[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 retained abroad; 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 Words above or below the 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.
54	13 & 4-9ths above the Cypher.
55	14 ditto, nearly.
57	15 2-9ths ditto, nearly.
59	16 4-9ths ditto, nearly.
60	17 1-9th ditto.
65	20 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]

	Pounds Averdupois.
Weight of the Aironaut	160
Provisions and Articles calculated at	20
Sand-Ballast prepared in Bags	44
Levity for Ascent	10
	——
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 Loop shoud have been furnished with a Swivel: 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 the Whetstone of 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]

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 a golden Colour: occasioned by a Haziness or Vapour which pervades the lower Regions 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 the Campania of Rome, for Instance, the Italians have a peculiar Name for such Kind of Weather, when the Sun is neither visible 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 in Imaginibus quâpropter Causa videtur Cernendi, neque posse sine his Res 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 of Epictetus, (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 they practically influenced 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 the apparent Height from the Balloon to the Ground was 7 Miles, viz. 4 to the Summit of the Clouds, and 3 below: and the barometric Height 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 Distance in Yards.		Greater Distance in Yards.
Thus 7, : 4 ::	2332	: 1332⁠4⁄7 Ans.
	4
	———
	7)9328
2332 the whole.	1332⁠4⁄7

Note. The Line A. B. here selected is the famous Measure of (half) a mathematical Rhinland and Roman Foot, according to Snellius. (See Geographia Generalis of Varenius, published by Newton. Lib. 1. Cap. 2. De variis Mensuris.)

[18]

To find the circular Boundary of the celestial Prospect 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.

Rule. To the Earth’s Diameter, equal to 7940 geographical Miles, add the Height of 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. (See Practical Navigator, by J. Moore, 7th Ed. Page 251.)

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.

	13932702(1⁄3)	Diameter of the Earth in Yards.
	2000	Addition to the Diameter.
	————
	13934702	Sum, to which add
	1332	the Height of the Eye or of the
	————	Balloon above the Floor of Cloud.
	13936034	Sum, which multiply into
	1332	the Height of the Eye above the
	————	Floor.
	27872068
	41808102
	41808102
	13936034
	——————
Extract the	.    .    .    .    .	1760) Yards in a Mile.
Square Root	18562797288	(136245 (77 Miles.
	1	12320
	—	———
	23) 85	13045
	69	12320
	——	———
	266) 1662	Yards 440) 725	(1 Quarter of a Mile.
	1596	440
	——	——
	2722) 6679	285	Yards.
	5444
	———	Ans. 77	Miles, 1 Qu. 285 Yards.
	27244) 123572
	108976
	———
	272485) 1459688
	1362425
	————
	97263

To find the circular Boundary of the terrestrial Prospect, 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	Yards,
	add 2332	the Height of the Eye or Balloon.
	————
	13935037	he Sum, multiply into
	2332	the Height of the Eye, &c.
	————
	27870074
	41805111
	41805111
	27870074
	——————
Extract the	.    .    .    .    .	1760) Yards in a Mile.
square Root	32496506284	(180267(102,
	1	1760 say 102​1⁄2 Miles, Ans.
	—	——
	28)224	4267
	224	3520
	——	——
	3602) 9650	747 Yards, Remainder.
	7204
	———
	36046) 244662
	216276
	————
	360527) 2838684
	2523689
	————
	314995	Remainder.

[19] See his “Minute Philosopher.”

[20] Ullòa in his voyage to South-America relates, that in passing over the Deserts, Írides are frequently seen by Travellers round their own Heads as the Center of the Iris; and visible only to themselves. But what Analogy the Balloon Iris bears to them, Time and future Experiments may discover. See his “Voyage to South America, Vol. 1. Pa. 442.”

[21]

As Sound travels	1142	Feet in a
Second, it must have moved in	30	Seconds
	———
Feet in a Yard	3)34260	= Feet
Yards in a Mile	1760)11420	(6 Miles
	10560
	——
Yards in a Quarter of a Mile	440)860	(1 Quarter
	440
	——
Answer 6 Miles, 1 Quarter, and	420	Yards.

