Question 9.4: Compound Gear Train Design to Approximate an Irrational Rati...
Compound Gear Train Design to Approximate an Irrational Ratio.
Find a pair of gearsets which when compounded will give a train ratio of 3.14159:1 with an error of < 0.0005%. Limit gears to tooth numbers between 15 and 100. Also determine the tooth numbers for the smallest error possible if the two gearsets must be reverted.
1 Input data to the algorithm are R = 3.141 59, N_{\text {low }}=15, N_{\text {high }}=100 , initial ε = 3.141 59 E-5.
2 The TKSolver file Compound.tk (see Table 9-9) was used to generate the nonreverted solu‑tions shown in Table 9-10.
3 The best nonreverted solution (7th row in Table 9-10) has an error in ratio of 7.849 9 E-06 (0.000 249 87%) giving a ratio of 3.141 582 with gearsets of 29:88 and 85:88 teeth.
4 The TKSolver file Revert.tk was used to generate the reverted solutions shown in Table 9-11.
5 The best reverted solution has an error in ratio of –9.619 8 E-04 (–0.030 62%) giving a ratio of 3.142 562 with gearsets of 22:39 and 22:39 teeth.
6 Note that imposing the additional constraint of reversion has reduced the number of possible solutions effectively to one (the two solutions in Table 9-11 differ by a factor of 2 in tooth numbers but have the same error) and the error is much greater than that of even the worst of the 11 nonreverted solutions in Table 9-10.
| TABLE 9-9 Algorithm for Design of Two-Stage Compound Gear Trains From Author’s downloadable TKSolver file Compound.tk. Based on Reference [2] |
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| ” Ratio is the desired gear train ratio and must be > 1. Nmin is the minimum number of teeth acceptable on any pinion. ” Nmax is the maximum number of teeth acceptable on any gear. eps1 is initial estimate of the error tolerance on Ratio. ” eps is the tolerance used in the computation, initialized to eps1 but modified (doubled) until solutions are found. ” counter indicates how many times the initial tolerance was doubled. Note that a large initial value on eps1 will cause long ” computation times whereas a too-small value (that gives no solutions) will quickly be increased and lead to a faster solution. ” pinion1, pinion2, gear1, and gear2 are tooth numbers for solution. |
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| eps = eps1 counter = 0 redo: S = 1 R_high = Ratio + eps R_low = Ratio – eps Nh3 = INT( Nmax^2 / R_high ) Nh4 = INT( Nmax / SQRT ( R_high)) For pinion1 = Nmin to Nh4 Nhh = MIN ( Nmax, INT (Nh3 / pinion1)) For pinion2 = pinion1 to Nhh Q = INT( pinion1 * pinion2 * R_low) + 1 P = INT( pinion1 * pinion2 * R_high) IF ( P < Q ) THEN GOTO np2 Nm = MAX ( Nmin, INT ( (Q + Nmax – 1) / Nmax )) Np = SQRT(P) For K = Q to P For gear1 = Nm to Np IF (MOD( K, gear1 ) <> 0 ) Then GOTO ng1 gear2 = K / gear1 error = ( Ratio – K / ( pinion1 * pinion2) )”check to see if is within current tolerance IF error > eps THEN GOTO ng1 ” else load solution into arrays pin1[S] = pinion1 ng1: Next gear1 ” test to see if any solution occurred with current eps value IF (Length (pin1) = 0 ) Then GOTO again ELSE Return |
” initialize error bound ” initialize counter” reentry point for additional tries at solution ” initialize the array pointer ” intermediate value for computation ” loop for 2nd pinion ” skip to next pinion2 if true ” loop for first gear ” increment array pointer ” have a solution ” double eps value and try again |
| TABLE 9-10 Nonreverted Gearsets and Errors in Ratio for Example 9-4 |
| \begin{array}{llllllll}N _{ 2 } & N _{ 3 } & \text {Ratio1 } & N _{ 4 } & N _{ 5 } & \text {Ratio2 } & m _{ V } & {\text { Error }} \\\hline 17 & 54 & 3.176 & 91 & 90 & 0.989 & 3.141564 & 2.5682 E -05 \\17 & 60 & 3.529 & 91 & 81 & 0.890 & 3.141564 & 2.5682 E -05 \\22 & 62 & 2.818 & 61 & 68 & 1.115 & 3.141580 & 1.0268 E -05 \\23 & 75 & 3.261 & 82 & 79 & 0.963 & 3.141569 & 2.0541 E -05 \\25 & 51 & 2.040 & 50 & 77 & 1.540 & 3.141600^{*} & 1.0000 E -05 \\28 & 85 & 3.036 & 86 & 89 & 1.035 & 3.141611 & 2.1296 E -05 \\29 & 88 & 3.034 & 85 & 88 & 1.035 & 3.141582^{†} & 7.8499 E -06 \\33 & 68 & 2.061 & 61 & 93 & 1.525 & 3.141580 & 1.0268 E -05 \\41 & 75 & 1.829 & 46 & 79 & 1.717 & 3.141569 & 2.0541 E -05 \\43 & 85 & 1.977 & 56 & 89 & 1.589 & 3.141611 & 2.1296 E -05 \\43 & 77 & 1.791 & 57 & 100 & 1.754 & 3.141575 & 1.5133 E -05\end{array} |
| TABLE 9-11 Reverted Gearsets and Errors in Ratio for Example 9-4 |
| \begin{array}{cccccccc}N _{ 2 } & N _{ 3 } & R \text { atio1 } & N _{ 4 } & N _{ 5 } & R a t i o 2 & m _{ V } & \text { Error } \\\hline 22 & 39 & 1.773 & 22 & 39 & 1.773 & 3.142562 & -9.6198 E -04 \\44 & 78 & 1.773 & 44 & 78 & 1.773 & 3.142562 & -9.6198 E -04 \\\hline\end{array} |