String length
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funkyXtina
Bracelet King
funkyXtina
1 month, 1 week ago by funkyXtina
How long does everyone cut their strings? I’m trying to find the perfect solution. I’ve had trouble with some string running low and having to add more. I’ve figured out how to add more but I just want input from y’all.
halokiwi
Super Moderator
halokiwi
1 month, 1 week ago by halokiwi
I have a formula for normal patterns:
45cm + percentage of knots a string makes of its own colour

I can explain further, if you want
funkyXtina
Bracelet King
funkyXtina
1 month, 1 week ago by funkyXtina
Please. Teach me your ways 😍
AlmaLlama
Bracelet King
AlmaLlama
1 month, 1 week ago by AlmaLlama
In adding of what @halokiwi said, yes, this formula works, I've tried it on the last 2 normal patterns I've recently made. @halokiwi said in a forum topic at "Newbies" section (https://www.braceletbook.com/forum/1_newbies/33745_what-is-a-good-length-for-normal-pattern-strings/) this following: " 3 weeks, 4 days ago by halokiwi
My formula is 45cm + % of rows in which a string makes a knot of its own colour.

Basically I look at each individual string. I follow it from the beginning of the pattern until the end of the pattern. If the string is in a different position then, I follow it again from beginning until end until it is in the original position again.

Let's say a pattern has 16 rows. A string makes knots in 8 of the rows. That's 50% of the rows. The string would need to be 45+50=95 cm long.

You have to calculate that for each individual string. The formula works for a bracelet of 14cm length with 10cm long ties on each side. "

(Copied from that forum topic I've linked above)
funkyXtina
Bracelet King
funkyXtina
1 month, 1 week ago by funkyXtina
Okay. Thank you all. I do need things explained like I’m 5 🙃 So I’m gonna give an example of a bracelet I’m working on right now. #140877. 32 strings done in a loop.
myabrclest
Skiller
myabrclest
1 month, 1 week ago by myabrclest
On a alpha do it just a wee bit longer than your bracelet should be or whatever your doing on a regular pattern just do it on how complicated the bracelet is
halokiwi
Super Moderator
halokiwi
1 month, 1 week ago by halokiwi
Ok, since you are doing a loop, you need double the length.

Strings for the straight edges (I):
2 × (0% + 45) = 90cm

The other string of colour I and H:
2 x (50% + 45) = 190cm

Outer string of colour B:
2 x (100% + 45) = 290cm

Next string of colour B:
(knots in 2 of 38 rows ≈ 5%)
2 x (5% + 45) = 100cm

String of colour D:
(there seems to be a mistake in row 17)
(knots in 13 of 38 rows ≈ 34%)
2 x (34% + 45) = 158cm

String of colour E:
(knots in 18 of 38 rows ≈ 47%)
2 x (47% + 45) = 184cm

String of colour C:
(mistake in row 17)
(knots in 21 of 38 rows ≈ 55%)
2 x (55% + 45) = 200cm

I will stop now, but I can continue, if you need me to.
Black is a little hard to figure out.
emp217
Skiller
emp217
1 month ago by emp217
So with this formula you don't count the knots, you count the rows? How are the rows counted? one by one horizontal? or diagonal?

I am wondering what's better/faster/easier? Counting through the pattern (sometimes you don't know how long it will exactly be, because the (diagonal) rows depend on the pattern - I have been counting diagonal rows so far) or adding in some string. Adding in some string is no problem, I just find it bad if there is more than 10 cm left at the end. I had that even with a length of 1 m each.

