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Word A	Word B	Word C	Word D	Occurrence	Frequency Band	MI	MI Band	Prominent
end_n	day_n	hour_n	minute_n	1,570	5	11.3251	5	false

Text snippets containing the quad

ID	Title	Author	Corrected Date of Publication (TCP Date of Publication)	STC	Words	Pages
A52264	Institutio mathematica, or, A mathematical institution shewing the construction and use of the naturall and artificiall sines, tangents, and secants in decimal numbers, and also of the table of logarithms in the general solution of any triangle, whether plain or spherical, with their more particular application in astronomie, dialling, and navigation / by John Newton.	Newton, John, 1622-1678.	1654 (1654)	Wing N1061; ESTC R20441	128,229	417

Artificiall SINES and TANGENTS as also of the LOGARITHMS IN Astronomie Dialling and Navigation By JOHN NEWTON LONDON Printed Anno Domini 1654. A Mathematicall Institution The second Part. CHAP. I. Of the Tables of the Suns motion and of the equation of time for the difference of Meridians WHereas it is requisite that the Reader should be acquainted with the Sphere before he enter upon the practise of Spherical Trigonometri the which is fully explained in Blundeviles Exercises or Ch●lmades translation of Hues on the Globes to whom I refer those that are not yet acquainted therewith that which I here intend is to shew the use of Trigonometrie in the actuall resolution of so me known Triangles of the Sphere And because the Suns place or distance from the next Equinoctial point is usually one of the three terms given in Astronomical Questions I will first shew how to compute that by Tables calculated in Decimal numbers according to the Hypothesis of Bullialdus and for the Meridian of London whose Longitude reckoned from the Canarie or Fortunate Islands is 21 deg and the Latitude North 51 deg 57 parts min. or centesms of a degree Nor are these Tables so confined to this Meridian but that they may be reduced to any other If the place be East of London adde to the time given but if it be West make substraction according to the difference of Longitude allowing 15 deg for an houre and 6 minutes or centesms of an houre to one degree so will the sum or difference be the time aequated to the Meridian of London and for the more speedy effecting of the said Reduction I have added a Catalogue of many of the chiefest Towns and Cities in diverse Regions with their Latitudes and difference of Meridians from London in time together with the notes of Addition and Substraction the use whereof is thus Suppose the time of the Suns enterance into Taurus were at London Aprill the 10th 1654 at 11 of the clock and 16 centesms before noon and it be required to reduce the same to the Meridian of Vraniburge I therefore seeke Vraniburge in the Catalogue of Cities and Places against which I finde 83 with the letter A annexed therefore I conclude that the Sun did that day at Uraniburge enter into Taurus at 11 of the clock and 99 min. or centesms before noon and so of any other Problem 1. To calculate the Suns true place THe form of these our Tables of the Suns motion is this In the first page is had his motion in Julian years compleat the Epochaes or roots of motions being prefixed which sheweth the place of the Sun at that time where the Epocha adscribed hath its beginning the Tables in the following pages serve for Julian Years Moneths Dayes Houres and Parts as by their Titles it doth appear The Years Moneths and Dayes are taken compleat the Houres and Scruples current After these Tables followeth another which contains the Aequations of the Eccentrick to every degree of a Semicircle by which you may thus compute the Suns place First Write out the Epocha next going before the given time then severally set under those the motions belonging to the years moneths and dayes compleat and to the hours and scruples current every one under his like onely remember that in the Bissextile year after the end of February the dayes must be increased by an unit then adding them all together the summe shall be the Suns mean motion for the time given Example Let the given time be 1654 May 13 11 hours 25 scruples before noon at London and the Suns place to be sought The numbers are thus   Longit. ☉ Aphel ☉ The Epocha 1640 291.2536 96.2297 Years compl 13 359.8508 2052 Moneth co April 118.2775 53 Dayes compl 12 11.8278 6 Hours 23 9444   Scruples 25 102         Sū or mean motiō 782.1643 96.4308 2. Substract the Aphelium from the mean Longitude there rests the mean Anomalie if it exceed not 360 degrees but if it exceed 360 degr 360 being taken from their difference as oft as it can the rest is the mean Anomalie sought Example The ☉ mean Longitude 782.1643 The Aphelium substracted 96.4308 There rests 685.7335 From whence deduct 360. There rests the mean Anomalie 325.7335 3. With the mean Anomalie enter the Table of the Suns Eccentrick Equation with the degree descending on the left side if the number thereof be lesse then 180 and ascending on the right side if it exceed 180 and in a straight line you have the Equation answering thereunto