B To its opposite side M O or D S So is the Radius that is the angle at O or D To its opposite side S B or M B. This equality being admitted if unto L B we add B S the sum is L S the Altitude for Summer Declination if from B L we take B M equal to B S the remainder M L is the Altitude for the like Declination towards the Depressed Pole being the Winter Altitude of that Azimuth But for Azimuths above 90 d from the Meridian it may be observed in the two Opposite Triangles E A N and E A I counting the Azimuth E A each way from the Vertical its Equinoctial Altitude A N in the Winter is equal to its Equinoctial Depression A I in the Summer and is to be found by the 1 Proportion The second Proportion varies not As the Sine of the Latitude N F equal to AE Z Is to N I equal to Z B the Cosine of the Equinoctial Altitude or Depression So is the Sine of the Declination I G equal to D S To Sine I K from which taking A I the Equinoctial Depression rests A K the Altitude sought To Calculate a Table of the Suns Altitude for all Azimuths and hours under the Equinoctial This will be two Cases of a Quadrantal Sphoerical Triangle 1. For the Altitudes on all Azimuths There would be given the side A B a Quadrant the angle at B the Azimuth from the Meridian and the side A D the Complement of the Suns Declination If the side B D be continued to a Quadrant the angle at C will be a right angled besides which in the Triangle A D C there would be given A D as before the Complement of the Suns Declination and A C the measure of the angle at B to find D C the Suns Altitude being the Complement of B D and so having the Hipotenusal and one of the Leggs of a right angled Sphoerical Triangle by the 7th Case we may find the other Legg the Proportion sutable to this question would be As the Sine of the Azimuth from East or West Is to the Radius So is the Sine of the Declination To the Cosine of the Altitude sought 2. For the Altitudes on all Hours There would be given the side A B a Quadrant A D the Complement of the Suns declination with the contained angle B A D the hour from noon to find the side B D the Complement of the Suns Altitude Here again if B D be continued to a Quadrant the angle at C is a right angle the side A D remains common the angle D A C is the Complement of the Angle B A D See Page 57 where it is delivered That if a Sphoerical Triangle have one right angle and one side a Quadrant it hath two right angles and two Quadrantal sides and therefore the angle B A C is a right angle this is coincident with the 8th Case of right angled Spherical Triangles the Proportion thereof is As the Radius Is to the Cosine of the Declination So is the Sine of the hour from six To the Sine of the Altitude sought Affections of Sphoerical Triangles BEcause the last Affection in page 57 is not Braced in the beginning and a mistake of lesser for greater in the last Brace but one I thought fit to recite it at large Any side of a Sphoerical Triangle being continued if the other sides together are equal to a Semicircle the outward angle on the side continued shall be equal to the inward angle on the said side opposite thereto If the sides are less then a Semicircle the outward angle will be greater then the inward opposite angle But if the said sides are together greater then a Semicircle the outward angle will be less then the inward opposite angle In the Triangle annexed if the sides A B and A C together are equal to a Semicircle then is the angle A C D equal to the angle A B C. If less then a Semicircle then is the said angle greater then the angle at B. But if they be greater then is the said angle A C D less then the angle at B. By reason of the first Affection in page 58 which wants a Brace in the first Line after the words two sides We require in the first second and other Cases of Oblique angled Sphoerical Triangles the sum of the two sides or angles given to be less then a Semicircle Before I finish the Trigonometrical part I think it not amiss to give a Determination of the certain Cases about Opposite sides and Angles in Sphoerical Triangles having before shewn which are the doubtful and the rather because that this was never yet spoke to Two sides with an Angle opposite to one of them to determine the Affection of the Angle opposite to the other 1. If the given angle be Acute and the opposite side less then a Quadrant and the adjacent side less then the former side The angle it subtends is acute because subtended by a lesser side for in all Sphoerical Triangles the lesser side subtends the lesser angle and the Converse 2. If the given angle be acute and the opposite side less then a Quadrant and the other side greater then the former side This is a doubtfull Case if it be less then a Quadrant it may subtend either an Acute or an Obtuse angle and so it may also do if it be greater then a Quadrant yet we may determine That when the given angle is acute and the opposite side less then a Quadrant but greater then the Complement of the adjacent side to a Semicircle which it cannot be unless the adjacent side be greater then a Quadrant the angle opposite thereto will be obtuse 3. The given angle Acute and the opposite side greater then a Quadrant and the other side greater If two sides be greater then Quadrants if one of them subtends an Acute angle the other must subtend an Obtuse angle by the 1st Affection in pag 58. 4. The given angle Acute and the opposite side greater then a Quadrant The other side cannot be lesser then the former by what was now spoken 5. If the given angle be Obtuse and the opposite side less then Quadrant the other side less subtends an Acute angle 6. If the given angle be Obtuse and the opposite side less then a Quadrant The other side greater then the former side must of necessity be also greater then a Quadrant otherwise two sides less then Quadrants should subtend two Obtuse angles contrary to the first Affection in p 58. 7. If the given angle be Obtuse and the Opposite side greater then a Quadrant If the other side be greater then the former it will subtend a more Obtuse angle 8. If the given angle be Obtuse and the Opposite side greater then a Quadrant If the other side be less then the former whether it be lesser or greater then a Quadrant it may either subtend an Acute or Obtuse
divided but the Author made his Table in page 5. without doubling to be graduated from a Quadrant divided into 45 equal parts Again If upon the Center C with a pair of Compasses each degree of the line of Sines be transferred into the Semicircle C G A it shall divide it into 90 equal parts the reason whereof is plain because the Sine of an Arch is half the chord of twice that Arch and therefore the Sines being made to twice the Radius of this circle shall being transferred into it become chords of the like Arch to divide a Semicircle into 90 equal parts Again upon the point A erect a line of Tangents of the same Radius with the former Sine which we may suppose to be infinitely continued here we use a portion of it A E. If from the point C the other extremity of the Diameter lines be drawn cutting the lower Semicircle as a line drawn from E intersects it at F through each degree of the said Tangent the said lower Semicircle shall be divided into 90 equal parts the reason is evident a line of Tangents from the Center shall divide a Quadrant into 90 equal parts and because an Angle in the circumference is but half so much as it is in the Center being transferred thither a whole Semicircle shall be filled with no more parts The chief use of this Circle is to operate Proportions in Tangents alone or in Sines and Tangents joyntly built upon this foundation that equiangled plain Triangles have their sides Proportional In streight lines it will be evident from the point D to E draw a streight line intersecting the Diameter at L and then it lies as C L to C D so is A L to A E it is also true in a Circle provided it be evinced that the points B L F fall in a streight line Hereof I have a Geometrical Demonstration which would require more Schemes which by reason of its length and difficulty I thought fit at present not to insert possibly an easier may be found hereafter As also an Algebraick Demonstration by the Right Honourable the Lord Brunkard whereby after many Algebraick inferences it is euinced that as L K is to K B ∷ so is L N to ◠F whence it will follow that the points B L F are in a right line If a Ruler be laid from 45d of the Semicircle to every degree of the Quadrant of Latitudes it will constitute upon the Diameter the graduations of the Line Sol whereby Proportions in Sines might be operated without the other supply From the same Scheme also follows the construction of the streight line of Latitudes from the point G at 90◠of the Quadrant of Latitudes draw a streight Line to C and transfer each degree of the Quadrant of Latitudes with Compasses one foot resting upon C into the said streight line and it shall be constituted To Calculate it The Line of Latitudes C G bears such Proportion to C A as the Chord of 90d doth to the Diameter which is the same that the Sine of 45d bears to the Radius or which is all one that the Radius bears to the Secant of 45 d which Secant is equal to the Chord of 90 d from the Diagram the nature of the Line of Latitudes may be discovered Any two Lines being drawn to make a right angle if any Ark of the Line of Latitudes be pricked off in one of those Lines retaining a constant Hipotenusal A C called the Line of Hours equal to the Diameter of that Circle from