But still I don't have much experience. I just always write it down how much string I used and how much string was left - just in case I want to make the same pattern again some day.
artsforbre
Skiller
artsforbre
1 week, 2 days ago by artsforbre
@emp217
This is how I understand and would use this method this method. In my example, I'm going to make 126 rows of a pattern. Horizontal rows are always numbered on the sides of the pattern so you don't usually have to count them.
Then I click on the A string, follow it down through the pattern, and count exactly how many knots that string is going to make on other strings. In my case it's 66 knots. 66 knots makes 52.3% of 126 rows.
Now you do 45+52=97cm
I used 80.5cm for the bracelet and was left with 16.5cm for the ties.
This formula is for 1 single string.
Then you do it with all the strings in the pattern.
artsforbre
Skiller
artsforbre
1 week, 2 days ago by artsforbre
Just gonna leave this here, maybe someone will find it useful 😅
I have to say that 45+% is a bit much for me sometimes and not enough other times. For example:
For the 126-row bracelet, I used 80.5 cm of string for 66 knots (52.3%). 45+52=97cm
If I used that formula I'd be left with 16.5cm for ties which sometimes is too much or not enough depending on the type of ties I'm going to make so my formula is 30+% and then add whatever length I'm gonna need for the ties:
30+53=83cm and 83+14=97cm
For how I knot I also use a different formula, amount of knots times 1.25. For that same example:
66×1.25=82.5cm and add the length of ties
Or the amount of knots times 1.5 for the standard bracelets that I sell. This gives me the length for both the bracelet and ties.
66×1.5=99cm and that's exactly what I needed.
For bracelets with loops I'd do
66×1.25= 82.5cm
Plus 5cm for the loop
82.5+=87.5cm
And 7cm for the tie
87.5+7=94.5
Then round it up to 95 or 96cm
halokiwi
Super Moderator
halokiwi
1 week, 2 days ago by halokiwi
@emp217 sorry, for not seeing your last question. You should really make a habit of tagging me 😉
I check back on threads when I know that people don't have a habit of tagging me, but sometimes I simply forget. So sorry.

The rows are easy to count. The rows are just the numbers you see on the left and right edge of the pattern. So rows are horizontal rows.

The knots are more difficult to count. You have to follow the string from the first to the last row and count in how many of the knots it is the ''leading string''.
If the string does not end in the same place as it started, you have to count again, but from the new position, until it is in the original position again.

I now learned that you don't do ties. If you don't do ties, you don't need to add 45 to the percentage. Of those 45, 25 are for the ties and 20 are what you need anyways. So you only need to add 20 plus whatever you need for the loop.
emp217
Skiller
emp217
1 week, 1 day ago by emp217
@artsforbre @halokiwi

I learned that I need 1 cm for 1 knot. I was recognizing that when doing alphas, but as I always write down how much string I cut and what I needed, with pattern #142398 it was exactly 60 cm per string for the knotted part. I knotted the pattern 2x (96 rows, 15 cm long). And I do a loop at the beginning and a button at the end, so I only need one string that is longer than it should be.

So the 66 knots would be 66 cm. In artsforbre's calculation it would be 52 cm, so it would be too less.

What I still don't know (but I will...haha, the last few days I cutted strings for about 10 bracelets and don't know where to start and when to end 😂) is how much string I need in a normal bracelet for only a passive string. Because there will be a huge difference if the string just runs through vertically like the string for the straight edge technique (have to learn this for normal bracelets...) or if it runs through diagonal. And when it runs through diagonal it depends on how many strings you have in the bracelet. To estimate that I have too less experience yet.

With black and white I don't really care how much string I need. But with the color strings, you know how quickly your favourite colors come to end...each cm counts! 😂 😇
halokiwi
Super Moderator
halokiwi
1 week, 1 day ago by halokiwi
For the passive string you need the square root of (2 x (the length of the knotted part squared)).

You can approximate that all the strings move in diagonals, no matter if the diagonal moves all the way from one edge of the bracelet to the other edge and back again or if the direction of the string switches every knot. An exception might be strings used for the straight edges technique. Those indeed move almost in a vertical line. But I think for a formula that is done by hand and not automated, if you consider this also, it would be too complex. If someone is ever willing to write a program for calculating string length, I think this would definitely be something to consider, but for now, to make it a little easier, I approximate all the strings as diagonals.

Because they are diagonals, you can calculate their length by using Pythagoras:
a^2 + b^2 = c^2
A is the diagonal you are looking for and b and c are the length of the knotted part.
halokiwi
Super Moderator
halokiwi
1 week, 1 day ago by halokiwi
@emp217 now I forgot to tag you
artsforbre
Skiller
artsforbre
1 week, 1 day ago by artsforbre
@emp217 I get it. Maybe try counting all the knots that other strings make on the passive string and calculate as you normally do then check how much string is left and calculate it from that. That's probably what I would do. I don't think the number of strings matters since the distance between base strings in normal patterns is always the same.
emp217
Skiller
emp217
1 week, 1 day ago by emp217
Who ever discovered this must have had quite a bit of phantasy! 😃 But it makes sense.
halokiwi
Super Moderator
halokiwi
1 week, 1 day ago by halokiwi
@emp217 it was me. I'm not sure, if I would consider myself having lots of fantasy, but I'd say I like to think mathematically and find solutions that make sense and make things easier 😄
emp217
Skiller
emp217
1 week, 1 day ago by emp217
😂 😂 as I said - it does make sense. but make things easier? 😉 😇 😘
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