Chapter 85

FOOTNOTES:[1]Ποιησον δ᾽ Αιθρην, δος δ᾽ Οφθαλμοῖσιν ιδεσθαι·Ἐν δε Φαει και ολεσσον, επει νυ τοι ευαδεν οὑτως.Homer’s Iliad, Book 17, Line 646.[2]Phil. Trans. Vol. LXVII, for 1777, Part II, Page 513, containing Sir G. Shuckburgh’s Rules for the Mensuration of Heights with the Barometer. Also Vol. LXVIII, for 1778, Part II, Page 681:[3]And Page 688.[4]It were to be wished that the Divisions of the Thermometer by Farenheit were become general throughout Europe, in preference to those by Reaumur yet retainedabroad; which Divisions of Reaumur are not sufficiently minute to mark the least sensible Change in the Temperature, are subject to frequent Mistakes, and the Inconvenience of adding in the Notation, the Wordsaboveorbelowthe Cypher, zero, or Point of Congelation: besides their being in Conversation not easily compared with those of Farenheit; each Degree of the latter having to that of the former nearly the Proportion of 18 to 11: since Farenheit from the freezing Point upwards to boiling Water has 212 − 32 = 180°, and Reaumur to the same Height, 110° Divisions: Mr. Saussure says as 4 to 9; in which there is an evident Oversight: see his curious and philosophic Investigation of the Atmosphere in“Essais fur L’Hygrometrie.”4to. A Neuchatel, 1783.Frequent Mention being made of the Thermometer graduated according to Farenheit’s Scale, in different Parts of the following Account; it may not be amiss to shew the corresponding Points according to Reaumur, taken from“Thermometre universel de Comparaison, extrait du Journal de Physique de M. L’Abbé Rozier.”Farenheit.Reaumur.5413 & 4-9ths above the Cypher.5514 ditto, nearly.5715 2-9ths ditto, nearly.5916 4-9ths ditto, nearly.6017 1-9th ditto.6520 1-9th ditto, nearly.[5]The Strength of the Rope, or Cable, if its Length does not exceed 10 or 12 Yards, ought to be such as to support a weight, greater than the Weight of the Balloon and it’s Appendages, for the Resistance made by the Grapple against the Balloon acted on by the Wind is immediate: The Rope ought therefore to be made of Indian-Gut, as most elastic, or Silk, as lightest. But if the Rope be half a Mile, or a Mile long; the Resistance is gradual: the Balloon descending for some Minutes; and having an open Space to move in through the Air: the Rope or Cable acting as a Radius, and the Levity of the Balloon and Opposition of the circumambient Air preventing it from falling with any Violence.The shorter Cable may be used at the Height of 10 Yards; in aid of the longer, to prevent it from rising; or to moor it, by winding the Reel, and hauling down the Balloon close to the Ground.[6]The Resistance being as the Square of the Velocity; therefore if the Velocity be increased 3 Times, the Resistance will be as 3 × 3 = 9, i. e. will be increased 9 Times.[7]PoundsAverdupois.Weight of the Aironaut160Provisions and Articles calculated at20Sand-Ballast prepared in Bags44Levity for Ascent10——Sum total,234[8]Ancient Warriors among the Arabs, Spaniards, Romans, Gauls, and Germans, being frequently obliged to pass deep Rivers, never undertook a Campaign without them. For the above Anecdote, and many curious Experiments on Air, see Sam. Reyheri,Dissertatio de Aëre, tertium edita. Kiliæ. 1673.[9]Equal Time with a Regulator corrected by an Observation.[10]Being a Dial-Compass, the Dipping of the Needle was frequently checked by the Glass at the Top. A Mariner’s Compass is the best.[11]The Defect of the Reel remediedThe Loop shoud have been furnished with aSwivel: or the Lath or Reel shoud have been a Kind of Pulley, a Foot in Diameter, and two Inches wide. The Hook of which having also a Swivel might have been held in the Hand: and thus the Twine woud have run off in a short Time with the greatest Readiness; the Swivel conforming to the circular Motion of the Balloon.[12]Slate(according to Cronstedt) is theWhetstoneof fine Particles, composed of Glimmer, Quartz; and, in some Species, of a martial argillaceous Earth, See“Essay on Mineralogy”by Mendes Da Costa, Sect. 264.[13]Method of discovering Haze round the Sun, in bright Weather.To know whether the Air is hazy, tho’ the Sun continues shining.The Method taken for that Purpose was by placing the Hand so as to cover his Disk or Body, and then observe the Glory blazing round him; which may, in general, be seen to issue in great Abundance, in Rays of agolden Colour: occasioned by a Haziness or Vapour which pervades thelowerRegions of the Air, most frequently in the hottest and calmest Weather, and in the hottest Climates. The Accumulation of these Vapours, before they are formed into Clouds, are often so great as to intercept the Sun’s Rays, or dye them the Colour of Blood: an Appearance frequent in Virginia, and also throughout the torrid Zone.In theCampaniaof Rome, for Instance, the Italians have a peculiar Name for such Kind of Weather, when the Sun is neithervisible nor invisible:Il Sole si vede, e’ non si vede.By Degrees the Hand is to be removed so as just to have a Glance of the Sun’s Limb. And it frequently happens that the Air is exceedingly hazy; tho’ not a Cloud appears above the Horizon.