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	rank_n	right_a	50,843	5	9.3193	5	true

Text snippets containing the quad

ID	Title	Author	Corrected Date of Publication (TCP Date of Publication)	STC	Words	Pages
A64224	Thesaurarium mathematicae, or, The treasury of mathematicks containing variety of usefull practices in arithmetick, geometry, trigonometry, astronomy, geography, navigation and surveying ... to which is annexed a table of 10000 logarithms, log-sines, and log-tangents / by John Taylor.	Taylor, John, mathematician.	1687 (1687)	Wing T534; ESTC R23734	190,995	514

the Fractional part of the given Mixt-number into a Decimal Fraction which shall consist of Tenaries of places Then to the whole number annex the Decimal Fraction and Extract the Cube-root of the Whole and observe that so many points as are over the Integers so many of the first places in the Quotient must be Esteemed Integers and the rest Expresseth the Fractional part of the Root in Decimal parts of a Fraction so the Cube-root of 2 ⅜ Decimal 2 375000000 c. will be found to be 1 334 or 1 334 1000 and is very near the true Root and so for any other Mixt-number of this nature CHAP. II. The Explication and use of the Tables of LOGARITHMS SECT I. The Explication of the Tables of the Logarithms and of parts proportional THE Logarithms were first invented found out and framed by that never to be forgotten and thrice Honourable Lord the Lord Nepeir which Numbers so found out and framed by his diligent industry he was pleased to call Logarithms which in the Greek signifies the Speech of Numbers I shall not here trouble you with the manner or the Construction of those Tables of Logarithms but shall first lay down some brief and general Rules that thereby the better you ma● Understand those Tables and then I shall e●plain their manifold uses in sundry Examples Arithmetical c. PROP. I. Any Number given under 10000 or 100000 to find the Logarithm corresponding thereunto 1. If the number propounded consist of one place whose Logarithm is required to be found as suppose 5 look for 5 in the top of the left hand Column under the Letter N and right against 5 and in the next Column under LOG you will find this number or rank of figures 0698970 which is the Logarithm of the number 5 required 2. If the number consisteth of two places as if it were 57 look 57 under N and opposite to it and under LOG you will find this number 1. 755875 which is the Logarithm of 57 the number propounded 3. If the number propounded consist of three places as 972 look for 972 under N and opposite to 972 and under ●o the Column you shall find this number 2. 987666 which is the Logarithm of 972 the number which was propounded 4. But if the number consist of four places as 685 look the three first figures 168 under the Column N and opposite to that and un●er 5 at the top of the page you will find this number 3. 226599 which is the Logarithm of 1685 the number propounded 5. But if the number given be above 10000 and under 100000 you may find its Logarithm by the Table of parts proportional printed at the latter end of this Book Thus if the Logarithm of 35786 be sought first seek the Log. of 3578 which will be 553649 and the common Difference under D is 121 with this difference 121 Enter the Table of parts proportional and finding 121 in the first Column under D you may then lineally under 6 find the number 72 which add to the Log. of 3578 that is 553649 it produceth 553712 which is the Log. of 35786 the number propounded now because the number propounded 35786 ariseth to the place of X. M. therefore there must be the figure 4 prefixed before its Logarithm and then it will be thus 4 553712 which 4 is called the Index as shall be hereafter shewed Now before we proceed to find numbers corresponding to Logarithms it will be necessary to explain the meaning of the first figure to the left hand of any Logarithm placed Mr. Briggs calleth it a Characteristick or Index which doth represent the distance of any the first figure of any whole number from Unity whose Index is 0 a Cypher so the