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A29761 The description and use of the trianguler quadrant being a particular and general instrument, useful at land or sea, both for observation and operation : more universally useful, portable and convenient, than any other yet discovered, with its uses in arithmetick, geometry, superficial and solid, astronomy, dyalling, three wayes, gaging, navigation, in a method not before used / by John Brown, philomath. Brown, John, philomath. 1671 (1671) Wing B5041; ESTC R15524 175,812 507

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shall be the sine of the Suns declination required for that distance from the next Equinoctial-point by the 1 st Rule abovesaid By the Sector Take 23-31 from the Sines make it a = in the sine of 90 then the = sine of the Suns distance from the next Equinoctial-point shall be the sine of the Suns declination Example as before Rule the 1 st Use XIII The Suns Declination being given to find his true place or distance from ♈ or ♎ the two Equinoctial Points Lay the Thred to the Declination counted in the degrees from 600 and in the Line of the Suns place is his true place required Example When the Suns declination is 12 degrees Northward the dayes increasing then the Sun will be 31 deg and 23 min. from ♈ or 1 deg 23 min. in ♉ his true place required As Sine of 23-31 the Suns greatest declination to Sine of 90 So Sine of 12-00 the Suns present declination to Sine of Suns distance from ♈ or ♎ 31-23 Which by considering the time of the year gives his true place by looking on the Months and Line of Suns place on the Quadrantal-side Take the Sine of the present declination make it a = Sine in the greatest declination laying the Thred to ND and on the degrees the Thred shall give the Suns distance from ♈ or ♎ required Example as before Make Sine of the given Suns declination a = Sine in the Suns greatest declination then = Sine of 90 measured from the Center is the = Sine of the Suns distance from ♈ or ♎ required or count 30 deg for one sign and the Center for the next Equinoctial-point and 90 for the two Tropicks of Cancer and Capricorn ♋ ♑ Use XIV The Suns place or Day of the Month and greatest Declination given to find his Right Ascention from the same Equinoctial Lay the Thred to the day of the Month or place given and in the Line of the Suns Right Ascention you have his Right Ascention in degrees or hours and minutes counting 4 minuts for every degree Example On the 9th of April near night the Sun being then entring ♉ the Suns Right Ascention will be 1 hour 52 min. or 28 degrees of Right Ascention distant from ♈ As the Sine of 90 to the Sine complement of the Suns greatest declination or C.S. of 23-31 counting backwards from 90 which will be at the Sine of 66-29′ So is the Tangent of the Suns distance from the next Equinoctial-point to the Tangent of the Suns Right Ascention from the same Equinoctial-point Take the co-sine of the greatest declination from the Center downwards being the sine of 66-29′ make it a = sine of 90 laying the Thred to ND and note what degree and minuit it cuts for this is fixed to this Proportion Then take the Tangent of the Suns distance from the next Equinoctial-point from the Center at 600 on the degrees toward the End and lay it on the sines from the Center downwards and note the Point where it stayeth for the ND from thence to the Thred shall be the Tangent of the Suns Right Ascention required Note That if the Suns distance from ♈ or ♎ be above 45 degrees then the Tangents on the loose-piece are to be used instead of the Tangents on the moveable-leg Or by Sines only thus Or Take Sine of the present Suns declination make it a = in the Sine of the Suns greatest declination and lay the Thred to ND then take = Co-sine of the Suns greatest declination and make it a = in Co-sine of the Suns present declination and lay the Thred to ND and in the degrees it cuts the Suns Right Ascention required Make Co-sine of 23-31 viz. the right Sine of 66-29 a = sine of 90 then the = Tangent of the Suns distance from ♈ or ♎ is the = Tangent of the Suns Right Ascention from the same Point of ♈ or ♎ as at 30 from ♈ it is 28 degrees or 1 hour and 52 minuts from ♈ neer Use XV. Having the Suns Right Ascention and greatest Declination to find the Angle of the Ecliptick and Meridian As Sine 90 to Sine 23-31 So is the Co-sine of the