[14]Esse inImaginibusquâpropterCausavideturCernendi, neque possesine hisRes ulla videri.Lucretius de Rerum Natura. L. 4. V. 238.[15]Notwithstanding what has been said;this, to the great and to the sordid Vulgar, woud still appear a solitary, helpless, and deplorable Situation. But such are not captivated with the golden Lines ofEpictetus, (Chap. 13. Line 3. see Mrs. Carter’s Translation.)“ΠΑΝΤΑ ΘΕΩΝ μεστα και ΔΑΙΜΟΝΩΝ·—Βλεπων τον ΗΛΙΟΝ και Σεληνην, και Ἀστρα, και ΓΗΣ απολαυων και ΘΑΛΑΣΣΗΣ, ἐρημος εστιν ου μαλλον ἠ και ἀβοηθητος·”Nor are theypracticallyinfluenced by the better Words of a much finer Writer: “The Earth is full,” &c. &c. And “If I take the Wings of the Morning,” &c. &c.[16]There being, at first, no Clouds, as usual, to occupy the Place of the lowest Stratum.[17]It has been said that theapparentHeight from the Balloon to the Ground was 7 Miles, viz. 4 to the Summit of the Clouds, and 3 below: and thebarometricHeight was about a Mile and half, viz. 2332 Yards,a Calculation of which will be given.If then we divide that Height or Distance into 2 such Parts, that the greater shall be to the less as 4 to 3; we obtain the Length of each Part; i. e. the barometric Height from the Balloon to the Summit of the Clouds, and thence to the Earth; which is done thus:Suppose the whole Distance to be any Line, as A. B. to be divided in C. Then, as 7 is the whole Line, and 4 the greater Part; say, as the whole 7 is to the greater Part 4, so is the whole Distance to a fourth Term proportional, which will be equal to the greater Distance sought:Whole Distancein Yards.Greater Distance in Yards.Thus 7, : 4 ::2332: 1332⁠4⁄7 Ans.4———7)93282332 the whole.1332⁠4⁄71332⁠4⁄7 being the greater Distance found; take the greater from the whole, and then will remain the lesser Distance wanted, viz. 999⁠3⁄7: the 1332⁠4⁄7 = the greater Distance, and 999⁠3⁄7 = the lesser Distance: and adding the Fractions 4⁄7 3⁄7 = 1 to the 999; we have 1332 Yards for the greater Distance, or Height of the Balloon above the Summit of the superior Clouds: and 1000 Yards for the less Distance, or Height from the Earth to the Summit of the superior Clouds.Note.The Line A. B. here selected is thefamous Measureof (half) amathematicalRhinland and RomanFoot,according to Snellius. (SeeGeographia Generalisof Varenius, published byNewton.Lib. 1. Cap. 2. De variis Mensuris.)[18]PROBLEM.To find the circular Boundary of thecelestialProspect over the Tops of the superior Clouds, from the Balloon at the Height of near a Mile and half above the Surface of the Earth, viz. 2332 Yards. The Height from the Earth to the upper Surface or Floor of Clouds being 1000 Yards; and the Height above the Floor to the Balloon being 1332 Yards.On the Curvature of the Earth and Clouds, and Elevation of the Eye above their circular Horizon.Rule. To the Earth’s Diameter, equal to 7940 geographical Miles, addthe Heightof the Eye above its Surface: multiply the Sum by that Height: then the square Root of the Product gives the Distance at which an Object on the Surface of the Earth can be seen by an Eye so elevated. Note the Diameter of the Earth, in Feet, is 41798117, according to Newton. (SeePractical Navigator, by J. Moore, 7th Ed. Page 251.)FIRST.Double 1000 Yards, the Height from the Earth to the Clouds for an Addition to the Diameter of the Earth, whose Surface is now considered, as extended to the concentric Floor of Cloud.10001000——2000SECOND.13932702(1⁄3)Diameter of the Earth in Yards.2000Addition to the Diameter.————13934702Sum, to which add1332the Height of the Eye or of the————Balloon above the Floor of Cloud.13936034Sum, which multiply into1332the Height of the Eye above the————Floor.27872068418081024180810213936034——————Extract the. . . . .1760) Yards in a Mile.Square Root18562797288(136245 (77 Miles.112320————23) 85130456912320—————266) 1662Yards 440) 725(1 Quarter of a Mile.1596440————2722) 6679285Yards.5444———Ans. 77Miles, 1 Qu. 285 Yards.27244) 123572108976———272485) 14596881362425————97263Circular Boundary of theterrestrialProspect from the Balloon on aclearDay.PROBLEM.To find the circular Boundary of theterrestrialProspect, on a clear Day, from the Balloon at the Height of near a Mile and half, viz. 2332 Yards: the Earth’s Diameter beingequal to13932705⁠2⁄3Yards,add 2332the Height of the Eye or Balloon.————13935037he Sum, multiply into2332the Height of the Eye, &c.————27870074418051114180511127870074——————Extract the. . . . .1760) Yards in a Mile.square Root32496506284(180267(102,11760 say 1021⁄2 Miles, Ans.———28)22442672243520————3602) 9650747 Yards, Remainder.7204———36046) 244662216276————360527) 28386842523689————314995Remainder.[19]See his “Minute Philosopher.”[20]Ullòa in his voyage to South-America relates, that in passing over theDeserts, Írides are frequently seen by Travellers roundtheir own Headsas the Center of theIris; and visible only to themselves. But what Analogy theBalloon Irisbears to them, Time and future Experiments may discover. See his“Voyage to South America, Vol. 1. Pa. 442.”