Selected quad for the lemma: ground_n

Word A	Word B	Word C	Word D	Occurrence	Frequency Band	MI	MI Band	Prominent
ground_n	distance_n	file_n	rank_n	1,730	5	11.1963	5	false

Text snippets containing the quad

ID	Title	Author	Corrected Date of Publication (TCP Date of Publication)	STC	Words	Pages
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

from Quit-rent and Taxes and the like the Rent is usually various according to the place and time where and wherein the Purchase shall happen to be then to find the quantity of the whole Purchase Say As 1 to 20 18 15 14 12 10 or 8 the number of years Purchase for Fee-simple or Coppy-hold Land or Houses Fee-simple or Coppy-hold For Leases of 60 50 40 or 30 years or 21 years So is the yearly Rent to the whole value Example A Parcel of Land worth 10 l. per annum Fee-simple valued at 20 years Purchase will amount to 200 l. For The Extent from 1 to 20 will reach the same way from 10 to 200 the whole price of 20 years Purchase at 10 l. per annum CHAP. XIV The Use of the Line of Numbers IN Military Questions Problem I. Any Number of Souldiers being propounded to order them into a Square Battel of Men that is as many in Rank as in File FInd the Square-root of the Number of Souldiers and that shall be the Number of Men in Rank and File required As suppose it were required to order 1770 Men in the order abovesaid you shall by the 8th Probl. of the 6th Chapt. find that the Square-root of 1770 is 42 and 6 over which here is not considerable Problem II. Any number of Souldiers propounded to order them into a double Battel of Men that is to say twice as many in Rank as File Find the Square-root of half the Number of Men and that is the Number of Men in File and the double the Number in Rank As for Example If 2603 were so to be placed the half of 2603 is 1301 whose Square-root by the 8th of 6th is 36 the number of Men in File and 72 the double thereof is the number in Rank For if you shall multiply 72 by 36 the Product is 2592 almost the number of Men propounded Problem III. Any Number of Souldiers being propounded to order them into a Quadruple Battel of Men viz. 4 times as many in Rank as File Find the Square-root of a 4th part of the Number of Men and that shall be the Number in File and 4 times so many the Number in Rank So the 4th part of 2603 is 650 whose Square-root is 25 ½ and 4 times 25 is 100 the Number in Rank Problem IV. Any Number of Souldiers being given together with their Distance one from another in Rank and File to order them into a Square Battel of Ground As suppose I would order 3000 Men so that being 7 foot asunder in File and 3 foot apart in Rank the Ground whereon they stood should be Square Extend the Compasses from 7 foot the distance in File to 3 foot the distance in Rank then that Extent applied the same way from 3000 the Number of Souldiers reaches to 1286 whose greatest Square-root is 35-7 that is 35 the Number of Men to be placed in File Then If you divide 3000 the whole Number by 35-7 the Quotient is 84 the Number in Rank to use and imploy a Square plat of ground to stand in As 7 to 3 so is 3000 to 1286 whose Square-root is 35-7 Then As ●5-7 to 1 so is 3000 to 84. Problem V. Any Number of Souldiers propounded to order them into Rank and File according to the ratio of any two Numbers given This Question is all one with the former For As the Number given for the distance in File is to that for the distance in Rank So is the whole Number of Souldiers to a 4th whose Square-root is the Number of Men in Rank Then The whole Number divided by the Number in Rank the Quotient is the Number to be placed in File Example Suppose 3000 Souldiers were to be ordered in Rank and File As 5 is to 10 or as 5 is to 9 that is to say that the Men in Rank might be in Proportion to them in File as 9 is to 5. Say thus As the Extent from 5 to 9 So is 3000 to 5400 whose Square-root is 73 ½ the Number of Men in Rank Then As 73 ½ to 1 So is 3000 to near 41 the Number in File Problem VI. There are 8100 Men to be ordered into a Square Body of Men and to have so many Pikes as to arm the main Square-Body round about with 6 Ranks of Pikes the Question is How many Ranks must be in the whole Square Battel And How many Pikes and Musquets First the Square-root of 8100 is 90 the Number of Men in and Number of Ranks and Files now in regard that there must be 6 Ranks of Pikes round about the Musquetiers there will be 12 Ranks less of them both in Front and Flank than in the whole Body therefore substracting of 12 from 90 rest 78 whose Square is 6084 the number of Musquetiers which taken from 8100 there remains 2061 the number of Pikes Problem VII To three Numbers given to find a fourth in a doubled Proportion For as much as like Squar●● are in double the Proportion of their answerable