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A56521 The description and use of an instrument called the dovble scale of proportion by which instrument, all questions in arithmetick, geometry, trigonometry, astronomy, geography, navigation, fortification, gunnery, gaging vessels, dialling may be most accurately and speedily performed without the assistance of either pen or compasses / by Seth Partridge. Partridge, Seth, 1603-1686. 1661 (1661) Wing P630A; ESTC R40063 73,667 202

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after the hour of six THe Latitude of the place is 51 degr 30 min. North and the Suns declination 11 deg 30 min. I demand the time of Sun-rising before six Set the Co-tangent of the Latitude 38 deg 30 min. on the first to the Tangent of 11 deg 30 min. the declination given on the second then on the other side against the Radius on the first is the sine 14 deg 50 min. This 14 deg 50 min. converted into time by accounting four minutes of an hour to every degree and one minute of an hour for ●… minutes of a degree and it will amount to 59 minutes of an hour So long the Sun riseth before six in the morning and sets after six at night in the Summer season and in the Winter season or Southern signes he riseth so long after six in the morning and sets so long before six at night where the Latitude is 51 deg 30 min. and when the Suns declination is 11 deg 30 min. South PROBLEM X. The Sun being in the Equinoctial by knowing his distance from the Meridian and the latitude of the place to find the Suns altitude at that time MArch the tenth the Sun being in the Equinoctial at ten a clock before noon or at two a clock after noon he being at that hour distant from the Meridian 30 degrees I would know his height in our Latitude of 51 deg 30 minutes Set the Radius on the first to the Co-sine of the Latitude 38 deg 30 min. on the second and then against the Co-sine of the Suns distance from the Meridian viz. against 60 degrees on the first is 32 deg 37 min. on the second So much is the Suns altitude on the tenth of March at ten a clock in the fore-noon and at two a clock after noon PROBLEM XI The latitude of the place and the Suns declination Northwards being known to find the Suns altitude at the hour of six LEt the Latitude of the place be 51 deg 30 min. the Suns declination 11 deg 50 min. North And the Suns height at six a clock is desired Set the Radius on the first to the sine of the declination 11 deg 50 min. on the second and then against 51 deg 30 min. the Latitude on the first is 8 deg 58 min. on the second So much is the Suns height at the hours of six at morning and night Or else thus Set the Radius to 51 deg 30 min. and then against 11 deg 50 min. on the first is 8 deg 58 min. on the second For note that the Radius on the first may be set to either of the other two terms given on the second and then against the other term on the first is the answer on the second PROBLEM XII The Latitude and declination of the Sun being known to find the Suns Azimuth at the hour of six from the North part of the Meridian IN the Latitude of 51 deg 30 min. the Suns declination given is 11 deg 30 min. What is his Azimuth at the hour of six Set the Radius on the first to the Co-sine of the Latitude 38 degr 30 min. on the second and then against the Co-tangent of the declination 11 deg 30 min. on the first is the Tangent 82 deg 47 min. on the second So much is the Suns Azimuth from the North towards the East which was required PROBLEM XIII The Latitude of the place and the Suns altitude at the point of his being due East or West to find the hour and minute when he will be so due East or West THe Latitude of the place is 51 deg 30 min. The Suns height found at his being on the true East point is 14 deg 45 min. I demand the hour of the day Set the Radius on the first to the Co-sine of the Latitude 38 deg 30 min. on the second and then on the Lines of Tangents right against the Tangent of the Suns altitude at his being due East viz. 14 deg 45 min. on that first is 9 deg 18 min. on the second the Tangent of time from the hour of six either before or after that the Sun will be due East or due West This 9 degr 18 minutes reduced into time by allowing to every degree four minutes of time and for fifteen minutes of a degree one minute of time