Selected quad for the lemma: reason_n

Word A	Word B	Word C	Word D	Occurrence	Frequency Band	MI	MI Band	Prominent
reason_n	angle_n	equal_a	side_n	2,221	5	9.5367	5	false

Text snippets containing the quad

ID	Title	Author	Corrected Date of Publication (TCP Date of Publication)	STC	Words	Pages
A80170	Navigation by the mariners plain scale new plain'd: or, A treatise of geometrical and arithmetical navigation; wherein sayling is performed in all the three kindes by a right line, and a circle divided into equal parts. Containing 1. New ways of keeping of a reckoning, or platting of a traverse, both upon the plain and mercators chart ... 2. New rules for estimating the ships way through currents, and for correcting the dead reckoning. 3. The refutation of divers errors, and of the plain chart, and how to remove the error committed thereby ... as also a table thereof made to every other centesm. 4. A new easie method of calculation for great circle-sayling, with new projections, schemes and charts ... 5. Arithmetical navigation, or navigation performed by the pen, if tables were wanting ... By John Collins of London, Pen-man, accomptant, philomathet.	Collins, John, 1625-1683.	1659 (1659)	Wing C5378A; Thomason E1002_1; ESTC R207824	177,819	300

What is here accomplished by the Tables may readily be performed by a Meridian-line out of which with Compasses take the distance between both Latitudes and prick it from L towards M at the end whereof raise a Perpendicular and therein prick down out of the Equinoctial degrees or equal parts the difference of Longitude drawing a line from it to L which shall pass through the former point O whatsoever be the Radius whereto the Meridian-line is fitted and after the same manner the error of the Plain Chart is to be removed when places are at first laid down in it according to their Longitudes and Latitudes which is most easily and suddenly done especially if a Meridian-line on a rod fitted to the degrees of Longitude on the Plain Chart and the maner of measuring a westwardly distance will be the same as in Mercators Chart. Before we proceed to the Demonstration of Mercators Chart it will be necessary to search into the Nature of the Rumbe-line on the Globe and to collect what we finde observeable in Forreign Authors to this purpose 1. They define it to be such a Line that makes the same Angles with every Meridian through which it passeth and therefore can be no Arch of a great Circle for suppose such a Circle to pass through the Zenith of some place not under the Equinoctial making an oblique Angle with the Meridian then shall it make a greater Angle with all other Meridians then with that through which it at first passeth therefore the Rumbe which maketh the same Angles with every Meridian must needs be a Line curving and bending which some call a Helix-Line and others because it is supposed to be on the Surface of the Sphere a Helispherical-line 2. Another Property of the Rumbe-line is that it leads nearer and nearer unto one of the Poles but never falleth into it and near the Pole it turneth often round the same in many Spires and turnings If by any Rumbe but not under the Meridian it were supposed possible to sayl directly under the Pole point it would follow that the Meridian-line of Mercators Chart should be finite and not infinite and that one and the same line should cut infinite other Lines as such at equal Angles in a meer point which is against the nature and definition of a line and then seeing that the Rumbe derives its definition from the Angle it makes with the Meridian the nature and use of a Meridia● under the Pole ceaseth and consequently the Rumbe ceaseth for under the Poles no star or other immoveable point of the heavens doth either rise or set in reference whereto the word Meridian hath its original being used to signifie the Mid-day time or high noon of the Sun or Star relating to their proper Motion moreover under the Pole point the sides of the Rumbe-triangles in the Sphere cease wherefo●e the Angles must needs do so too Of the Nature of the Triangles made upon the Globe by the Rumbe-Line And in the whole Rumbe-triangle AE t m the whole distance is AE m the whole difference of Latitude is AE t but the whole Departure from the Meridian is not given in any one Triangle but in divers Triangles and is by supposition as much in one Triangle as