whence the Line of Latitudes is constituted if the said Hipotenusal from the Point formerly pricked off be made the Hipotenusal to the Legs of the right angle formerly pricked off the said Legs or sides including the right angle shall bear such Proportion one to another as the Radius doth to the sine of the Ark so prickt off and this is evident from the Schem for such Proportion as A C bears to C D doth A B bear to B C for the angle at A is Common to both Triangles and the angle at B in the circumference is a right angle and consequently the angle A C B will be equal to the angle A D C and the Legs A C to C D bears such Proportion by construction as the Radius doth to the Sine of an Ark and the same Proportion doth A B bear to B C in all cases retaining one and the same Hypotenusal A C the Proportion therefore lies evident As the Radius the sine of the angle at B To its opposite side A C the Secant of 45d So is the sine of the angle at A To its opposite side B C sought Now the quantity of the angle at A was found by seeking the natural Sine of the Ark proposed in the Table of natural Tangents and having found what Ark answers thereto the Sine of the said Ark is to become the third Tearm in the Proportion But the Cannon prescribed in the Description of the small Quadrant is more expedite then this which Mr Sutton had from Mr Dary long since for whom and by whose directions he made a Quadrant with the Line Sol and two Parrallel Lines of Sines upon it as is here added to the backside of this Quadrant Of the Line of Hours alias the Diameter or Proportional Tangent This Scale is no other then two Lines of natural Tangents to 45 d each set together at the Center and from thence beginning and continued to each end of the Diameter and from one end thereof numbred with 90 d to the other end This Line may fitly be called a Proportional Tangent for whersoever any Ark is assumed in it to be a Tangent the remaining part of the Diameter is the Radius to the said Tangent So in the former Schem if C L be the Tangent of any Ark the Radius thereto shall be A L. In the Schem annexed let A B be the Radius of a Line of Tangents equal to C D and also parralel thereto and from the Point B to C draw the Line B C and let it be required to divide the same into a Line of Proportional Tangents I say Lines drawn from the Point D to every degree of the Tangent A B shall divide one half of it as required from the similitude of two right angled equiangled plain Triangles which will have their sides Proportional it will therefore hold As C F To C D So F B To B E and the Converse As the second Tearm C D To the fourth B E So is the first C F To the third F B and therefore C F bears such Proportion to F B as C D doth to B E which is the same that the Radius bears to the Tangent of the Ark proposed If it be doubted whether the Diameter wil be a double Tangent or the Line here described such a Line a Proportion shall be given to find by Experience or Calculation what Line
this latter Case resolve the Opposite Triangle Example In the former Triangle given ☉ P Comple Declination 66d 29′ Z P Comple Latitude 38 28 Angle ☉ Z P the Azimuth 63 54 To find the hour Z P ☉ 105 The first operation wil find the angle of Position as before 37 d 32′ The second Operation half difference of the given angles 13 d 11 m Tangent 936966 half sum of the side 52 28′ 30″ Sine 989931 1926897 half difference of the sides 14 d 00′ 30″ Sine 938393 Tangent 37 d 30′ 988504 Comple is 52 30 doubled makes 105 d the Angle sought 7. Two Angles with a side opposite to one of them being given To find the third Angle the kind of the side opposite to the other Angle being foreknown First find the side opposite to the other Angle by 4th Case And then we have two angles and their opposite sides to find the third angle by transposing the order of either of the Proportions used in the first Case the latter will be As the Cosine of halfe the difference of the sides To the Tangent of halfe the sum of the angles So the Cosine of halfe the sum of the sides To the Cotangent of half the contained angle Example In the Triangle Z ☉ P Data angle ☉ 37 d 32′ Angle P 105 00 Side ☉ Z 80 31 To find the angle Z 63 54 The first Operation will find Z P 38 28 The second Operation half sum of the angles 71 d 16′ Tangent 1046963 half sum of the sides 59 d 29′ 30″ Sine Compl 970558 2017521 half difference of the sides 21 d 1′ 30″ Cosine 997007 Tangent 58d 3′ 1020514 Compl 31d 57′ doubled is 63◠54′ the angle sought 8. Two angles with a side Opposite to one of them being given To find the Interjacent side the kind of the side opposite to the other angle being fore known First find the side opposite to the other angle by 4 Case And then you have two sides and their opposite angle given to find the 3 side by tranposing the Order of either of the Proportions used in the 2d Case the latter will be As the Cosine of halfe the difference of the two angles To the tangent of halfe the sum of the two sides So the Cosine of halfe the sum of the two given angles To the Tangent of halfe the third side Example In the former Triangle given the Hour angle at P 105d 00 Azimuth angle at Z 63 54 Compl Altitude Side Z ☉ 80 31 To find the Compl. of the Latitude the side Z P 38 28 The first Operation will find the side P ☉ 66◠29′ Second Operation half the sum of the two sides 73d 30′ Tangent 1052839 half the sum of the two angles 84 27 Cosine 898549 1951388 half the difference of the two angles 20d 33′ Cosine 997144 Tangent of 19d 14′ 954244 Doubled is 38 28 the side sought These 6 last precedent Cases may be called the Doubtful Cases because that three given terms are not sufficient Data to find one single answer without the quality of a fourth which is demonstrated by Clavius in Theodosium and seeing it passes without due caution in our English Books I shall insert it from him LEt A D and A C be two equal sides including the angle D AC and both of them less or greater then a Quadrant Draw through the Points C and D the arch of a great Circle C D continue it and draw thereunto another Arch or Side from A namely A B neither through the Poles of the Arch C D nor through the Poles of the Arch A D so that the angles B and B A D may not be right angles nor the angle A D B if then each of these sides A D A C be less then a Quadrant the two angles C and A D C will be Acute and if these Arks be greater respectively then a Quadrant the two angles C and A D C will be Obtuse whence it comes to pass that the angle A D B is Obtuse when the angle A D C is Acute and the contrary Now forasmuch as the sides A C and A D are equal to each other the other Data viz. the side A B and the angle at Bare common to both for in each Triangle A B D and A B C there is given two sides with the angle at B opposite to one of them Now this is not sufficient Data to find the angle opposite to the other side which may be either the acute angle at C or the Obtuse angle ADB the Complement thereof to a Semicircle Nor to find the third side which may be either B D or the whole side B C nor the angle included which may be either B A D or B A C therefore in these 3 Cases we have required the quality of the angle opposite to the other given side A B and though it be not so much observed in the other Trigonometry by Perpendiculars let fall without the knowledge of the said angle it could not be determined whether the Perpendicular would fall with in or without the Triangle nor whether the angle found in the first Case be the thing sought or its Complement to 180◠nor whether the angles or Segments found by 1st and 2d Operation in the other Cases are to be added together or substracted from each other to obtain the side or angle sought So also two angles with a side opposite to one of them are not sufficient Data to obtain a fourth thing in the said Triangle without the affection of the side opposite to the other given angle LEt A B and A C be two unequal sides containing the angle B A C both together equal to a Semicircle one being greater the other less then a Quadrant Draw through the Points B and C the arch of a great Circle B C continue it and draw thereto from A another side AD but not through the Poles of A C nor through the Poles of B C so that the angles D and C A D may not be right angles nor the angle A C D a right angle for if it were a right angle the angle A B C whereto it is equal should be also a right angle and so the two sides A B and A C by reason of their right angles at B and C should be equal and be Quadrants contrary to the Supposition Now the angles A C D and A B C being equal which is thus proved Suppose the two sides A B and B D to be continued to a Semicircle at E then will the said angle be equal to its opposite angle at B the side A C by supposition is equal to the side A E the Complement of the side A B to a Semicircle but equal sides subtend equal angles therefore the angle at C is equal to the angle at B or at E which being admitted retaining the side A D and angle at D we have another angle opposite thereto either C or B