amounteth to 37 minutes of time So then the Sun is due East on the day given at 37 minutes after six a clock in the morning and also will be due West at 37 minutes before six at night which is at five a clock 23 minutes PROBLEM XIV The elevation of the Pole the Suns or a Stars declination and distance from the Meridian for any time being given to find the Suns altitude at that time LEt the elevation of the Pole be 51 deg 30 min. whose Complement to 90 is 38 deg 30 min. The Suns declination 11 deg 30 minutes whose Complement is 78 deg 30 min. his distance from the Pole his distance from the Meridian 30 degrees or which is all one two hours that is either at ten or two a clock and the Complement of this distance to 90 degrees is 60 degrees And let the Suns Altitude for that time be required 1 Set the Radius on the first to 60 deg the Co-sine of the Suns distance from Noon on the second and then on the lines of Tangents against 38 deg 30 minutes the Co-tangent of the elevation on the first is the Tangent 34 degr 33 min. on the second for the fourth term which taken from the Suns distance from the Pole the Remainder is 43 degrees 56 minutes 2 Set the Co-sine of that 34 deg 33 min. the fourth term viz. 55 deg 27 min. on the first to 46 deg 4 min. the Co-sine of that Remainder on the second and then against the sine of the Elevation 51 deg 30 min. on the first is the sine 43 deg 18 min. on the second So much is the altitude of the Sun at that time the thing required This is when the Sun hath Northern declination but when he hath Southern declination then add his declination to 90 deg and it makes 101 deg 30 min. the Suns distance from the Pole from which take the fourth term found and the Remain is 66 deg 27 min. This done the second work would be thus Set the Co-sine of the fourth term found on the first viz. 55 deg 27 min. to the Co-sine of that Remain which is 23 deg 33 min. on the second and then against 51 deg 50 min. the Elevation on the first is 21 deg 52 min. on the second The Suns height at ten or two a clock when he hath 11 deg 30 minutes of South declination If the Sun be just in the Equinoctial having no declination then to find his height on that day at ten or two a clock the work will be done at one working in
this manner Set the Radius on the first to 60 degrees the Co-sine of the distance from the Meridian on the second and then against 38 degr 50 min. the Co-sine of the Elevation on the first is the sine 32 deg 62 min. So much is the Suns altitude at the hours of ten and two a clock on the Equinoctial day PROBLEM XV. Having the Poles elevation and the Suns declination to find the ascensional difference LEt the Poles elevation be 51 deg 30 min. and the Suns declination 11 deg 30 min. And by them the ascensional difference required to be given Set 38 deg 30 min. the Co-tangent of the elevation on the first to the Tangent of the declination 11 deg 30 min. on the second and then on the Lines of Sines against the Radius on that first is the sine of 14 deg 49 min. on the second So much is the ascensional difference required at that time as the Sun hath 11 deg 30 min. declination I could shew by this ascensional difference converted into time How to find the time of Sun-rising and setting with the length of day and night c. But these are things done without the Lines and therefore I passe them by having limited my self to shew onely the use of the double Lines PROBLEM XVI Having any Planets declination and Latitude with its distance from the next Equinoctial point to find its right ascension LEt a Planets declination be 26 degrees its Latitude 4 degrees and its distance from the neerest Equinoctial point 70 degrees it being in the 10 deg of Gemini I demand the right ascension thereof which to attain Set 64 deg the Co-sine of the declination to 20 deg the Co-sine of the Planets distance from the next Equinoctial point and then against 86 deg on the first the Co-sine of the Planets latitude is 22 deg 17 min. on the second the Co-sine of the right ascension Therefore 67 deg 43 min. being the Complement of 22 deg 17 min. is the right ascension of that Planet which was demanded PROBLEM XVII The Suns greatest declination with his distance from the next Equinoctial point being had to find the Meridian angle that is the intersection of the