. American engineer . ent fromChurchs Mechanics, page 23, the values for Xi and x. are (AB) and X2 (AB)^ 12F. N = — F=.r: Substituting the values for x, and .v- in the last equation, i - r(AB)-l r(AB) -1 F3 .r, = F, ~ F J N I L 12Fi J L 12F, J F„.r., = NF. Tor equilibrium of forces in this case Wa Fs = Wa or Fs .ra = tqCombining the two equations and solving for t we find Wa ~ qNF; In using this formula, it will be found more convenient to as-sign values for M and Rj, increasing the value of M if the thick-ness of counterbalance is found to be too large for clearancepast t
. American engineer . ent fromChurchs Mechanics, page 23, the values for Xi and x. are (AB) and X2 (AB)^ 12F. N = — F=.r: Substituting the values for x, and .v- in the last equation, i - r(AB)-l r(AB) -1 F3 .r, = F, ~ F J N I L 12Fi J L 12F, J F„.r., = NF. Tor equilibrium of forces in this case Wa Fs = Wa or Fs .ra = tqCombining the two equations and solving for t we find Wa ~ qNF; In using this formula, it will be found more convenient to as-sign values for M and Rj, increasing the value of M if the thick-ness of counterbalance is found to be too large for clearancepast the side rods, etc. The area F2 of segment ABD may bedetermined by the aid of the planimeter or by using the follow-ing formula. The area of the segment is equal to the area of the sectorADBO2 minus the area of the triangle ABO,. r,„..^ -1 Tab -1. Area Fj = 2R. — (N -f .r) in which the value for x is found from the two right-angled tri-angles BEO, and BEO2. In triangle BEO,, x + (BE) = R,=, and in triangle BEO=. (x 4- N)= + (BE)= = R=:. Solving both equations for (BE) =and combining: R2 — R-i — N= SEGMENT COUNTERBALANCE. Using the same notation, as above For equilibrium of forces. FtSolving for x Wa Ftq Combining the two equations and solving for AB we find AB = \ tq SECTOR BALANCE. Let M =: Number of spaces filled by = Number of spokes in wheel.
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Keywords: ., bookcentury1900, bookdecade1910, booksubjectrailroa, bookyear1912