using the part proportional if need require Lastly according to the title Add or Substract this Equation found to or from the mean longitude so have you the Suns true place Example The Suns mean longitude 782.1643 Or deducting two circles 720. The Suns mean longitude is 62.1643 The Suns mean Anomalie 325.7335 In this Table the Equation answering to 325 degrees is 1.1525 The Equation answering to 326 degrees is 1.1236 And their difference 289. Now then if one degree or 10000 Give 289 What shall 7335 Give the product of the second and third term is 2119815 and this divided by 10000 the first term given the quotient or term required will be 212 fere which being deducted from 1.1525 the Equation answering to 325 degr because the Equation decreased their difference 1.1313 is the true Equation of this mean Anomalie which being added to the Suns mean longitude their aggregate is the Suns place required Example The Suns mean longitude 62.1643 Equation corrected Add 1.1313 The Suns true place or Longitude 63.2956 That is 2 Signes 3 degrees 29 minutes 56 parts The Suns Equation in this example corrected by Multiplication and Division may more readily be performed by Addition and Substraction with the help of the Table of Logarithmes for As one degree or 10000 4.000000     Is to 289 2.460898 So is 7335 3.865400     To 212 fere 2.326298 The Suns mean Motions Epochae Longitud ☉ Aphelium ☉   ° ′ ″ ° ′ ″ Per. Jul. 242 99 61 355 85 44 M●●di 248 71 08 007 92 42 Christi 278 98 69 010 31 36 An. Do. 1600 290 95 44 095 58 78 An. Do. 1620 291 10 41 095 90 39 An. Do. 1640 291 25 36 096 21 97 An. Do. 1660 291 40 33 096 53 56 1 356 76 11 0 01 58 2 359 52 22 0 18 17 3 359 28 30 0 04 74 B 4 000 03 00 0 06 30 5 359 79 11 0 07 89 6 359 55 19 0 09 47 7 359 31 30 0 11 05 B 8 000 05 97 0 12 64 9 359 82 08 0 14 22 10 359 58 19 0 15 78 11 359 34 30 0 17 36 B 12 000 08 97 0 18 94 13 359 85 08 0 20 52 14 359 00 19 0 22 11 15 359 37 30 0 23 69 B 16 000 11 97 0 25 25 17 359 88 08 0 26 83 18 359 64 19 0 28 41 19 359 40 28 0 30 00 B 20 000 14 97 0 31

fractions is in the calculation very tedious besides here no fractions almost are exquisitely true therefore the Radius for the making of rhese Tables is to be taken so much the more that there may be no errour in so many of the figures towards the left hand as you would have placed in the Tables and as for the numbers superfluous they are to be cut off from the right hand towards the left after the ending of the supputation Thus to finde the numbers answering to each degree and minute of the Quadrant to the Radius of 10000000 or ten millions I adde eight ciphers more and then my Radius doth consist of sixteen places This done you must next finde out the right Sines of all the arches lesse then a Quadant in the same parts as the Radius is taken of whatsoever bignesse it be and from those right Sines the Tangents and secants must be found out 21. The right Sines in making of the Tables are either primary or secondary The primarie Sines are those by which the rest are found And thus the Radius or whole Sine is the first primary Sine the which how great or little soever is equall to the side of a six-angled figure inscribed in a circle that is to the subtense of 60 degrees the which is thus demonstrated Out of the Radius or subtense of 60 degrees the sine of 30 degrees is easily found the halfe of the subtense being the measure of an angle at the circumference opposite thereunto by the 19 of the second if therefore your Radius consists of 16 places being 1000.0000.0000.0000 The sine of 30 degrees will be the one half thereof to wit 500.0000.0000.0000 22. The other primary sines are the sines of 60 45 36 and of 18 degrees being the halfe of the subtenses of 120 90 72 and of 36 degrees 23. The subtense of 120 degrees is the side of an equilateral triangle inscribed in a circle and may thus be found The Rule Substract the Square of the subtense of 60 degrees from the Square of the diameter the Square root of what remaineth is the side of an equilateral triangle inscribed in a circle● or the subtense of 120 degrees The reason of the Rule The subtense of an arch with the subtense of the complement thereof to 180 with the diameter make in the meeting of the two subtenses a right angled triangle As the subtense AB 60 degrees with the subtense AC 120 degrees and the diameter CB make the right angled triangle ABC right angled at A by the 19 of the second And therfore the sides including the right angle are equal in power to the third side by the 〈◊〉 of the second Therefore the square of AB being taken from the square of CB there remaineth the square of AC whose squar root is the subtense of 〈◊〉 degrees or the side of an equilateral triangle inscribed in a circle Example Let the diameter CB be 2000.0000 0000.0000 the square thereof is 400000. 