The Mechanical Principles of Engineering and ArchitectureJ. Wiley, 1875 - 699 стор. |
З цієї книги
Результати 1-5 із 44
Сторінка xiii
... intersect the base of such a wall at a point whose distance from its extrados ie ths the distance between the extrados at the base and the vertical through the centre of gravity . have called it the modulus of stability ; conceiving ...
... intersect the base of such a wall at a point whose distance from its extrados ie ths the distance between the extrados at the base and the vertical through the centre of gravity . have called it the modulus of stability ; conceiving ...
Сторінка 17
... distance from A , mea- sured along the line AC , at which R intersects that line , we have , since R is the moment of R , ≈R = 2 m2 P , where the sign of R indicates the direction in which R tends 2 OF PARALLEL PRESSURES . 17.
... distance from A , mea- sured along the line AC , at which R intersects that line , we have , since R is the moment of R , ≈R = 2 m2 P , where the sign of R indicates the direction in which R tends 2 OF PARALLEL PRESSURES . 17.
Сторінка 18
... intersect the line joining the points P , and P , in the point R ,; produce the line P , P. , to intersect any plane given in position . in the point L. Through the points P , P. , and R ,, draw P , M ,, P , M ,, and R , N ...
... intersect the line joining the points P , and P , in the point R ,; produce the line P , P. , to intersect any plane given in position . in the point L. Through the points P , P. , and R ,, draw P , M ,, P , M ,, and R , N ...
Сторінка 24
... intersect the plane ADC in AM , then by similar triangles DM : MC :: dm : mc , but DM MC ; therefore dm = -me . Also by similar triangles GM : BM :: gm : bm , but GM = BM ; therefore_gm = } bm . Since then dm de and gm- bm , therefore g ...
... intersect the plane ADC in AM , then by similar triangles DM : MC :: dm : mc , but DM MC ; therefore dm = -me . Also by similar triangles GM : BM :: gm : bm , but GM = BM ; therefore_gm = } bm . Since then dm de and gm- bm , therefore g ...
Сторінка 25
Henry Moseley. therefore the lines AG and CH intersect , and the centre of gravity is at their intersection K. Now GK is one - fourth of GA ; for suppose equal weights to be placed at the angles A , B , C , and D of the pyramid ( the ...
Henry Moseley. therefore the lines AG and CH intersect , and the centre of gravity is at their intersection K. Now GK is one - fourth of GA ; for suppose equal weights to be placed at the angles A , B , C , and D of the pyramid ( the ...
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Загальні терміни та фрази
a₁ angle of resistance angular velocity axes axis beam body bordering upon motion centre of gravity circumference co-efficient compression conical surfaces corresponding crank crank arm cubic foot curve cylinder deflexion determined displaced fluid distance ditto driven wheel driving epicycloidal equal equation equilibrium evident exceedingly small extrados forces formula friction given horizontal hypocycloidal inclination inertia intersect involute lamina length limiting angle line of centres line of resistance load machine modulus moment of inertia moving n₁ n₂ neutral line observing obtain oscillation P₁ P₂ parallel passing perpendicular pitch circle plane point of application point of contact portion position radius repre respect resultant revolution revolve rupture space stability Substituting supposed surfaces of contact taken to represent tion tooth U₁ unguent vertical vis viva viva voussoirs wall weight whence it follows whilst whole