consider not fully because the center and transverse diameter of the Ellipsis lies within and of the Hyperbola without the Section And if h or any point within a Section be given and required through it to draw an Ordinate that may be easily done because it must be parallel to a tangent at the Vertex a. Any Section given to find that diameter thereof which shall make an angle with the Ordinate to it equall to an angle given If first the Section given be a Parabola finde any diameter and from the end or vertex thereof draw a right line to the Section making an angle with the said diameter equall to the angle given to which if a parallel through the middle of the other right line be drawn that parallel is the diameter required Let there be given therefore the Hyperbola bac and the angle z to finde the diameter eg which with the Ordinate af shall make the angle ega = z. Finde the transverse axis ad and the center e and upon ad describe by the 33. of the 3. of Euclide a portion of a Circle dfa capable of an angle equall to z then draw df and af and through the middle of af draw eg the diameter required The work is altogether the same in an Ellipsis only the lesser axis is to be used Midor 3.67 Any Hyperbola being given to finde the Asymptoti Because ah toucheth the Section it is equidistant to the Ordinates per Coroll 2 ad 17 primi But to the Rectangle or Parallelogram mag that is to the figure comprehended of the two sides ma and ag is made equal the Square or Rhombus of ah and an is half of ah therefore the square or Rhombus of an is equal to a fourth part of the Square or Rhombus of ah that is to the quadrant of the figure mag and therefore by the 38. of the first and Coroll to it by conversion it may be shewed that the right line en drawn from the Center and produced how far soever shall never meet with the Section bac and by the same reason and because an = ao eo drawn from the Center shall doe the like c. From hence it appears that the Asymptotes are lines drawn from the center of the Section and produced so as that inclining toward the section still more shall never be coincident therewith More for the Parabola Numerically Let the base be given in Numbers 20 that is of what lenght soever let it be parted into 20 equal parts And at any inclination to it let there also be given a diameter which divide into 100 parts And through all the other 9 divisions of the Semi-base draw lines equidistant to the Diameter shortening them in this proportion viz. Of such parts as the Diameter is 100 let the next be 99 the next 96 the next 91 the fourth 84 the fift 75 the sixt 64 the seventh 51 the eighth 36 the ninth 19. A line drawn with an even hand by the ends of these lines shall be a Semiparabola The said Numbers are made thus 10 in 10. 11 in 9. 12 in 8. 13 in 7. 14 in 6. 15 in 5. 16 in 4. 17 in 3. 18 in 2. and 19 in 1. Prop. 62. lib. 2. And they differ just as the square Numbers immediately succeeding to Unity viz. 1 4 9 16 25 36 49 64 81 100. c by the quantity of the odd numbers intercepted as 1 3 5 7 9 11 13 15 17 c. Which is the same proportion by which the degrees of Velocity of the falling of any thing toward the center of the earth are increased as Galileo hath sufficiently proved in his Dialogues And therefore the course of every Projectile or thing shot from Gun or Bow may easily be proved to be a Parabolical line And the making a Rectilone figure equal to a Parabola might be facilitated from hence if it were not needlesse the thing being already often done Moreover it is to be noted that the equidistant lines thus drawn may represent squares because they differ as the Square numbers doe For an Hyperbola Numerically The burning points and transverse axis being given the Vertex is also given Let the transverse axis be 80 the distance of each burning point 20 of the same parts the said points a and b the center a space 23 and the other center b and space 103 describe arches which shall cut where the Section is to passe and so describing from the center a other arches 34 57 100 and from the center b with distances 114 137 180 other arches so as the distances from b may exceed 100 as much as the distances from a exceed 20. Those arches of Circles shall intersect and thereby give points by which the Hyperbola is to passe by the 26. of the 2. of Midorgius For an Ellipsis Numerically The burning points and Vertices being given as they were before the Ellipsis also may be described by numbers as followeth let the one burning point be at a the other at b and let the diameter be z the distance betwixt a and b let that be x equall to 100 and let it be x″ 16′ 100″ Therefore also z = 232 wherefore making the center b at severall spaces more then 16 and lesse then 116 of such parts as z is 132 as 110 97 81 c. describe arches Again making the center a with distances 22 35 51 and others still the correspondent complements of the former distances to 132 draw other arches which shall cut the former in points whereby the Ellipsis desired must passe by the said 26 of the second And it is plain from the generation of an Ellipsis shewed in the instrumentall way before in this Book for the string which describes it is alwayes equall to z + x that is 232 and so is 100 + 110 + 22 and 100 + 97 + 35 c. wherefore this is evident And thus they that like this last way better may accomplish the Section by number Moreover put the diameter of a Parabola of an inch ferè And let the whole base inclined to the Diameter at angle 84 ferè be c = 150 64 Lastly Let the perpendicular from the Vertex to the base be d = 64 64 Multiply 150 64 by 64 64 the Product is 9600 4096 Of which two thirds is equall to the supersicies of the Parabola and is 6400 4099 Of these parts the middle Parallel which was before 75 when the diameter was supposed 100 is 50 64 which doubled is 100 64 that is 6400 4096 as before So that in this case the residue of the Rectangle or Parallelogram when the superficiall content of the Parabola is taken from it and the length of the middle parallel are both denominated by the same number but this is left to the Reader to try by a Figure delineated by himselfe But what use might be made of this if it were further urged either in naturall or artificiall numbers I