Index o● 10 is 1 and so to 100 whose Index is 2 and s● to 1000 whose Index is 3 and so to 10000 whose Index is 4 and so if you persist furthe● the Characteristick is always one less in dignity than the places or figures os the number propounded PROP. II. To find the Logarithm belonging to a Vulgar Fraction and a Mixt number First as is before shewed if it be a Vulgar Fraction find the Log. of the Numerator and the Log of the Denominator then substract the Log of the Numerator from the Log of the Denominator the remainder is the Log of the Fraction propounded Now if you would find the Logarithm of 5 7 do as is prescribed whose Log. I find to be 0. 146121 Now to find the Log. of a Mixt Number reduce it into an Improper Fraction and then do as before so the Log of 15 ⅖ Improper 77 5 is 1 187 52 and so do for any other Mixt number PROP III. A Logarithm propounded to find the whole or Mixt number corresponding thereunto For the more speedy finding the number answering unto the Logarithm propounded observe that if the Index be 0 then the Number sought may be found between 1 and 10 If 1 between 10 and 100 if 2 between 100 and 1000 if 3 between 1000 and 10000 and so on still observing the Rules of the Characteristick or Index therefore loo in the Table untill you find the Logarithm proposed and against it in the Margent according to the aforegoing directions under N you shall find the number belonging thereunto This Rule holds in force in Mixt Numbers also Thus. 0. 845098 1. 556302 2. 130334 3. 980276 Are the Logarithms of 7 36 135 9556 NOTE But if you cannot find the Logarithm exactly in the Table as in many operations it so hapneth you must then take the nearest Logarithm Number to the Logarithm propounded and so take the number belonging thereto ●or the desired number SECT II. Of the Admirable use of the Logarithms in Arithmetick PROP. I. To Multiply one number by another Admit 90 be to be multiplied by 42 what is the product To find which first find the Log. of the Multiplicand 90 whos 's Log. is 1. 95424 Then find the Log. of the Multiplier 42 whose Log is 1 62324 then add these two Log together viz. the Log of the Multiplicand and Multiplier their sum is 3 57748 which is the Log of 3780 the product of 90 and 42 Multiplied together PROP. II. To Divide one number by another Admit the Dividend or number to be divided be 648 and the Divisor 72 what is the number that the Quotient shall consist off To find which first write down the Logarithm of the Dividend 648 which is 2. 81157 and also write down the Logarithm of the Divisor 72 which is 1. 85733. Now substract the Log of the Divisor out of the Log of the Dividend the remainder is 0. 95424 which is the Logarithm of 9 so I conclude that the Divisor 72 is contained in the Dividend 648 9 times and so do for any other PROP. III. To find the Square-Root of a Number Admit it be required to Extract the Square-Root of the Number 144

as in the former Proposition PROP. III. In a Rectangled Triangle the Base and Cathetus given to find the other parts thereof In the Triangle ABC the Base AB is 736 parts and the Cathetus BC is 467 4 10 parts and the Angle B between them is a right angle o● 90° And here you may make either side of the Triangle Radius but I shall make BC the Cathetus Radius and then to find the Angle at the Cathetus C this is the Analogy or ●●●portion As Log. Cathet BC 467 4 10 parts To Radius or S 90° So Log. Base AB 736 parts To T. V. Cathe C 57° 35 ' as required Secondly I find the other Angle at A to be 32° 25 ' it being the Complement to C 57 35 ' unto 90° Thirdly To find out the Hypothenuse AC this is the analogy or proportion As S. V. Cathe C. 57° 35 ' To Log. Base AB 736 parts So Radius or S. 90° To Log. Hypothenuse AC 871 8 10 parts 〈…〉 quired But making the Base AB Radius yo● may find the Hypothenuse AC by this anal 〈…〉 or proportion Plate 1 Page 65 As Radius or S. 90° To Log. Base AB 736 parts So Sc. V. Base A 32° 25 ' To Log. Hypothenuse AE 871 8 10 parts required and thus you have all the parts of the Triangle propounded PROP. IV. The Base and