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Application of "CD-4" theory for determining the width of implant in breast augmentation.

Cai J, Zhou Y - Chin. Med. J. (2015)

Bottom Line: From January 2006 to June 2014, the authors have found and applied "CD -4" theory to determine the width of breast implant (W) in dual plane I or II breast augmentation cases through transaxillary or periareolar incision for 560 patients. "CD" is defined as the curved distance on skin from the midline of the sternal bone to the anterior axillary line (AAL) on the lateral chest wall through the horizontal level on inferior mammary fold.Their new intermammary cleavages without bras are between 1 cm and 2.5 cm, and lateral borders of the breast are on the area of the AAL.For the very thin patient, 4 should be 3.5.

View Article: PubMed Central - PubMed

Affiliation: Department of Plastic Surgery, China-Japan Friendship Hospital, Beijing 100029, China.

ABSTRACT

Background: The determination of the width of the implant is the first key step to select shape and volume of the implant in breast augmentation. The aim of this study was to introduce a new method to determine the width of the implant (W) and explain the reasons to do so in details.

Methods: From January 2006 to June 2014, the authors have found and applied "CD -4" theory to determine the width of breast implant (W) in dual plane I or II breast augmentation cases through transaxillary or periareolar incision for 560 patients. "CD" is defined as the curved distance on skin from the midline of the sternal bone to the anterior axillary line (AAL) on the lateral chest wall through the horizontal level on inferior mammary fold. W = CD - 4 (or 3.5) cm.

Results: The 560 patients used both round and anatomic implants with W from 10.5 cm to 12.5 cm. Their CDs are from 14.5 cm to 17 cm. About 78% of the patients have got followed up from 1 month to 5 years postoperatively. Except for four patients who got unilateral capsular contractions, all the other patients are satisfied with their nature new breast shapes and volumes. Their new intermammary cleavages without bras are between 1 cm and 2.5 cm, and lateral borders of the breast are on the area of the AAL.

Conclusions: W (width of the implant) = CD - 4 (cm) when doing dual plan I or II breast augmentation. For the very thin patient, 4 should be 3.5.

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Related in: MedlinePlus

W (implant width) = CD − (a + b + c + d) = CD − (1.5 + 0.5 + 1 + 1 or 0.5) = CD – 4 or 3.5 (cm). The serratus anterior muscle (flat red “c”) becomes upright because of the implant projection. The “b” is “death zone that implant edge could not reach to”. The “d” is lateral thickness of soft tissue, which is about 1 cm for regular patients and about 0.5 cm for very thin patients.
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Figure 5: W (implant width) = CD − (a + b + c + d) = CD − (1.5 + 0.5 + 1 + 1 or 0.5) = CD – 4 or 3.5 (cm). The serratus anterior muscle (flat red “c”) becomes upright because of the implant projection. The “b” is “death zone that implant edge could not reach to”. The “d” is lateral thickness of soft tissue, which is about 1 cm for regular patients and about 0.5 cm for very thin patients.

Mentions: Why W = CD − 4 (or 3.5) (cm). Figures 4 and 5 (a, b, c, d are defined) shows the reasons. “a” (1.5 cm) is the NTZ. As we mentioned before, any sharp dissection with electric cautery within the NTZ could have a high possibility of damaging the vessel perforators and uncontrolled bleeding, especially through the transaxillary incision with endoscope. Because the broken artery will shrink into the deep muscles and electric coagulation could not work probably, ligation of the artery is sometimes needed. Dissection of NTZ could also induce the palpation of the implant edge, even the symmastia.[8910] Although some experienced doctors could reach out of the margin of NTZ a little bit without complications, as a rule, the NTZ should be highly respected.


Application of "CD-4" theory for determining the width of implant in breast augmentation.

Cai J, Zhou Y - Chin. Med. J. (2015)

W (implant width) = CD − (a + b + c + d) = CD − (1.5 + 0.5 + 1 + 1 or 0.5) = CD – 4 or 3.5 (cm). The serratus anterior muscle (flat red “c”) becomes upright because of the implant projection. The “b” is “death zone that implant edge could not reach to”. The “d” is lateral thickness of soft tissue, which is about 1 cm for regular patients and about 0.5 cm for very thin patients.
© Copyright Policy - open-access
Related In: Results  -  Collection

License
Show All Figures
getmorefigures.php?uid=PMC4836252&req=5

Figure 5: W (implant width) = CD − (a + b + c + d) = CD − (1.5 + 0.5 + 1 + 1 or 0.5) = CD – 4 or 3.5 (cm). The serratus anterior muscle (flat red “c”) becomes upright because of the implant projection. The “b” is “death zone that implant edge could not reach to”. The “d” is lateral thickness of soft tissue, which is about 1 cm for regular patients and about 0.5 cm for very thin patients.
Mentions: Why W = CD − 4 (or 3.5) (cm). Figures 4 and 5 (a, b, c, d are defined) shows the reasons. “a” (1.5 cm) is the NTZ. As we mentioned before, any sharp dissection with electric cautery within the NTZ could have a high possibility of damaging the vessel perforators and uncontrolled bleeding, especially through the transaxillary incision with endoscope. Because the broken artery will shrink into the deep muscles and electric coagulation could not work probably, ligation of the artery is sometimes needed. Dissection of NTZ could also induce the palpation of the implant edge, even the symmastia.[8910] Although some experienced doctors could reach out of the margin of NTZ a little bit without complications, as a rule, the NTZ should be highly respected.

Bottom Line: From January 2006 to June 2014, the authors have found and applied "CD -4" theory to determine the width of breast implant (W) in dual plane I or II breast augmentation cases through transaxillary or periareolar incision for 560 patients. "CD" is defined as the curved distance on skin from the midline of the sternal bone to the anterior axillary line (AAL) on the lateral chest wall through the horizontal level on inferior mammary fold.Their new intermammary cleavages without bras are between 1 cm and 2.5 cm, and lateral borders of the breast are on the area of the AAL.For the very thin patient, 4 should be 3.5.

View Article: PubMed Central - PubMed

Affiliation: Department of Plastic Surgery, China-Japan Friendship Hospital, Beijing 100029, China.

ABSTRACT

Background: The determination of the width of the implant is the first key step to select shape and volume of the implant in breast augmentation. The aim of this study was to introduce a new method to determine the width of the implant (W) and explain the reasons to do so in details.

Methods: From January 2006 to June 2014, the authors have found and applied "CD -4" theory to determine the width of breast implant (W) in dual plane I or II breast augmentation cases through transaxillary or periareolar incision for 560 patients. "CD" is defined as the curved distance on skin from the midline of the sternal bone to the anterior axillary line (AAL) on the lateral chest wall through the horizontal level on inferior mammary fold. W = CD - 4 (or 3.5) cm.

Results: The 560 patients used both round and anatomic implants with W from 10.5 cm to 12.5 cm. Their CDs are from 14.5 cm to 17 cm. About 78% of the patients have got followed up from 1 month to 5 years postoperatively. Except for four patients who got unilateral capsular contractions, all the other patients are satisfied with their nature new breast shapes and volumes. Their new intermammary cleavages without bras are between 1 cm and 2.5 cm, and lateral borders of the breast are on the area of the AAL.

Conclusions: W (width of the implant) = CD - 4 (cm) when doing dual plan I or II breast augmentation. For the very thin patient, 4 should be 3.5.

Show MeSH
Related in: MedlinePlus