Suns Right Ascention to the Co-sine of the Angle of the Ecliptick and Meridian Lay the Thred to 23-31 counted on the degrees from the Head then count the Co-sine of the Right Ascention from the Center downward or the Sine from 90 upwards and take the ND from thence to the Thred and measure it from the Center and it shall reach to the Co-sine of the Angle required Example The Right Ascention being 30 degrees or 2 hours the Angle shall be 69-50 Make the right sine of 23-31 a = sine of 90 then the = co-sine of 30 viz. = 60 shall make the sine of 69-50 the Angle of the Ecliptick and Meridian Use XVI Having the Latitude and Declination of the Sun or Stars to find the Suns or Stars Amplitude at rising or Setting Take the Suns declination from the particular Scale of Sines and lay it from 6 in the hour or Azimuth-line and it shall give the Amplitude from South as it is figured or from East or West counting from 90 observing to turn the Compasses the same way from 90 or 6 as the declination is Northward or Southwards Example The Suns declination being 10 degrees Northward the Suns Amplitude or Line is 106-12 from the South or 16-12 from the East-point As co-sine of the Latitude to S. 90 So is S. of the Suns declination to S. of the Amplitude Take the Sine of the Suns declination make it a = in the co-sine of the Latitude and lay the Thred to the nearest distance and on the degrees the Thred shall shew the true Amplitude required Make the right Sine of the Suns declination a = in co-sine latitude then = 90 taken and measured from the Center gives the Amplitude or Line Use XVII Having the same Amplitude and Declination to find the Latitude As S. of the Suns Amplitude to S. the Suns Declination So is S 90 to Co-sine Latitude Take the sine of the Suns declination set one Point in the Sine of the Suns Amplitude lay the Thred to ND and on the degrees it sheweth the complement of the Latitude required Example The Declination being 20 degrees and the Amplitude 33-15 the complement of the Latitude will be 38-28 counting from the Head toward the End Make the right Sine of the Suns Declination a = sine in the Suns Amplitude then the = sine of 90 shall be the co-sine of the Latitude required Use XVIII Having the Latitude and Suns Declination to find his Altitude at East or West commonly called the Vertical-Circle or Azimuth of East or West Take the Suns Declination from the particular Line of Sines set one Point in 90 on the
Azimuth-line and lay the Thred to the ND and on the degrees it sheweth the Altitude required counting from 600 toward the End As S. latitude S. of 90 So S. of Suns declination to S. Suns height at East or West Take the sine of the Suns declination make it a = in the sine of the latitude and lay the Thred to ND and on the degrees it shall shew the Suns Altitude at East and West required Example Declination 10. Latitude 51-32 the Altitude is 12 degrees and 50 minuts As S. of the Suns Declination to = S. of Latitude So is the = S. of 90 to S. of Vertical Altitude Use XIX Having the Latitude and Suns Declination to find the time when the Sun will be due East or West Having gotten the Altitude by the last Rule take it from the particular Sine then lay the Thred to the Suns declination counted on the degrees then setting one Point in the Hour-line so as the other turned about shall but just touch the Thred and the Compass-point shall stay at the hour and minuit of time required As Tangent latitude to Sine 90 So is the Tangent of the Suns declination to Co-sine of the hour Or As sine 90 to Tangent Suns declination So is Co-tangent-latitude to Co-sine of the hour from noon Example Latitude 51-32 declination 10 the Sun will be due East at 6-32 and West at 5-28 Take the Tangent of the Latitude on the loose-piece counting from 60 towards the moveable-leg or else from 600 on the moving-leg or degrees according as the Latitude is above or under 45 degrees and lay it from the Center downwards and note the Point where it ends Then take from the same Tangent the Tangent of the Suns declination and setting one foot in the Point last noted lay the Thred to ND then the = sine of 90 shall be the sine of the hour from 6. Or by the Sines only work thus Take the sine of the Suns declination make it a = in sine of the latitude lay the Thred to ND then take ND from the Co-sine latitude to the Thred then set one foot in the Co-sine of