Hypothenuse with the Angle between them given to find the other parts of a Rect-angled Triangle In the Triangle ABC the Base AB is 736 parts and the Hypothenuse AC is 871 8 10 parts and the Angle A included between them is 32° 25 ' First to find the Angles and first remember that the Angle B is a right Angle or 90° Secondly that the Angle at C is the Complement to the Angle at A 32° 25 ' unto 90° and therefore is 57° 35 ' Now these being known you may find the Cathetus by this analogy or proportion As S. V. Cathe C. 57° 35 ' To Log. Base AB 736 parts So S. V. Base A 32° 25 ' To Log. Cathe BC 467 4 10 parts required Thus I have sufficiently explained all the Cases of Plain Rect-angle Triangles for to these rules they may be all reduced SECT III. Of Oblique-Angled Plain Triangles PROP. I. Two Angles and a side opposite in an Oblique-Angled Triangle given to find the other parts thereof IN the Triangle ABC the Angleat A is 50° and at C is 37° and the side AB is 30 parts and opposite to the Angle C First to find the Angle B remember that as 't is said in the third Maxim aforegoing 't is the Complement to the Angles A 50° and C 37° to 180° and therefore is the Angle at B 93° Secondly having thus found the Angles the two unknown sides may be found by the proportion they bear to their opposite Angles for that proportion holds also in these thus to find the side BC this is the analogy or proportion As S. V. C 37° 00 ' To Log. side AB 30 parts So S. V. A. 50° 00 ' To Log. side BC 38 19 100 parts required to be found But it may be more readily found and performed in such case as this where you have a Sine or Tangent in the first place by the Arithmetical Complement thereof and so save the Substraction Now the readiest way to find the Arithmetical Complement is that of Mr. Norwood in his Doctrine of Triangles which is thus begin with the first Figure towards the left hand of any Number and write down the Complement or the remainder thereof unto 9 And so do with all the rest of the Figures as you see here done Saying 9 wants of 9 0 and again 9 wants 0 6 wants 3 2 wants 7 3 wants 6 9 wants 0 only when you come to the last Figure to the right hand take it out of 10 so 8 wants 2 of 10 Thus you may readily find the Co-Ar of any Sine almost as soon as the Sine it self But if you want the Complement Arithmetical of any Tangent you may take the Co-tang which is exactly the Co-Arith of the double Radius so that the Tangent and Co-tangent of an Arch makes exactly 20. 000000. Now if the Radius be in the first place then there is no need of taking the Co-Arith of the first Number only you must cut off the first I to the left hand thus X and you will have the Logarithm of the Number desired Thirdly now to find the side AC by the opposite Angle B which is 93° 00 ' And see 〈…〉 ng the Angle B exceeds 90° you must work 〈…〉 y the Complement to 180° as in the seventh 〈…〉 ork in page 61 is taught Thus having found all the parts of the Triangle propounded Viz. The Angle B to be 93° 00 ' the side AC to be 49 78 100 parts and the side BC to be 38 19 100 parts as was required to be found PROP. II. Two sides and an Angle opposite to one of them in an Oblique-angled Triangle given to find the other parts thereof In the Triangle ABC the side AB is 30 parts and the side AC is 49 78 100 parts and the opposite Angle C is 37° 00 ' First To find the Angle at B this is the Analogy or Proportion As Log. cr AB 30 parts To S. V. at C 37° 00 ' So Log. cr AC 49 78 100 parts To Sc. V. B 93° 00 ' as was required to be found Now seeing that the Angle C is 37° 00 ' and the Angle B is 93° 00 ' which makes 120° 00 ' therefore must the Angle A be 50° 00 ' the Complement to 180° so having found all the three Angles you may find the other side CB 38 19 100 parts as afore in the first proposition by his opposite Angle PROP. III. Two Sides of an Oblique-angled Triangle with the Angle included between them given to find the other parts thereof In the Triangle ABC the side AC is 49 78 100 parts the side DB is 30 parts and the Angle A between them is 50° 00 ' and 't is required