the tangent of the present declination To the right ascension required Onely you must regard to give it a right account by considering the time of the year and how many 90s past PROR 14. To find an altitude by the length and shadow of any perpendicular object Lay the hair on one legg to the length of the shadow found on the line of numbers and the hair of the other leg to the length of the object that caused the shadow found on the same line of the numbers then observe the lines between and which way when the legs are so set bring the first of them to the tangent of 45 and the other leg shall â—hew on the line of tangents so many turns between and the same way the tangent of the altitude required Thus may you apply all manner of quest to the Serpentine-line work them by the same Canons that you use for the Logarithms in all or most Authors PROP. 15. To square and cube a number and to findethe square root or cube roat of a number The squaring of a number is nothing else but the multiplying of the number by it self as to square 12 is to multiply 12 by 12 and then the cubing of 12 is to multiply the square 144 by 12 that makes 1728 and the way to work it is thus Set the first leg to 1 and the other to 12 then set the first to 12 and then the second shall reach to 144 then set the first to 144 and the second shall reach to 1728 the cube of 12 required but note the number of figures in a cube that hath but one figure is certainly found by the line by the rule aforegoing but if there be more figures then one so many times 3 must be added to the cube and so many times two to the square To find the square root of a number do thus Put a prick under the first the third the 5th the 7th the number of pricks doth shew the number of figures in the root and note if the figures be even count the 100 to be the unit if odde as 3 5 7 9 c. the 10 at the beginning must be thâ— unit as for 144 the root consists of two figures because there is two pricks under the number and if you lay the index to 144 in the numbers it cuâ—s on the line of Logarithms 15870 the half of which is 7915 whereunto if you lay the index it shall shew the 12 the root required but if you would have the root of 14+44 then divide the space between that number and 100 you shall finde it come to 8 4140 that is four turnes and 4140 for which four turnes you must count 80000 the half of which 8,4140 is 4,2070 whereunto if you lay the index and count from 1444 â—r 100 at the end you shall have it cut at 38 lack four of a 100. To extract the cubique root of a number set the number down and put a point under the 1 the 4th the 7th and 10th and look how many pricks so many figures must be in the root but to finde the unity you must consider if the prick falls on the last figure then the 10 is the unit at the beginning of the line as it doth in 1728 for the index laid on 1728 in the Logâ—rithms sheweth 2,3760 whose third part 0,7920 counted from 10 falls on 12 the root but in 17280 then you must conceive five whole turnes or 1000 to be added to give the number that is to be divided by three which number on the outermost circle in this place is 12 +3750. by conceiving 10000 to be added whose third part counted from 10 viz. two turnes or 4.125 shall fall in the numbers to be near 26. But if the prick falls of the last but 2 as in 172800 then 100 at the end of the line must be the unit and you must count thus count all the turnes from 172830 to the end of the line and you shall finde them to amount to 7,6250 whose third part 2 5413 counted backward from 100 will fall on 55,70 the cubique root required PROP. 16. To work questions of interest or progression you must use the help of equal parts as in the extraction of roots as in this question if 100 l. yield 106 in one year what shall 253 yield in 7 year Set the first leg to 10 at the beginning in this case representing a 100 and the other to 106 and you shall finde the legs to open to 253 of the small divisions on the Logarithms multiply 253 by 7 it comes to 1771 now if you lay the hair upon 253 and from the place where the index cuts the Logarithms count onwards 1771 it shall stay on 380 l. 8 s. or rather thus set one leg to the beginning of the Logarithms and the other to 1771 either forward or backward and then set the same first leg to the sum 253 and the second shall fall on 380. 8 s. according to estimation the contrary work is to finde what a sum of money due at a time to â—ome is worth in ready money this being premised here is enough for the ingenious to apply it to any question of this nature by the rules in other Authors However you may shortly expect a more ample treatise in the mean time take this for a taste and farewell The Use of the Almanack Having the year to finde the day of the week the first of March is on in that year and Dominical letter also First if it be a Leap-year then look for it in the row of Leap-year and in the column of week-days right over it is the day required and in the row of dominical letters is the Sunday letters also but note the Dominical letter changeth the first of Ianuary but the week day the first of March so also doth the Epact Example In the year 1660 right over 60 which stands for 1660 there is G for the Dominical or Sunday letter beginning at Ianuary and T for thursday the day of the week the first of March is on and 28 underneath for the epact that year but in the year 1661. being the next after 1660 the Leap-year count onwards toward your right hand and when you come to the last column begin again at the right hand and so count forwards till you come to the next Leap-year according to this account for 61 T is the dominical letter and Friday is the first of March But to finde the Epact count how many years it is since the last Leap-year which can be but three for every 4th is a Leap-year and adde so many times 11 to the epact in the Leap-year last past and the sum if under 30 is the Epact if above 30 then the remainder 30 or 60 being substracted is the Epact for that year Example for 1661.28 the epact for 1660 and 11 being added makes 39 from which take 30 and there remaineth 9 for the Epact for the year 1661 the thing required Note that in orderly counting the years when you come to the Leap-year you must neglect or slip one the reason is because every Leap-year hath two dominical letters and there also doth the week day change in the first of March so that for the day of the month in finding that the trouble of remembring the Leap-year is avoided To find the day of the Month. Having found the day of the week the first of March is on the respective year then look for the month in the column and row of months then all the daies right under the month are the same day of the week the first of March was on then in regard the days go round that is change orderly every seven days you may find any other successive day sought for Example About the middle of March 1661 on a Friday what day of the month is it First the week day for 1661 is Friday as the letter F on the next collumn beyond 60 sheâ—etâ— then I look for 1 among the months and all the days right under viz. 1 8 15 22 29. in March and November 61 are Friday therefore my day being Friday and about the middle of the month I conclude it is the 15th day required Again in May 1661. on a Saturday about the end of May what day of the month May is the third month by the last rule I find that the 24 and 31 are Fridays therefore this must needs be the 25 day for the first of Iune is the next Saturday FINIS ERRATA PAge 23. l. 4. adde 1660 p. 24. l. 6. for 5 hours r. 4. l. 9. for 3. 29. r. 4. 39. 1. 12 for 5. 52. r. 4 52. l. 13. for 3. 39. r. 4. 39. l. 17 for 5 hours 52. r. 4. 52. p. 27. l. ult dele or 11. 03. p 31. l 4. for sun r. sum p. 50. l. 8. for B r. A. p. 50 d CHAP. XII p. 51. r. 16. for 6. 10. 1. 6 to 10. p. 71. l. 6. for 7 4 r. 1 4. l. penult for 2 afternoon r. 1. p. 74 l. ult for 1. r. 1 2. p. 83. l. 18. for BC r. BD. p. 69 l. 17. add measure p. 129 l. 24. for right of r. right sine of p. 114 l. 9 for 18 3. r 18 13. p. 147 1. 2 for 20 r. 90 p. 163. l. 16. for of r. on