angled Triangle C G F right angled at G As the sine of the Latitude the angle at F To its Opposite side C G the sine of the Declination So the Radius the angle at G To the Secant C Z F. Again in Summer As the Cosine of the Latitude the angle at ☉ To its opposite side D Z F the difference between the former Secant and the sine of the Altitude So is the sine of the Latitude the angle at F To its opposite side D ☉ the sine of the Azimuth from the Vertical in the Parralel of Altitude In Winter As Cosine Latitude angle at A To B E the sum of the former Secant equal to E M and of the sine of the Altitude M B So is the sine of the Latitude the angle B To A E equal to B D the sine of the Azimuth in a Parallel as before to be reduced to the common Radius From this Schem may be observed the reason why the Sun in those Latitudes upon some Azimuths hath two Altitudes because the Parralel of his Declination F R intersects and passeth through the Azimuth namely the prickt Ellipsis in the two points S ☉ I now proceed to the Vse in Calculating a Table of Hours For those that have occasion to Calculate a Table of Hours to any assigned Altitude and parralel of Declination it will be the readiest way to write down all the moveable Tearms first as the natural sines of the several Altitudes in a ruled sheet of Paper and then upon a peice of Card to write down the natural sine of the Suns Altitude at 6 and removing to every Altitude get the sum or difference accordingly which being had seek the same in the natural sines and write down the Log m that stands against it then upon the other end of the piece of Card get the sum of the Arithmetical Complements of the Logarithmical Cosine of the Declination and of the Logarithmica Cosine of the Latitude and add this fixed Number to the Logme before wrote down by removing the Card to every one of them and the sum is the Logme of the sine of the Hour from 6 if the Logmes be well proportioned out to the differences which may be sufficiently done by guess Example Comp Latitude 38d 28′ Ar Comp 0,2061683 Comp Delinat 66 29 Ar Comp 0,0376572 fixed Number ,2438255 Let the Altitude be 36◠42′ Nat Sine 5976251 Natural sine Altitude at 6 3124174 difference 2852077 Log against 9 4550441 Sine of 30d the hour from 6 towards noon 9,6988696 Another Example N S Altitude at 6 3124174 Let the Altitude be 13d 46′ N S 2379684 difference 744490 Logm 8,8715646 The former fixed Number 0,2438255 Sine of 7d 30′ the hour from 6 towards midnight because the Altitude is less then the Altitude of 6. 9,1153901 This method of Calculation will dispatch much faster then the common Canon when three sides are given to find an angle the Azimuth may in like manner be Calculated but will be more troublesome not having so many fixt Tearms in it and having got the hour the Azimuth will be easily found in this Case we have two sides and an angle opposite to one of them given to find the angle opposite to the other and the Proportion will hold As the Cosine of the Altitude To the sine of the hour from the Meridian So the Cosine of the Declination To the sine of the Azimuth from the Meridian And in this Case the three sides being given we may determine the affection of any of the angles If the Sun or Stars have declination towards the depressed Pole the Azimuth is always Obtuse and the hour and angle of position-Acute If the Sun c. have declination towards the Elevated Pole but less then the Latitude of the place the angle of Position is always acute the hour before 6 obtuse the hour and Azimuth between the Altitude of 6 and the Vertical Altitude both acute afterwards the hour acute and the Azimuth obtuse But when the Sun or Stars come to the Meridian between the Zenith and the Elevated Pole as when their declination is greater then the Latitude of the place the Azimuth is always acute the hour before 6 obtuse afterwards acute The angle of position from the time of rising to the remotest Azimuth from the Meridian is acute afterwards obtuse Another General Proportion for the Hour As the Radius To the Tangent of the Latitude So the Tangent of the Suns declination To the sine of the hour of rising from six Again To the Rectangle of the Cosine of the Latitude and of the Cosine of the declination Is to the Square of the Radius So is the sine of the Altitude To the difference of the Versed sines of the Semidiurnal Ark and of the hour sought Having got the Logarithm of this difference take the natural number out of the Sines or Tangents that stands against it accordingly as the Logme is sought and in Winter add it to the natural sine of the hour of rising from 6 the sum is the natural sine of the hour from 6 towards noon In Summer get the difference between this fourth and the sine of rising from 6 the said difference is the natural sine of the hour from 6 towards noon when the Number found by the Proportion is greater then the sine of rising towards midnight when less The Canon is the same without Variation as well for South declinations as for North and therefore we may by help thereof find two hours to the same Altitude Example Comp Lat 38d 28′ Ar Comp Sine 2061683 Comp declin 77d Ar Comp 0112761 fixed number 2174444 Let the Altitude be 14d 38′ Sine 9,4024889 Natural sine against it 41660 00 9,6199333 Nat sine of rising from 6 29058 79 Sum 70718 79 N sine of 45d the hour from six in Winter Difference 12601 21 Sine of 7d 14′ the hour from 6 in Summer towards noon to the former Altitude and like declination towards Elevated Pole Another Example for the same Latitude and Declination Logme Let the Altitude be 20d 25′ Sine 9,5426321 The former fixed Logme 2174444 Natural sine against it 5754811 Sum 9,7600765 Natural sine of rising 2905879 Sum 9,7600765 Sum 8660690 sine 60d the ho from 6 in Wint Difference 2848932 sine 16d 33′ the hour from 6 in Summer towards noon And thus may two hours be found at one operation for all Altitudes less then the Winter Meridian Altitude to be converted into usual Time by allowing 15d to an hour and 4◠to a degree To Calculate a Table of the Suns Altitudes on all Hours As the Secant of the Latitude To the Cosine of the Declination Or which is all one As the Square of the Radius To the Rectangle of the Cosines both of the Latitude and of the Declination So is the sine of the hour from 6. To a fourth namely in Summer the difference of the sines of the Suns
angle But we may determine That when the given angle is Obtuse and the Opposite side greater then a Quadrant If the Complement of the adjacent side to a Semicircle be greater then the said opposite side the angle subtended by the said adjacent side is Acute Two Angles with a side Opposite to one of them to determine the Affection of the side opposite to the other 1. If the given angle be Acute and the opposite side less then a Quadrant If the other angle be less the side opposite thereto shall be less then a Quadrant because it subtends a lesser angle 2. If the given angle be Acute and the Opposite side less then a Quadrant If the other angle be greater the Case is ambiguous yet we may determine If the given angle be Acute and the opposite side lesser then a Quadrant if the Complement of the other angle to a Semicircle be less then the Acute angle which it cannot be but when the latter angle is Obtuse the side subtending it shall be greater then a Quadrant 3. If the given angle be Acute and the opposite side grnater then a Quadrant if the other angle be lesser Then by 12 of 4 book of Regiomontanus if two Acute angles be unequal the side opposite to the lesser of them shall be less then a Quadrant 4. If the given angle be Acute and the opposite side greater then a Quadrant If the other angle be greater It must of necessity be Obtuse because otherways two Acute unequal angles the side opposite to the lesser of them should not be lesser then a Quadrant contraty the former place of Regiomontanus 5. If the given angle be Obtuse and the opposite side lesser then a Quadrant If the other angle be lesser It must of necessity be Acute and the side subtending it less then a Quadrant otherways two sides less then Quadrants should subtend two Obtuse angles contrary to 1st Affection in page 58. 6. If the given angle be Obtuse and the opposite side less then a Quadrant If the other angle be greater By 13 Prop of 4 h of Regiomontanus if a Triangle have two Obtuse unequal angles the side opposite to the greater of them shall be greater then a Quadrant 7. If the given angle be Obtuse and the opposite side greater then a Quadrant If the other angle be greater or more obtuse then the former it is subtended by a greater side 8. If the given angle be Obtuse and the opposite side greater then a Quadrant If the other angle be less then the former the Case is ambiguous yet we may determine That when the given angle is Obtuse and the opposite side greater then a Quadrant if the other angle be less then the Complement of the said Obtuse angle to a Semicircle the side subtending it shall be less then a Quadrant The former Cases that are still and alwais will be doubtful may be determined when three sides are given A Sphoerical Triangle having two sides less then Quadrants and one greater will always have one Obtuse angle opposite to that greater side and both the other angles Acute A Sphoerical Triangle having three sides less then Quadrants can have but one obtuse angle and many times none and that obtuse angle shall be subtended by the greatest side But whether the greatest side subtend an Acute or Obtuse angle cannot be known unless given or found by Calculation and that may be found several ways First by help of the Leggs or Sides including the angle sought by 15â— Case of right angled Sphoerical Triangles As the Radius To the Cosine of one of those Leggs So is the Cosine of the other Legg To the Sine of a fourth Arch. If the third side be greater then the Complement of the fourth Arch to 90d the angle included is Obtuse if equal to it a right angle if less an Acute angle Secondly By help of one of the Leggs