$1911

I think it would be great if the pattern generator could put out estimated string lengths based on an average bracelet length (maybe 15 or 16cm?), average row height (my bracelets vary between 8 and 10 rows per cm) and the number of knots and rows without knots each string has to pass plus 20cm for braided ends.
For this we'd have to find out average string use, of course. But that can't be too hard. When I calculate with 0.3cm for each row without a knot and 0.95cm for each knot I get in the right ballpark.

Might be even cooler if there was a calculator on each pattern site that the user could put his or her variables into (with average values as a preset for inexperienced users)...

[COLOR=purple]
I totaly agree!!!

[COLOR=purple]
lol

i thinks thats a very good idea!

That would be awesome !
Calculate
number of rows for the pattern R,
number of times the specified string is used for a knot K
number of times the pattern must be replicated to complete the bracelet P,

minimal yarn length = P*(K*0.95+(R-K)*0.3)+length for braiding, etc.

YESSS bracelet math!! :)

@Bretzelbuth Can you give us an example with any pattern?

Pattern #51499
For the first blue string
You repeat it about 12 times to make 5 inches. So p=12. You make 5 knots with the blue string in one repeat so k=5. There are 8 rows so r=8. And say you need 10 inches for braids on both sides. So if you plug everything in
12*(5*.95+(8-5)*.3)+10
So all together that equals 77.8. You will need about 77.8 inches for that one blue string.

I tried it and it works but you must make sure that strings will start and end at the same position!
@wildcattz98: I think you made a mistake - if you want calculate this in inches you MUST convert centimeters to inches so you will have P*(K*0,38+(R-K)*0,12)+braids = 12*(5*.38+(8-5)*.12)+10 = 37.12 inches

oh ok. that makes more sense lele97. i thought the number i got seemed too big.