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the Velocity of the other Body, being to the Velocity of the firft, as the Cofine of (52°) the given Latitude to Radius, it will be expreffed by

84'43"

38′

X

Cof. 52°
Rad. ›
Unity: Now let

the circular Velocity being denoted by

84' 43'
38'

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X

Cof. 52° Rad. then m2 being lefs than 2, the Body projected from the Parallel of 52° will defcribe an Ellipfis ABFDA, whereof the Semi-tranfverfe AO =

by m, and the Earth's Radius AC by a,

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A

√2

-

2 m

Therefore, of different Bodies revolving round the fame Center, the Squares of their periodic Times, being as the Cubes of the transverse Diameters, of the Section described, let the other Diameters be what they will, AC'

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B

Ellipfis ABFDA; therefore

H

G.

D

F

P

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E

is the perio

dic Time required, which divided by 38′ leaves

84′ 43′′

2

; therefore as 38' is to the faid Re

38′ x 2-m mainder, fo is 360° to the Difference of Longitudes of the two Points of the fame Parallel from whence the Body is projected and where it falls.

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Again, with regard to the other Body projected from under the Equator, the Curve AE which it defcribes, because n is greater than 2, will be an

Hyperbola whose Semi-tranfverfe is

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nz

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a

2

and

;' (fee Simpfon's Flux

ions, Page 154) and the Hyperbolical Sector ACEA defcribed in 6 Hours, will be to the Area of the Circle AHGA described in 38', as 6 Hours to 38', and therefore will be expreffed by

x

ra

AHGA x 360'

38

a2 × 360′ × 3.141592, &c. ="a2; but the Area

38

2

ACEA, fuppofing the Ordinate PE to be expref

fed by az, will be alfo expreffed by

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n2 — 1x a2 Z

2 x n2

2

n2 a2

2 xn

2

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taz

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2

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I =d, it will become bdz-cdx Hip. Log.

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2

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=r, or z= baix Hip. Log.

√x2+c2+%; whence the Distance of the Body

;

from the Earth's Center may be determined.

This Problem was, also, answered by Tycho Oxonienfis, who makes the periodic Time of the Body projected from the Parallel of 52°, to be 1 Day 9 Hours 44′ 52", and that projected from under the Equator, to be 186947 Miles diftant from the Earth's Center at the End of the propofed Time.

A

Ꭺ .

COLLECTION

O F

PROBLEMS

TO BE

Answered in the next NUMBER.

I

PROBLÈM XXI. L.

F to enjoy the Benefit of an Estate for 23 Years after the Expiration of eight Years, be worth 400l. prefent Money; what will the faid Eftate be worth for 21 Years after the Expiration of 10 Years, at the Rate of 51. for every 100 Yearly.

PROBLEM XXII. L.

A Traveller benighted, fees before him two Lights, and looking back difcovers three others, in a right Line with the former; and judges the Quantity of Light received from the former Place,

to

to be to that received from the latter, in the Ratio of 3 to 4; then proceeding forwards 400 Yards, he finds the Ratio of Light there to become as 5 to 3: The Queftion is, fuppofing all the Lights to be equal among themselves, what Distance he was from each Set of Lights, at the two Placesof Obfervation.

PROBLEM XXIII. A Theorem to be demonftrated.

As the leffer of the two Sides, including any propofed Angle, of a Triangle, is to the greater, fo is Radius, to the Tangent of an Arch or Angle: And as Radius, is to the Tangent of the Excefs of the faid Angle above half a right Angle, fo is the Tangent of half the Sum of the oppofite Angles, to the Tangent of half the Difference of the fame Angles.

PROBLEM XXIV. Thomas Hulme, London.

The Line bifecting the Base, the Difference of the Sides, and the Difference of the Angles at the Base, of any plane Triangle being given; to determine the Triangle.

PROBLEM XXV. John Turner, London.

The Base of any plane Triangle, the vertical: Angle and the Side of the infcribed Square being given; to construct the Triangle.

PROBLEM XXVI. R.

To defcribe a Circle, through a given Point, which fhall touch a right Line given in Pofition, and also another Circle given in Magnitude and Pofition.

· PRO

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