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Changing hook & yarn size


Horsy

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Hi.  Maybe someone here can help me.  I have poor spatial/mathematical aptitude (er, maybe I'm just plain thick) but here goes:

If a pattern specifies Lace or Fingering weight (3.50mm for Lace & 3.75mm for Fingering) for a project, but I want to use Worsted weight with a smaller-than-average hook (maybe 4.5mm) to get a stiffer fabric, do I need more or less yarn than if I followed the recommended yarn & hook sizes?   

Resulting size of project does not matter that much.   It's a small throw. 

Also, I want to  use a chainette yarn, which I know behaves a little different.  I just love that stuff. 

Sorry to bother you and wish I could figure these things out for myself...😅

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You are going to be making a LOT more stitches per square foot with the smaller yarn, but less yardage used per stitch.  Since it's a blanket, it ought to be easier to figure out than say a garment because of the shaping (I assume it's a rectangle).  Warning, some math ahead, but it's simple arithmetic.  

Do a 6" square swatch, in the yarn weight, stitch pattern, and hook size you want to use.  Mark the end of the yarn at the last stitch, unravel the square, and measure the yardage that you used (it's not wasted, you can use it in the blanket).  Let's say it is 5 yards to make the math easy, it is probably really more than that.

A 6" square = 36 square inches, and a 12" square contains 4 six" squares, so that would be 4 squares @ 5 yards per 12" square, or 20 yards per 12" square. 

What you need to do with that information is to figure out the dimension of YOUR intended blanket in square feet.  Example, if you want to make a sofa throw that is 4'x6', that is 24 square feet, times 20 yards per foot, or 480 yards.  (then, you need to look at the yardage per skein of your yarn and to a bit more math of how many skeins you need.

Does that make sense?  I'm making up an example to explain the concept, don't use the actual numbers I arrived at unless your swatch and your intended blanket match what I described, which is unlikely (the guesses I used to arrive at the yardage sounds like my guess for how much might be used for a 6" swatch is probably too small). 

 

Edited by Granny Square
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Thank you so much for your formula!  I am on my way to trying this out.  You really go out of your way, all the time, to help.  ☺️

 Warning, some math ahead, but it's simple arithmetic.  

By the way, a fellow who studied engineering agrees with me about math VS arithmetic.  You don't learn mathematics til you are at the university level.  Even 12th grade stuff is still "arithmetic".  FWIW.  Your first paragraph brought that memory back for me.

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