00000.00000.00000.00000.00000 The subtense of AB is 100000.00000.00000 The square thereof is 100000.00000.00000 00000.00000.00000 which being substracted from the square of CB the remainder is 300000.00000.00000.00000.00000.00000 whose square root 173205.08075.68877 the subtense of 120 degrees CONSECTARY Hence it followeth that the subtense of an arch lesse then a Semicircle being given the subtense of the complement of that arch to a Semicirc●e is also given 24. The Subtense of 90 degrees is the side of a square inscribed in a circle and may thus be found The Rule Multiply the diameter in it self and the square root of half the product is the subtense of 90 degrees or the side of a square inscribed in a circle The reason of this Rule The diagonal lines of a square inscribed in a circle are two diameters and the right angled figure made of the diagonals is equal to the right angled figures made of the opposite sides by the 20 th of the second now because the diagonal lines AB and CD are equal it is all one whether I multiply AC by it self or by the other diagonal CD the p●oduct will be still the same then because the sides AB AC and BC do make a right angled triangle right angled at C by the 〈◊〉 of the second that the 〈◊〉 AC and ●B are equal by the work the half of the square of AB must needs be the square of AC or CB by the 17 th of the second whose square rootes the subtense of CB the side of a square or 90 degree Example Let the diameter AB be 200000.00000 00000 the square thereof is 400000.00000 00000.00000.00000.00000 the half whereof is 200000.00000.00000.00000.00000 00000. whose square root 14142● 356●3 73095. is the subtense of 90 degrees or the side of a square inscribed in a Circle 25. The subtense of 36 degrees is the side of a decangle and may thus be found The Rule Divide the Radius by two then multiply the Radius by it self and the half thereof by it self and from the square root of the summe of these two products substract the half of Radius what remaineth is the side of a decangle or the subtense of 36 degrees The reason of the rule For example Let the Radius EB be 100000.00000.00000 then is BH or the half thereof 500000. 00000.00000 the square of EB is 100000 00000.00000.00000.00000.00000 and the square of BH 250000.00000.00000.00000 00000.00000.00000 The summe of these two squares viz 125000.00000.00000 00000 00000. 00000 is the square of HE or HK whose square root is 1118033● 887●9895 from which deduct the halfe Radius BH 500000000000000 and there remaineth 618033988749895 the right line KB which is the side of a decangle or the subtense of 36 degrees 26 The subtense of 72 degrees is the side of a Pentagon inscribed in a circle and may thus be sound The Rule Substract the side of a decangle from the diameter the remainer multiplied by the Radius shall be the square of one side of a Pentagon whose square root shall be the side it self or subtense of 72 degrees The Reason of the Rule In the following Diagram let AC be the side of a decangle equal to CX in the diameter and let the rest of the semicircle be bisected in the point E then shall either of the right lines AE or EB represent the side of an equilateral pentagon for AC the side of a decangle subtends an arch of 36 degrees the tenth part of a circle and therefore AEB the remaining arch of a semicircle is 144 degrees the half whereof AE or EB is 72 degrees the fift part of a circle or side of an equilateral pentagon the square whereof is equal to the oblong made of DB and BX Demonstration Draw the right lines EX ED and EC then will the sides of the angles ACE and ECX be equal because CX is made equal to AC and EC common to both and the angles themselves are equal because they are in equal segments

to swell equally as much in longitude as in latitude till it joyn it self unto the concavity of the cylinder so as hereby no part thereof is any way distorted or displaced out of his true and natural situation upon his meridian parallel or rumbe but onely dilated and enlarged the meridians also parallels and rumbes dilating and enlarging themselves likewise at every point of latitude in the same proportion Now then let us diligently consider of the Geometrical lineaments that is the meridians rumbes and parallels of this imaginary Nautical planisphere that we may in like manner expresse the same in the Mariners Chart for so undoubtedly we shall have therein a true Hydrographical description of all places in their longitudes latitudes and directions or respective situations each from other according to the points of the compasse in all things correspondent to the Globe without either sensible or explicable errour First therefore in this planisphere because the parallels are every where equal each to other for every one of them is equal to the Equinoctiall or circumference of the circumscribing cylinder the meridians also must needs be parallel streight lines and consequently the rumbes making equall angles with every meridian must likewise be streight lines Secondly because the spherical superficies whereof this planisphere is conceived to be made swelleth in every part thereof equally that is as much in Latitude as in Longitude till it apply it self round about to the concavity of the cylinder therefore at every point of Latitude in this planisphere a part of the Meridian keepeth the same proportion to the like part of the parallel that