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41)
[[margin notes not related to text]]
To place the first Hair in a Telescope in the Center of the Glasses.
To fix the second Hair at Right-angles with the first.
Of fixing a telescope to a Quadrant.
Another way.
[[/margin]]

Upon the NONIUS.
Let ABCDEF (Fig 11.) be a right-line divided into any Number of equal parts, n; and abcdefg another right line equal in length to the former, divided likewise into a number of equal parts, n+1; let these two lines be contiguous, and even at their extremities; which then are the only divisions that will concide; and if the position of one ag be altered by sliding along AF, the extremities A and a, F and f will no longer concur; but some one of the others may, as D and e, in which situation no other division upon AF can concur with any division upon ag. For, putting P = one division AB, and p = ab; Then Bb = P-p = P/(n+1) = P-p = p/n, when the extremities A & a, F & g. concur, and Cc = twice, Dd = three times, Ee = four times, &c. the distance Bb: Now the two Lines AF, ag, being contiguous, & moving parallel so as the two divisions at D and e may concur; then it is evident ^ [[insertion]] no one of [[/insertion]] the divisions upon the [[strikethrough]] right [[/strikethrough]] ^ [[insertion]] left [[/insertion]] -hand of D is so far from [[strikethrough]] as [[/strikethrough]] its corresponding division from a, upon ag. it will have moved over a space greater than its distance from its corresponding division, and therefore will not concur with it, and not hiving moved over space sufficient to reach the next, it concurs with no division: so likewise those on the right hand of D, as E, being farther from its corresponding division e, than D is from e, and not so far from f, E will move over f, but not reach g; and therefore conjoined with none, the same may be said of all the othe divisions on the right; or when any other two divisions are conjoined. After the same manner may it be proved, that no one of the divisions may be conjoined. 
When ag, which has the most divisions, is the moveable arch, and AF the fixed limb, it is then a [[underline]] Nonius of the first kind [[/underline]], & is mostly prefer[e]d as in [[underline]] Hadley's Octant [[/underline]]: but when AF moves upon the fixed limb ag, it then is a [[underline]] Nonius of the second kind [[/underline]]. --- V. page 40. [[underline]] Tycho Brahe [[/underline]], in subdividing his Quadrant with diagonal Lines, says the space included between the exterior & interior concentric circles, should never be more than 1/48 of the Radius; [[insertion]] and [[/insertion]] [[strikethrough] but [[/strikethrough]] by how much less than 1/48, the more exact will ^ [[insertion]] be [[/insertion]] the subdivisions; and that all the concentric circles must be accurately [[underline]] equidistant. [[/underline]] In all which the moderns agree, though for what reasons, I am ignorant. Because, in fig [[insertion]] -s [[/insertion]]. 21. ^ [[insertion]] & 22. [[/insertion]] let AB be the Arch of one degree to the Radius AC or CB, [[margin]] See p. 159 [[/margin]] and let AE or BD e 1/48 of the radius, draw the concentric arc EGD, then are the arcs AFB, EGD the exterior and interior concentric circles; and AOD, the Digonal Line, is cut by the arch IPK, which is equidistant both from EGD & AFB, in Q; thro' which if HL a part of the radius, pass it will divide the Arc AB unequally in L. Becasue the Sides AE, BD being not parallel, as AE & BM are, but E & D incline to each other, [[strikethrough]] the Diagonal [[/strikethrough]] & therefore D lying nearer E than M does, the diagonal AQD will lye [[strikethrough]] able [[/strikethrough]] above the diagonal AM, and consequently Q above P; So that the Arch ROS passing thro' the intersection of GF, (which bisects ED & AB) and AD, is that [[strikethrough]] will [[/strikethrough]] which cuts the Diagonal AD at the proper place to divide AB into 2 equal parts; & yet this Arc ROS is nearer EGD than to AFB. and therefore the concentric circles cannot be [[underlined]] equidistant [[/underlined]], as Tycho Brahe, asserts.  Again, Let ελδ & αμβ be the interior & exterior concentric circles, draw the Diagonal αδ, & Quinquisect the L C by the lines Cμ, Cμ, Cμ, Cμ, and transfer their intersections, η, η, η, η with the diagonal αδ, to the side δβ, & they will be all [[underline]] unequal [[/underline]]. and nowhere conincide with the equal divisions θ, which are also transferred to the diagonal,(Continue on page 43.)    

Transcription Notes:
Not sure how to type the Greek letters, so I spelled them out. Check for accuracy, too! Quinquisect has been googled and it is a real word.