Meridian with the Ecliptick LEt the Suns greatest declination be 23 degr 30 min. and his place just entering Taurus which is distant from the neerest Equinoctial point 30 degrees I would know the Meridian angle Set the Radius on the first to 60 degrees the Co-sine of the Suns distance from the neerest Equinoctial point on the second and then on the lines of Tangents against 23 deg 30 min. the Tangent of the Suns greatest declination on the first is 20 deg 38 min. on the second which is the Co-tangent of the angle sought Wherefore the Meridian angle is 69 degr 22 min. being the Complement of 20 degr 38 min. to 90 degrees PROBLEM XVIII The Suns declination and amplitude given to find the heigth of the Pole LEt the Suns declination given be 14 degr 51 min. and his amplitude 19 degr 7 min. And the Poles elevation demanded Set 19 deg 7 min. the amplitude on the first to 14 deg 51 min. the declination on the second and then against the Radius on the first is 51 deg 30 min. on the second which is the sine of the latitude Wherefore 51 degr 30 min. is the Latitude sought PROBLEM XIX The amplitude of the Sun and time of his rising being known to find thereby his declination LEt the Suns amplitude be 33 degr 38 min. and the time of his rising be at 4 a clock 10 minutes by which two things known I would find the declination of the Sun Here first convert the 4 hours 10 minutes into degrees and minutes and they make 62 degr 30 minutes Now set this 62 degr 30 min. the time of Sun-rising converted into degrees on the first to 56 deg 22 min. the Co-sine of the amplitude on the second and then against the Radius on the first is 69 degr 50 min. on the second the Co-sine of the declination Whereby it followeth that 20 deg 10 min. is the declination desired PROBLEM XX. Having the elevation of the Pole and the Suns amplitude to find his declination LEt the Elevation of the Pole be 51 degr 30 min. and the Suns amplitude 19 deg 7 min. By which two things given the declination is required Set the Radius on the first to 19 deg 7 min. the Suns amplitude on the second and then against 51 deg 30 min. the sine of the elevation on the first is 14 deg 51 min. the sine of the declination upon the second The thing desired PROBLEM XXI Having the hour of the day the Suns nltitude and declination to find the Azimuth LEt the Suns altitude be 25 degr 56 min. whose Complement is 64 deg 4 min. his declination 11 deg 30 min. whereof the Complement is 78 degr 30 min. The hour of the day 7 hours 56 minutes fore noon which time is distant from noon 4 hours 4 minutes This time converted into degrees giveth 61 degrees Now by these things known the Azimuth is desired to be found Set 64 degrees 4 minutes the Co-sine of the Suns altitude on the first to 61 degrees the Suns distance from the Meridian on the second and then against 78 degrees 30 minutes the Co-sine of the declination on the first is 72 degrees 22 minutes on the second So much is the Suns Azimuth from the South Eastwards PROBLEM XXII The Suns declination altitude and azimuth being known to find the hour of the day SVppose the Suns declination be 11 degr 30 min. his altitude 25 deg 56 min. and the Azimuth 72 deg 22 min. Hereby to find the hour of the day Set 78 deg 30 min. on the first which is the Complement of the declination to 64 deg 4 min. on the second which is the Complement of the Altitude and then against 72 deg 22 min. the Azimuth on the first is 61 grees on the second This 61 degrees converted into time giveth 4 hours 4 minutes which taken from 12 hours the Remain is 7 hours 56 minutes before noon the time of the day required PROBLEM XXIII The altitude of the Equator and the Suns or a Stars declination being given to find the angle of the Meridian with the Horizon LEt the altitude of the Equator be 51 degrees 30 minutes and the declination 22 degrees And the angle of the Meridian with the Horizon sought for Set 68 degrees the Co-sine of the declination on the first to the Radius on the second and then against 51 deg 30 min. the Co-sine of the Equators altitude on the first is 57 degrees 34 minutes on the second So much is the angle of the Meridian or circle of Declination with the Horizon CHAP. IX The Vse of the double Scales in Geography PROBLEM I. Two places lying without the Equinoctial and having both one Latitude