another to wit the whole Departure from the Meridian is the sum of a b c d e f g h i k l m each of which Portions are supposed to be equal each to other and the whole difference of Longitude is the Arch AE Q the Segments whereof Q L Q s s r cannot be equal to each other because the Arks or Departures l m i k c. are supposed to be equal each to other not in the same but in different Parallels of Latitude Now then for all uses in Navigation we suppose the Segments of the Rumbe m k k h and the rest to be a right Line and if we give the Rumbe to wit the Angle l k m and the Side k l we may by the Doctrine of Plain Triangles finde the Side k m the distance by the common Proportion As the Cosine of the Rumbe from the Meridian Is to the Radius ∷ So is the difference of Latitude To the distance ∷ And from hence we may observe another property in the Rumbe as namely that the Segments or Pieces thereof contained between two Parallels having the like difference of Latitude are also equal to each other so that the like distance being sayled in the same Rumbe in several parts of the earth shall cause the like difference or alteration of Latitude in each of those parts For Example If a Ship sayl from the Latitude of 10d North-East till she be in the Latitude of 30d and then sayl from thence on the same Rumbe till she be in the Latitude of 50d the latter distance shall be equal to the former distance and the like shall hold from thence to 70d and the same also should hold from 70d to the Pole if it were possible to sayl thither by any Rumbe and the reason is because in the Triangle k l m the Angles at k and l with the Side l k are equal by construction to the Angles at h and i and the Side i h in the Triangle i h k and the like in any other Triangle wherefore if these were given to finde the Side m k or k h it must needs be found the same in both Again if in the former Triangle we should give the Rumbe to wit the Angle l k m and the difference of Latitude l k we might finde the Departure from the Maridian l m by this Proportion As the Radius Is to the Tangent of the Rumbe viz. Tang. l k m So is the difference of Latitude 1 minute to wit k l To the Departure from the Meridian l m ∷ Now such Proportion as one Circle hath to another such Proportion have their Degrees Semi-diameters and Sines of like Arches one unto another As in the former Scheme A C is the Radius of the Equinoctial and E D is the Radius of a Parallel or lesser Circle whose Latitude from the Equinoctial is A E but E D is the Sine of the Ark E B which is the complement of the Latitude A E. It therefore holds As the Cosine of the Latitude Is to the Radius ∷ Or rather with other tearmes in the same Proportion As the Radius Is to the Secant of the Latitude ∷ So is a Mile or any part thereof or number of Miles To the difference of Longitude answering thereto ∷ And because the two first tearms of the Proportion vary not it will hold after the manner of the Compound rule of three As the Radius Is to the Sum of the Secants of all the parallels or Latitude between any two places and we allow a parallel to pass through every minute or Centesme of a degree So is a Mile or any number or part thereof allotted to every Parallel To the whole difference of

Semicircle which may sometimes happen when both places are in one Hemisphere if the sum of the Complements exceed 90 degrees Secondly the Latitudes of the great Ark are found by this Proportion which is in a manner the same as for a Parallel Course As the Cosine of either of the Vertical Angles but rather the lesser Is to the tangent of the latitude of that place to which it is adjacent ∷ So is the Cosine of the Ark of difference of Longitude from the Perpendicular To the Tangent of the Arks Latitude sought ∷ Thus by this excellent Method of Calculation we dispatch that at two Operations which Master Norwood and others do not attaine under seven or eight which rendred the Sayling by the great Arch so difficult and laborious that none cared to practise it And this latter Proportion having two fixed tearms in it will be performed on a Serpentine-line or Logarithmical Ruler without altering the Index or Compasses which Proportion being varied is carried on in the former Scheme As the Secant of either of the Vertical Angles Is to the Cotangent of the Latitude ad●acent thereto ∷ So is the Secant of the difference of