Selected quad for the lemma: hand_n

Word A	Word B	Word C	Word D	Occurrence	Frequency Band	MI	MI Band	Prominent
hand_n	left_a	right_a	turn_v	28,868	5	7.6673	4	true

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

Zenith and therefore needlesse to put them on In the North reclining more then the Equator the plane in our example must be elevated 120 degr above the horizon and the stiles of both must point to the North pole Lastly as all other planes have two faces respecting the contrary parts of the heavens so these recliners have opposite sides look downwards the Nadir as those do towards the Zenith and may be therefore made by the same rules or if you will spare that labour and make the same Dials serve for the opposite sides turn the centers of the incliners downwards which were upwards in the recliners and those upwards in the incliners which were downwards in the recliners and after this conversion let the hours on the right hand of the meridian in the recliner become on the left hand in the incliner and contrarily so have you done what you desired and this is a general rule for the opposite sides of all planes Probl. 11. To draw the hour-lines upon a declining reclining or declining inclining plane DEclining reclining planes have the same varieties that were in the former reclining North and South for either the declination may be such that the reclining plane will fall just upon the pole and then it is called a declining Equinoctial or it may fall above or under the pole and then it is called a South declining cast and west recliner on the other side the declination may be such that the reclining plane shall fall just upon the intersection of the Meridian and Equator and then it is called a declining polar or it may fall above or under the said intersection and then it is called a North declining East and West recliner The three varieties of South recliners are represented by the three circles AHB falling between the pole of the world and the Zenith AGB just upon the pole and AEB between the pole and the horizon and the particular pole of each plane is so much elevated above the horizon upon the azimuth DZC crossing the base at right angles as the plane it self reclines from the Zenith noted in the Scheme with I K and L. 1. Of the Equinoctiall declining and reclining plane This plane represented by the circle AGB hath his base AZB declining 30 degrees from the East and West line EZW equal to the declination of the South pole thereof 30 degrees from S the South part of the Meridian Easterly unto D reclining from the Zenith upon the azimuth CZD the quantity ZG 34 degrees 53 min. and Passeth through the pole at P. Set off the reclination ZG from D to K and K shall represent the pole of the reclining plane so much elevated above the horizon at D as the circle AGB representing the plane declineth from the Zenith Z from P the pole of the world to K the pole of the plane draw an arch of a great circle PK thereby the better to informe the fancie in the rest of the work And if any be desirous to any declination given to fit a plane reclining just to the pole or any reclination being given to finde the declination proper to it this Diagram will satisfie them therein for in the Triangle ZGP we have limited First the hypothenusal PZ 38 degrees 47 min. Secondly the angle at the base PZG the planes declination 30 degrees Hence to finde the base GZ by the seventh case of right angled spherical triangles the proportion is As the Radius 90 10.000000 To the co-sine of GZP 30 9.937531 So the tangent of PZ 38.47 9.900138 To the tangent of GZ 34.53 9.837669 the reclination required If the declination be required to a reclination given then by the 13 case of right angled spherical triangles the proportion is As the Radius 90 10.000000 To the tangent of ZG 34.53 9.837669 So the co-tangent of PZ 38.47 10.099861 o the co-sine of GZP 39. 9.937530 And now to calculate the hour-lines of this Diall you are to finde two things first the arch of the plane or distance of the meridian and substile from the horizontal line which in this Scheme is PB the intersection of the reclining plane with the horizon being at B. And secondly the distance of the meridian of the place SZPN from the meridian of the plane PK which being had the Diall is easily made Wherefore in the triangle ZGP right angled at G you have the angle GZP given 30 degrees the declination and ZP 38 degr 47 min. the complement of the Pole to finde GP and therefore by the eighth case of right angled spherical triangles the proportion is As the Radius 90 10.000000 To the sine of ZP 38.47 9.793863 So is the sine of GZP 30 9.698970 To the sine of GP 18.12 9.492833 Whos 's complement 71 deg 88 min. is the arch PB desired The second thing to be found is the distance of the Meridian of the place which is the houre of 12 from the substile or meridian of the plane represented by the angle ZPG which may be found by the 11 Case of right angled sphericall Triangles for As the Radius 90 10.000000 Is to the sine of GP 18.12 9.492833 So is the co-tang of GZ 34.53 10.162379 To the co-tang of GPZ 65.68 9.655212 Whos 's complement is ZPK 24 deg 32 min. the arch desired Now because 24 deg 32 min. is more then 15 deg one houres distance from the Meridian and lesse then 30 deg two houres distance I conclude that the stile shall fall between 10 and 11 of the clock on the West side of the Meridian because the plain declineth East if then you take 15 deg from 24 deg 32 min. there shall remain 9 deg 32 min. for the Equinoctiall distance of the 11 a clock houre line from the substile and taking 24 deg 32 min. out of 30 deg there shall remain 5 deg 68 min. for the distance of the houre of 10 from the substile the rest of the houre distances are easily found by continual addition of 15 deg Unto these houre distances joyn the naturall tangents as in the East and West Dials which will give you the true distāces of each houre from the substile the plane being projected as in the 5 Pro. for the east west dials or as in the 8 Prob. for the Equinoctial according to which rules you may proportion the length of the stile also which being erected over the substile and the Diall placed according to the declination 30 deg easterly and the whole plain raised to an angle of 55 deg 47 min. the complement of the reclination the shadow of the stile shall give the houre of the day desired 2. To draw the houre lines upon a South reclining plain declining East or West which passeth between the Zenith and the Pole In these kinde of declining reclining plains the South pole is elevated above the plane as is clear by the circle AHB representing the same which falleth between the Zenith and the North pole and therefore hideth