and the Base or Side subtending the angle sought by 7th Case of right angled Sphoerical Triangles As the Cosine of the adjacent side being one of the lesser sides Is to the Radius So the Cosine of the opposite side To the sine of a fourth Arch. If the third side be greater then the Complement of the 4th Arch to 90d the angle subtended is Acute if equal to it a right angle if less an Obtuse angle All other Cases need no determination if a Triangle have two sides given bigger then Quadrants make recourse to the opposite Triangle and it will agree to these Cases If three angles were given to determine the Affection of the sides if they were all Acute the three sides subtending them will be all less then Quadrants But observe that though a Triangle that hath but one side greater then a Quadrant can and shall always have but one Obtuse angle yet a Triangle that hath but one Obtuse angle may frequently have two sides greater then Quadrants In this and all other Cases let the angles be changed into sides and the former Rules will serue I should have added a brief Application of all the Axioms that are necessary to be remembred and have reduced the Oblique Cases to setled Proportions with the Cadence of Perpendiculars only to shew how they arise whereby they will be rendred very facil as also the Demonstration of the Affections which may be hereafter added to some other Treatise to be bound with this Book Of working Proportions by the Lines on the Quadrant BEfore I come to shew how all Proportions may in some measure be performed upon the Lines of this Quadrant it is to be intimated in general That the working of a Proportion upon a single natural Line was the useful invention of the late learned Mathematician Mr Samuel Foster and published after his decease as his In the use of his Scale a Book called Posthuma Fosteri as also by Mr Stirrup in a Treatise of Dyalling in which Books though it be there prescribed and from thence may be learned yet I acknowledge I received some light concerning it from some Manuscripts lent me by Mr Foster in his life time to Transcribe for his and my own use touching Instrumental Applications Yet withal be it here intimated that there are no ways used upon this Quadrant for the obtaining the Hour and Azimuth with Compasses and the Converse of the 4th Axiom but what are wholly my own and altogether novel though not worth the owning for Instrumental Conclusions not being so exact as the Tables are of small esteem with the learned as in Mr Wingates Preface to the Posthuma besides the taking off of any Line from the Limb to any Radius the Explanation of the reason of Proportions so wrought the supply of many Defects and the inscribing of Lines in the Limb I have not seen any thing of Mr Fosters or of any other mans tending thereto Of the Line of equal Parts THis Line issueth from the Center of the Quadrant on the right edge of the foreside and will
sine and enter the former extent between the Scale and the Thread and the foot of the Compasses will on the Line of equal parts shew the fourth Proportional The Proportion for finding the Altitude of a Tower at one Station by the measured distance may also be wrought in in equal parts and Sines For As the Cosine of the Ark at first Station To the measured distance thereof from the Tower So is the Sine of the said Ark To the Altitude of the Tower In that former Scheme the measured distance B H is 85 and the angle observed at H 48 d 29′ Wherefore I lay the Thread to the Sine of the said Ark in the Limb counted from the right edge and from the measured distance in the equal parts take the nearest extent to the Thread then laying the Thread to the Cosine of the said Ark in the Limb and entring the former extent between the Thread and the Scale I shall find the foot of the Compasses to fall upon 96 the Altitude sought So also in the Triangle A C B if there were given the side A C 194 the measured distance between two Stations on the Wall of a Town besieged and the observed angles at A 25 d 22′ at C 113 d 22′ if B were a Battery we might by this work find the distance of it from either A or C for having two angles given all the three are given it therefore holds As the Sine of the angle ot B 41d 16′ To its opposite side A C 194 So the Sine of the angle at C 66d 38′ the Complement To its Opposite side B A 270 the distance of the Battery from A Such Proportions as have the Radius in them will be more easily wrought we shall give some few Examples in Use in Navigation 1. To find how many Miles or Leagues in each Parralel of Latitude answer to one degree of Longitude As the Radius To the Cosine of the Latitude So the number of Miles in a degree in the Equinoctial To the Number of Miles in the Parralel So in 51 d 32′ of Latitude if 60 Miles answer to a degree in the Equinoctial 37 ‑ 3 Miles shall answer to one degree in this Parralel This is wrought by laying the Thread to 51 d 32′ in the Limb from the left edge towards the right then take the nearest distance to it from 60 in the equal parts which measured from the Center will be found to reach to 37 ‑ 3 as before The reason of this facil Operation is because the nearest distance from the end of the Line of equal parts to the Thread is equal to the Cosine of the Latitude the Scale it self being equal to the Radius and therefore needs not be taken out of a Scale of Sines and entred upon the first Tearm the Radius as in other Proportions in Sines of of the greater to the less when wrought upon a single Line only issuing from the Center where the second Tearm must be taken out of a Scale and entred upon the first Tearm 2. The Course and Distance given to find the difference of Latitude in Leagues or Miles As the Radius To the Cosine of the Rumb from the Meridian So the Distance sailed To the difference of Latitude in like parts Example A Ship sailed S W by W that is on a Rumb 56 d 15′ from the Meridian 60 Miles the difference of Latitude in Miles will be found to be 33 ‑ 3 the Operation being all one with the former Lay the Thread to the Rumb in the Limb and from 60 take the nearest distance to it which measured in the Scale of equal parts will be found as before 3. The Course and Distance given to find the Departure from the Meridian alias the Variation As the Radius To the Sine of the Rumb from the Meridian So the distance Sailed To the Departure from the Meridian In the former Example to find the Departure from the Meridian Lay the Thread to the Rumb counted from the right edge towards the left that is to 56d 15′ so counted and from 60 in the equal parts being the Miles Sailed take the nearest distance to it this extent measured in the said Scale will be found to be 49 ‑ 9 Miles and so if the converse of this were to be wrought it is evident that the Miles of Departure must be taken out of the Scale of equal parts and entred Parralelly between the Scale and the Thread lying over the Rumb Many more Examples and Propositions might be illustrated but these are sufficient those that would use a Quadrant for this purpose may have the Rumbs traced out or prickt upon the Limb Now we repair to the backside of the Quadrant Of the Line of on the right Edge of the Backside THe Uses of this Line are manifold in Dyalling in drawing Projections in working Proportions c. 1. To take of a Proportional Sine to any lesser Radius then the side of the Quadrant or which is all one to divide any Line shorter in length then the whole Line of Sines in such manner as the same is divided Enter the length of the Line proposed at 90 d the end of the Scale of Sines and to the other foot lay the Thread according to nearest Distance or measure the length of the Line proposed on the Line of Sines from the Center and observe to what Sine it is equal then lay the Thread over the like Arch in the Limb and the nearest distances to it from each degree of the Line of Sines shall be the Proportional parts sought And if the Thread be laid over 30 d of the Limbe the nearest distances to it will be Sines to half the Common Radius 2. From a Line of Sines to take off a Tangent the Proportion to do it is As the Cosine of an Arch To the Radius of the Line proposed So the Sine of the said Arch To the Tangent of the said Arch. Enter the Radius of the Tangent proposed at the Cosine of the given Arch and to the other foot lay the Thread then from the Sine of that Arch take the nearest distance to the Thread this extent is the length of the Tangent sought thus to get the Tangent of 20 d enter the Radius proposed at the Sine of 70 d then take the nearest distance to the Thread from the Sine of 20 d this extent is the Tangent of the said Arch in reference to the limited Radius Otherways by the Limb. Lay the Thread to the Sine of that Arch counted from the right edge whereto you would take out a Tangent and enter the Radius proposed down the Line of Sines from the Center and take the nearest distance to the Thread then lay the Thread to the like Arch from the left edge and enter the extent between the Scale and the Thread the distance of the Foot of the Compasses from the Center shall be the length of the