Longitude from the Perpendicular To the Cotangent of the Arks Latitude sought ∷ In the former Scheme making the Perpendicular P A Radius P T becomes the Secant of the Vertical Angle T P A to that Radius and is also by construction the Tangent of the Complement of the Latitude next that Angle to another Radius then by reason of the Proportion of Equality which we have formerly handled the Secant of any other Angle from the Perpendicular to the former Radius as is P e shall be also the Cotangent of the Arks Latitude to the latter Radius Also the former Scheme shews us how to delineate Proportions in Sines and Tangents by framing of right-lined oblique Triangles for as the Sine of the Angle at L is to its opposite Side T P a Tangent so is the Sine of the Angle at T to its opposite Side P L another Tangent and by the like reason Proportions in Sines alone or in Equal parts and Sines c. may be carried on in the Angles and Sides of Plain Triangles Of the six parts of a right-angled Spherical Triangle no more but four at a time can be laid down in right Lines and Angles As in the right-angled Triangle P A T to wit the right Angle at A the Hipotenusal P T the Perpendicular P A and the Vertical Angle between T P A so that the Angle at T being the Complement of the Vertical Angle is not the Angle of Position in the Sphaere nor is the Side A T the measure of the Distance on that side the Perpendicular however both these Arks may be easily found Example for the Distance Prick the Radius of the Tangents from A to B and place the extent B P being the Secant of the Perpendicular from A to E from whence lines drawn to L and T shall contain an Angle equal to the Distance between the Lizard and Trinity Harbour in the great Arch To measure it with 60d of the Chords setting one foot at E draw K M which extent measured in the Chords sheweth the Distance to be 50d 9′ at 20 leagues to a degree is 1003 leagues the Rumbe between these two places is 74d 17′ from the Meridian and the Distance of the Rumbe 1034 leagues and after the same manner any part of it may be measured and so likewise might the parallel Distance I V which if we had room would be found to be 55 degrees and a half Also the Distance I T would be found to be 74d 7′ the Complement whereof to a Semicircle being 105d 53′ would be the Distance between the Lizard and Trinity Harbour as we supposed them the one in South the other in North Latitude The Proportion carried on to finde the Distance is As the Radius Is to the Sine of the Perpendicular ∷ So is the Tangent of the Vertical Angle To the Tangent of the Distance ∷ The Perpendicular A P is a Tangent to the Radius A B and if we make the extent B P Radius it then becomes a Sine as is evident if you describe an Ark from P therewith which R●dius if we should place from P outward beyond C and suppose a Tangent erected thereon parallel to A T then doth the extent A T become the Tangent of the fourth Proportional to that Radius which that it might be measured the Radius was pricked from A to E. To finde the Angle of Position If the Hipotenusal and one Legg be given the Proportion to finde the Angle next that Legg would be As the Radius Is to the Tangent of the Hipotenusal ∷ So is the Tangent of the given Legg To the Cosine of its adjacent Angle ∷ And so in the former Scheme if we make T P Radius which is also the Tangent of the Hipotenusal then doth P A being the Tangent of the given Leg become the Sine of the Angle P T A which is the Complement of the Angle T P A whence we may observe that if a Perpendicular be raised at the end of the Tangent of the given Leg the Tangent of the Hipotenusal being from the other end of the said Leg made the Hipotenusal opposite to the right Angle the Angle included shall be the Angle sought therefore having the Distance or Leg A T given with its Radius A E you may proportion out the Hipotenusal to that Radius and with it setting one foot in T cross A E with the other from whence drawing a line to T the Angle between it and A T shall be the Angle of Position required Otherwise more readily Take the nearest distance from C to T P and it shall be the Cosine of the Angle of Position if we make E A Radius wherefore upon A describe the prickt Ark n a ruler from the Center E just touching it cuts the Ark x n at n which Ark measured on the Chords is 38d 49′ the complement whereof being 