to find the other parts of the Triangle propounded To resolve this Conclusion let fall a Perpendicular DB from the Angle B on the side AC by prop. 3. § 1. chap. 4 and then proceed thus First Seeing the Oblique-angled Triangle ABC is divided into two Rectangled Triangles Viz. ADB and BDC Now I will begin with the Triangle ADB in which is given the Angle A 50° 00 ' and the Angle D is a right Angle or 90° and the side AB 30 parts and the sides AD and DB and the Angle at B are required First to find the Angle at B remember that it is the Complement unto the Angle A 50° 00 ' unto 90° 00 ' and therefore must the Angle B be 40° 00 ' Now for to find the Cathetus BD as in prop. 1. and 2 §

2. chap. 5. by the Rule of opposition the Analogy or Proportion holds thus As Radius or S. 90° To Log. Hypoth AB 30 parts So S. V. at A 50° 00 ' To Log. Cath. BD 22 98 100 parts sought And AGAIN say As Radius or S. 90° To Log. Hypoth AB 30 parts So S. V. at B 40° 00 ' To Log. Base AD 19 28 100 parts sought Thus in the Triangle ADB you have found the Angle B to be 40° 00 ' the Cathetus BD to be 22 98 100 parts and the Base AD to be 19 28 100 parts as was so required Now for the other Triangle which is BDC in which there is given the side BD 22 98 100 parts and the Angle at D is a Right-angle or 90° and the sides DC and CB and the Angles B and C are required First to find the side DC substract AD 19 28 100 parts out of AC 49 78 100 parts there remains the Base DC 30 50 100 parts Thus have you the two sides of the Triangle to wit the Base DC 30 50 100 parts and the Cathetus BD 22 98 100 parts and the Angle D between them is a Right-angle or 90° Now you may find the Angle at B by the Tangent as in prop. 3. § 2. chap. 5. thus As Log. Cath. BD 22 98 100 parts To Radius or S. 90° So Log. Base CD 30 50 100 parts To T. V. B. 53° 00 ' Secondly For the Angle C remember 't is the Complement of the Angle B 53° to 90° and therefore is the Angle C 37° 00 ' required Thirdly To find the Hypoth BC this is the Analogy or Proportion As S. V. B. 53° 00 ' To Log. Base DC 30 50 100 parts So Radius or S. 90° To Log. Hypoth BC 38 19 100 parts Thus have you found all the required parts of the Triangle ABC propounded viz. the Angle C to be 37° 00 ' the Angle B to be 93° 00 ' and the Side BC 38 19 10● parts as required to be found Another way to perform the same Take the Sum of the two sides and the difference of the two sides and work as followeth Now to find the two Angles B and C this is the Manner and by this Analogy or Proportion they are found out and known As Log. Z. cr s. AB and CA 79 78 100 parts To Log. X. cr s. AB and CA 19 78 100 parts So T. of ½ VV unknown 65° 00 ' To T. ½X of VV 28° 00 ' This difference of Angles 28° 00 ' add unto 65° 00 ' half the difference of the unknown Angles and it shall produce 93° 00 ' which is the greater Angle and substracted from it leaves 37° 00 ' which is the lesser Angle C so have you the required Angles PROP. IV. The three sides of an Oblique-angled Triangle given to find the Angles In the Triangle ABC the side AC is 49 78 100 parts the side AB is 30 parts and the side BC is 38 19 100 parts and the three Angles of the Triangle are required The resolution of this Conclusion is thus Take the Summ and Differ of the two sides AB and BC And then work as follows To find a Segment of the Base AC to wit CE say As Log. Base AC 49 78 100 parts To Z. cr s. AB and BC 68 19 100 parts So X. cr s. AB and BC 8 19 100 parts To Log of a Segment of the Base AC to wit C E 11 22 100 parts This Segment of the Base CE 11 22 100 parts being substracted from the whole Base AC 49 78 100 parts the remainder is EA 38 56 100 parts in the middle of which as at D the Perpendicular DB will fall from the Angle B and so divide it into two Rectangled Triangles to wit ADB and CDB whose Base DA is 19 28 100 parts which taken from AC 49 78 100 parts leaves the Base of the greater Triangle CD 30 50 100 parts Now having the two Bases of these two Triangles and their Hypothenuses to wit CD 30 50 100 parts DA 19 28 100 parts CB 38 19 100 parts and BA 30 