sides therefore you must work by their Squares and Square-root But by the Line of Numbers in this manner If a Fathom of Rope of 6 inches compass about weigh 6 pound 2 ounces or 6-●25 1000 what shall a Fathom of Rope of 12 inches compass weigh Here Note alwayes That when the two Numbers of like denomination which are given are of Lines or sides of Squares or Diameters of Circles then the Extent of the Compasses upon the Line of Numbers from one Line to the other or from one side to the other side that Extent turned twice the same way from the given Area or Content shall reach to the other required So here the Extent of the Compasses from 6 to 12 being turned two times the same way from 6-125 shall reach to 24-50 for 24 pound and a half the weight required But if the two terms given of one denomination are of Squares or Superficies or Areas then the half distance on the Line of Numbers between one Area and the other being turned the same way on the Line from the given Line or Side it shall reach to the Side or Line required For the half-distance between 24-50 and 6-125 shall reach from 12 to 6 or the contrary from 6 to 12. An Example whereof you have in the 4th and 5th Problems of the 12th Chapter Also in the 6th and 7th Problems of the 8th Chapter which treates of Superficial-measure in measuring of Land Note also That if you have three Lines of Numbers viz. a Great a Mean and a Less after Mr. Windgates way then these Questions are wrought without doubling or halving and very neatly and speedily As thus The Extent on the mean Line from 24-50 to 6-125 the weight of the two Ropes shall reach on the great Line from 12 to 6 or from 6 to 12 the inches in compass about of each Rope Problem VIII To three Numbers given to find a fourth in a tripled Proportion For as much as like

Extent shall reach in the same Line from the true hour of the Night to the Stars hour from the Meridian then laying the Thred to the Stars Declination the ND from the Stars hour in the Line of hours to the Thred measured on the particular Scale of Altitudes gives the Stars Altitude then by his Declination and Altitude you may soon find his Azimuth by Use 27. And if the Instrument be neatly fixed to a Foot to set North and South and turn to any Azimuth and Altitude you may find any Star at any time convenient and visible Use XLVII The Altitude and Azimuth of any Star being given to find his Declination Lay the Thred to the Altitude on the degrees counted from 600 toward the end then setting one Point on the Stars Azimuth counted in the Azimuth Line and take the ND from thence to the Thred which distance measured from the beginning of the particular Scale of Altitudes shall give the Declination If the Compasses stand on the right-side of the Thred then the Declination is North if on the left it is South according as you work for the Suns Azimuth in a particular Latitude Use XLVIII The Altitude and Declination of any Star with the Right Ascention of the Sun and the true Hour of the Night given to find the Right Ascention of that Star First by the 43d Use find the Stars Hour viz. How many hours and minuts it wants of coming to or is past the Meridian then the Extent of the Compasses on the Line of 24 hours on the Head-leg from the Stars hour to the true hour shall reach the same way from the Suns Right Ascention to the Stars Right Ascention on the Line of twice 12 or 24 hours Use XLIX To find when any Fixed-Star cometh to South by the Line of twice 12 or 24 hours In Use 42 Section 4 you have the way by Substraction with its Cautions But by the Line of twice 12 or 24 hours work thus Count the Suns Right Ascention on that Line and take the distance from thence to the next 12 backward viz. that at ♈ at the beginning of the Line when the Suns Right Ascention is under 12 hours or to the next 12 in the middle of the Rule at ♎ when the Suns Right Ascention is above 12 hours which is nothing but a rejecting 12 for more conveniency Then The same Extent laid the same way from the Stars Right Ascention shall reach to the Stars coming to South Or The Extent from the Sun to the Stars Right Ascention shall reach the same way from 12 to the Stars coming to South Example for the Lyons-Heart August 20. The Suns Right Ascention the 20th of August is 10 hours 36 minuts the Right Ascention of the Lions-Heart is 9 hours and 50 min. Therefore The Extent from 10 hours 35 min. to the beginning shall reach the same way from 9 hours 50 min. by borrowing 12 hours because the Suns Right Ascention is more than the Stars to 11 hours 13 min. of the next day viz. at a quarter past 11 or at 11 hours and 13 min. the same day where you may observe that the remainder being above 12 if you add 24 hours the time of Southing is between mid-night and mid-day next following Use