51d 11′ is the Angle of Position required and so may all the other Angles of Position be found if there were any need of them The Proportion carried on is As the Radius Is to the Sine of the Vertical Angle So is the Cosine of the Perpendicular To the Cosine of the Angle of Position ∷ If we make B P Radius then is A B the Cosine of the Perpendicular to that Radius and the nearest distance from C to T P is the Cosine of the Angle of Position to the Radius E A. Thus we have shewed the use of this Pro●ection in more particulars then the Sea-man shall have occasion to use Lastly it may be noted that in stead of pricking down a new Line of Tangents from the Perpendicular in every question that shall be put that if the Perpendicular be placed in the Side P L

given I have shewed in a Treatise Entituled The Sector on a Quadrant page 139 140. And how to find the points I or K without drawing the lines e F or O G and that by help of a cross or Intersection like that at e which may either happen within or without the outward Circle the Reader may attain from the last Scheme for finding the Amplitude The Converse of the former Scheme for finding the Hour will finde the Suns Altitudes on all Hours and the Distances of Places in the Arch of a great Circle Example Latitude 51d 32′ Declination 23d 31′ North. Hour 75d from Noon that is either 7 in the Morning or 5 in the Afternoon Having drawn the Semicircle its Diameter and by a Perpendicular from the Center divided it into two quadrants and therein having prickt off A L the Latitude and thorough the same drawn L M produced and parallel to D C therein from L to M and Q prick off the Sine of the Declination Then prick off the Hour from Noon from A to R and laying the Ruler from the Center draw the line R E and with the Co-sine of the Declination namely the nearest distance from F to D C draw the Arch G E and transfer the distance between the points M and E from M to H. Lastly the distance between the points N and H is the Chor● of the Suns distance from the Zenith for that Hour namely 62d 37′ the Complement whereof 27d 23′ is the Altitude sought Moreover the distance between H and Q is the Chord of the Suns distance from the Zenith for the winter declination namely 99d 30′ which being greater then a quadrant argues the Sun to have 9d 30′ of Depression under the Horizon and so much is his Altitude for the hours of 5 in the Morning or 7 in the Afternoon when his Declination is 23d 31′ North. Another Example for the same Latitude and Declination Let the Hour from Noon be either 10 in the Morning or 2 in the Afternoon prick off 30d from A to K and from the Center draw the line K G place the distance M G from M to O so is the distance O N the Chord of 36d 16′ the Complement whereof 53d 44′ is the Summer Altitude for that Hour and the distance O Q is the Chord of 79d 32′ the Complement whereof 10d 28′ is the Winter Altitude for that Hour Also for the Distances of places in the ark of a great Circle the Case in Sphaerical Triangles is the same with that here resolved So if there were two places the one in North Latitude 51d 32′ the other in North Latitude 23d 31′ the difference of Longitude between them being 75d their distance by the former Scheme will be found to be 62d 37′ but if the latter place were in as much South Latitude then their distance would be 99d 30′ Another Example for finding the Distances of Places in the Arch of a great Circle Example Let the two Places be according to the Sea-mans Calendar Isle of Lobos Longitude 307d 41′ Latitude 40d 21′ South Lizard 18 30. Latitude 50 10 North. Difference of Longitude 289 11. Complement 70d 49′ Having described a Circle make AE M 70d 49′ M I the Latitude of the Island B I the Sine thereof falling Perpendicularly on C M AE L the Latitude of the Lizard L A the Sine thereof make A E equal to the extent A B and prick B I from L upwards to H when the places are in different Hemispheres but downwards to K when in the same Hemisphere and the extent H E or K E is the Chord of the Ark of distance between both places in this Example H E is 109d 41′ K E is 48d 57′ Demonstration This Scheme I first met with in a Map made in Holland the foundation whereof was long since laid by Copernicus and Regiomontanus who from a right lined plain Triangle happening at the Center of the Sphere have prescribed a Method of Calculation for