[21]As Sound travels1142Feet in aSecond, it must have moved in30Seconds———Feet in a Yard3)34260= FeetYards in a Mile1760)11420(6 Miles10560——Yards in a Quarter of a Mile440)860(1 Quarter440——Answer 6 Miles, 1 Quarter, and420Yards.[22]Equal to 2085 Yards; or 1 Mile, 325 Yards.[23]Long’s Astronomy. Pages 227, 229.[24]Also called theHorsham Stone, from a Place so named, in Surrey, where great Quantities are found.[25]PROBLEM.To find theLengthof theShadowfrom a Person ofmiddleStature, (five Feet and a half High) viz. at XII o’Clock, on the 8th Day of September, 1785, at Chester, whose North Latitude is 53° 12′; (and 3° 11′ West Longitude from London.)FIRST,To find the Sun’s Altitude at XII.From90°. 00′′SubtractThe Latitude53. 12———The Remain.36. 48is the Complement of Latitude,to which add (from the Tables)Sun’s N. Decl.5. 29———The Remain.42. 17is the Sun’s Altitude (viz. at XII.)SECOND,For the Shadow say,As the Sine of the Sun’s Altitude 42° 17′To the Person’s Height, viz. 66 Inches,So is the Co-Sine of the Sun’s Altitude,To the Length of the Shadow.For the Sine of the Sun’s Altitude 42° 17′ in the Table of artificial Sines, is the Logarithm 9.82788, which, subtracted from the arithmetic Complement, viz. 9.99999 (supposing the last Figure a 10) becomes,.17212Then for the Person’s Height, viz. 66 Inches: in the Table of Logarithms is the corresponding Number,1.81254And for the Co-Sine (had by subtracting the Altitude 42.17 from 90.00) viz. 47.43: among the artificial Sines is the Logarithm,9.86913————The above Sums added, are11.86079which logarithmic Number (deducting theInitial1 as useless) viz. 1.86079, in the Table of Logarithms, corresponds to 72.57, equal to 72 Inches, for the Length of the Shadow at XII.Reducing then the Numbers 66 and 72, to the lowest Denomination, thus 6)66⁄72 = 11⁄12 the Proportion which theLengthof theShadowbears to theHeightof theObjectis thereby obtained: that is[26]If theLengthof the Shadow be divided into 12 Parts, the Height of the Object would be 11 of those Parts.See Moore’sPractical Navigator.PROBLEM.AneasyWay to find the Proportion which theLengthof the Shadow bears to theHeightof an Object is,at any time when the sun shines, to fix a Plummet Line andframeuprightin the Ground; measure theLengthof itsShadow, and compareitwith theHeightof theframe.[27]Equal to 3 Quarters of a Mile and 121 Yards.[28]i. e. When the Barometerbelowis at 30 Inches, and Thermometerbelowat 60° viz. about 1000 Yards high infineWeather, and 500 inchangeable.[29]Being 1083 Yards, i. e. half a Mile, and 203 Yards.[30]It was High Water at Chester and Frodsham-Bridge, at 38 Minutes past I.[31]Articles parted with, to check thefirstDescent at Bellair, near Frodsham: and to ascend thesecondTime.To check thefirstDescent.Pounds.Ounces.Ballast, at twice:240To clear Trees and Hedges, andre-ascend:Barometer and Frame,0121⁄2Basket with Tunning Dish and Bottles (except the Flask with Brandy and Water)410Half Mile of Twine on the Reel10Speaking Trumpet081⁄2Woollen Gloves01—————310240—————Remains for Re-ascent70[32]The Sun’s Azimuth from the North PointWestward, being 118.26′: its Supplement to 180° is 61°.34′ South westerly: i. e. South West by West, half Westnearly.[33]TheLengthof the Shadows being more thandoublethe Height of theObjects: see[34].[34]To find the Length of the Shadow at half past III.(See Section 84,Note [25].)Given{Lat. of Chester,53°12′{To find Sun’s Alt.Sun’s Dec.529Hour III, 30M.5230This is the Case of an oblique spheric Triangle, wherein are two Sides and one Angle between them given, to find the Sun’s Azimuth, and the Sun’s Co-Alt.Side84.31{Sum of Sides121.19Side36.48Diff. of Sides47.43(31⁄2 Hour) Angle contained52.30Half ditto26.15{Co.63.45Half Sum of Sides60.3929.22Half Difference ditto23.5166.9THE FIRST PREPARATIVE PROPORTION.As Sine of 1⁄2 Sum of Sides60.390.05966Co-Ar.To Sine of 1⁄2 Difference of Sides23.519.60675So Co-Tangent 1⁄2 contained Angle63.4510.30703———————To T. of 1⁄2 Diff. of the other two Angles43.159.97344SECOND PREPARATIVE PROPORTION.As Co-Sine 1⁄2 Sum of Sides29.210.30968Co-Ar.To Co-Sine 1⁄2 Diff.66.99.96123So Co-Tangent 1⁄2 contained Angle63.4510.30703————To T. 1⁄2 Sum of other Angles75.1110.57794Half Diff. before found43.15———Sum, is greater Angle118.26= Sun’s Azim.Diff. is lesser Angle31.56= S’s right Asc.Then by first Axiom in Trigonometry, to know the Sun’s Altitude say,As Sine Sun’s right Asc.31.560.27659To Sine Co-Lat.36.489.77744So Sine of the contained Angle52.309.89947————To Co-Sine of the Sun’s Alt.63.579.95350from90.———Sun’s Alt.26.3Having Sun’s Alt. to find the Shadow,As Sine Sun’s Alt.26.30.35738Co-Ar.To Person’s Height,66Inches,1.81954So Co-Sine of the Sun’s Alt.63.579.95350————To Length of Shadow,135Inches,2.13042Then 6(66⁄135 = 11⁄22 − | − 3⁄6 or 1⁄2, i. e. as 22 to 45: supposing the Length of the Shadow divided into 45 Parts; theHeightof the Object woud be 22 of those Parts; or not quitehalftheLengthof the Shadow, at half past III.

