. American engineer and railroad journal . angle ] F E, which the bearing face of the lock makeswith a line drawn from the point i*^ perpendicular to thecenter line L L, be given also FZ, Z E and O E, then , O E F O = \^ O E + {F Z + Z Ef and Sin OFE=-^yj^- In the triangle O F A the angle O F A = t)o + VFOVF O = O F E - VF O FA = cp + OFF- V the triangle F A Z we get F Z FA = —^— Sm F A Z and FA Z = FA C = V FE. The angle FA O may now be found. In the triangle FA O {FO ■>r FA)-AFO~ F A) .l»n^{FA (>+ )):tan l(F A O — F O A)!i{F A 0 +FO A) ~ HiSo - O FA). (a) AH otherquant
. American engineer and railroad journal . angle ] F E, which the bearing face of the lock makeswith a line drawn from the point i*^ perpendicular to thecenter line L L, be given also FZ, Z E and O E, then , O E F O = \^ O E + {F Z + Z Ef and Sin OFE=-^yj^- In the triangle O F A the angle O F A = t)o + VFOVF O = O F E - VF O FA = cp + OFF- V the triangle F A Z we get F Z FA = —^— Sm F A Z and FA Z = FA C = V FE. The angle FA O may now be found. In the triangle FA O {FO ■>r FA)-AFO~ F A) .l»n^{FA (>+ )):tan l(F A O — F O A)!i{F A 0 +FO A) ~ HiSo - O FA). (a) AH otherquantilies entering the equation (a) being known,the value oi i {F A O — F O A) is at once determined,and FAO = k(EAO + FOA) + i{FAO-FOA),and CAO = FAO-FAZ. In the triangle A /> C, the angles A /! C = iSo - (B A C + B C A)B C A = C: O. and B A C = F A C. A B = A D Sin B C ASm A B C XCA S\t\B A C ^ ,S^ATTc ^ 1IJ. I. I gives the values of the forces acting on the lock and atthe pivotal point of knuckle, which result from a pulling. Fig. II force, .1 C, when frictional resistance at the bearing sur-face of lock is neglected. If, however, the coefficient of friction equals i, the fric-tional resistance at the bearing face of the lock equals(f X A B), and the resultant force at the lock acts in the direction A F and tan A FA = —, from which the angle A FA is the triangle FA A the angle FA A = 180° - FA Cand FA A = 180° - (FA A + A FA) = FA Z. From the triangle Z we get FZ FA Sin FAZ The angle F A O may now be found, for in the triangleF A- O ( FA + FO): (.FA - FO) - tan J (FOA+FAO): tuniiFOA - FAO) (/■)1 (FOA f FAO) = J (180° — OFA) = ^ (i8o° — (OFA -f A FA). All other quantities entering the equation (b) beingknown, the value of ^ (F O A — FA O) may be deter-mined, and FA 0= i(B O A + FA O)and CA O = FA O - FA Z. i (F O A - FA O) In the triangle A B C the angles A B C = 180 - (/) A C + B C A) B C A C A D , and B A C = B A Z. ,. ,., Sin B C A
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