Winter Quilt Patch Fifteen
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
 |
Wednesday, February 7, 2001
1. We have 15 p. so far.
2. We have 30 s.t. so far.
3. We have 18 big triangles so far.
4. We have 3 windmills. The 3rd p. in the 3rd r is in 2 windmills.
5.We have a pattern of fabrics by rows that goes: [1234 2143] [5313 5313] [4312 3421] [3135 31??]
6. So far, the fourth row reverses the patch pattern of the first row and reverses the fabric patterns in each patch becauseÉ the second row is C(5+3) D(1+3) C(5+3) D(1+3) and the fourth row is D(3+1) C(3+5) D(3+1)
7. We have a prediction that the next patch will be C (3+5) because our pattern for this row will go DCDC and because the fourth row will keep being the reverse of the second row.
Our Math Ideas:
# s.t. = # p. x 2 (Number of small triangles equals the number of patches times two)
1 f.g. = 2 plg (one flying geese equals two parallelograms)
1 wl = 8 s.t. (one windmill equals 8 small triangles)
1 dmd = 4 s.t. (one diamond equals 4 small triangles)