FOOTNOTES:

[1]Ποιησον δ᾽ Αιθρην, δος δ᾽ Οφθαλμοῖσιν ιδεσθαι·Ἐν δε Φαει και ολεσσον, επει νυ τοι ευαδεν οὑτως.Homer’s Iliad, Book 17, Line 646.

[2]Phil. Trans. Vol. LXVII, for 1777, Part II, Page 513, containing Sir G. Shuckburgh’s Rules for the Mensuration of Heights with the Barometer. Also Vol. LXVIII, for 1778, Part II, Page 681:

[3]And Page 688.

[4]It were to be wished that the Divisions of the Thermometer by Farenheit were become general throughout Europe, in preference to those by Reaumur yet retainedabroad; which Divisions of Reaumur are not sufficiently minute to mark the least sensible Change in the Temperature, are subject to frequent Mistakes, and the Inconvenience of adding in the Notation, the Wordsaboveorbelowthe Cypher, zero, or Point of Congelation: besides their being in Conversation not easily compared with those of Farenheit; each Degree of the latter having to that of the former nearly the Proportion of 18 to 11: since Farenheit from the freezing Point upwards to boiling Water has 212 − 32 = 180°, and Reaumur to the same Height, 110° Divisions: Mr. Saussure says as 4 to 9; in which there is an evident Oversight: see his curious and philosophic Investigation of the Atmosphere in“Essais fur L’Hygrometrie.”4to. A Neuchatel, 1783.Frequent Mention being made of the Thermometer graduated according to Farenheit’s Scale, in different Parts of the following Account; it may not be amiss to shew the corresponding Points according to Reaumur, taken from“Thermometre universel de Comparaison, extrait du Journal de Physique de M. L’Abbé Rozier.”Farenheit.Reaumur.5413 & 4-9ths above the Cypher.5514 ditto, nearly.5715 2-9ths ditto, nearly.5916 4-9ths ditto, nearly.6017 1-9th ditto.6520 1-9th ditto, nearly.

Frequent Mention being made of the Thermometer graduated according to Farenheit’s Scale, in different Parts of the following Account; it may not be amiss to shew the corresponding Points according to Reaumur, taken from“Thermometre universel de Comparaison, extrait du Journal de Physique de M. L’Abbé Rozier.”

[5]The Strength of the Rope, or Cable, if its Length does not exceed 10 or 12 Yards, ought to be such as to support a weight, greater than the Weight of the Balloon and it’s Appendages, for the Resistance made by the Grapple against the Balloon acted on by the Wind is immediate: The Rope ought therefore to be made of Indian-Gut, as most elastic, or Silk, as lightest. But if the Rope be half a Mile, or a Mile long; the Resistance is gradual: the Balloon descending for some Minutes; and having an open Space to move in through the Air: the Rope or Cable acting as a Radius, and the Levity of the Balloon and Opposition of the circumambient Air preventing it from falling with any Violence.The shorter Cable may be used at the Height of 10 Yards; in aid of the longer, to prevent it from rising; or to moor it, by winding the Reel, and hauling down the Balloon close to the Ground.

The shorter Cable may be used at the Height of 10 Yards; in aid of the longer, to prevent it from rising; or to moor it, by winding the Reel, and hauling down the Balloon close to the Ground.

[6]The Resistance being as the Square of the Velocity; therefore if the Velocity be increased 3 Times, the Resistance will be as 3 × 3 = 9, i. e. will be increased 9 Times.

[7]PoundsAverdupois.Weight of the Aironaut160Provisions and Articles calculated at20Sand-Ballast prepared in Bags44Levity for Ascent10——Sum total,234

PoundsAverdupois.

160

——

Sum total,

234

[8]Ancient Warriors among the Arabs, Spaniards, Romans, Gauls, and Germans, being frequently obliged to pass deep Rivers, never undertook a Campaign without them. For the above Anecdote, and many curious Experiments on Air, see Sam. Reyheri,Dissertatio de Aëre, tertium edita. Kiliæ. 1673.

[9]Equal Time with a Regulator corrected by an Observation.

[10]Being a Dial-Compass, the Dipping of the Needle was frequently checked by the Glass at the Top. A Mariner’s Compass is the best.

