A Treatise on Mensuration: Both in Theory and PracticeSaint, 1770 - 646 pages |
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Page 108
... versed fines ( marked V. S. ) in the table of circular fegments at the end of the book . 3. Multiply the number immediately on the right of the verfed fine , in the table , by the fquare of the diameter , and the product will be the ...
... versed fines ( marked V. S. ) in the table of circular fegments at the end of the book . 3. Multiply the number immediately on the right of the verfed fine , in the table , by the fquare of the diameter , and the product will be the ...
Page 112
... versed fine ; the tabular area for which is 07134954 . Then 07134954 × 25 × 25 = 44 ° 5934625 = the feg- ment AGDRA . DO + PA But D 2 · x P Q = 35 × 35 = 1225 = 153 * 125 = the trapezoid APQD . Therefore 153125 + 44′5934625 = 1977184625 ...
... versed fine ; the tabular area for which is 07134954 . Then 07134954 × 25 × 25 = 44 ° 5934625 = the feg- ment AGDRA . DO + PA But D 2 · x P Q = 35 × 35 = 1225 = 153 * 125 = the trapezoid APQD . Therefore 153125 + 44′5934625 = 1977184625 ...
Page 147
... versed fine , to which in the table of circu- lar fegments correfponds the area 08604117 ; hence 08604117 × 9 ( AB2 ) = 77437053 = the area of the feg . BDE , and 78539816-08604117 × 9 • = * 69935699 × 96′29421291 = the fegment ADE ...
... versed fine , to which in the table of circu- lar fegments correfponds the area 08604117 ; hence 08604117 × 9 ( AB2 ) = 77437053 = the area of the feg . BDE , and 78539816-08604117 × 9 • = * 69935699 × 96′29421291 = the fegment ADE ...
Page 172
... versed fine of the bafe of the hoof divided by the diameter of the base of the fruftum ; multiply the former feg- ment by the cube of the diameter of the end , by the quotient of the verfed fine of the base of the hoof divided EBF- BD ...
... versed fine of the bafe of the hoof divided by the diameter of the base of the fruftum ; multiply the former feg- ment by the cube of the diameter of the end , by the quotient of the verfed fine of the base of the hoof divided EBF- BD ...
Page 245
... versed fine , whose segment is 1851669 . · Then * 1851669 × 35 × 25 = 162 ° 0210375 = the area of the lefs fegment . If the greater fegment had been required . Then 78539816-1851669 = 60023126 . And 60023126 × 25 × 35 = 525 * 2023525 ...
... versed fine , whose segment is 1851669 . · Then * 1851669 × 35 × 25 = 162 ° 0210375 = the area of the lefs fegment . If the greater fegment had been required . Then 78539816-1851669 = 60023126 . And 60023126 × 25 × 35 = 525 * 2023525 ...
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Common terms and phrases
abfciffa againſt alfo Alnwick alſo altitude angle area Verf bafe baſe becauſe breadth bung cafe cafk circle whofe circumference cofine cone confequently conftruction conjugate COROLLARY defcribe difference dimenfions diſtance ditto divided divifion ellipfe equal expreffed expreffion faid fame example fecond fection feet fegment feries fhall fide figure fimilar fince find the area firſt fixed axe fluxion folid fome fpheroid fpindle fquare fruftum fubtract fuch fuppofing furface gallons girt greateſt half head diameter hence hoof hyperbola inches interfecting laft laſt laſt problem lefs length meaſure method middle moſt multiply muſt Newcaſtle oppofite ordinate parabola paraboloid parallel perpendicular plane prob quotient radius rule Schoolmafter ſhall Sliding Rule ſphere ſquare ſtation tangent theſe thickneſs thofe thoſe tranfverfe trapezium triangle uſed vertex Wherefore whofe whofe height whole whoſe zone ΙΟ