Learn how to create a pattern of direct dress by yourself

Learn how to create a pattern of direct dress by yourself

To create a drawing of the pattern of a straight dress, such measurements are necessary, as shown below. For a better understanding of the construction, let's take their approximate values ​​for 46 sizes:

  • dress length 112 centimeters;
  • hem of the neck 18 centimeters;
  • a semicircumference of a breast of 48 centimeters;
  • semi-grip waist 40 centimeters;
  • half-girth of hips 52 centimeters;
  • length of the back to the waist 40 centimeters;
  • the length of the shelf is 44 centimeters;
  • the back width is 18 centimeters.

This article will not be detaileddrawing drawing patterns of direct dress. Let's consider the basic moments describing this process by means of already constructed drawings (alphabetic and numerical) Draw a rectangle AA1HH1. The width of the straight dress. AA1 = HH1 = 52 cm. It is calculated as follows: half-breast (48 cm) + 4 cm (regardless of size). The length of the dress. AH = A1H1 = dress length according to measure (112 cm). Calculate the depth of the armhole. AG constitutes the third part of the half-breast (48 cm: 3 = 16 cm) + 3 cm (regardless of size) = 19 cm. From G, we construct a line parallel to AA1 up to the line A1H1. We obtain Γ1. Define the waist line. AT = length of back to waist, (38 cm). From T we construct a line parallel to AA1 up to the line A1H1. We get T1. We mark the line of the thighs. ТЯ = 20 cm (standard size). From point H construct a line parallel to AA1 up to line A1H1. We obtain the point X1. The track AN determines the line of the middle of the back, and the segment A1H1 - the line of the middle of the transmission. Calculate the width of the back. AC constitutes the third part of the waist semicircle (48 cm: 3 = 16 cm) + 3 cm = 19 cm. Similarly, we determine the position of the point C1, ГС1 = 19 cm. Define the width of the armhole. C1B corresponds to the fourth part of the semicircular breast (48 cm / 4 = 12 cm). From point B we construct a line perpendicular to ГГ1, before crossing with АА1. We will denote the point of intersection E. We make a depression for the armhole C1C2 = BB1 = 2 cm. Building a back Let's make a neckline. AR is equal to the third part of the semicircle of the neck 18 cm / 3 = 6 cm. PP1 = 2 cm (regardless of size). A and P1 are connected according to the template. The inclination of the shoulder. SP = 3 cm. Connect P1 and P, continue the line 1 cm to the right, point P2. Line of armholes. C2g2 = 3cm, C2C3 = 6cm, divide the angle C2 in half, C2C4 = 2.5 cm. The data are used for all sizes. P2, C3, C4, and G2 are connected according to the form. Side seam. From Γ2 down we lower the line, until the intersections with HH1 and TT1, we obtain points H2 and T2, respectively. From T2 lay 2 cm to the left. We denote the point H2, where Hl2 = 24 cm. From H2, drop the perpendicular down to the intersection with HH1. Raise this point 1 cm. We smoothly connect it to the point H2. Construction of a dart on the back. From T we postpone to the right along the TT2 segment 6 cm, perpendicularly up and down, from the obtained point we lay off respectively 10 and 12 cm. The tuck width should be 3 cm. G1B = ½ × semicircle of the breast. We get 48cm / 2 = 24 cm. Г1Б = ВЕ0. Let's fill the neck. BB1 is the third part of the semicircle of the neck. According to the available data, we get 18 cm / 3 = 6 cm. BB0 = BB1 + 2 cm. B1 and B0 are connected according to the pattern. Breast recess. The point Γ3 is the midpoint of the segment 1Γ1. T3P1 is built perpendicular to B1T1. P1n2 = 6 cm. Г3е = Г3П1 + 1. PP3 = 5cm. Define the length of the shoulder. P2P4 = measured shoulder width of the backrest (14 cm) - B1P1 (4.5 cm as measured by the segment) = 9.5 cm. Draw the armhole line. BB2 = 4cm (for any size). Point a is obtained by deepening the middle of B2p4 by 0.5 cm. Angle B1 is divided in half, B1O = 2cm. A4, a, B2, 0, and T2 can be joined by a pattern. Waistline. T1T3. = 2 cm. Thigh line. Я1Я3. = 32 cm. The outline of the bottom of the dress. Н2Н3 = 10 cm. It is necessary to make a bevel along the line Я3Н3 in 2 cm. The pattern of a straight dress is ready! To this dress the sleeve of the classical form will suit. The construction of such a sleeve, as well as a collar to the dress are shown in the figures.

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