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

[23] Long’s Astronomy. Pages 227, 229.

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

[25]

To find the Length of the Shadow from a Person of middle Stature, (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.)

To find the Sun’s Altitude at XII.
From	90°. 00′′	Subtract
The Latitude	53.  12
	———
The Remain.	36.  48	is the Complement of Latitude, to which add (from the Tables)
Sun’s N. Decl.	5.  29
	———
The Remain.	42.  17	is the Sun’s Altitude (viz. at XII.)

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,	.17212
Then for the Person’s Height, viz. 66 Inches: in the Table of Logarithms is the corresponding Number,	1.81254
And 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, are	11.86079
which logarithmic Number (deducting the Initial 1 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 the Length of the Shadow bears to the Height of the Object is thereby obtained: that is

[26] If the Length of the Shadow be divided into 12 Parts, the Height of the Object would be 11 of those Parts. See Moore’s Practical Navigator.

An easy Way to find the Proportion which the Length of the Shadow bears to the Height of an Object is, at any time when the sun shines, to fix a Plummet Line and frame upright in the Ground; measure the Length of its Shadow, and compare it with the Height of the frame.

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

[28] i. e. When the Barometer below is at 30 Inches, and Thermometer below at 60° viz. about 1000 Yards high in fine Weather, and 500 in changeable.

[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 the first Descent at Bellair, near Frodsham: and to ascend the second Time.

To check the first Descent.	Pounds.	Ounces.
Ballast, at twice:	24	0
To clear Trees and Hedges, and re-ascend:
Barometer and Frame,	0	12​1⁄2
Basket with Tunning Dish and Bottles (except the Flask with Brandy and Water)	4	10
Half Mile of Twine on the Reel	1	0
Speaking Trumpet	0	8​1⁄2
Woollen Gloves	0	1
	—————
	31	0
	24	0
	—————
Remains for Re-ascent	7	0

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

[33] The Length of the Shadows being more than double the Height of the Objects: see [34].

[34] To find the Length of the Shadow at half past III.

Given	{	Lat. of Chester,	53°	12′	{	To find Sun’s Alt.
		Sun’s Dec.	5	29
		Hour III, 30M.	52	30

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.

Side	84.	31	{	Sum of Sides	121.	19
Side	36.	48		Diff. of Sides	47.	43

(3​1⁄2 Hour) Angle contained	52.	30
Half ditto	26.	15	{	Co.	63.	45
Half Sum of Sides	60.	39			29.	22
Half Difference ditto	23.	51			66.	9

As Sine of 1⁄2 Sum of Sides	60.	39	0.05966	Co-Ar.
To Sine of 1⁄2 Difference of Sides	23.	51	9.60675
So Co-Tangent 1⁄2 contained Angle	63.	45	10.30703
	———		————
To T. of 1⁄2 Diff. of the other two Angles	43.	15	9.97344

As Co-Sine 1⁄2 Sum of Sides	29.	21	0.30968	Co-Ar.
To Co-Sine 1⁄2 Diff.	66.	9	9.96123
So Co-Tangent 1⁄2 contained Angle	63.	45	10.30703
			————
To T. 1⁄2 Sum of other Angles	75.	11	10.57794
Half Diff. before found	43.	15
	———
Sum, is greater Angle	118.	26	= Sun’s Azim.
Diff. is lesser Angle	31.	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.	56	0.27659
To Sine Co-Lat.	36.	48	9.77744
So Sine of the contained Angle	52.	30	9.89947
			————
To Co-Sine of the Sun’s Alt.	63.	57	9.95350
from	90.
	———
Sun’s Alt.	26.	3

Having Sun’s Alt. to find the Shadow,

As Sine Sun’s Alt.	26.	3	0.35738	Co-Ar.
To Person’s Height,	66	Inches,	1.81954
So Co-Sine of the Sun’s Alt.	63.	57	9.95350
			————
To Length of Shadow,	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; the Height of the Object woud be 22 of those Parts; or not quite half the Length of the Shadow, at half past III.