Altitude be well given These Scales in their Use presuppose the Hour and Azimuth of the Sun to be nearer the noon Meridian then 60d. Operation to find the Hour Take the distance between the Altitude and the Declination proper to the season of the year out of the Hour Scale and enter one foot of this Extent at the Cosine of the Declination in the Line of sines and laying the Thread to the other foot according to nearest distance it shews the hour from noon in the Versed sines Quadrupled Example When the Sun hath 23d 31′ of North Declination and 60d of Altitude the hour from noon will be 13d 58′ to be Converted into time When the hour is found to be less then 40d from Noon the former extent may be doubled and entred as before and it shews the hour in the Versed sines Octupled And when the hour is less then 30d from Noon the former extent may be tripled and entred as before and after this manner it is possible to make the whole Limb give the hour next Noon the Versed Sine Duodecupled lies on the other side of the Quadrant and in this case an Ark must first be found in the Limb and the Thread laid over the said Ark counted from the other edge will intersect the said Versed Sine at the Ark sought To find the Suns Azimuth TAke the distance in the Azimuth Scale between the Altitude and the Declination proper to the season of the year and entring it at the Cosine of the Altitude laying the Thread to the other foot according to nearest distance it will shew the Azimuth in the Versed Sines quadrupled or when the Azimuth is near Noon according to the former restrictions for the hour the extent may be doubled or tripled and the answer found in the Versed Sines Octupled or Duodecupled as was done for the hour Example So when the Sun hath 23d 31′ of North Declination his Altitude being 60d. The Azimuth will be found to be 26d 21′ from the South By the like-like-reason when we found the Hour and Azimuth in the equal Limb by the Diagonal Scale if those extents had been doubled the Hour and Azimuth near six or the Vertical might have been found in a line of Sines of 30d put thorow the whole Limb but that we thought needless FINIS THE DESCRPITION AND VSES Of a Great Universal Quadrant With a Quarter of Stofters particular Projection upon it Inverted Contrived and Written by John Collins Accomptant and Student in the MATHEMATIqUES LONDON Printed in the Year 1658. The DESCRIPTION Of the Great Quadrant IT hath been hinted before that though the former contrivance may serve for a small Quadrant yet there might be a better for a great one The Description of the Fore-side On the right edge from the Center is placed a line of Sines On the left edge from the Center a line of Versed Sines to 180d. The Limb the same as in the small Quadrant Between the Limb and the Center are placed in Circles a Line of Versed Sines to 180d another through the whole Limb to 90d. The Line of lesser Sines and Secants The line of Tangents The Quadrat and Shaddows Above them the Projection with the Declinations Days of the moneth and Almanack On the left edge is placed the fitted Hour and Azimuth Scale Within the Projection abutting against the Sines is placed a little Scale called The Scale of Entrance being graduated to 62d and is no other but a small line of Sines numbred by the Complements At the end of the Secant is put on the Versed Sines doubled that is to twice the Radius of the Quadrant and at the end of the Tangents tripled to some few degrees to give the Hour and Azimuth near Noon more exactly The Description of the Back-side On the right edge from the Center is placed a Line of equal part being 10 inches precise decimally subdivided On the out-side next the edge is placed a large Chord to 60â— equal in length to the Radius of the Line of Sines On the left edge is placed a Line of Tangents issuing from the Center continued to 63d 26′ and again continued apart from 60d to 75d The equal Limb. Within it a Quadrant of Ascensions divided into 24 equal hours and its parts with Stars affixed and Letters graved to refer to their Names Between it and the Center is placed a Circle whereof there is but three Quadrants graduated The Diameter of this Circle is no other then the Dyalling Scale of 6 hours or double Tangents divided into 90d. Two Quadrants or the half of this Circle beneath the Diameter is divided into 90 equal parts or degrees The upper divided Quadrant is called the Quadrant of Latitudes From the extremity of the said Quadrant and Perpendicular to the Diameter is graduated a Line of Proportional Sines M Foster call it the Line Sol. Diagonal-wise from one extremity of the Quadrant of Latitude to the other is graduated a line of Sines that end numbred with â—0 d that is next the Diameter being of the same Radius with the Tangents Opposite and parallel thereto from 45d of the Semicircle to the other extremity of the Diameter is placed a Line of Sines equal to the former Diagonal-wise from the beginning of the Line Sol to the end of the Diameter is graduated a Line of 60 Chords From the beginning of the Diameter but below it towards 45d of the Semicircle is graduated the Projection Tangent alias a Semi-tangent to 90d being of the same Radius with the Tangents The other Quadrant of this Circle being only a void Line there passeth through it from the Center a Tangent of 45d for Dyalling divided into 3 hours with its quarters and minutes Below the Diameter is void space left to graduate any Table at pleasure and a Line of Chords may be there placed Most of these Lines and the Projection have been already treated upon in the use of the small Quadrant those that are added shall here be spoke to Of the Line of Versed Sines on the left Edge issuing from the Center THis Line and the uses of it were invented by the learned Mathematician M. Samuel Foster of Gresham Colledge deceased from whom I received the uses of it applyed to a Sector I shall and have added the Proportions to be wrought upon it and in that and other respects diversifie from what I received wherein I shall not be tedious because there are other ways to follow since found out by my self The chief uses of it are to resolve the two cases of the fourth Axiom of Spherical Trigonometry as when three sides are given to find an Angle or two sides with the Angle comprehended to find the third side which are the cases that find the Hour and Azimuth generally and the Suns Altitudes on all hours For the Hour the learned Author thought meet to add a Zodiaque of the Suns-place annexed to it both in the use of his Sector as also
doubtful whether the angle opposed to the greater side be Acute or Obtuse yet a true Sine of the 4 â—h Proportional The Sine found will give the angle opposite to the other given side if it be Acute and it will always be Acute when the given angle is Obtuse But if it be fore known to be Obtuse the Arch of the Sine found substract from a Semicircle and there will remain the Angle sought In this Case the quality or affection of the angle sought must be given and fore-known for otherwise it is impossible to give any other then a double answer the Acute angle found or its Complement to 180d yet some in this Case have prescribed Rules to know it supposing the third Side given which is not if it were then in any Plain Triangle it would hold That if the Square of any Side be equal to the Sum of the Squares of the other two Sides the angle it subtends is a right angle if less Acute if greater Obtuse but if three Sides are given we may with as little trouble by following Proportions come by the quantity of any angle as by this Rule to know the affection of it Two Sides with an Angle opposite to one of them to find the third Side In this Case as in the former the affection of the angle opposite to the other given side must be fore-known or else the answer may be double or doubtful In the Triangle annexed there is given the sides A B and A C with the angle A B C to find the angle opposite to the other given side which may be either the angle at C or D and the third side which may be either B C or B D the reason hereof is because two of the given tearms the Side A B and the angle at B remain the same in both Triangles and the other given side may be either A C or A D equal to it the one falling as much without as the other doth within the Perpendicular A E. The quality of the angle opposite to the other given side being known by the former Case get the quantity and then having two angles the Complement of their Sum to 180d is equal to the third angle the Case will be to find the said side by having choice of the other sides and their opposite angles as follows 2. Two Angles with a Side opposite to one of them given to find the Side opposite to the other As the Sine of the angle opposed to the given Side Is to the given Side So is the Sine of the angle opposite to the side sought To the side sought Ricciolus in the Trigonomerical part of his late Almagestum Novum suggests in the resolution of this Case that if the side opposite to the Obtuse angle be sought it cannot be found under three Operations as first to get the quantity of the Perpendicular falling from the Obtuse angle on its Opposite side and then the quantity of the two Segments of the side on which it falleth and so add them together to obtain the side sought But this is a mistake and if it were true by the like reason a side opposite to an Acute angle could not be found without the like trouble for in the Triangle above As the Sine of the angle at B Is to its Opposite side A C or A D So is the Sine of the angle B C A or B D A To its opposite Side B A where the Reader may perceive that the same side hath opposite to it both an Acute and an Obtuse angle the one the Complement of the other to 180d the same Sine being common to both for the Acute angle A C E is the Complement of the Obtuse angle B C A but the angle at C is equal to the angle at D being subtended by equal Sides and the