How to knit the armhole, neckline, and shoulders.

Back.

1) Armhole.

Determine the number of stitches for the width of the armhole. Let the width of the back armhole be 6cm, with a knitting density of 2.2 stitches and 2.4 rows per 1 cm.

6 cm x 2.2 st. = 13 st. Divide 13 stitches into 3 equal parts.

13 : 3 = 4 (1 remainder).  The calculation resulted in  a remainder. Add it to the first part.

First part.  4 + 1 stitch remainder = 5 stitches. Close in 2 steps: 3 stitches and 2 stitches.

Second part. 4 stitches — decrease 1 stitch every second row.

Third part. 4 stitches —  decrease 1 stitch every fourth row.

This is the classic calculation for knitting the armhole. There is also another  «grandmother’s» method. Stitches for the armhole are decreased in the following order: 4-3-2-1. For example, close 13 stitches for the armhole like this:  in every second row, decrease 4 stitches once, 3 stitches once, 2 stitches twice, and 1 stitch twice.

After decreasing the stitches for the armholes, continue knitting the back  straight to the shoulder.

One tip. Before knitting the armhole, check the length of the back by placing it against yourself.

2) Shoulders and neckline.

Perform calculations on half of the pattern. Measure the width of the neckline and shoulder. For example: the width of the half neckline is 8 cm, the width of the shoulder is 11 cm.

Number of stitches 8 cm x 2.2 st. = 18 st.

11 cm x 2.2 st. = 24 st.

The shoulder slope is usually 2.5 cm.

2.5 cm x 2.4 r. = 6 rows. In six rows, there are 3 edge stitches (every 2 rows equal 1 edge stitch).

The number of edge stitches in the shoulder slope corresponds to the number of groups into which the shoulder stitches should be divided during decreasing: 24 st. : 3 = 8 st.

If there is a remainder, add it to the first group. Thus, to form the shoulder, you need to decrease 8 stitches 3 times every second row.

Simultaneously with the shoulder decrease, execute the neckline. Usually, the height of the back neckline is approximately equal to the shoulder slope, i.e. 2.5 cm x 2.4 r. = 6 rows = 3 edge stitches (3 groups into which the stitches should be divided during decreasing).

18 st. : 3 = 6. If there is a remainder, add it to  1 group.  Thus, half of the back neckline needs to be decreased 3 times by 6 stitches.

For convenience, mark the results of the calculations on the pattern.

Front.

1) Armhole.

Classic calculation. The width of the front armhole is greater than the width of the back armhole and is equal to 8cm in our example.

8 cm x 2.2 st. = 18 st. Let’s divide this number into 4 equal parts. 18 st. : 4 = 4 st. + 2 remainder.

It resulted in 4 groups of stitches,  4 in each. If you have a remainder from the calculation, add it to the second part.

First part — 4 stitches.

Second part. Divide into triples — 3 stitches + 1 remainder + 2 stitches remainder from dividing.

Third part. Divide into pairs — 2 + 2.

Fourth part. Divide into single stitches: 1 + 1 + 1 + 1.

Mark the calculations on the pattern. Thus, decrease the stitches in the following order: in every second row - 1 time of 4 stitches, 2 times of 3 stitches, 2 times of 2 stitches  and in every 4th row  4 times of 1 stitch.

In modern knitting magazines, the armholes of the back and front are knitted the same way.

2) Shoulders of the front are knitted the same as  the shoulders of the back. The neckline of the front depends on the model.