L. To find what two dayes in the year are of equal length and the Suns Rising and Setting Lay the Thred on any one day in the upper Line of Months and Dayes and at the same time the Thred cuts in the lower-Line of Months the day that is answerable to it in length rising setting and declination and other requisites Example The 1st of April and the 21 of August are dayes of equal length and the Suns Rising and Setting is the same on both those dayes only in the upper-Line the dayes are increasing in length and in the lower-Line they are decreasing Use LI. To find how many degrees the Sun is under the Horizon at any Hour the Declination and Hour being given Count the Suns Declination on the degrees the contrary way viz. for North Declination count from 600 toward the end and count for Southern Declination toward the Head and thereunto lay the Thred then take the nearest distance from the hour given to the Thred this distance measured in the particular Scale of Altitudes shall shew the Suns Depression under the Horizon at that hour Example January the 10th at 8 at Night how many degrees is the Sun under the Horizon On that Day and Hour the Suns Declination is about 20 degrees South then if I lay the Thred to 20 degrees of Declination North and take the nearest distance from 8 to the Thred that distance I say measured in the particular Scale gives 34 degrees and 9 min. for the Suns Depression under the Horizon of 8 afternoon To do this in other Latitudes you are to find the Suns Altitude at 8 in Northern Declination by Use 37. CHAP. XVII The use of the Trianguler Quadrant in finding of Heights and Distances accessable or inaccessable Use I. To find an Altitude at one Station FIrst The Trianguler Quadrant being rectified and fixed to a Ball and Socket and three-legged-staff being necessary in these Operations to perform them exactly especially for Distances look up to the object as you would to a Star and observe what degree and minut the Thred cuts and set it down Also observe the place where you stand at the time of Observation and the distance from your Eye to the ground and the place on the object that is level with your eye also as the playing of the Thred and Plummet will plainly shew Also you must have the measure from the place where you stood observing to the Point exactly right under the object whose height you would have in Feet Yards Perch or what you please to Integers and Fractions in Decimals if it may be Also Note That in all Right-Angle-Triangles one Acute Angle is alwayes the complement of the other so that observing or finding one by Observation by consequence you have the other by taking that from 90. These things being premised the Operation followes by the Artificial Numbers Sines and Tangents and also by the Natural Note also by the way That in regard the complement of the Angle observed is frequently used if you count the degrees the contrary way that is to say from the Head you shall have the complement required as hath been oftentimes hinted before Then As the sine of the Angle opposite to the measured side is to the measured side counted on the Numbers So is the sine of the Angle found to the Altitude or Height required on Numbers Example at one station Standing at C I look up to B the object whose Height is required and I find the Thred to fall on 41 degrees and 45 minuts but if you count from the Head it is 48-15 the complement thereof as in the Figure you see Also the measure from C to A is found to be

218 foot Then As the sine of 48-15 the Angle at B being the complement of the Angle at C is to 218 on the Line of Numbers So is the sine of the Angle at C 41-45 to 195 the Altitude of AB the height required found on the Line of Numbers A second Example standing at D. But if I were standing at D 129 foot and a half from A and would find the height AB the complement of the Angle at D that is to say the Angle at B is 33-30 This being prepared then say As the sine of 33-30 the Angle at B to the measured-side DA 129 ½ counted on the Numbers So is 56-30 the sine of the Angle at D to 195 the Altitude required AB and 5 foot more the usual height of the eye from the Level to the ground makes 200 the whole height required To work this by the Trianguler Quadrant say thus As 129 ½ taken from any Scale is to the = sine of 33 deg 30 min. laying the Thred to the nearest distance So is the = sine of 56-30 the Angle at D to the measure of 195 on the Scale you took 129 ½ from The like manner of work is by the Sector as thus in the foregoing Example As 218 taken from the Line of Lines to the = sine of 48 deg 15 min. So is the = sine of 41-45 to 195 on the Line of Lines latterally And yet further So is the = sine of 90 to 291 the Line CB. Use II. To find an Altitude at two stations But if you cannot come