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Some naive comments/questions:

First, how much of the discrepency between Wright's calculated  
distance and the modern digital calculator is due to the elliptical  
shape of the earth, or were you using the same assumptions?

Second, one could guess that a graphical method would be good to 3  
decimal places (about what you got for question 1).  Five-decimal- 
place precision is needed to get 0.1 arcminute accuracy, more or  
less, so a graphical method would only be good to 10 arcminutes, more  
or less.  Perhaps it's the lack of precision that led to Wright's  
method not being adapted to standard sight reduction.  Certainly back  
in his time, simple reduction of noon sights for altitude was easy  
enough.  By the period when time sights for longitude became  
prevalent, and especially by the point when intercept methods took  
over, 3 decimal places wasn't close enough anymore.

Fred

On Dec 4, 2007, at 4:18 AM, Gary J. LaPook wrote:

> Gary J. LaPook wrote:
>
>
> It is not surprising that nobody ever noticed this before  
> (considering that Wright published in 1599 almost 300 years prior  
> to Marc St. Hilaire) that Wright's method of calculating the great  
> circle distance on the earth using only a strait edge and a compass  
> could just as easily be used to calculate the altitude of a  
> celestial body. The great circle distance is simply 60 NM times the  
> number of degrees of the great circle between two points and this  
> is exactly the same as the zenith distance to a body having the  
> geographical position represented by the second point.     The  
> formula is 90� minus zenith distance equals altitude.
>
> Wright's example of calculation of the great circle distance  
> between London and Jerusalem resulted in his calculated distance of  
> 2325 NM and a modern digital calculator comes up with 2316.8 NM a  
> difference of  only 8.2 NM or minutes of zenith distance or of  
> computed altitude for those coordinates! Using his method Wright  
> could compute altitudes to a precision of 8.2'. It is surprising in  
> light of the many devices invented later in an attempt to find a  
> mechanical method for this calculation that none (that I am aware  
> of) attempted to use Wright's method, a method that would seem  
> easily adapted to a mechanical device and that could provide much  
> greater accuracy using a larger scale and precise machining of the  
> parts.
>
> I would really like it if someone could explain why Wright's method  
> works since I have not been able to find such an explanation  
> anywhere. I am attaching pages 45-52 of "Certaine Errors" in which  
> he lays out his method. I am also including the errata sheet  
> showing that the corrections of typos I identified in my previous  
> posts were correct.
>
> gl
>
>
> >
> 
> 



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