[11]The Defect of the Reel remediedThe Loop shoud have been furnished with aSwivel: or the Lath or Reel shoud have been a Kind of Pulley, a Foot in Diameter, and two Inches wide. The Hook of which having also a Swivel might have been held in the Hand: and thus the Twine woud have run off in a short Time with the greatest Readiness; the Swivel conforming to the circular Motion of the Balloon.

The Defect of the Reel remedied

[12]Slate(according to Cronstedt) is theWhetstoneof fine Particles, composed of Glimmer, Quartz; and, in some Species, of a martial argillaceous Earth, See“Essay on Mineralogy”by Mendes Da Costa, Sect. 264.

[13]Method of discovering Haze round the Sun, in bright Weather.To know whether the Air is hazy, tho’ the Sun continues shining.The Method taken for that Purpose was by placing the Hand so as to cover his Disk or Body, and then observe the Glory blazing round him; which may, in general, be seen to issue in great Abundance, in Rays of agolden Colour: occasioned by a Haziness or Vapour which pervades thelowerRegions of the Air, most frequently in the hottest and calmest Weather, and in the hottest Climates. The Accumulation of these Vapours, before they are formed into Clouds, are often so great as to intercept the Sun’s Rays, or dye them the Colour of Blood: an Appearance frequent in Virginia, and also throughout the torrid Zone.In theCampaniaof Rome, for Instance, the Italians have a peculiar Name for such Kind of Weather, when the Sun is neithervisible nor invisible:Il Sole si vede, e’ non si vede.By Degrees the Hand is to be removed so as just to have a Glance of the Sun’s Limb. And it frequently happens that the Air is exceedingly hazy; tho’ not a Cloud appears above the Horizon.

Method of discovering Haze round the Sun, in bright Weather.

To know whether the Air is hazy, tho’ the Sun continues shining.

The Method taken for that Purpose was by placing the Hand so as to cover his Disk or Body, and then observe the Glory blazing round him; which may, in general, be seen to issue in great Abundance, in Rays of agolden Colour: occasioned by a Haziness or Vapour which pervades thelowerRegions of the Air, most frequently in the hottest and calmest Weather, and in the hottest Climates. The Accumulation of these Vapours, before they are formed into Clouds, are often so great as to intercept the Sun’s Rays, or dye them the Colour of Blood: an Appearance frequent in Virginia, and also throughout the torrid Zone.

In theCampaniaof Rome, for Instance, the Italians have a peculiar Name for such Kind of Weather, when the Sun is neithervisible nor invisible:Il Sole si vede, e’ non si vede.

By Degrees the Hand is to be removed so as just to have a Glance of the Sun’s Limb. And it frequently happens that the Air is exceedingly hazy; tho’ not a Cloud appears above the Horizon.

[14]Esse inImaginibusquâpropterCausavideturCernendi, neque possesine hisRes ulla videri.Lucretius de Rerum Natura. L. 4. V. 238.

[15]Notwithstanding what has been said;this, to the great and to the sordid Vulgar, woud still appear a solitary, helpless, and deplorable Situation. But such are not captivated with the golden Lines ofEpictetus, (Chap. 13. Line 3. see Mrs. Carter’s Translation.)“ΠΑΝΤΑ ΘΕΩΝ μεστα και ΔΑΙΜΟΝΩΝ·—Βλεπων τον ΗΛΙΟΝ και Σεληνην, και Ἀστρα, και ΓΗΣ απολαυων και ΘΑΛΑΣΣΗΣ, ἐρημος εστιν ου μαλλον ἠ και ἀβοηθητος·”Nor are theypracticallyinfluenced by the better Words of a much finer Writer: “The Earth is full,” &c. &c. And “If I take the Wings of the Morning,” &c. &c.

“ΠΑΝΤΑ ΘΕΩΝ μεστα και ΔΑΙΜΟΝΩΝ·—Βλεπων τον ΗΛΙΟΝ και Σεληνην, και Ἀστρα, και ΓΗΣ απολαυων και ΘΑΛΑΣΣΗΣ, ἐρημος εστιν ου μαλλον ἠ και ἀβοηθητος·”Nor are theypracticallyinfluenced by the better Words of a much finer Writer: “The Earth is full,” &c. &c. And “If I take the Wings of the Morning,” &c. &c.

[16]There being, at first, no Clouds, as usual, to occupy the Place of the lowest Stratum.

[17]It has been said that theapparentHeight from the Balloon to the Ground was 7 Miles, viz. 4 to the Summit of the Clouds, and 3 below: and thebarometricHeight was about a Mile and half, viz. 2332 Yards,a Calculation of which will be given.If then we divide that Height or Distance into 2 such Parts, that the greater shall be to the less as 4 to 3; we obtain the Length of each Part; i. e. the barometric Height from the Balloon to the Summit of the Clouds, and thence to the Earth; which is done thus:Suppose the whole Distance to be any Line, as A. B. to be divided in C. Then, as 7 is the whole Line, and 4 the greater Part; say, as the whole 7 is to the greater Part 4, so is the whole Distance to a fourth Term proportional, which will be equal to the greater Distance sought:Whole Distancein Yards.Greater Distance in Yards.Thus 7, : 4 ::2332: 1332⁠4⁄7 Ans.4———7)93282332 the whole.1332⁠4⁄71332⁠4⁄7 being the greater Distance found; take the greater from the whole, and then will remain the lesser Distance wanted, viz. 999⁠3⁄7: the 1332⁠4⁄7 = the greater Distance, and 999⁠3⁄7 = the lesser Distance: and adding the Fractions 4⁄7 3⁄7 = 1 to the 999; we have 1332 Yards for the greater Distance, or Height of the Balloon above the Summit of the superior Clouds: and 1000 Yards for the less Distance, or Height from the Earth to the Summit of the superior Clouds.Note.The Line A. B. here selected is thefamous Measureof (half) amathematicalRhinland and RomanFoot,according to Snellius. (SeeGeographia Generalisof Varenius, published byNewton.Lib. 1. Cap. 2. De variis Mensuris.)

