ards to the number before the start number...
i.e. 9, 0, 1, 2, 3, 4, 5, 6, 7, 8
then it will need to repeat this pattern (continuing to count upwards) and the repeat number is based on a slider (for example 3 in the case illustrated below):
9, 0, 1, 2, 3, 4, 5, 6, 7, 8
19, 10, 11, 12, 13, 14, 15, 16, 17, 18,
29, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29,
if anyone has any ideas on how to create this series it would be great
M.…
e! I do not have good ideas today!
The end result of the list would be:
5, 10, 15, 20, 21, (21 + 5), (21 + 10), (21 + 15), (21 + 20), (21 + 21), (42 + 5), (42 + 10), (42 + 15), (42 + 20), (42 + 21), etc …
1. Duplicate the first list.(24-->48) and Graft(48 branchs of 1 item each)
2. Graft the second list.(48 branchs of 1 item each)
3. merge two list
4. Join Curves
5. Fillet Curve
jaja, I tried a brute force method using Mathematica...lol I give it the list of values and it gave out instantly:
(1/9) * (57 + 48 * n + 15 * Cos((2 * n * Pi)/3) + Sqrt(3) * Sin((2 * n * Pi)/3) )
Added by Jesus Galvez at 7:42am on November 27, 2012
dont get you, i am saying sleect numbers in range 1 to 10, starting from 1 with a step of 2.
1 to 10 by 3 = 1 4 7 10
1 to 10 by 5 = 1 6
1 to 10 by 1 = 1 to 10 = 1 2 3 4 5 6 7 8 9 10
Added by Steve Lewis at 3:15pm on November 11, 2013