finding an Angle when three Sides are given Here we shall illustrate the Converse how from two Sides and the Angle comprehended to find the third Side From any two points in the Sphere suppose Perpendiculars to fall on the plain of the Equator here represented by AE L Q M which Perpendiculars are the Sines of the Latitudes of those two Points and the distance of the points in the Plain of the Equinoctial from the Center of the Sphere are the Cosines of those Latitudes the angle at the Center between those points in the plain of the Equator is equal to the arch of the Equinoctial between the two Meridians passing through the supposed Points in the Sphere now a right line extended in the Sphere between any two Points is the Chord of the Ark of distance between those Points Understand then that the three Points A C B limit the Sides of a righ● lined Triangle in the Plain of the Equator whereof the Angle A C B is at the Center then the extent A B is placed from A to E if then we draw D E G perpendicular to A Q and place B I from E to G and D the extent L G shall be the Chord of the third Side when the places are in different Hemispheres and the extent L D is the Chord of their distance when they are in the same Hemisphere And if the extent E D E G be placed from L to H and K the line D E G need not be drawn because the extent L G and L D if it be rightly conceived are the two very Points at first supposed in the Sphere the extent A B as to matter of Calculation being one Side of a plain Triangle right angled The sum or difference of the Sines of both Latitudes the other Side and the Chord of the distance is the Hipotenusal or third Side sought Thus the Ancients by Calculation and we by Protraction having two Sides and the Angle comprehended given find the third Side or having three Sides given find the Angle opposite to that Side which in the Scheme is measured by a Chord as by result from the three Sides there will be got the two extents A E and C B and consequently the Intersection at B and thence the Angle AE M which before was insisted on in finding the Azimuth and Hour by the like reason the Distances of Stars may be found from their Longitudes and Latitudes or from their Declinations and right Ascensions Divers other Schemes from other Proportions might be added for finding the Hour and Azimuth c. which which I am loth to trouble the Reader withal I shall adde another Scheme for this purpose which carries on the same Proportion by which this Case is usually Calculated The Proportions are expressed in a Treatise The Sector on a Quadrant page 127. Example Latitude 51d 32′ Declination 23 31. Hour 60 from Noon Having drawn a Semicircle and the Radius Z D prick the Latitude from C to L

which Snellius asserts is That the Tangent cut off to wit I B is somewhat shorter then the Arch I A though near it in length when the Arch is not above 1 12 part of a Quadrant and this Hugenius demonstrates in his Book De Magnitudine Circuli where he finds fault with Snellius his Demonstration thereof Now the Proportion for finding an Angle raised from that Scheme lyes thus As R E the sum of the double of the Hipotenusal C A and of the side C E Is to E A the lesser side ∷ So is R I the triple of the Radius or Hipotenusal ∷ To I B the length of the Arch required ∷ propé verum This Proportion finds the length of the Arch making the Hipotenusal always Radius whereas in Calculation we always retain such a Radius whereof the Circumference of the Circle is 360d now the Diameter of such a Circle will be found by the numbers in page 112 in which the Proportion of the Circumference to the Diameter is expressed to be 114 5915 wherefore the Radius of such a Circle is 57 2957 and the triple thereof is 171 8871 then retaining the two first tearms of the former Proportion we may make this number the thi d tea●m and by one single work finde the Angle or rather taking the halfs of all four tearms the Proportion will hold As the sum of the Hipotenusal and of half the greater Leg of a Plain right Angled Triangle Is to the lesser Leg thereof ∷ So is 86 To the Angle opposed to the lesser Leg The half of 171 8871 is 85 9435 which because we have taken it to be 86 the Proportion if the Angle be less then 30d finds the Angle to be about one Centesimal part of a degree too much but if the Angle be above 35d by reason the Scheme is not absolutely