An Encyclopaedia of Architecture, Historical, Theoretical, and PracticalLongmans, Green, 1876 - 1395 pages |
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Page 21
... feet from the portals , southward , is another stair- case of two flights ( lettered C ) , one west and the other east . On the top of the ramp of the steps are some foliages , and a lion tearing to pieces a bull , in bas - relief , and ...
... feet from the portals , southward , is another stair- case of two flights ( lettered C ) , one west and the other east . On the top of the ramp of the steps are some foliages , and a lion tearing to pieces a bull , in bas - relief , and ...
Page 329
... feet , and height 8.5 feet . 12.25 x 8.5 = 104-125 feet , or 104 feet 11 inches . Example 2. Required the content of a piece of land in the form of a rhombus whose length is 6 20 chains , and perpendicular height 5.45 . Recollecting ...
... feet , and height 8.5 feet . 12.25 x 8.5 = 104-125 feet , or 104 feet 11 inches . Example 2. Required the content of a piece of land in the form of a rhombus whose length is 6 20 chains , and perpendicular height 5.45 . Recollecting ...
Page 334
... feet . Here we have six sides ; therefore 20 × 20 × 6 = 2400 feet , the area required . Example 2. Required the surface of a triangular prism whose length is 20 feet and each side of its end or base 18 inches . Here we have , for the ...
... feet . Here we have six sides ; therefore 20 × 20 × 6 = 2400 feet , the area required . Example 2. Required the surface of a triangular prism whose length is 20 feet and each side of its end or base 18 inches . Here we have , for the ...
Page 335
... feet . Here , 8.5 x 3 · 1416 × 50 ÷ 2 = 667 5 , the convex surface required . 1233. PROBLEM III . To find the ... feet , each side of the base being 3 feet 4 inches and each side of the top 2 feet 2 inches . Here , 3 feet 4 inches x 4 ...
... feet . Here , 8.5 x 3 · 1416 × 50 ÷ 2 = 667 5 , the convex surface required . 1233. PROBLEM III . To find the ... feet , each side of the base being 3 feet 4 inches and each side of the top 2 feet 2 inches . Here , 3 feet 4 inches x 4 ...
Page 336
... feet and the circum- ference of its base 9 feet . Here , 07958 ( Prob . 9. 1222. ) × 81 = 6 · 44598 area of base , And 3-5 being of 10 feet , 6 · 44598 × 3.5 = 22 · 56093 is the content required . 1236. PROBLEM VI . To find the solidity ...
... feet and the circum- ference of its base 9 feet . Here , 07958 ( Prob . 9. 1222. ) × 81 = 6 · 44598 area of base , And 3-5 being of 10 feet , 6 · 44598 × 3.5 = 22 · 56093 is the content required . 1236. PROBLEM VI . To find the solidity ...
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15th century aisles arch architect architecture axis base beam breadth bricks building built called cast iron Castle cathedral cement centre of gravity chapel choir church circle circumference colour columns construction Corinthian order cube cubic foot curve decorated depth describe diameter dome Doric order draw edifices ellipsis employed entablature equal erected examples extrados feet girder given Gothic granite half horizontal inches intercolumniations joints length lime limestone marble mortar mouldings nave oolite ornaments palace parallel parallelogram perpendicular piece piers placed plane plates portico Portland stone Prop proportion pyramid quarries radius rectangle ribs right angles right line Roman Roman architecture Rome roof sandstone side similar sofite solid square stone strength style surface tangent temple thickness tiles timber tons tower transepts triangle upper vault vertical Vitruvius voussoirs walls weight whereof width wood