to measure to the foot of the object then you must observe at two places As thus for Example First as before find the Angle at D or rather the complement thereof viz. 33-30 then go further backward in a right Line with the object and first station any competent Number of feet as suppose 88 ½ to C there also observe the Altitude or Complement viz. the Angle ABC 48-15 Then Find the difference between 48-15 and 33-30 and it is 14-45 Then As the sine of the difference last found viz. the Angle CBD 14-45 to 88 ½ on the Line of Numbers So is the sine of the Angle at C 41-45 to the measure of the side DB 233 on the Line of Numbers Again for the second Operation As the sine of 90 the Angle at A to the Hypothenusa DB 233 So is the sine of 56-30 the Angle at D to 195 the Altitude required The same by the Trianguler Quadrant or Sector As 88 ½ the measured distance CD to the = sine of 14-45 CBD So is = sine of 41-45 to the measure of 233 the opposite-side DB. Again As 233 taken from the Line of Lines to = sine of 90 So is the = sine of 56-30 the Angle at D to 195 on the Line of Lines the height required Use III. Another way to save one Operation from IC First observe the complement of the Angle at D and also the complement of the Angle at C then count these two complements on the Line of Natural Tangents on the loose-piece or moving-leg and take the distance between them and measure it on the same Tangent-line from the beginning thereof and note what Tangent the Compass-point stayeth at and count that for the first term in degrees and minuts Then As the Tangent of this first term to the measured distance CD 88 ½ on the Line of Numbers So is the Tangent of 45 to the Altitude required Thus in our Example The distance measured is 88 ½ the two complements 33-30 and 48-15 the distance between them makes the Tangent of 24-34 to be used as a first term Then As the Tangent of ●4-34 the first term last found to 88 ½ on the Numbers So is the Tangent of 45 to 195 ferè on the Numbers the height required But if the distance from D or C to A the foot of the Object were required then the manner of Calculation runs thus As the Tangent of the difference of the Co-tangents first found 24-34 is to the distance between D and C 88 ½ So is the Co-tangent of the greater Ark 48-15 to the greater distance CA 218. Or So is the Co-tangent of the lesser Ark 33-30 to the lesser distance DA 129 ½ But if the Hypothenusaes be required then reason thus As the Tangent of the difference first found is 24-34 to the distance between the stations D and C 88 ½ So is the Secant of the Angle at B the greater viz. 48-15 counted beyond 90 to CB 291. Or So is the Secant of 33-30 the lesser Angle at B to 233 the lesser distance DB the Hypothenusa required To work these two last by the Trianguler Quadrant First prick off the Tangents and Secants to be used parallelly from the loose-piece on the greater general Scale and note those Points for your present use As thus The Tangent of 24-34 taken from the loose-piece from 60 counted as 00 will reach to the sine of 10-40 on the general Scale Secondly The Secant of 33-30 being the measure from the Tangent of 33-30 on the loose-piece counting from 60 to the Center will reach on the general Scale from the Center to 28-50 Thirdly The measure from the Tangent of 48-15 on the loose-piece to the Center being the Secant of 48-15 will reach from the Center to 32-5 on the general Scale This being prepared the work is thus As distance between the two stations to = Tangent of the first term at 10-40 So is = Tangent of 45 to the Altitude required Again for the Distance As distance between the two stations to the = Tangent of the first term So is the = Tangent of the greater Angles complement at 26-36 to the greatest distance CA 218. Or So is the = Tangent of the lesser Angles complement at 15-25 to the lesser distance DA 129 ½ Or So is the = Secant of the greater Angles complement at 32-5● to the greater Hypothenusa CB 291. Or So is the = Secant of the lesser Angles complement at 28-50 to the lesser Hypothenusa DB 233. Use IV. Another way for Altitudes by the Line of Shadows either accessable or unaccessable by one or two stations If this way be desired it may be put on this as well as any other Quadrants Then the use is thus Figure II. Suppose that AB be the height of a Tree or other Object to be found go so far back from it as suppose to C till looking up by the two Pins put for sights the Thred falls on 45 degrees on the Quadrant or on 1 on the Line of Shadows then I say that the height AB is equal to the distance CA more by the height of your eye from the ground But if you go further back still to D till the Thred falls on 2 on the Line of Shadows that is to say at 26 deg 34 min. the Altitude will be but half the distance