Proportion of the Sines of Angles to their opposite Sides is already demonstrated in every Book of Trigonometry 3. Two Sides with the angle comprehended to find either of the other Angles Substract the angle given from 180d and there remains the Sum of the two other Angles then As the Sum of the Sides given To tangent of the half sum of the unknown Angles So is the difference of the said Sides To Tangent of half the difference of the unknown Angles If this half difference be added to half sum of the angles it makes the greater if substracted from it it leaves the lesser Angle 4. Two Sides with the Angle comprehended to find the third side By the former Proposition one of the Angles must be found and then As the Sine of the angle found is to its Opposite side So is the Sine of the angle given To the Side opposed thereto If two angles with a Side opposite to one of them be given to find the side opposite to the third Angle is no different Case from the former because the third angle is by consequence given being the Complement of the two given angles to 180d and the like if two angles with the side between them were given 5. Three Sides to find an Angle The Side subtending the Angle sought is called the Base As the Rectangle or Product of the half Sum of the three sides and of the difference of the Base therefrom Is to the Square of the Radius So is the Rectangle of the difference of the Leggs or conteining Sides therefrom to the Square of the Tangent to half the Angle sought And by changing the third tearm into the place of the first As the Rectangle of the Differences of the Leggs from the half Sum of the 3 sides Is to the Square of the Radius So is the Rectangle of the said half sum and of the difference of the Base there from To the Square of the Tangent of an angle which doubled is the Complement of the angle sought to 180d Or the Complement of this angle doubled is the angle sought To Operate the former Proportion by the Tables From the half Sum of the three Sides substract each Side severally then from the Sum of the Logarithmes of the Square of the Radius which in Logarithms is the Radius doubled and of the differences of the Sides containing the angle sought Substract the Sum of the Logarithms of the half sum of the three sides and of the difference of the Base therefrom the half of the remainder is the Logarithm of the Tangent of half the angle sought Example In the Triangle A B C let the three sides be given to find the Obtuse angle at C. Differences from the half sum B C 126 Leggs 101 2,0043214 A C 194 Leggs 169 2,2278867 A B 270 Base 25 Logarithms  1,3979400 Sum with double Radius 24,2322081 Sum 590   half sum 295 2,4698220    3,8677620   3,8677620   20,3644461 Tangent of 56d 41′ which doubled is 113d 22′ the angle sought 10,1822230 half For the 2d Proportion for finding an Angle the
equidistant one from another but having determined the distance between the two Extream Latitudes to which they are fitted for the the larger sine it will hold As the difference of the Secants of the two extream Latitudes It to the distance between the Lines fitted thereto So is the difference of the Secants of the lesser extream Latitude and any other intermediate Latitude To the distance thereof from the lesser extream And so for the lesser sine continued the other way having placed the two Extreams under the two former Extreams to place the imtermediate Lines the Canon would be As the difference of the sines of the two extream Latitudes Is to the distance between the Lines fitted thereto So is the difference of the sines of the lesser extream Latitude and of any other intermediate Latitude To the distance thereof from the lesser Extream Having fitted the distances of the greater sine streight Lines drawn through the two extream sines shall divide the intermediate Parralels also into Lines of sines proper to the Latitudes to which they are fitted Now for the lesser sines they are continued the other way at the ends of the former Parralells the Line proper to each Latitude should be divided into a Line of sines whose Radius should be equal to the sine of the Latitude of the other sine whereto it is fitted and so Lines traced through each degree to the Extreams but by reason of the small distance of these Lines the difference is so exceeding small that it may not be scrupled to draw Lines Diagonal wise from each degree of the two outward extream Sines for being drawn true they will not be perceived to be any other then streight Lines Whereas these Lines by reason of the latter Proportion should not fall absolutely to be drawn at the ends of the former Lines whereto they are fitted and then they would not be so fit for the purpose yet the difference being as we said so insensible that it cannot be scaled they are notwithstanding there placed and crossed with Diagonals drawn through each degree of the Extreams The Vses of the Diagonal Scale 1. To find the time of Sun rising or setting In the Parralel proper to the Latitude take out the Suns Declination out of the lesser continued sines and enter one foot of this extent at the Complement of the Declination in the Line of sines and in the equal Limb the Thread being laid to the other foot will shew the time sought In the Latitude of York namely 54d if the Sun have 20d of Declination Northward he rises at 4 and sets at 8 Southward he rises at 8 and sets at 4 2. To find the Hour of the Day or Night for South Declination In the Parralel proper to the Latitude account the Declination in the lesser continued sine and the Altitude in the greater sine and take their distance which extent apply as before to the Cosine of the Declination in the Line of sines on the Quadrant and laying the Thread to the other foot according to nearest distance it shews the time sought in the equal Limbe Thus in the Latitude of York when the Sun hath 20d of South declination his Altitude being 5d the hour from noon will be found 45 minutes past 8 in the morning or 15 minutes past 3 in the afternoon feré For North Declination The Declination must be taken out of the lesser sine in the proper Parralel and turned upward on the greater sine and there it shews the Altitude at six for the Sun or any Stars in the Northern Hemispere the distance between which Point and the given Altitude must be entred as before at the Cosine of the declination laying the thread to the other foot and it shews the hour in the Limb from six towards noon or midnight according as the Sun or Stars Altitude was greater or lesser then its Altitude at six So in the Latitude of York when the Sun hath 20d of North declination if his Altitude be 40d the hour will be 46 minutes past 8 in the morning or 14 minutes past 3 in the afternoon 4. The Converse of the former Proposition will be to find the Altitude of the Sun at any hour of the day or of any Star at any hour of the night I need not insist on this having shewn the manner of it on the small quadrant only for these Scales use the Limb instead of the lesser sines for Stars the time of the night must first be turned into the Stars hour and then the Work the same as for the Sun 5. To find the Amplitude of ehe Sun or Stars Take out the Declination out of the greater sine in the Parralel proper to the Latitude and measure it on the Line of sines on the lesser Quadrant and it shews the Amplitude sought So in the Latitude of York 54d when the Sun hath 20d of Declination his Amplitude will be 35d 35′ 6. To find the Azimuth for the Sun or any Stars in the Hemisphere For South Declination Account the Altitude in the lesser sine continued in the proper Parralel and the Declination in the greater sine and take their distance enter one foot of this extent at the Cosine of the Altitude on the Quadrant and lay the Thread to the other according to nearest distance and in the Limbe it shews the Azimuth from East or West Southwards So in the Latitude of York when the Sun hath 20d of South Declination his Altitude being 5d the Azimuth will be found to be 44d 47′ to the Southwards of the East or West For North Declination Account the Altitude in the lesser sine continued and apply it upward on the greater sine and it finds a Point thereon from whence take the distance to the declination in the said greater sine in the Parralel proper to the Latitude of the place and enter one foot of this Extent at the Cosine of the Altitude on the Line of sines and the Thread being laid to the other foot according to nearest distance shews the Azimuth in the Limbe from East or West So in the Latitude of York when the Sun hath 20d of North Declination and 40d of Altitude his Azimuth will be 23d 16′ to the Southwards of the East or West When the Hour or Azimuth falls near Noon for more certainty you may lay the Thread to the Complement of the Declination for the Hour or the Complement of the Altitude for the Azimuth in the Limbe and enter the respective extents Parralelly between the Thread and the Sines and find the answer in the sines We might have fitted one Scale on the quadrant to give both the houre and Azimuth in the Equall Limb by a Lateral entrance and have enlarged upon many more Propositions which shall be handled in the great Quadrants Mr Sutton was willing to add a Backside to this Scale and therefore hath put on particular Scales of his own for giving the requisites of an upright Decliner
