A Practical System of Mensuration of Superficies and Solids ... |
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Page xii
... pound . 20 cwt . 28 pounds 1 quarter . 1 lb .: = 14 oz 11 dwt . 16 gr . troy . 7000 troy grains = 5760 175 " pounds 175 ounces 437 " grains 1 lb. avoirdupois . 1 lb. troy . 144 lbs . avoirdupois . 1 " pound 192 oz . 1 oz . 6 .8228 lb ...
... pound . 20 cwt . 28 pounds 1 quarter . 1 lb .: = 14 oz 11 dwt . 16 gr . troy . 7000 troy grains = 5760 175 " pounds 175 ounces 437 " grains 1 lb. avoirdupois . 1 lb. troy . 144 lbs . avoirdupois . 1 " pound 192 oz . 1 oz . 6 .8228 lb ...
Page xiii
... pound . NOTE . - An ounce of Gold is divided into 24 equal parts called carats , and an ounce of silver into 20 parts called pennyweights ; therefore to distinguish fineness of metals , such Gold as will abide the fire without loss , is ...
... pound . NOTE . - An ounce of Gold is divided into 24 equal parts called carats , and an ounce of silver into 20 parts called pennyweights ; therefore to distinguish fineness of metals , such Gold as will abide the fire without loss , is ...
Page 73
... pounds avoirdupois to the cubic foot ? Ans . 3375 pounds . NOTE . The weight of water which a vessel of any given dimen- sions will contain , may be found by calculating the capacity in cubic feet , and multiplying by 62 lbs . or 1000 ...
... pounds avoirdupois to the cubic foot ? Ans . 3375 pounds . NOTE . The weight of water which a vessel of any given dimen- sions will contain , may be found by calculating the capacity in cubic feet , and multiplying by 62 lbs . or 1000 ...
Page 74
... pound , the thickness of the lead being such as to require 7 lbs . for each square foot of surface , the inner dimensions of the cistern being as follows ; viz : the length 3 feet 2 inches , the breadth 2 feet 8 inches , and the depth 2 ...
... pound , the thickness of the lead being such as to require 7 lbs . for each square foot of surface , the inner dimensions of the cistern being as follows ; viz : the length 3 feet 2 inches , the breadth 2 feet 8 inches , and the depth 2 ...
Page 90
... pounds of water ? Ans . 8 ft . PROBLEM X. To find the Solidity of a Spherical Segment . ART . 60. Rule . - To three times the square of the radius of its base , add the square of its height 90 MENSURATION To find the Solidity of a ...
... pounds of water ? Ans . 8 ft . PROBLEM X. To find the Solidity of a Spherical Segment . ART . 60. Rule . - To three times the square of the radius of its base , add the square of its height 90 MENSURATION To find the Solidity of a ...
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A Practical System of Mensuration of Superficies and Solids: Designed ... J. M. Scribner No preview available - 2017 |
A Practical System of Mensuration of Superficies and Solids J. M. Scribner No preview available - 2022 |
A Practical System of Mensuration of Superficies and Solids: Designed ... J. M. Scribner No preview available - 2017 |
Common terms and phrases
12 feet 18 inches 20 feet 9 feet ABCD ABFD assumed cube avoirdupois axis base 26 breadth centre chord circle whose diameter Circular Sector circular segment circumfer circumference circumscribed contained convex surface Cube Root cubic feet cubic ft cubic inches cylinder cylindrical ring decimal diff divide the product ellipse ends entire surface equal extract the square feet 6 inches feet long find the area find the Solidity frustrum gallon half hypotenuse inner diameter inscribed square lateral surface length miles multiply the sum Nonagon number of degrees number of sides number of square OPERATION parabola parallel sides parallelogram pentagonal pyramid perpendicular distance perpendicular height plane prism PROBLEM radius regular pentagonal regular Polygon Required the area Required the solidity rhombus right angled triangle Rule Rule.-I slant height solid contents sphere spherical segment spheroid square feet square rods square root square yards thickness trapezium zoid zone
Popular passages
Page 53 - RULE. Find the area of the sector which has the same arc, and also the area of the triangle formed by the chord of the segment and the radii of the sector. Then...
Page 35 - Now, since the areas of similar polygons are to each other as the squares of their homologous sides...
Page 79 - A sphere is a solid terminated by a curved surface, all the points of which are equally distant from a point within called the centre.
Page 80 - A zone is a portion of the surface of a sphere included between two parallel planes.
Page 90 - ... to three times the square of the radius of the segment's base, add the square of its height ; then multiply the sum by the height, and the product by .5236 for the contents.
Page 49 - From 8 times the chord of half the arc subtract the chord of the whole arc, and ' of the remainder will be the length of the arc nearly.
Page 72 - RULE.* To the sum of the areas of the two ends add four times the area of a section parallel to and equally distant from both ends, and this last sum multiplied by £ of the height will give the solidity.
Page 51 - As 360 degrees is to the number of degrees in the arc of the sector, so is the area of the circle to the area of the sector.
Page 91 - From three times the diameter of the sphere subtract twice the height of the segment; multiply this remainder by the square of the height and the product by 0.5236.
Page 39 - A Circle is a plane figure bounded by a curve line, called the Circumference, which is everywhere equidistant from a certain point within, called its Centre.