true must have these additions made to the Angle found thereby from 35d to 38d adde one Centesm from 38 to 40d adde two Centesmes to it and afterwards to 45d for every degree it exceed 40d adde one Centesm more besides the former two Centesms and thus we may always finde the lesser acute Angle and consequently the greater being the Complement thereof within one Centesm of the truth which is nearer then any Mechanick way Example In the former Triangle let there be given the Sides C E 4 17 A E 3 93 by extracting the square root of the sum of the squares of these two Numbers we shall finde the side C A to be 5 73 to which adding the half of C E the sum is 782 the Divisor then multiplying 393 the lesser Leg by 86 the Product is 33798 to which you may annex Ciphers at pleasure to finde the Decimal parts of a degree and dividing by 782 you will finde the quotient to be 43d 22 Centesmes to which if you adde 5 Centesmes error the Angle sought is 43d 27 Centesms Readily to finde what allowance must be made in respect of the Arch found you may repair to a Table of Natural Sines and take the two Legs of the right Angled Triangle to be the sine and cosine of any Arch and by the last Proportion finde how near you can recover the Arch again whereby you will find what allowance must be made The Example here used is that mentioned in page 16 and 109 so that hereby we finde a Course and Distance on the Plain Chart without the help of Tables and by the like reason the height of a Gnomon and the length of its shadow being given the Suns height may be got without Tables Hugenius not thinking this way of Snellius to be exact enough propounds another of his own upon this consideration that the Chord of an Arch being increased by one third part of the difference between the Sine and the Chord of the said Arch shall be very near equal in length to the Arch it self yea so near that in an Arch of 45d it shall not erre or fall short above 1 18000 part of a degree and in an Arch of 30d not above the 1 21600 part of a degree whereby the Sines may be examined and an Angle found without Tables as if the sides of the former Triangle be given by taking C E out of C A there remains E I the square whereof added to the square of E A the Sine is equal to the square of the line A I the Chord whereby may be found the length of the Arch I A to the given Radius C A and then by another Proportion the measure of the said Arch to the Radius of such a Circle whose Circumference is 360d. In like manner if the sides of an oblique Plain Triangle were given the Angles thereof might be found if you first reduce that oblique Triangle to two right angled Plain Triangles which is performed in every Book of Trigonometry without Tables But for such Cases of Plain Triangles in which but one Side with two Angles are given to finde the other Sides in regard the Proportions for such Cases require Sines and that we have not attained any ready way to make the Sine of any arch at command forbearing to mention such ways as are both troublesom and uncertain we must suppose that the Reader is furnished with a Table of Sines which most Mariners have in their Sea-mans Kalendar FINIS ERRATA PAge 1 line 31 for would read should p. 48. l. 30. for 48 degrees r. 84 degrees p. 49 l. 3. for Angle r Angles l. 4 for Sine r. Line p. 51. l. 25. for following r former p. 59. l. 14. for or r. of p. 60. l. 21. for as every r. or ever p. 81 l. 16. for one is r. one in page 88. l. 26. for Tangent r. Tangents A Table of Meridionall parts D 0 1 2 3 4 5 6 7 8 9 0 0 000 1 000 2 000 3 001 4 003 5 006 6 011 7 017 8 02● 9 037 2 020 020 2 020 0●1 023 029 031 037 046 057 4 040 040 04● 041 043 046 051 057 066 077 6 060 060 060 061 063 066 071 077 086 097 8 080 080 080 081 083 086 091 097 106 117 10 100 1 100 2 100 3 101 4 103 5 106 6 111 7 118 8 127 9 138 12 120 120 120 121 123 126 131 138 147 158 14 140 140 140 141 143 146 151 158 167 178 16 160 160 160 161 163 166 171 178 187 198 18 180 180 180 181 183 186 191 198 207 218 20 200 1 200 2 200 3 201 4 204 5 207 6 212 7 219 8 228 9 239 22 220 220 220 221 224 227 232 239 248 259 24 240 240 240 241 244 247 252 259 268 279 26 260 260 260 261 26● 267 272 279 288 300 28 280 280 280 281 284 287 292 299 308 320 30 300 1 300 2 300 3 301 4 304 5 307 6 312 7 319 8 329 9 341 32 320 320 320 321 324 327 332 339 349 361 34 340 340 340 341 344 347 352