the Suns declination lay the Thred to ND and on the degrees it gives the hour from noon as it is figured or the hour from 6 counting from the head counting 4 minuts for every degree Make the small Tangent of the Latitude if above 45 taken from the Center a = sine of 90 then the Tangent of the Suns declination taken from the same small Tangent and carried Parallely till it stay in like Sines shall be the Sine of the hour from 6. Or as before by Sines only Make sine Declination a = sine Latitude then take = Co-sine Latitude and make it a = Co-sine of the Suns Declination then take = 90 and lay it from the Center it gives the Sine of the hour from 6. Use XX. Having the Latitude and Suns Declination to find the Ascentional Difference or the Suns Rising and Setting and Oblique Ascention Lay the Thred to the Day of the Month or to the Suns Declination or true Place or to his Right Ascention for the Thred being laid to any one of them is then also laid to all the rest then in the Azimuth-line it cuts the Ascentional difference if it you count from 90 or the Suns Rising as you count the morning hours or his Setting counting the afternoon hours The Oblique Ascention is found out for the six Northern signs or Summer half-year by substracting the Suns difference of Ascentions out of the Suns Right Ascention But for the other Winter-half year or six Southern signs it is found by adding the Suns difference of Ascentions to his Right Ascention this sum in Winter and the remainder as above-said in Summer shall be the Suns Oblique Ascention required As Co-tangent Lat. to sine 90 To is the Tangent of the Suns declination to the sine of the Suns Ascentional difference Take the co-tangent latitude from the loose or moveable-piece as it is above or under 45 degrees make it a = in sine 90 lay the Thred to ND then take the Tangent of the Suns declination from the same Tangents and carry it = till it stay in the parts that the other foot turned about will but just touch the Thred which parts shall be the Sine of the Suns Ascentional difference required Or ●hus by Sines only Make the sine of Declination a = Co-sine of the Latitude lay the Thred to ND then take the = sine of Latitude make it a = in Co-sine of the declination and lay the Thred to ND and on the degrees it shall cut the Suns Ascentional-difference required which being turned into time by counting 4 minuts for every degree and added to or taken from 6 gives the Suns Rising in Summer or Winter Make the Co-tangent Latitude a = sine of 90 then take Tangent of the Suns declination and carry it = till it stay in like parts viz. the Sine of the Suns Ascentional difference required Example otherwise As sine 90 to = Tangent 38-28 So is = Tangent of 23-31 the Suns greatest declination to the sine of the Suns greatest Ascentional difference 33 deg and 12 min. Use XXI The Latitude and Suns Declination given to find the Suns Meridian Altitude When the Latitude and Declination is both alike viz. both North or both South then substract the Declination out of the Latitude or the less from the greater and the remainder shall be the complement of the Suns Meridian Altitude But when they be unlike then add them together and the sum shall be the complement of the Meridian Altitude The contrary work serves when the complement of the Latitude and Declination is given to find the Meridian Altitude Lay the Thred to the Declination counted on the degrees from 600 the right way toward the Head for North and toward the End for South declination Then Take the nearest distance from the Center-prick at 12 in the Hour-line to the Thred this distance measured on the Particular-line of Sines shall shew the Suns Meridian-Altitude required Use XXI The Latitude and Hour from the midnight Meridian given to find the Angle of the Suns Position viz. the Angle between the Hour and Azimuth-lines in the Center of the Sun As Sine 90 to Co-sine of the Latitude So is the Sine of the Hour from Midnight to the sine of the Angle of Position Example As Sine 90 to Co-sine Latitude 38-28 So is the Co-sine of the Hour from midnight 120 for which you must use 60 to 32-34 the Angle of Position Take the distance from the Hour to the 90 Azimuth on the Hour-line and measure it in the particular sines and it shall shew the Angle of Position required This holds in the Equinoctial Take Co-sine Latitude make it a = in sine 90 then take