If then we divide that Height or Distance into 2 such Parts, that the greater shall be to the less as 4 to 3; we obtain the Length of each Part; i. e. the barometric Height from the Balloon to the Summit of the Clouds, and thence to the Earth; which is done thus:

Suppose the whole Distance to be any Line, as A. B. to be divided in C. Then, as 7 is the whole Line, and 4 the greater Part; say, as the whole 7 is to the greater Part 4, so is the whole Distance to a fourth Term proportional, which will be equal to the greater Distance sought:

2332

: 1332⁠4⁄7 Ans.

———

7)9328

1332⁠4⁄7

1332⁠4⁄7 being the greater Distance found; take the greater from the whole, and then will remain the lesser Distance wanted, viz. 999⁠3⁄7: the 1332⁠4⁄7 = the greater Distance, and 999⁠3⁄7 = the lesser Distance: and adding the Fractions 4⁄7 3⁄7 = 1 to the 999; we have 1332 Yards for the greater Distance, or Height of the Balloon above the Summit of the superior Clouds: and 1000 Yards for the less Distance, or Height from the Earth to the Summit of the superior Clouds.

Note.The Line A. B. here selected is thefamous Measureof (half) amathematicalRhinland and RomanFoot,according to Snellius. (SeeGeographia Generalisof Varenius, published byNewton.Lib. 1. Cap. 2. De variis Mensuris.)

[18]PROBLEM.To find the circular Boundary of thecelestialProspect over the Tops of the superior Clouds, from the Balloon at the Height of near a Mile and half above the Surface of the Earth, viz. 2332 Yards. The Height from the Earth to the upper Surface or Floor of Clouds being 1000 Yards; and the Height above the Floor to the Balloon being 1332 Yards.On the Curvature of the Earth and Clouds, and Elevation of the Eye above their circular Horizon.Rule. To the Earth’s Diameter, equal to 7940 geographical Miles, addthe Heightof the Eye above its Surface: multiply the Sum by that Height: then the square Root of the Product gives the Distance at which an Object on the Surface of the Earth can be seen by an Eye so elevated. Note the Diameter of the Earth, in Feet, is 41798117, according to Newton. (SeePractical Navigator, by J. Moore, 7th Ed. Page 251.)FIRST.Double 1000 Yards, the Height from the Earth to the Clouds for an Addition to the Diameter of the Earth, whose Surface is now considered, as extended to the concentric Floor of Cloud.10001000——2000SECOND.13932702(1⁄3)Diameter of the Earth in Yards.2000Addition to the Diameter.————13934702Sum, to which add1332the Height of the Eye or of the————Balloon above the Floor of Cloud.13936034Sum, which multiply into1332the Height of the Eye above the————Floor.27872068418081024180810213936034——————Extract the. . . . .1760) Yards in a Mile.Square Root18562797288(136245 (77 Miles.112320————23) 85130456912320—————266) 1662Yards 440) 725(1 Quarter of a Mile.1596440————2722) 6679285Yards.5444———Ans. 77Miles, 1 Qu. 285 Yards.27244) 123572108976———272485) 14596881362425————97263Circular Boundary of theterrestrialProspect from the Balloon on aclearDay.PROBLEM.To find the circular Boundary of theterrestrialProspect, on a clear Day, from the Balloon at the Height of near a Mile and half, viz. 2332 Yards: the Earth’s Diameter beingequal to13932705⁠2⁄3Yards,add 2332the Height of the Eye or Balloon.————13935037he Sum, multiply into2332the Height of the Eye, &c.————27870074418051114180511127870074——————Extract the. . . . .1760) Yards in a Mile.square Root32496506284(180267(102,11760 say 1021⁄2 Miles, Ans.———28)22442672243520————3602) 9650747 Yards, Remainder.7204———36046) 244662216276————360527) 28386842523689————314995Remainder.

PROBLEM.

To find the circular Boundary of thecelestialProspect over the Tops of the superior Clouds, from the Balloon at the Height of near a Mile and half above the Surface of the Earth, viz. 2332 Yards. The Height from the Earth to the upper Surface or Floor of Clouds being 1000 Yards; and the Height above the Floor to the Balloon being 1332 Yards.

On the Curvature of the Earth and Clouds, and Elevation of the Eye above their circular Horizon.