Arithmeticall proportion the summe of the two middle termes is equal to the summe of the two extreames And so here the fine of the middle part plus Radius is equal to the tangent of the of the adjacent part known plus the tangent of the part required I hope the word plus needs no interpretation Note 1. It is notwithstanding to be ever remembred that every of the five circular parts must be considered two wayes that is whether it be contiguous to the right angle or quadrant if so then all before is right and unalterable But if not so that is if some other part lie betwixt them then all that hath been said of their sines and tangents which were then supposed contiguous must be performed by the sines complements and tangents of the complements of such of the parts respectively as are remote from the right angle or quadrant 4 If now the two parts given be remote and the part required lie betwixt them then make the part required the middle part and it may be found as easily as in the former case 5 If the two known parts be contiguous and the part required adjacent to neither of them but opposite to one of them then working stil by Logarithmes make the part required the middle part and then the sine of the middle part plus Radius is equal to the sines complements of the opposite parts given if therefore from those two sines complements added be taken Radius the rest is the sine of the thing required Note 2. It is further to be noted that the sines complements of those parts which by the former note are accompted complements are the sines of the things themselves Example In the triangle zps let p be the Pole s the Sun and z the Zenith zps the hour from noon in Winter or the hour from Midnight in Summer pzs the Azimuth from the North and zsp the angle of position and sz be 90 Deg. as at Sun-rise Then first sine z plus Radius is equal to tangent complement zp plus tang s. Secondly sine compl p plus Radius is equal to tang compl pz plus tang compl ps per Not. 1. Lastly by the 5 direction sine compl sp plus Radius is equal to the sine of pz plus the sine complement of z because pz is accompted a compl and z not so Note 1. And if instead of a quadrant as zs there were a right angle at p or at z or at s the aforegoing directions serve If there be neither right angle nor quadrant there must be two operations to do this having first supposed a circle to passe from one of the angles to cut the opposite side produced when need is at right angles This perpendicular circle shall fall sometimes within the triangle sometimes without Within when the other angles are both obtuse Or both acute as at the hours between six and noon are the angles at p and s. In letting fall the perpendicular Mr. Norwoods advice is to do it 1 From the end of a side given being adjacent to an angle given let it fall opposite to that angle 2 And touching in some part the side required 3 And opposite if it may be to the angle required One of the most difficult cases in oblique angled sphericall triangles is this In the triangle pzs Let there be given comp Elevation pz = 38 28′ comp Declinati ps = 66 29′ comp Azimuth pze = 70 0′ To find the complement of the Suns Altitude zs = 75 45′ having produced sz and from p let fall pe perpendicular to it then First making pze the middle part s.c. pze plus Radius 195340517 From which taking the tangent of the Elevation t.c. zp 100999135 Remains tang of ze = 15 12′t ze 94343382 Secondly to sine compl ze 99845347 Adde sine of Declination s.c. ps 96009901 The summe of them is 195855248 From which taking sine Eleva s.c. pz 98937412 Remains the sine of 29 27′ s.c. 60 33′ 96917696 To whose comple 60 33′ adding ze = 15 12′ the sum is 75 45′ the thing required Whereas if the Azimuth it selfe pzs were 70d then ze being taken from 60 33′ the rest 45 21′ is the Suns Altitude CHAP. I. An explanation of the Characters and Symbols used in this Work FIrst One single letter of the Alphabet is usually put for any quantity whatsoever as well Line as Number whether known or unknown But for the most part where any quantity is sought there a or some other Vowel is put for it and the other Quantities known are signified by Consonants These letters are multiplyed one into another by joyning them together without any pricke or comma between nor doth it import at all which is first or last written for bcd bdc and cbd are all one So a multiplyed by a produceth aa And a multiplyed by b produceth ab And ab multiplyed by c produceth abc The like of all others whatsoever except Fractionall quantities as ab+ fg d and bc − fb+ rc b+ c If the first of these were to be multiplyed by d it is done by taking away the d under the line and the product is ab + fg If the second were to be multiplyed by b + c it is done by taking away the Denominator b + c and the product will be bc − fb + rc For all Fractions aswell in Plaine as in Figurate Arithmetick are nothing else but Quotients of one number divided by another and are multiplied again by taking away their Divisor and line of Separation Division is done in Figurative Arithmetique most commonly by applying some line of separation between the Dividend and the Divisor So a b is a divided by b and abc f signifies that abc is divided by f. But yet if the letter f had been found in the Dividend the application of this line had not been necessary for it might have been better done by taking away that letter out of the Dividend So afc divided by f quotient is ac and ffcc divided by fc quotient is fc by ff quotient is cc by cc quotient is ff by ffc quotient is c by fcc quotient is f by f quotient is fcc by c quotient is ffc And the like may easily be understood of all the rest Symbols of Majority > Minority <
