The Mechanical Principles of Engineering and ArchitectureJ. Wiley, 1875 - 699 стор. |
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Результати 1-5 із 83
Сторінка 24
... repre- sented in amount by A , a weight equal to 2A will then be collected in D , and a weight equal to A at A , and the centre of gravity of these is in G ; therefore DA x A = DG × ( 2A + A ) , ... DA = 3 DG , or DG DA . * 26. THE ...
... repre- sented in amount by A , a weight equal to 2A will then be collected in D , and a weight equal to A at A , and the centre of gravity of these is in G ; therefore DA x A = DG × ( 2A + A ) , ... DA = 3 DG , or DG DA . * 26. THE ...
Сторінка 25
... repre- sented by A , and let the three weights B , C , and D be supposed to be collected in their centre of gravity G ; the four weights will then be reduced to two , viz . 3A at G , and A at A , whose common centre of gravity is K ...
... repre- sented by A , and let the three weights B , C , and D be supposed to be collected in their centre of gravity G ; the four weights will then be reduced to two , viz . 3A at G , and A at A , whose common centre of gravity is K ...
Сторінка 29
... repre- sent , respectively , the distances of the element AM of the Poisson , Journal de l'Ecole Polytechnique , 18me cahier , p . 320 , or Art . 2 , in the Treatise on Definite Integrals in the Encyclopædia Metropolitana by the author ...
... repre- sent , respectively , the distances of the element AM of the Poisson , Journal de l'Ecole Polytechnique , 18me cahier , p . 320 , or Art . 2 , in the Treatise on Definite Integrals in the Encyclopædia Metropolitana by the author ...
Сторінка 33
... repre- sented by a , and the are AP by 8. Then x = -DP cos . PDMa cos . PDM = a sin ... 2xy = 2a ' sin . .. 2 S1 = Sxyds = na S2 = a ' cos . 1 + cos . = 1 πα ( 1 S , = πα cos . a -- 252 a 2S , 2 8 DM y = PM = DP sin . a 8 COS . 2 = a2 ...
... repre- sented by a , and the are AP by 8. Then x = -DP cos . PDMa cos . PDM = a sin ... 2xy = 2a ' sin . .. 2 S1 = Sxyds = na S2 = a ' cos . 1 + cos . = 1 πα ( 1 S , = πα cos . a -- 252 a 2S , 2 8 DM y = PM = DP sin . a 8 COS . 2 = a2 ...
Сторінка 36
... repre- sented by 8 , then the area FQPK equals the sector FQM - the sector KPM MQ × 6- MP2 × ê = K ¿ ¿ ( MQ2 — MP® ) = ¿ ( MQ + MP ) 2 × ( MQ - MP ) = 8 ( MT × PQ ) . Now the solid ring generated by PRSQ is evidently equal to the ...
... repre- sented by 8 , then the area FQPK equals the sector FQM - the sector KPM MQ × 6- MP2 × ê = K ¿ ¿ ( MQ2 — MP® ) = ¿ ( MQ + MP ) 2 × ( MQ - MP ) = 8 ( MT × PQ ) . Now the solid ring generated by PRSQ is evidently equal to 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