While for train [42, -18, 18, 12, 6] = 6*[7, -3, 3, 2, 1] the polynomial is 400*X^6 - 2148*X^5 + 22425*X^4 - 38960*X^3 + 19680*X^2 - 3840*X + 256 --- so 3x, 4x, 6x yield polynomials identical except for X^5 coefficient. Weird! WFL On 8/25/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:

I have been developing a procedure for computing the polynomial associated with a given eccentric planetary gear train. In outline, it generates sunsets by solving inverse trigonometric equations of form belt-length = integer , numerically to high precision; then deduces an integer relation between successive powers of some chosen sunset, via LLL lattice basis reduction.

The process is not currently fully automated: the user remains obliged to estimate suitable precision and polynomial degree, and to select a sunset (often but not always the minimum) not subsumed by some smaller train.

As an example, consider the (symmetric) train [r, s; p, t, q] = [28, -12; 12, 8, 4] = 4*[7, -3; 3, 2, 1] .

Computing possible sunsets h , then processing h -> h^2 * 2/(r-s) = h^2/20 (to reduce eventual polynomial degree and coefficient size) yields values to 6 decimal digits [0.164185, 0.177674, 0.203751, 0.250000, 0.334135, 0.504112, 0.937348, 4.00000] . (Maximum value 4 represents sun and ring touching externally: an overlapping, numerically pathological case best excluded!)

Because this train is (concentric with a multiple of) 4x a smaller train, alternate values merely yield polynomials associated with 2x or 1x multiples. But applying LLL to powers of (say) the first value yields polynomial 400*X^6 + 768*X^5 + 22425*X^4 - 38960*X^3 + 19680*X^2 - 3840*X + 256 , with just 4 positive real roots, corresponding to the 4 remaining values. Its coefficients are highly composite: [2^4 5^2, 2^8 3, 3 5^2 13 23, -2^4 5 487, 2^5 3 5 41, -2^8 3 5, 2^8] .

Substituting all 8 values into the polynomial yields syndrome [2.5 E-179, -1.0, -3.6 E-179, 5.7, -1.3 E-179, -1.9 E2, 2.5 E-177, 6.0 E6] .

Working single precision was 180 digits, double that for solving equations and accumulating inner products; degree assumed was m = 12 ; polynomial factors of degrees 1,5 were dropped from the raw result.

A remarkable feature: the smaller multiple [21, -9; 9, 6, 3] = 3*[7, -3; 3, 2, 1] yields processed values h^2/15 = [0.167544, 0.193318, 0.250000, 0.377104, 0.736875, 4.00000] (4 relevant), and polynomial 400*X^6 + 3684*X^5 + 22425*X^4 - 38960*X^3 + 19680*X^2 - 3840*X + 256 --- identical to that above except for a single coefficient!

QUERY: Given dimension (ie. presumed polynomial degree) m , what precision is necessary to ensure successful lattice basis reduction?

Fred Lunnon

So I have this algebraic number x , a root of the polynomial, irreducible over the integers, F(X) = \sum_{0 <= i <= m} c_i X^{m-i} . If it happens that some b divides each coefficient, I can divide c_i -> c_i/b , and the root remains unchanged: x -> x . Or if c_i is divisible by b^i , I can divide c_i -> c_i/2^i , and the root is halved: x -> x/2 . But suppose c_i divisible by b^|i-k| for some 0 < k < m : I wish to dispose of these superfluous factors as well, but what appropriate transformation should apply to x ? This behaviour is displayed by sunset polynomials of dilations of train [7, -3; 3, 2, 1] , for which b = 4 and k = m/3 . For example, the 8 novel sunsets of [r, s, p, t, q] = [84, -36; 36, 24, 12] = 12*[7, -3; 3, 2, 1] , after reduction h -> h' = (h/50)^2 [note typo in my earlier post!] , are the positive real roots, approximately 0.160458, 0.172003, 0.184713, 0.231571, 0.272437, 0.432110, 0.601110, 1.85694, of F(X) = 160000*X^12 + 614400*X^11 - 6979344*X^10 + 3276800*X^9 + 458782065*X^8 - 1720199520*X^7 + 2394836160*X^6 - 1705296384*X^5 + 697996800*X^4 - 171089920*X^3 + 24821760*X^2 - 1966080*X + 65536 --- note c_4 is odd. Fred Lunnon