Rule. To the Earth’s Diameter, equal to 7940 geographical Miles, addthe Heightof the Eye above its Surface: multiply the Sum by that Height: then the square Root of the Product gives the Distance at which an Object on the Surface of the Earth can be seen by an Eye so elevated. Note the Diameter of the Earth, in Feet, is 41798117, according to Newton. (SeePractical Navigator, by J. Moore, 7th Ed. Page 251.)

FIRST.

Double 1000 Yards, the Height from the Earth to the Clouds for an Addition to the Diameter of the Earth, whose Surface is now considered, as extended to the concentric Floor of Cloud.

10001000——2000

SECOND.

13932702(1⁄3)

2000

————

13934702

1332

————

13936034

1332

————

27872068

41808102

13936034

——————

. . . . .

18562797288

12320

—

———

23) 85

13045

12320

——

———

266) 1662

Yards 440) 725

1596

440

——

2722) 6679

285

5444

———

Ans. 77

27244) 123572

108976

———

272485) 1459688

1362425

————

97263

Circular Boundary of theterrestrialProspect from the Balloon on aclearDay.

PROBLEM.

To find the circular Boundary of theterrestrialProspect, on a clear Day, from the Balloon at the Height of near a Mile and half, viz. 2332 Yards: the Earth’s Diameter being

equal to

13932705⁠2⁄3

add 2332

————

13935037

2332

————

27870074

41805111

27870074

——————

. . . . .

32496506284

—

28)224

224

——

3602) 9650

7204

———

36046) 244662

216276

————

360527) 2838684

2523689

————

314995

[19]See his “Minute Philosopher.”

[20]Ullòa in his voyage to South-America relates, that in passing over theDeserts, Írides are frequently seen by Travellers roundtheir own Headsas the Center of theIris; and visible only to themselves. But what Analogy theBalloon Irisbears to them, Time and future Experiments may discover. See his“Voyage to South America, Vol. 1. Pa. 442.”

[21]As Sound travels1142Feet in aSecond, it must have moved in30Seconds———Feet in a Yard3)34260= FeetYards in a Mile1760)11420(6 Miles10560——Yards in a Quarter of a Mile440)860(1 Quarter440——Answer 6 Miles, 1 Quarter, and420Yards.

1142

———

Feet in a Yard

3)34260

Yards in a Mile

1760)11420

10560

——

Yards in a Quarter of a Mile

440)860

440

——

Answer 6 Miles, 1 Quarter, and

420

[22]Equal to 2085 Yards; or 1 Mile, 325 Yards.

[23]Long’s Astronomy. Pages 227, 229.

[24]Also called theHorsham Stone, from a Place so named, in Surrey, where great Quantities are found.

[25]PROBLEM.To find theLengthof theShadowfrom a Person ofmiddleStature, (five Feet and a half High) viz. at XII o’Clock, on the 8th Day of September, 1785, at Chester, whose North Latitude is 53° 12′; (and 3° 11′ West Longitude from London.)FIRST,To find the Sun’s Altitude at XII.From90°. 00′′SubtractThe Latitude53. 12———The Remain.36. 48is the Complement of Latitude,to which add (from the Tables)Sun’s N. Decl.5. 29———The Remain.42. 17is the Sun’s Altitude (viz. at XII.)SECOND,For the Shadow say,As the Sine of the Sun’s Altitude 42° 17′To the Person’s Height, viz. 66 Inches,So is the Co-Sine of the Sun’s Altitude,To the Length of the Shadow.For the Sine of the Sun’s Altitude 42° 17′ in the Table of artificial Sines, is the Logarithm 9.82788, which, subtracted from the arithmetic Complement, viz. 9.99999 (supposing the last Figure a 10) becomes,.17212Then for the Person’s Height, viz. 66 Inches: in the Table of Logarithms is the corresponding Number,1.81254And for the Co-Sine (had by subtracting the Altitude 42.17 from 90.00) viz. 47.43: among the artificial Sines is the Logarithm,9.86913————The above Sums added, are11.86079which logarithmic Number (deducting theInitial1 as useless) viz. 1.86079, in the Table of Logarithms, corresponds to 72.57, equal to 72 Inches, for the Length of the Shadow at XII.Reducing then the Numbers 66 and 72, to the lowest Denomination, thus 6)66⁄72 = 11⁄12 the Proportion which theLengthof theShadowbears to theHeightof theObjectis thereby obtained: that is

PROBLEM.

To find theLengthof theShadowfrom a Person ofmiddleStature, (five Feet and a half High) viz. at XII o’Clock, on the 8th Day of September, 1785, at Chester, whose North Latitude is 53° 12′; (and 3° 11′ West Longitude from London.)

FIRST,

From

90°. 00′′

The Latitude

53. 12

———

The Remain.

36. 48

Sun’s N. Decl.

5. 29

———

The Remain.

42. 17

SECOND,

For the Shadow say,As the Sine of the Sun’s Altitude 42° 17′To the Person’s Height, viz. 66 Inches,So is the Co-Sine of the Sun’s Altitude,To the Length of the Shadow.

.17212

1.81254

9.86913

————

11.86079

Reducing then the Numbers 66 and 72, to the lowest Denomination, thus 6)66⁄72 = 11⁄12 the Proportion which theLengthof theShadowbears to theHeightof theObjectis thereby obtained: that is

[26]If theLengthof the Shadow be divided into 12 Parts, the Height of the Object would be 11 of those Parts.See Moore’sPractical Navigator.PROBLEM.AneasyWay to find the Proportion which theLengthof the Shadow bears to theHeightof an Object is,at any time when the sun shines, to fix a Plummet Line andframeuprightin the Ground; measure theLengthof itsShadow, and compareitwith theHeightof theframe.