the like parts of the Meridian and parallel have each to other in the Globe without any explicable errour And because like parts of wholes keep the same proportion that their wholes have therefore the like parts of any parallel and Meridian of the Globe have the same proportion that the same parallel and meridian have For example sake as the meridian is double to the parallel of 60 degrees so a degree of the meridian is double to a degree of that parallel or a minute to a minute and what proportion the parallel hath to the meridian the same proportion have their diameters and semidiameters each to other But the sine of the complement of the parallels latitude or distance from the Equinoctial is the semidiameter of the parallel As here you see AE the sine of AH the complement of AF the latitude or distance of the parallel ABCD from the Equinoctial is the semidiameter of the same parallel And as the semidiameter of the meridian or whole sine is to the semidiameter of the parallel so is the secant or hypothenusa of the parallels latitude or of the parallels distance from the Equinoctial to the semidiameter of the meridian or whole sine as FK that is AK to AE that is GK so is LK to KF Therefore in this nautical planisphere the Semidiameter of each parallel being equal to the semidiameter of the Equinoctial that is to the whole sine the parts of the Meridian at every point of Latitude must needs increase with the same proportion wherewith the secants of the ark conteined between those points of Latitude and the Equinoctial do increase Now then we have an easie way laid open for the making of a Table by help of the natural Canon of Triangles whereby the meridians of the Mariners Chart may most easily and truly be divided into parts in due proportion and from the Equinoctial towards either Pole For supposing each distance of each point of latitude or of each parallel from other to contein so many parts as the secant of the latitude of each point or parallel conteineth by perpetual addition of the secants answerable to the latitudes of each point or parallel unto the summe compounded of all the former secants beginning with the secant of the first parallels latitude and thereto adding the secant of the second parallels Latitude and to the summe of both these adjoyning the secant of the third parallels Latitude and so forth in all the rest we may make a Table which shall truly shew the sections and points of latitude in the Meridians of the Nautical Planisphere by which sections the parallels must be drawn As in the Table of meridional parts placed at the end of this Discourse we made the distance of each parallel from other to be one minute or centesm of a degree and we supposed the space between any two parallels next to each other in the Planispere to contain so many parts as the secant answerable to the distance of the furthest of those two parallels from the Equinoctial and so by perpetual addition of the secants of each minute or centesm to the sum compounded of all the former secants is made the whole Table As for example the secant of one centesm in Master Briggs 's Trigonometrica Britannica is 100000.00152 which also sheweth the section of one minute or centesm of the meridian from the Equinoctial in the Nautical Planisphere whereunto adde the secants of two minutes or centesmes that is 100000. 00609 the sum is 200000.00761 which sheweth the section of the second minute of the meridian from the Equinoctial in the planisphere to this sum adde the secant of three minutes which is 100000.01371 the sum will be 3000●0 02132 which sheweth the section of the third minute of the meridian from the Equinoctial and so ●orth in all the rest but after the Table was thus finished it being too large for so small a Volume we have contented our selves with every tenth number and have also cut off eight places towards the right hand so that in this Table the section of 10 minutes is 100 of one degree 1000 and this is sufficient for the making either of the generall or any particular Chart. I call that a general Chart whose line AE in the following figure represents the Equinoctial as here it doth the parallel of 50 degrees and so containeth all the parallels successively from the Equinoctial towards either Pole but they can never be extended very near the Pole because the distances of the parallels increase as much as secants do But notwithstanding this it may be remed general because a more general Chart cannot be contrived in plano except a true projection of the Sphere it self And I call that a particular Chart which is made properly for one particular Navigation as if a man were to sail betwen the Latitude of 50 and 55 degrees and his difference of Longitude were not to exceed six degrees then a Chart made as this figure is for such a Voyage may be called particular and is thus to be projected Probl. 3. The Latitudes of two places being known to finde the Meridional difference of the same Latitudes IN this Proposition there are three varieties First when one of the places is under the Equinoctial and the other without and in this case the