remote from the said Center NOTE This Aequation − aaa + 3 a − b = 0 is naturally without the second terme aa which is the cause that it hath the false Root not discerned by twice + or twice − succeeding as hath been spoken of Chap. 4. If therefore one would have it so he must fill up the second term by augmenting the root never so little putting e − x = a. The Demonstration of this Problem is as followeth It is to be proved that kg in the Section is equal to be the subtense of the third part of the angle given Put kg = y. Then because of the Section ag = yy From the center m draw mk and ma which are equal because of the circle And draw kn parallel to ae and produce me to it in n. Then it is kn = ge = 2 − yy The square therfore of kn is 4 − 4 yy + yyyy And mn being equall to y plus halfe the subtense bg call bg by the single letter b as before then mn = y + ½ b the square of which being yy + by + ¼ bb add to it the former square of kn that is 4 − 4 yy + yyyy it makes + 4 − 4 yy + yyyy + yy + by + ¼ bb equall to the square of the Hypothenusal mk Againe the square of ae is 4 and the square of me is ¼ bb which two squares are equall also to the square of mk because mk = ma. Therfore 4 − 4 yy + yyyy + yy + by + ¼ bb = = 4 + ¼ bb That is − 3 yy + yyyy + by = 0. That is by adding on each part 3 yy and subtracting yyyy + 3 yy − yyyy = by Or lastly dividing all by y + 3 y − yyy = b. But this aequation is alike graduated and like affected as the first aequation + 3 a − aaa = b. Wherefore y = a. But a = be and y = kg. And therefore kg = be Which was to be proved In like sort it might be proved that fd is a true root of the aequation 3 a − aaa = b in the first figure and the subtense of the third part of the complement of the angle given bag to a Circle And by such working one may finde it evident that when a Circle cuts a Parabola in points how many soever the Vertex excepted perpendiculars let fall from all those points to the axis are all the severall roots of one and the same aequation Nor hath that aequation any more roots then those perpendiculars to the axis NOTE 1. In the aequation + aaa − bca = bbd the construction differs somwhat from the former for b being reputed unity if c as here be signed with − the axis of the Parabola must be produced from the point c in the axis within the Section distant from a by ½ beyond the vertex till the continuation be equal to ½ c and at the end thereof raise a perpendicular equal to ½ d at the end of that is the center of the circle desired And according to this method may any aequation not above biquadratical be resolved after by taking away the second term if there be any by the second Rule of Chapter the fourth it is reduced to such a form as this aaa * bca = bbd if the quantity unknown hath but three dimensiōs or if it have four then thus aaaa * bcaa * bbda = = bbbf Or else taking b for Unity then thus aaa * caa = d nd thus aaaa * caa * da = f the signes + and − are here omitted for they must be supplied as the nature of the Aequation requireth NOTE 2. Note that in this breviate the line b is that which was ba in the example of trisection and that which was r or unity in the example of two Meanes Also the line c is that which in the former example of trisection was 2 ce or 3. And if this quantity be nothing then the perpendicular equal to half d is to be erected at the end of half b or ½ set off from the vertex upon the Axis within but if c have any length then at the distance of ½ c from that end upon the axis And this which hath been said is enough for all Cubiques Prob. 3. But where the equation is aaaa − caa + da = f so placed as here if there be + f and the Probleme be to find the value of the root a then producing ma towards a Make as equall to the right parameter of the Section and make ax = f and upon the diameter xs describe the Circle xhs cut by a perpendicular to ma namely ah in h then making the center m and the space mh describe the Circle desired But if it be − f as in this Example I put it then after ah is found as before upon the diameter am describe a circle and in it from a apply a line ai = ah and making the center m and the space mi describe the circle fik which is the circle sought for Now this Circle fik may cut or touch the Parabola in 1 2 3 or 4 points from all which perpendiculars let fall to the axis give all the roots of the Aequation as well the true as false ones Namely if the quantity d be marked − then those Perpendiculars which are on that side the Parabola where the center m is are the true Roots but if it be + d as here the true roots are those of the other side as gk and no and those of the center side as fz pq are the false DEMONSTRATION Put ce = c 2 and draw me perpendicular to ag and gl equall and parallel to it lastly Put gk = a then ag = aa and taking from it ae that is ½ c + ½ then ge = aa − − ½ c − ½ whose square is aaaa − caa − − aa + ¼ cc + ½ c + ¼ And because by construction gl = ½ d therefore kl = a + ½ d and the square of it is aa + da + ¼ dd which added to the former square of ge it gives the square of km that is a4 − caa + ¼ cc + da + ¼ dd + ½ c + ¼ Again the square of ae is ¼ cc + ½ c + ¼ To which adding the square of me that is ¼ dd the whole is the spuare of ma to wit ¼ cc + ¼ dd + ½ c + ¼ But the square of ah that is ai is equal to f because sa = 1 and xa = f between which ah or ai is a mean Therefore the square of mi is ¼ cc + ¼ dd + ½ c + ¼ − f But mi = mk Therefore their squares are equal that is aaaa − caa + ¼ cc + ¼ dd + da + ½ c + ¼ = ¼ cc + ¼ dd + ½ c + ¼ − f. That is aaaa − caa + da = − f. or else aaaa =