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159)

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The Subject of a Quadrant being divided by Diagonal lines & equidistant concentric circles farther prosecuted. Scep. 41, 43. 

M.[[superscript]] r [[/superscript]] Jones's Quadrant to the best of my remembrance was 5 feet 10 Inches from the center to the exterior circle of the limb, had 30 concentric circles about, 15 Inch distant from each other, and each upon the diagonal line was 5 seconds of a degree: from whence the limb of the quadrant or the whole angular extent of one Diagonal line reaching from the exterior to the interior concentric circle was 2 1/2 Minutes; which gives only ,051 Inch for the length of the arc between each division on the exterior limb; whereas I think it was not much less than 1/10 Inch. And if so, it was divided into every 5.' & 60 concentric circles. which I take to be the case upon farther consideration & that the breadth of the limb was about 4 1/2 or 5 Inches. - I think each 5." was about this measure, vis. [[Image: two horizontal lines crossing two vertical lines]]
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[[right margin]] [[Image - drawing of an angle ABC, projection of D, perpendicular EF]] [[/right margin]]

   Let C be the center of a quadrant, AB so much of the limb as is divided diagonally by the line AE, and [[strikethrough]] the [[/strikethrough]] equidistant concentric arches. Now
  
    1. To find the error of a quadrant so divided; continue [[strikethrough]] [[?]] [[/strikethrough]] the diagonal AE indefinitely, from C let fall the perpendicular CD, & from E let fall the perpendicular EF. Put R = AC, the Radius of the quadrant; r = GC, the Rad. exclusive of the breadth of the limb; d = R-r = GA, the said breadth; s & v, the sine & versed sine of the given