parts you may find all their Angles by the Rule of Opposite sides to their Angles as afore I. In the Triangle CDB To find the Angles this is the Analogy or Proportion As Log. BC 38 19 100 parts To Radius or S. 90° So Log. DC 30 50 100 parts To S. V. B 53° 00 ' whose Complement is the Angle at C 37° 00 ' unto 90 or a Quadrant II. In the Triangle ADB To find the Angles this is the Analogy or Proportion As Log. AB 30 parts To Radius or S. 90° So Log. AD 19 28 100 parts To S. V. B 40° 00 ' The Complement whereof unto 90° 00 ' is the Angle at A 50° 00 ' Now in the first Triangle CDB there is found the Angle C to be 37° 00 ' and the Angle B to be 53° 00 ' In the second Triangle ADB there is found the Angle A to be 50° 00 ' and the Angle B to be 40° 00 ' Now the two Angles at B to wit 53° 00 ' and 40° 00 ' makes 93° 00 ' which is the Angle of the Oblique-angled Triangle ABC at B Thus the three Angles of the said given Triangle ABC are found as was required viz. the Angle A to be 50° 00 ' the Angle B to be 93° 00 ' and the Angle C to be 37° 00 ' as sought Thus I have sufficiently fully and plainly explained all the Cases of Plain Right-lined Triangles both Right and Oblique-angled I shall now fall in hand with Spherical Triangles both Right and Oblique-angled SECT IV. Of Spherical Rectangled Triangles And here first it will be necessary also to understand those few general Maxims or Rules that are of special Moment in the Doctrine of Spherical Triangles 1. THat a Spherical Triangle is comprehended and formed by the Conjunction and Intersection of three Arches of a Circle described on the Surface of the Sphere or Globe 2. That those Spherical Triangles consisteth of six distinct parts viz. three Sides and three Angles any of which being known the other is also found out and known 3. That the three Sides of a Spherical Triangle are parts or Arches of three great Circles of a Sphere mutually intersection each other and as plain or Right-lined Triangles are measured by a Measure or Scale of equal parts So these are measured by a Scale or Arch of equal Deg●ees 4. That a Great Circle is such a Circle that doth bessect the Sphere dividing it into two equal parts as the Equinoctial the Ecliptick the Meridians the Horizon c. 5. That in a Right-angled Spherical Triangle the Side subtending the Right-angle we call the Hypothenuse the other two containing the Right-angle we may simply call the Sides and for distinction either of them may be called the Base or Perpendicular 6. That the Summ of the

edge shall shew the Hour of the day Now to draw the Hour-lines with the Radius of your line of Chords on M strike the Arch QN which divide into 5 equal parts in the points ● ● ● c. Then lay a Ruler from M unto each of those points and it will cut the line JK in the points * * * c. through which points by prop. 4. § 1. chap. 4. draw Parallels to 6 I 6 as the lines 77 88 c. which shall be the true Hour-lines of an East Plane from 6 in the Morning till 11 before Noon Then for the Hour-lines of 4 and 5 you must prick off 5 as far from 6 as 6 is from 7 and 4 as far as 6 is from 8 and draw the Hour-lines 55 and 44 as before Thus is your Dial compleated and in the forming of which you have made both an East and a West Dial which is the same in all respects only whereas the Arch H G through which the Equinoctial passed in the East Dial was described on the right hand of the Plane in the West it must be drawn on the left hand and the Hour-lines 4 5 6 7 8 9 10 and 11 in the Forenoon in the East Dial must be 8 7 6 5 4 3 2 and 1 in the West in the Afternoon as in the Figure plainly appeareth Now you may find the distance of the Hour-lines from the Substile by this Analogy or Proportion As the Radius To the Height of the Stile So is the Tangent of any Hours distance from 6 To the distance thereof from the Substile PROP. IV. How to draw the Hour-lines on a direct South and North Plane This Plane or Dial must stand upright having his face or Plane if it be a South Dial directly opposite unto the South but if a North Plane directly opposite unto the