this whole extent being entred at 12 deg 13′ in the Scale of entrance lay the thread to the other foot according to nearest distance and it will intersect the equal Limbe at 40 deg and so much is the Suns Azimuth from the East or West Because the Scale of entrance could not be continued by reason of the Projection the residue of it is put on an little Line neare the Amanack the use whereof is to lay the thread to the Altitude in it when the Azimuth is sought and in the Limbe it shewes at what Arke of the Sines the point of entrance will happen which may likewise be found by pricking downe the Co-altitude on the line of Sines out of the fitted houre Scale on the right edge How to find the houre and Azimuth generally in the equal limb either with or without Tangents or Secants hath been also shewed and how that those two points for any Latitude might be there prickt and might be taken off either from the Limbe or from a line of Sines or best of all by Tables for halfe the natural Tangent of the Latitude of London is equal to the sine of 〈◊〉 39 deg And half the Secant thereof equal to the sine of 〈◊〉 53d 30 Against which Arkes of the Limbe the Tangent and Secant of the Latitude are graduated but of this enough hath been said in the Description of the small quadrant Of the Quadrat and Shadowes THe use thereof is the same as in the small quadrant onely if the thread hang over any degree of the Limb lesse then 45d to take out the Tangent thereof out of the quadrat count the Arch from the right edge of the quadrant towards the left and lay the thread over it the pricks are repeated in the Limbe to save this trouble for those eminent parts Of the equal Limbe WE have before shewed that a Sine Tangent and Secant may be taken off from it and that having a Sine or Secant with the Radius thereof the correspondent Arke thereto might be found that a Chord might be taken off from Concentrick Circles or by helpe of a Bead but if both be wanting enter the Semidiameter or Radius whereto you would take out a Chord twice downe the right edge from the Center and laying the thread over halfe the and laying the thread over halfe the Arch proposed take the nearest distance to it and thus may a chord be taken out to any number of degrees lesse then a Semicircle It hath been asserted also that the houre and Azimuth might be found generally without Protraction by the sole helpe of the Limb with Compasses and a thread Example for finding the houre THe first work will be to find the point of entrance take out the Cosine of the Latitude by taking the nearest distance to the thread laid over the said Arke from the concurrence of the Limbe with the right edge and enter it down the right edge line and take the nearest distance to the thread laid over the complement of the Declination counted from the right edge this extent entred down the right edge finds the point of entrance let it be noted with a mark Next to find the sine point take out the sine of the Declin enter it dowh the right edge from the point of termination take the nearest distance to the thread laid over the ark of the Latit counted from the right edge this extent enter from the Center and it finds the sine point let it be noted with a marke Thirdly take out the sine of the Altitude in Winter add it in lenght to the sine point in Summer enter it from the Center take the distance between it the sine point which extent entred upon the point of entrance if the thread be laid to the other foot shewes the the houre from 6 in the equal limb before or after it as the Sine of the Altitude fell short or beyond the sine point Example In the latitude of 39 d. the Sun having 23d 31′ of North Declination and Altitude 51 deg 32′ the houre will be found to be 33 deg 45′ from six towards noon Note the point of entrance and sine point Vary not till the Declination Vary After the same manner may the Azimuth be found in the limb by proportions delivered in the other great quadrant Also both or any angle when three sides are given may be found by the last general Proportion in the small quadrant which finds the halfe Versed sine of the Arke fought which would be too tedious to insist upon are more proper to be Protracted with a line of Chords To find the Azimuth universally THe Proportion used on the smal quadrant for finding it in the equal limbe wherein the first Operation for the Vertical Altitude was fixed for one day by reason of its Excursions will not serve on a quadrant for the Sun or Stars when they come to the Meridian between the Zenith and the elevated Pole but the Proportion there used for finding the houre applyed to other sides will serve for the Azimuth Universally and that is As the Radius Is to the sine of the Latitude So is the sine of the Altitude To a fourth sine Again As the Cosine of the Altitude Is the Secant of the Latitude Or As the Cosine of the Latitude Is the Secant of the Altitude So In Declinations towards the Elevated Pole is the difference but towards the Depressed Pole the summe of the fourth sine and of the sine sine of the Declination To the sine of the Azimuth from the Vertical In Declinations towards the Depressed Pole the Azimuth is alwayes obtuse towards the elevated Pole if the Declination be more then the fourth Arch it is acute if lesse obtuse Example for the Latitude of the Barbados 13 deg Altitude 27 deg 27′ Declination 20 deg North. Lay the thread to 27 deg 27′ in the Limbe and from the sine of 13 deg tahe nearest distance to it which enter on the line of Sines from the Center and take the distance between the limited point and the sine of 20 deg the Declination this latter extent enter twice downe the line of the Sines from the Center and take the nearest distance to the thread laid over the Secant of 27 deg 27′ this extent enter at the sine of 77 deg the Complement of the Latitude and laying the thread to the other foot it will lye over 16 deg in the equal Limbe the Suns Azimuth to the Northwards of the East or West Otherwaies Another Example for the same Latitude and Declination the Altitude being 52 deg 27′ lay the thread to it in the Limbe and take the nearest distance to it from the sine of 13 deg as before and enter it downe the line of sines from the Center and from the point of the limitation take the distance to the sine of 20 deg the Suns Declination this
particular sine on the foreside by the intersection of the thread and for this Example will be 42 deg 53′ 4 The Angle of 12 and 6. Account the Complement of the Reinclination in the peculiar hour Scale as a sine and just against it in the annexed Tangent stands the Complement of the Angle sought in this Example the Angle of 12 and 6 is 68 deg 20′ In other Latitudes the Operations must be performed by Proportional worke with the Compasses Of the Lines derived from Mr. Gunters Sector Such are the Lines of superficies Solids c. Of the Line of Superficies or Squares THe chiefe uses of this Line joyntly with the Line of Lines in the Limbe is when a square number is given to find the Root thereof or a Root given to find the square number thereto these Lines placed on a quadrant will perform this some what better then a Sector because it is given by the Intersection of the thread without Compasses the properties of the quadrant casting these lines large where on a Sector they would be narrow To find the square Root of a number The Root being given to find the Square Number of that Root IN extracting the square Root pricks must be set under the first third fift and seventh figure and so forward and as many pricks as fall to be under the square number given so many figures shall be in the Root and accordingly the line of lines and superficies must vary in the number they represent I am very unwilling to spend any time about these kind of Lines as being of small performance and by my self and almost by all men accounted meere toyes If a number be given in the superficies the thread in the lines sheweth the Root of it and the contrary if a number be given in the lines the thread laid over it intersects the Square thereof The performance thereof by these lines is so deficient that I shall give no Example of it When a number is given to find the square thereof if not to large the Reader may correct the last figure of it by multiplying it in his memory To three numbers given to find a fourth in a Duplicated Proportion That is to worke a Proportion between Numbers and Squares Example If the Diameter of a Circle whose Area is 154 be 14 what shall the Diameter of that Circle be whose Area is 616. Example Lay the thread over 616 in the superficies and from 14 in the equal parts take the nearest distance to it then lay the thread to 154 in the superficies and enter the former extent between the thread and the Scale and the foot of the Compasses will rest upon 28 the diameter sought To find a Proportion between two or more like superficies ADmit there be two Circles and I would know what Proportion their Areas bear to each other in this case the proper use of a Line of superficies would be to have it on a ruler and to measure the lengths of their like sides for Circles the lengths of their Diameters upon it and then I say the numbers found on the superficies beare such Proportion each to other as the Areas or superficial contents and for small quantities may be done on the quadrant by entring downe the larger extent of the Compasses on the Line of Lines from the Center and mind the point of limitation enter then the other extent on the point of limitation and lay the thread to the other foot find what number it cuts in the superficies and the greater shall beare such Proportion to the lesser as 100 c. the length of the whole