require shall give respective points enough in each hour to draw each parallel by Example In the latitude 51. 32′ the Sun being in Pisces the beginning thereof the degrees of the Suns height above the horizon at every hour being as followeth that is 25. 37′ at one of clock 21. 49′ at two 15. 57′ at three 8. 32′ at foure and the same for eight nine ten and eleven respectively if the perpendicular stile being Radius the tangents of the complements of 25. 37′ 21. 49′ 15. 57′ 8. 32′ be applyed from the foot of the stile to the respective hour that is the co-tangent of 25. 37′ from the foot of the stile to the hours of 1. and 11. and so the others they shall give points in every hour-line one by which a line being drawn with an even hand shall be the Parallel at the beginning of Pisces And the like of all the rest And therefore generally in verticals as also in all recliners that is to say upon all planes whatsoever draw a horizontal dial proper to the plane and inscribe the signes or parallels upon it by setting off from the foot of the perpendicular stile the tangents complements of the Suns height at every hour in the beginning of every such signe above that plane taken as an horizon the perpendicular stile being ever Radius and at the end of these tangents so set off upon every respective hour-line will be a point by which points lines drawn with an even hand shall give the parallels desired This horizontal Dial being drawn in obscure lines the Dial for the plane may be drawn afterwards The Parallels serving which were drawn before Example Suppose as M. Wells doth pag. 185 a plane declining 30 degrees and reclining 55 degrees the height of the Pole above the plane 19 degrees 25 minutes the Suns height at the beginning Of Taurus to be at the hours of 12h 82d 5′ Of Taurus to be at the hours of 1 73 30 Of Taurus to be at the hours of 2 60 3 Of Taurus to be at the hours of 3 46 1 Of Taurus to be at the hours of 4 31 53 Of Taurus to be at the hours of 5 17 47 The tangents of the complements of 82 5′ and 73 30′ and 60 3′ c. set off from the foot of the perpendicular stile the said stile being the Radius to those tangents to the obscure horizontal hours of 12 1 2 c. give the true distances between the foot of the stile and those auxiliary hours for the parallel of Taurus And so the other Parallels may be found It is true the height of the Sun at every hour of the day at the beginning of every signe in any latitude is not easily found out without Trigonometrical Calculation by Logarithms of the sines Tangents or by trusting to Tables already Calculated if any happen to be done for that latitude already the way of making a table shall be shewed towards the end Of the Vertical Circles These are vulgarly called Azimuths and are great Circles whose Poles lye in the horizon and intersecting one another in the Zenith and Nadir of the Place The whole Horizon being divided into 32 parts equal these circles shewing those divisions are called points of the Compasse and marked S. SbE.SSE c. Every one distant from other by 11¼ degrees But the better way of accompting them is 10 20 30 40 50 60 70 c. degrees from the Meridian 1 In all horizontal dials the Perpendicular stile being chosen making the foot thereof the center at any convenient distance describe a circle and accompt from the meridian both ways arches equal to 10 20 30 c. degrees from which divisions right lines drawn to the-foot of the stile aforesaid shall represent those Azimuths upon that dial 2 Upon a Prime Vertical or South Dial through the foot of the perpendicular stile draw a right line parallel to the horizon and making the said stile radius upon the parallel line set off both ways from the Meridian tangents of 10 20 30 40 c. degrees through which divisions right lines drawn all at right angles with the parallel line shall be the Azimuths 3 Upon any declining Vertical the same being done shall give the Azimuths of 10 20 30 c. from the meridian of the plane or from the Meridian of the place just allowance being made for the distance of Meridians 4 In South Declining Reclining Planes the perpendicular stile being chosen and made the Radius the tangent complement of the Reclination applyed from the foot of the said stile to the meridian of the place shall determine the Zenith of the place through which and the foot of the stile that is the Zenith of the plane a right line drawn shall be a perpendicular to the horizontal line which shall concurre with the aequator in the hour of 6 and the therefore if from the foot of the stile upon the said perpendicular towards the North for the former application is made towards the South be set off the tangent of the reclination a line drawn from the end thereof at right angles with it shall be the horizontal line upon which the tangents of 10 20 30 c the secant of the reclination being now made Radius set from the said right angle lines drawn from them to the Zenith of the place shall be the Azimuths 5 The distance betwixt the Meridians being known upon the horizontal line the Azimuths which were accompted from the meridian of the plane may be fitted for accompt from the meridian of the place with easie For example let that distance be the tangent of 20 deg then that Azimuth which is 10 from the one is 10 from the other also and that which is 30 on the same side of the substile is 10 on the other side of the Meridian of the place the like Method serves for any distance Note 1. It may be noted that although I have shewed the construction of a South reclining plane at the beginning hereof in a figure proper only to those planes which recline not further then the Pole whereas in those that doe and although there be some variation of the Scheme as you may see by comparing this with the former at the first beginning of this subject for the point h which there fell on that side of the vertical meridian zx towards q here falls on the other side towards s likewise the hour of 12 that is bc did there fall betwixt the axis and the substile but falls here betwixt the substile and the horizontal meridian ba yet this notwithstanding the construction is the very same in both NOTE 2. It may be further Noted that as the Reclination may increase the points n c r all approach still neerer to a and when the reclination is 90 they are all coincident and this vanisheth into an horizontal dial whose substile will be ba. Also if the Declination be still increased at last
