# Numerical solution of differential eccentric gear based on unequal division model

According to the above model, the actual rotation angle of passive eccentric gear φ The numerical solution steps of 2 are as follows:

① Set the module of eccentric gear as m and the number of teeth as Z, calculate the standard tooth profile of involute eccentric gear, and the reference tooth is marked as 0 (Fig. 1).

② Geometric center angle of driving eccentric gear θ Determine the teeth that rotate relative to the reference teeth at any rotation angle, and judge which two pairs of teeth or one pair of teeth have the working tooth profile on the inner common tangent N1N2.

③ Obtain the common tangent and tangent point in the two base circles under the fixed coordinate system xo1y.

④ In the initial solution φ~ 2 find the intersection K1 (K2) and K’1 (K’2) of the working tooth profile of the paired teeth of the two base circles with the common tangent N1N2 and the driving eccentric gear 1 and the passive eccentric gear 2.

⑤ With φ 1 and φ’ 2. The determined eccentric gear state is to search the initial position, fix the driving gear 1, and determine the intersection K1 (K2) and K’1 (K’2) as the pairing points according to the pairing rule that the sum of the labels of the teeth at the intersection of the driving eccentric gear and the passive eccentric gear is equal to round (Z / 2) or Z + round (Z / 2).

Judge that the driven eccentric gear 2 will be spaced at a small angle γ Whether to rotate forward or reverse along wheel 2, and then calculate the distance D (I) = lk1k’1 of the mating point of the front pair of teeth. If the distance is less than the given distance d0, update the mating point. Round is the rounding function. The termination condition is set as follows according to whether the passive eccentric gear 2 rotates forward or reverse:

Difference of longitudinal coordinates of intersection of front pair of teeth Δ YK1 (0) = YK1 － YK’1, if Δ YK1 (0) ＞ 0, when 0 ＜ Δ When YK1 (n) ≤ d0 and D (n) ≤ d0, it is considered that K1 and K’1 are close enough, then the search is terminated. At this point, Δφ ( n) = － n γ。 If Δ YK1 (0) ＜ 0, when 0. 000 ＜ 1 Δ When YK1 (n) ≤ d0 and D (n) ≤ d0, it is also considered that K1 and K’1 are close enough to terminate the search. At this point, Δφ ( n) = n γ， N is the number of times the wheel 2 is slightly rotated. After the iteration is terminated, judge whether the rear pair of teeth matching points meet the termination conditions of meshing point search. If so, it is double tooth meshing, otherwise it is single tooth meshing, as shown in Fig. 2 and Fig. 3.

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