PROBLEM.

AneasyWay to find the Proportion which theLengthof the Shadow bears to theHeightof an Object is,at any time when the sun shines, to fix a Plummet Line andframeuprightin the Ground; measure theLengthof itsShadow, and compareitwith theHeightof theframe.

[27]Equal to 3 Quarters of a Mile and 121 Yards.

[28]i. e. When the Barometerbelowis at 30 Inches, and Thermometerbelowat 60° viz. about 1000 Yards high infineWeather, and 500 inchangeable.

[29]Being 1083 Yards, i. e. half a Mile, and 203 Yards.

[30]It was High Water at Chester and Frodsham-Bridge, at 38 Minutes past I.

[31]Articles parted with, to check thefirstDescent at Bellair, near Frodsham: and to ascend thesecondTime.To check thefirstDescent.Pounds.Ounces.Ballast, at twice:240To clear Trees and Hedges, andre-ascend:Barometer and Frame,0121⁄2Basket with Tunning Dish and Bottles (except the Flask with Brandy and Water)410Half Mile of Twine on the Reel10Speaking Trumpet081⁄2Woollen Gloves01—————310240—————Remains for Re-ascent70

121⁄2

81⁄2

—————

Remains for Re-ascent

[32]The Sun’s Azimuth from the North PointWestward, being 118.26′: its Supplement to 180° is 61°.34′ South westerly: i. e. South West by West, half Westnearly.

[33]TheLengthof the Shadows being more thandoublethe Height of theObjects: see[34].

[34]To find the Length of the Shadow at half past III.(See Section 84,Note [25].)Given{Lat. of Chester,53°12′{To find Sun’s Alt.Sun’s Dec.529Hour III, 30M.5230This is the Case of an oblique spheric Triangle, wherein are two Sides and one Angle between them given, to find the Sun’s Azimuth, and the Sun’s Co-Alt.Side84.31{Sum of Sides121.19Side36.48Diff. of Sides47.43(31⁄2 Hour) Angle contained52.30Half ditto26.15{Co.63.45Half Sum of Sides60.3929.22Half Difference ditto23.5166.9THE FIRST PREPARATIVE PROPORTION.As Sine of 1⁄2 Sum of Sides60.390.05966Co-Ar.To Sine of 1⁄2 Difference of Sides23.519.60675So Co-Tangent 1⁄2 contained Angle63.4510.30703———————To T. of 1⁄2 Diff. of the other two Angles43.159.97344SECOND PREPARATIVE PROPORTION.As Co-Sine 1⁄2 Sum of Sides29.210.30968Co-Ar.To Co-Sine 1⁄2 Diff.66.99.96123So Co-Tangent 1⁄2 contained Angle63.4510.30703————To T. 1⁄2 Sum of other Angles75.1110.57794Half Diff. before found43.15———Sum, is greater Angle118.26= Sun’s Azim.Diff. is lesser Angle31.56= S’s right Asc.Then by first Axiom in Trigonometry, to know the Sun’s Altitude say,As Sine Sun’s right Asc.31.560.27659To Sine Co-Lat.36.489.77744So Sine of the contained Angle52.309.89947————To Co-Sine of the Sun’s Alt.63.579.95350from90.———Sun’s Alt.26.3Having Sun’s Alt. to find the Shadow,As Sine Sun’s Alt.26.30.35738Co-Ar.To Person’s Height,66Inches,1.81954So Co-Sine of the Sun’s Alt.63.579.95350————To Length of Shadow,135Inches,2.13042Then 6(66⁄135 = 11⁄22 − | − 3⁄6 or 1⁄2, i. e. as 22 to 45: supposing the Length of the Shadow divided into 45 Parts; theHeightof the Object woud be 22 of those Parts; or not quitehalftheLengthof the Shadow, at half past III.

(See Section 84,Note [25].)

53°

12′

This is the Case of an oblique spheric Triangle, wherein are two Sides and one Angle between them given, to find the Sun’s Azimuth, and the Sun’s Co-Alt.

84.

121.

36.

47.

52.

26.

63.

60.

29.

23.

66.

THE FIRST PREPARATIVE PROPORTION.

60.

0.05966

23.

9.60675

63.

10.30703

———

————

43.

9.97344

SECOND PREPARATIVE PROPORTION.

29.

0.30968

66.

9.96123

63.

10.30703

75.

10.57794

43.

118.

31.

Then by first Axiom in Trigonometry, to know the Sun’s Altitude say,

31.

0.27659

36.

9.77744

52.

9.89947

————

63.

9.95350

from

90.

———

26.

Having Sun’s Alt. to find the Shadow,

26.

0.35738

Inches,

1.81954

63.

9.95350

————

135

Inches,

2.13042

Then 6(66⁄135 = 11⁄22 − | − 3⁄6 or 1⁄2, i. e. as 22 to 45: supposing the Length of the Shadow divided into 45 Parts; theHeightof the Object woud be 22 of those Parts; or not quitehalftheLengthof the Shadow, at half past III.

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