Nassau New Providence July 4th. 1789
	Sir.
		Your established Character in the World as a Natural Philosopher
		and Master of Science, being a Native of America myself (tho’
		expelled by the Fortune of War from South Carolina) induces me to
		address a Line to you on a Subject interesting to the World at
		large but particularly to all maritime States. I shall esteem the
		acknowledgment of the Receipt of this with your candid and free
		Opinion thereon as an Honor conferred on Sir Your most Obedient
		and most humble Servant
	
 
	The Honble. Benjamin Franklin Esquire
		By measuring Space by Time The Moon appears to Me to be at the
		Distance of 36,313.845 Statute Miles from the Earths Center—Radius
		of her Orbit  2 = 72627.69 Diameter of Ditto x 3.1426968 (The
		true Proportion of Circumference to Diameter) = 228246.809 Miles
		for the Circumference of Orbit  29.530851 Days to one
		Revolution (= 235 migh?) = 7729.09689 Miles as the Diurnal
		Motion of the Moon and supposing her to revolve round her own Axis
		 3.1426968 = 2459.383574 Miles for her Diameter. I go on the
		Data that the Earths Radius is = 4000 Miles and Revolution in 24
		Hours of our Time.
	
		The Use I would recommend the above to would be to ascertain the
		true Longitude of Places East and West of any fixt Meridian (for
		which Purpose I have begun (tho a most tedious Business to
		calculate the sines and Co-sines to 1¹, for 45º the 8th
		part of Earths Circumference. As the lines of 45°, become the
		Co-sines for the Remainder of Quadrant by squaring the former,
		deducting the same from  of Radius and taking the Roots of
		Remainders for Co-sines which are the Sines for other Part. The
		Sines by Logarithms are I think very untrue tho’ they answer for
		small Distances. Such Table being compleat for real Time, If an
		Observation be make on any fixt day when the Sun is in the
		Meridian of Observer and the Moons Angle be ascertained to the
		Earth’s Center by referring to the Table for Angle of Moon at fix
		Meridian for that day The difference of Angle of Observation from
		Tabulat Angle will be the Difference of Time and Distance And by
		simple Proportion. As suppose from 0 in Scheme, you find the Angle
		of Moon, as at Moon observed, and by your Table it should be in
		your Zenith on due South The difference between the Time being
		known Say As 10631.10636 Minutes in Quadrant of Moons Orbit :
		3600' (=6 Hours Earths Quadrant) :: Difference of Time in Angle
		: Min on Earths Surface Then As 3600' : 6285.3936 Miles in
		Quadrant :: difference of Time : Miles in Longitude.
	
 
		Suppose a Person at the Equator on any day when the Moon at any
		fixt Meridian should be due South at Noon but by Observation
		should be found to have an Angle which being corrected for Earths
		Center as in the Scheme below as OE Moon observed = 65° from
		Observer’s Zenith = 7678.021293 Minutes of Time They say As
		10631.10636 : 3600' :: 7678.021293 : 2600' Time As 3600 :
		6285.3936 Miles in Quadrant :: 2600' : 4539.45093 Miles But
		Suppose the Observer to be in Latitude 25° North and find same
		Angle of Difference look into Table for Cosine and you’ll find
		3678.4313  1.5713484 = 5680.0971 Miles for Quadrant in that
		latitude then say as 90° : 5780.0971 :: 65° : 4174.514572
		Miles Quare if not the true Distance of Nassau in New Providence
		West of such fixt Meridian.
		Please excuse the want of Scholastic Terms as the Ideas are my own
		without such Help, but from Mechanic Principles.
	
 
		Quadrant of Moons Orbit at 25°. taken as a Globe of that Size
		as a Proof for this Globe
		Earths Radius 4000  1.5713484 = 6285.3936 Miles Quadrant And
		the Proportion is easily demonstrated tho differing from all
		before it.
		Quare How can any Person know exactly in what Latitude He is in by
		the usual method of taken the declination at a fixt Meridian when
		perhaps she be from 50° to 90° West of same when the
		Declination is varying constantly and at a considerable Rate and 6
		Hours carries a Quadrant of this Globe to Meridian fixt.