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

appear by that which followes 6. § When of four numbers given the second exceeds the first as much as the fourth exceeds the third the summe of the first and fourth is equal to the summe of the second and third and contrarily As 8 5 6 3. here 8 exceeds 5 as much as 6 exceeds 3 therefore the summe of the first and fourth namely of 8 and 3 is equall to the summe of the second and third namely of 5 and 6 from whence necessarily followes this Corollary When four numbers are proportionall the summe of the Logarithmes of the mean numbers is equal to the summe of the Logarithmes of the extreams Example Let the four proportional numbers be those exprest in the first column of the first Table in this Chapter viz. 4 16 32 128 in which Table the Logarithme of 4 under the letter A is 3 the Logarithme of 16 5 the Logarithme of 32 6 and the Logarithme of 128 is 8. Now as the summe of 5 and 6 the Logarithmes of the mean numbers do make 11 so the summe of 3 and 8 the Logarithmes of the extreames do make 11 also 7. § When four numbers be proportional the Logarithme of the first substracted from the summe of the Logarithmes of the second and third leaveth the Logarithme of the fourth Example Let the proportion be as 128 to 32 so is 16 to a fourth number here adding 5 and 6 the Logarithmes of the second and third the sum is 11 from which substracting 8 the Logarithme of 128 the first proportional the remainer is 3 the Logarithm of 4 the fourth proportional 8. § If instead of substracting the aforesaid Logarithme of the first we adde his complement arithmetical to any number the totall abating that number is as much as the remainer would have been The complement arithmetical of one number to another as here we take it is that which makes that first number equall to the other thus the complement arithmetical of 8 to 10 is 2 because 8 and 2 are 10. Now then whereas in the example of the last Proposition substracting 8 from 11 there remained 3 if instead of substracting 8 we adde his complement arithmeticall to 10 which is 2 the totall is 13 from which abating 10 there remains 3 as before both the operations stand thus As 128 is to 32 So is 16 Logar 8 compl arithmetical 2 6   6 5   5 The aggreg of 1.2   11 Their aggregate is 13 To 4   3     from which abate 10 there remaines 3 and the like is to be understood of any other The reason is manifest for whereas we should have abated 8 out of 11 we did not onely not abate it but added moreover his complement to 10 which is 2 wherefore the total is more then if should be by 8 2 that is by 10 wherefore abating 10 from it we have the Logarithme desired which rule although it be generall yet we shall seldome have occasion to use any other complements then such as are the complements of the Logarithmes given either to 10,000000 or to 20,000000 the ● complement arithmetical of any Logarithme to either of these numbers is that which makes the Logarithme given equal to either of them Thus the complement arithmetical of the Logarithme of 2 viz. 0301030 is 9698970 because these two numbers added together do make 10.000000 and thus the complement thereof to 20 000000is 19698970 if therefore 0301030 be substracted from 10.000000 the remainer is his complement arithmetical But to finde it readily you may instead of substracting the Logarithme given from 10.000000 write the complement of every figure thereof unto 9 beginning with the first figure toward the left hand and so on till you come to the last figure towards the right hand and thereof set down the residue unto 10. Thus for the complement arithmetical of the aforesaid Logarithme 0301030 I write for 0 9 for 3 6 for 0 9 for 1 8 for 0 9 for 3 again I should write 6 but because the last place of the Logarithme is a cipher and that I must write the complement thereof to 10 instead of 6 I write 7 and for 0 0 and so have I this number 9698970 which is the complement arithmetical of 0301030 as before 9. § Every Logarithme hath his proper Characteristick and the Character or Characteristicall root of every Logarithme is the first figure or figures towards the left hand distinguished from the rest by a point or comma Thus the Character of the Logarithmes of every number lesse then 10 is 0 but the Character of the Logarithme of 10 is 1 and so of all other numbers to 100 but the Character of the Logarithme of 100 is 2 and so of the rest to 1000 and the Character of the Logarithme of 1000 is 3 and so of the rest to 10000 in brief the Characteristick of any Logarithme must consist of a unite lesse then the given number consisteth of digits or places And therefore by the Character of a Logarithme you may know of how many places the absolute number answering to that Logarithme doth consist 10. § If one number multiply another the summe of their Logarithme is equal to the Logarithme of the product As let the two numbers multiplied together be 2 and 2 the products is 4 I say then that the summe of the Logarithmes of 2 and 2 or the Logarithme of 2 doubled is equal to the Logarithme of 4 as here you may see 2. 0.301030 2. 0.301030     4. 0.602060 Again let the two numbers multiplied together be 2 and 4 the product is 8 I say then that the summe of the Logarithmes of 2 and 4 is equall to the Logarithme of 8 as here you may also see 2. 0.301030 4. 0.602060     8. 0.903090 And so for any other The reason is for that by the ground of multiplication as unit is in proportion to the multiplier so is the multiplicand to the product therefore by the sixth of this Chapter the sum of the Logarithmes of a unit and of the product is equall to the summe of the Logarithmes of the multiplier and multiplicand but the Logarithme of a unit is 0 therefore the Logarithme of the product alone is equal to the summe of the Logarithmes of the multiplier and multiplicand And by the like reason it three or more numbers be multiplied together the summe of all their Logarithmes is equall to the Logarithme of the product of them all 11. § If one number divide another the Logarithme of the Divisor being substracted from the Logari●hme of the Dividend leaveth the Logarithme of the Quotient As let 10 be divided by 2 the quotient is 5. I say then if the Logarithme of 2 be substracted from the Logarithme of 10 there will remain the Logarithme of 5 as here is to be seen 10. 1.000000 2. 0.301030     5. 0.698970 For seeing that the quotient multiplied by the divisor produceth the dividend therefore