The New Practical Builder and Workman's Companion, Containing a Full Display and Elucidation of the Most Recent and Skilful Methods Pursued by Architects and Artificers ... Including, Also, New Treatises on Geometry ..., a Summary of the Art of Building ..., an Extensive Glossary of the Technical Terms ..., and The Theory and Practice of the Five Orders, as Employed in Decorative ArchitectureThomas Kelly, 1823 - 596 sider |
Inni boken
Resultat 1-5 av 35
Side 6
... cutting stone , illustrated by several plates , answering to most purposes which present themselves . And since the principles laid down in this Work are every where of a general tendency , the judgement of the work- man will enable him ...
... cutting stone , illustrated by several plates , answering to most purposes which present themselves . And since the principles laid down in this Work are every where of a general tendency , the judgement of the work- man will enable him ...
Side 62
... cutting BA at g , and BC at h ; from the point E , with the same radius , describe an arc , ik , cutting ED ati : make ik equal to gh , and through the point k draw EF : then the angle DEF will be equal to the given angle ABC . PROBLEM ...
... cutting BA at g , and BC at h ; from the point E , with the same radius , describe an arc , ik , cutting ED ati : make ik equal to gh , and through the point k draw EF : then the angle DEF will be equal to the given angle ABC . PROBLEM ...
Side 63
... cutting each other in A and B : join AB , and this line will bisect AB perpendicularly . PROBLEM 6 . 179. From a given point C , ( fig . 40 , pl . I , ) in a given straight line , AB , to erect a perpendicular . In the straight line ...
... cutting each other in A and B : join AB , and this line will bisect AB perpendicularly . PROBLEM 6 . 179. From a given point C , ( fig . 40 , pl . I , ) in a given straight line , AB , to erect a perpendicular . In the straight line ...
Side 64
... cutting AB at e and f ; from the points e and f , as centres , with any equal radius greater than the half of A , describe arcs cutting each other in D , and draw CD , which will be the perpendicular required . PROBLEM 9 . 182. To ...
... cutting AB at e and f ; from the points e and f , as centres , with any equal radius greater than the half of A , describe arcs cutting each other in D , and draw CD , which will be the perpendicular required . PROBLEM 9 . 182. To ...
Side 65
... cutting each other in S ; from S , with the radius AS , BS , or CS , describe a circle ABCDE , then carry the side AB or BC round the remaining part of the arc , which will be found to contain the remaining sides of the number required ...
... cutting each other in S ; from S , with the radius AS , BS , or CS , describe a circle ABCDE , then carry the side AB or BC round the remaining part of the arc , which will be found to contain the remaining sides of the number required ...
Vanlige uttrykk og setninger
ABCD abscissa adjacent angles altitude angle ABD annular vault axes axis major base bisect called centre chord circle circumference cone conic section conjugate contains COROLLARY 1.-Hence cutting cylinder describe a semi-circle describe an arc diameter distance divide draw a curve draw lines draw the lines edge ellipse Engraved equal angles equal to DF equation equiangular figure GEOMETRY given straight line greater groin homologous sides hyperbola intersection join joist latus rectum less Let ABC line of section meet multiplying Nicholson opposite sides ordinate parallel to BC parallelogram perpendicular PLATE points of section polygon PROBLEM produced proportionals quantity radius rectangle regular polygon ribs right angles roof segment similar triangles square straight edge subtracted surface Symns tangent THEOREM timber transverse axis triangle ABC vault vertex wherefore
Populære avsnitt
Side 27 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Side 20 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Side 51 - The area of a parallelogram is equal to the product of its base and its height: A = bx h.
Side 15 - AXIOMS. 1. Things which are equal to the same thing are equal to one another. 2. If equals be added to equals, the wholes are equal. 3. If equals be taken from equals, the remainders are equal.
Side 15 - LET it be granted that a straight line may be drawn from any one point to any other point.
Side 28 - ... angles of another, the third angles will also be equal, and the two triangles will be mutually equiangular. Cor.
Side 81 - C' (89) (90) (91) (92) (93) 112. In any plane triangle, the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference.
Side 80 - The sine of an arc is a straight line drawn from one extremity of the arc perpendicular to the radius passing through the other extremity. The tangent of an arc is a straight line touching the arc at one extremity, and limited by the radius produced through the other extremity.
Side 28 - After remarking that the mathematician positively knows that the sum of the three angles of a triangle is equal to two right angles...
Side 22 - The perpendicular is the shortest line that can be drawn from a point to a straight line.