line doth to the parts cut The Proportion that two superficies beare each to other is the same that the squares of their like sides and therefore their sides may be measured either in foot or inch measure and then the Squares taken out as before shewed The line of superficies serves for the reducing of Plots to any proportion ADmit a Plot of a piece of ground being cast up containes 364 Acres and it were required to draw another Plot which being cast up by the same Scale should containe but a quarter so much and let one side of the said Plot be 60 inches against 60 in the lines the square of it will be found to be 3600 and the fourth part hereof would be 900 which account in the superficies and you will find the Square Root of it to be 30 and so many inches must be the like side of the lesser Plot if being cast up by the same Scale it should containe but ¼ of what it did before If the line of Superficies were on a streight ruler then to perform such a Proposition as this would be to measure therewith the side of the Plot given minding what number it reaches to in the Superficies the fourth part of the said Number being reckoned on the Superficies and thence taken shall be the length of the side in the Proportion required Of the Line of Solids IF a number be duly estimated in the said line and the thread laid over it it will in the line of lines shew the cube Root of that number and the converse the Root being assigned the Cube may be found but by reason of the sorry performance of these Lines I shall spend no time about it if this line be placed on a loose Ruler and the like sides of two like Solids be measured therewith those Solids shall beare such Proportion in their contents each to other as the measured lengths on the Solids Three Numbers being given to find the fourth in a Duplicated Proportion Example IF a Bullet of 4 inches Diameter weigh 9 pound what shall a Bullet of 8 inches Diameter weigh Answer 72 pounds In this case let the whole line of Solids represent 100 alwayes the Solid content whether given or sought must be accounted in the line of Solids and the Sides or Diameters in the Equall parts Lay the thread to 9 in the line of Solids and from 8 in the inches take the nearest distance to it enter one foot of that extent at 4 in the inches and lay the thread to the other foot and it will lye over 72 in the Solids for the weight of the Bullet sought An Example of the Converse If a Bullet whose Diameter is 4 Inches weigh 9 pound another Bullet whose weight is 40 pound what shall be the Diameter of it Lay the thread to 40 in the Solids and from 4 Inches in the lines take the nearest distance to it Then lay the thread to 9 in the Solids and enter the said extent at the equal Scale so that the other foot turned about may but just touch the thread and it it will rest at 6½ Inches nearest which is the Diameter sought Of the Line of inscribed Bodies This Line hath these letters set to it D Signifying the Sides of a Dodecahedron S Signifying the Sides of a _____ I Signifying the Sides of a Icosahedron
the Boul at the same time the shade in the Boul is the hour line And if the Boul be full of water or any other liquor you may draw the hour-lines which will never shew the true hour unlesse filled with the said Liquor again Reflected Dialling To draw a Reflected Dial on any Plain or Plains be they never so Gibous and Concave or Convex or any irregularity whatsoever the Glass being fixed at any Reclination at pleasure provided it may cast its Reflex upon the places proposed Together with all other necessary lines or furniture thereon viz. the Parallels of Declination the Azimuth lines the Parallels of Altitude or proportions of shadows the Planetary Hour-lines and the Cuspis of those Houses which are above the Horizon c. 1. If the Glasse be placed Horizontal upon the Transome of a window or other convenient place How upon the Wall or Cieling whereon that Glasse doth reflect to draw the Hour-lines thereon although it be never so irregular or in any form whatsoever CONSTRUCTIO FIrst draw on Pastboard or other Material an Horizontal Dial for the Latitude proposed Then by help of the Azimuth or at the time when the Sun is in the Meridian or by knowing the true hour of Day whereby may be drawn several lines on the Cieling Floor and Walls of the Room so as in respect of the center of the Glasse they may be in the true Meridian-circle of the World For if right lines were extended from the center of the said Glasse by any point though elevated in any of those lines so drawn it would be directly in the Meridian Circle of the World Now all Reflective Dialling is performed from that principle in Opticks which is That the angle of Incidence is equal to the angle of Reflection And as any direct Dial may be made by help of a point found in the direct Axis so may any Reflected Dial be also made by help of any point found in the Reflected Axis And in regard the reflected Axis for the most part will fall above the Horizon of the Glasse without the window so that no point there can be fixed therefore a point must be found in the said Reflected Axis continued below the Horizontal of the said Glasse until it touch the ground or floor of the Room in some part of the Meridian formerly drawn which point will be the point in the reversed Axis desired and may be found as followeth One end of the thread being fixed at or in the center of the said Glasse move the other end thereof in the meridian formerly drawn below the said Glasse until the said reversed Axis be depressed below the Horizon as the direct Axis was elevated above the Horizon which may be done by applying the side or edge of a Quadrant to the said thread and moving the end thereof to and fro in the said meridian until the thread with a plummet cut the same degree as the Pole is above the Horizontal Glasse and then that point where the end of the thread toucheth the Meridian either on the floor or wall of the room is the point in the reflected reversed Axis sought for Now if the Reversed Axis cannot be drawn from the Glasse by reason of the jetting of the window or other impediment that point in the reverse Axis may be found by a line parallel thereto by fixing one end of it on the Glasse and the other end in the meridian so as that it may be parallel to the floor or wall in which the reversed Axis-point will fall and finde the Axis point from that other end of the lath so if the same Distance be set from that point backward in the Meridian on the floor as is the Lath the point will be found in the Reversed Axis desired Thus having found a point in the reflected reversed Axis it is not hard by help whereof and the Horizontal Dial to draw the reflected hour-lines on any Cieling or Wall be it never so concave or convex To do which First note that all straight lines in any projection on any Plain do always represent great Circles in the Sphere such are all the hour-lines Place the center of this Horizontal Dial in the center of the Glasse the hour-lines of the said Dial being horizontal and the Meridian of the said Dial in the Meridian of the world which may be done by plumb lines let fall from the meridian on the Cieling Then fix the end of a thread or silk in the said center of the Dial or Glasse and draw it directly over any hour-line on the Dial which you intend to draw and at the further side of the room and there let one hold or fasten that thread with a small nail Then in the point formerly found on the reversed Axis on the â—oor fix another thread there as formerly was done in the center of the Diall then take that thread and make it just touch the thread on the hour-line of the Horizontal Dial extended in any point thereof it matters not whereabouts and mark where the end of that thread toucheth the Wall or Cieling and there make some mark or point Then again move the same thread higher or lower at pleasure till it as formerly touch the said same hour thread and mark again whereabouts on the wall or Cieling the end of the said thread also toucheth In like manner may be found more points at pleasure but any two will be sufficient for the projecting or drawing any hour-line on any plain how irregular soever For if you move a thread and also your eye to and fro until you bring the said thread directly between your eye and the points formerly found you may project thereby as many points as you please at every angle of the Wall or Cieling whereby the reflected hour-line may be exactly drawn Again in like manner remove the said thread fastned in the center of the Horizontal Dial which also is the center of the Glasse on any other hour-line desired to be drawn and as before fasten the other end of the thread by a small nail or otherwise at the further side of the room but so that the said thread may lie just on the hour-line proposed to be drawn on the Horizontal Dial. Then as before take the thread fastened in the point on the reflected Axis and bring it to touch the thread of the hour-line in any part thereof and mark where the end of that thread toucheth the said Wall or Cieling Then again as before move the said thread so as that it only touch the said thread of the hour-line in any other part thereof and also mark where the end of that thread toucheth the said Wall or Cieling So is there found two points on the Wall or Cieling being in the reflected hour-line desired by help of which two points the whole hour-line may be drawn for if as before a thread be so scituated that it may interpose between the eye and the said