I just bought my Kromski rigid heddle loom. It came with an 8 dent heddle.
1st project - kitchen towel
Yarn - 5/2 mercerized cotton
Get the sett. Wrapped 1 inch around ruler. 24 divide by 2 = 12 ends per inch
Balanced weave would be warp 12 epi (size 12 heddle) and weft 12 picks per inch.
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Can I calculate my warp based on the number 8 and use the 8 heddle?
Size 8 heddle (8 ends per inch). Fabric density would be more drapey; 4 less threads per inch. Would that work?
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Warp width
Towel width 17.50 inches
Draw (10%) +1.75
Total width =19.25 (rounded to 20)
Yarn Sett x 8 (ends per inch)
Total # ends = 160 ends
Warp length
Towel length 23.00 inches counting fringe
Take up 10% + 2.30
Loom waste +18.00
Total length = 43.00 inches (1 1/4 yard)
Wrap each end 43 inches on warp board.
Stripe warp (160 ends)
White: 23 | xx | 22 | xx | 22 | xx | 22 | xx | 23 = 112 ends
Color: xx | 12 | xx | 12 | xx | 12 | xx | 12 | xx = 48 ends