMONSTRATION.
The lines gI, fK, and hC, being parallel to each other, it follows {from similar triangles) that the ratio of gI to GI, and of fK to FK, will be the same as the ratio of hC to HC; that is (by construction) as the ratio of the greater axis of the given ellipsis to the lesser, therefore (by conics) the points F and G, will be in the periphery of the given ellipsis BGFC, Q. E. D. 

[[right margin]] Luna Eclipses retard the moon in her periodical motion [[/margin]]

CALCULATION. 
The Angle DLC and ELB, being given (by the question) we shall have the angle DAL = [[undelrine]] (DLC - ELB)/2* [[underline]]; therefore in the triangle ADL, there is given all the angles, together with the Side DL, from which AL becomes known; but the tangent of the Angle FAK, is to the tangent of the angle fAK, as FK to fK, that is, as the lesser axis of the given ellipsis, is to the greater; whence in the triangle AfL there is given the angle fAL, together with the sides AL, Lf, by which the angle AfL will be found; but the sum of the angles fAL, AfL is equal to the angle fLK, and the tangent of the angle fLK, will be to the Tangent of the angle FLK, as fK to FK, that is, as the greater axis of the 6[[given]] Ellipsis, is to the lesser, therefore the angle FLK will be known, from which, taking the given angle DLC, there will remain the angle FLD; but fL is to FL, as the secant of the angle fLK, is to the secant of the angle FLK (or, which amounts to the same, reciprocally as their co-sines) therefore FL will become known; lastly, in the triangle FLD, there is given, the two sides FL, DL, with the included angle FLD; whence FD will be easily found: and by a similar process, EG may be determined.

[[left margin]] * This appears from lines being drawn from B to D, and from E to C: for then DEC = 1/2DLC, & ECA = 1/2ELB, but DEC = DAL + ECA; DAL = DEC - ECA = DLC/2 - ELB/2. [[/margin]]

[[right margin]] Thence arises an Error in the Tables. [[/margin]]

ALGEBRAIC SOLUTION. 
Let ABCD (fig. 18.) represent the given Ellipsis, and M, N, the two given points in the circumscribing circle, draw the ordinates MK, LI, and NP, then will PH, HN, and HK, HM, be given. To find the segments ML, and NE, put OA = OC = t, OB = OD = C, HC = a, HK = b, KM = d, CI = x. Then, [[underline]] per [[/underline]] property of the ellipsis, IL = c/t [[square root]] 2tx - x^2 [[/square root]], and [[underline]] per [[/underline]] similar triangles, as HK = b:KM = d :: HI = a-x: IL = c/t [[square root]] 2tx - x^2 [[/square root]]; therefore td x (a-x) = bc [[square root]] 2tx - x^2 [[/square root]], from which equation the value of x may be easily found, and in consequence of that the segment MI may be found also: and according to the same method of reasoning, the other segment NE may be found. Contind. On p.g.

[[right margin]] Water, exhalations [[vc?]]. & not Air, is the cause of Refraction, w.ch proves no Solid Orbits in the expanse. The argument of rays passing thro' a vacuum is false. Pure Air neither reflects nor refracts light, but water does both, hence the refraction of the

Transcription Notes:
The parentheses are not found around equations as they are literally drawn as DLC - ELB over 2; however, I have inserted them to preserve the intent of the equation.