the Cone all along from b to a and make right angles with bc the diameter of the base and again another plain fd parallel to eboaz cut the semi-cone bac the section ir in the superficies of the Cone is halfe a Parabola the other halfe underneath if the Cone be supposed entire and is not to be projected in plano 2 Again if the Semi-cone bac be cut by another plain gkz parallel to the Axis ax the section in the superficies of the semicone to wit gk shall be halfe an Hyperbola and the like for the other halfe underneath if the Cone were supposed intire and further whatsoever plain cutting the Semicone as aforesaid being produced shall concur with the plaine ba produced towards z. Thirdly If the said Semi-cone be cut by a plain nph neither of the former wayes nor parallel nor subcontrary to the base the line in the superficies namely nh is a Semiellipsis Subcontrary position is that where two like triangles are joyned at an equall and then verticall angle yet have not their bases parallel Lastly if it be cut by a plain lorq parallel to the plain of the base the section or is a Semicircle Definition 1. Opposite Sections are two Hyperbola's in opposite superficies cut by the same plain Definition 2. The Vertex of a Section is a point in the greatest curvature thereof but more generally the point where any diameter cuts the Section and where the Axis cuts is called the highest Vertex Definition 3. Any two lines applyed within the Section and equidistant are called Ordinately applied in respect of some diameter which divides them into two equall parts Definition 4. Any line drawn so as it cuts the section and divides the Ordinates into two equall parts is called the Diameter of the Section and if it divide them as aforesaid and at right angles it is the Axis and so much of the Axis or diameter as lies betwixt the Vertex and any ordinate is called in respect of that ordinate the intercepted Axis or intercepted Diameter and those two diametets which mutually divide lines applied in the Section and parallel to the Diameters into two equall parts are called Conjugate diameters of which as likewise of the oppsite Sections I intend to say no more in this Tract Definition 5. The transverse Diameter of an Hyperbola is a right line in the intercepted diameter continued without the Section and is equall to the double of that line intercepted betwixt the Vertex and the center and connects the Vertices of opposite Sections In an Ellipsis or Circle it is any whole Diameter in the Hyperbola and Ellipsis if it be the continuation of the Axis or the Axis in the later it is called the transverse Axis But the Parabola whose Diameters are all equidistant hath no transverse Diameter nor any center Definition 6. The Center is a point where all the Diameters meet Definition 7. The Figures of Hyperbola's and Ellipses and Circles are paralleligrams included between the transverse Diameter and the contiguous Parameter of whicb those are called transverse sides and these Coefficients by some Definition 8. The said Parameter is a right line drawn to touch the Section at the end of the intercepted Diameter to which all the Ordinates are parallel and according to which they are compared and valued which is therefore called juxta quam possunt and if it be contiguous to the Axis it is called the right Parameter Definition 9. The umbilicius focus or burning point in the Parabola is a point in the Axis distant from the Vertex by a fourth part of the right Parameter But in the other two Sections the burning points are assigned in the Axis of either Section distant from either end of the transverse axis by the space of a right line that is the square root of the fourth part of the figure produced by the said transverse axis and the right Parameter which applied to the transverse axis is in the Hyperbola excedent in the Ellipsis deficient The same points in any Ellipsis whose diameters or diameter are given may easily be found by the mechanique way of describing an Ellipsist a little before shewed Wherein also it is plain that these points are as it were Centers proper to the generation of the Section CHAP. XII Of the description of the Sections MAny are the methods Generall and Speciall which Midorgius shews to describe these three Sections I will only mention one or two 1. To describe a Parabola about any Diameter given with one of the ordinate lines Let the Diameter given be ab and let bc be one of the ordinate lines applied unto it by which the angle abc being given joyn a and c by the right line ac And let ab be divided into as many parts as you please and through every such division draw right lines parallel to bc and produce them and make dk = √ bc in dg likewise el = √ bc in eh and fm = √ bc in fi and so of all the rest and the points c k l m a c. shall be all in the same Section so that a line drawn with an even hand by all the said points shall be by the first Prop. of the second of Midorgius the Parabola required And bc on the one side is equall to bc on the other side because by supposition that and all the parallels to it kd le mf c. ate those lines which are called Ordinates or Ordinately applyed and so ab in respect of bc also ae in respect of le c. are the intercepted Diameters or if the angle abc were a right angle the intercepted Axes Def. 4. And if you make ad′ dk″ aq′″ and draw aq parallel to dk then aq shall be the contiguous Parameter in respect of the intercepted diameter ad and so may the Parameter by ab or any other diameter given be found and therefore the Parameter aq only being given the Parabola by points may easily be described 2. About any Diameter and one Ordinate line to describe an Hyperbola known in kinde in a plain by points Let ab be a diameter of the Hyperbola and bc an Ordinate to it comprehending the angle given abc and let the Section be of such a kinde as that the transverse diameter to the contiguous parameter may be as r to s. Make ab′ bc″ bd′″ And s′ r″ bd′ be″ and joyn the points d and e and in the line ab take points how many soever and by them points f g t c. draw lines parallel to bc as fh gn ti c. the more the better and making the triangle ded compleat produce these parallels both wayes to the sides de in the points h n i c. Lastly making fk gd tl c. the square roots of the rectangles afh agn ati c. the points k o and l shall be in the Hyperbola required per 5. of 2. Midorg And therefore a line drawn with an
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
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 =