North now admit it be required to make a Direct South Dial for the Latitude of 51° 32 ' To make which first describe the Circle ABCD to represent an E●ect direct South Plane cross it with the Diameters CB and AD then out of your Line of Chords take 38° 28 ' the Complement of the Latitude and set it from A unto a and from B unto b Then lay a Ruler from C unto a and it will cut the Meridian ARD in P the Poles of Plate V Page 261 As Radius or S. 90° To Sc. of the Latitude So is T. of the Hour from Noon To T. of the Hour-line from the Meridian PROP. V. How to draw the Hour-lines on an Horizontal Plane This Horizontal Plane or Dial is one of the best and most usefull Dials in our Oblique Hemisphere Admit it be required to make an Horizontal Dial for the Latitude of 51° 32 ' To make which first describe the Circle AB CD which representeth your Horizontal Plane Then cross it with the two Diameters ARC and BRD Then take 51° 32 ' out of your Line of Chords and set it from B to a and from C to b Then lay a Ruler from A unto a and it will cut the Meridian BD in P the Pole of the World Then lay a Ruler from A unto b and it will cut ABD the Meridian in the point AE where the Equinoctial cutteth the Meridian then through the three points A AE and C draw the Equinoctial Circle whose Center is at H and found as in the former proposition Then divide the Semicircle ADC into 12 equal parts in the points ● ● ● c. Then lay a Ruler to R the Center of the Plane and on those points so shall the Equinoctial Circle AAeC be by it divided into 12 unequal parts in the points * * * * c. Then a Ruler laid unto P the Pole of the World and those Points shall cut the Semicircle CDA in those Points I I I c. Lastly from the Center R and through those Points let there be drawn right lines which shall be the true Hour-lines of such an Horizontal Plane from 6 in the Morning untill 6 at Night but for the Hours of 4 and 5 in the Morning and 7 and 8 in the Evening they are delineated by producing 4 and 5 in the Evening through the Center R and 7 and 8 in the Morning extending them out unto the other side of the Plane so shall you have those Hour-lines also on your Plane delineated as you see in the Figure The Stile of this Plane may be a thin Plate of Brass cut exactly unto the Quantity of an Angle of 51° 32 ' and set Perpendicular on the Meridian line for the forming of this Stile take out of your Line of Chords 51° 32 ' and set it from D unto e and draw Re which shall be the Axis of the Stile you may also prefix the Halves and Quarters of Hours in the very same manner as the Hours themselves were drawn Now to find out the distance of the Hour-lines from the Meridian say As the Radius or S. 90° To the S. of the Latitude So is the T. of the Hour from Noon To the T. of the Hour-line from the Meridian Line These kinds of Dials being so frequently used with us in this Oblique Sphere for the help of the speedy delineating of them I have annexed hereunto the Table of Longomontanus wherein the Hour-lines for many Latitudes are calculated A Table shewing the Distance of the Hour-lines from the Meridian in these Degrees of Latitude An Horizontal Dial Latitude The Hours from the Meridian A South Erect Dial Latitude xi i. x. ii ix iii viii iv vii v. vi D M D M D M D M D M D M 30 7 38 16 6 26 34 40 54 61 49 90 00 60 31 7 51 16 34 27 14 41 42 62 28 90 00 59 32 8 4 17 1 27 53 42 30 63 6 90 00 58 33 8 17 17 27 28 34 43 17 63 45 90 00 57 34 8 30 17 54 29 13 44 5 64 42 90 00 56 35 8 43 18 20 29 49 44 46 64 56 90 00 55 36 8 56 18 45 30 25 45 28 65 27 90 00 54 37 9 9 19 9 31 1 46 9 65 58 90 00 53 38 9 21 19 34 31 37 46 50 66 29 90 00 52 39 9 33 19 57 32 9 47 26 66 55 90 00 51 40 9 46 20 20 32 40 48 1 67 20 90 00 50 41 9 58 20 43 33 14 48 37 67 45 90 00 49 42 10 10 21 7 33 47 49 13 68 11 90 00 48 43 10 22 21 29 34 17 49 44 68 32 90 00 47 44 10 24 21 50 34 46 50 14 68 52 90 00 46 45 10 43 22 12 35 15 50 45 69 14 90 00 45 46 10 54 22 33 35 44 51 16 69 37 90 00 44 47 11 5 22 33 36 10 51 43 69 53 90 00 43 48 11 16 23 12 36 35 52 9 70 10 90 00 42 49 11 26 23 32 37 1 52 35 70 28