1️⃣ What is mm² in cable?

mm² = Square millimeter
It tells the cross-sectional area of the copper conductor inside the cable — not the outer diameter.

🔹Example:

• 2.5 mm² cable means the copper area is 2.5 mm²
• 4 mm² cable → copper area is 4 mm²
  (Bigger mm² = more current flow)

🌟 2️⃣ Why mm²? (Not diameter)

Because some cables are:

• Round
• Stranded (many small wires)
• Shaped differently

So area is fixed… but diameter changes depends on insulation thickness.

🌟 3️⃣ How to calculate diameter from mm²?

Assume the conductor is round:

📌 Formula:

$$\text{Diameter} = \sqrt{\frac{4 \times \text{Area}}{\pi}}$$Diameter=π4×Area​​

Example: 2.5 mm² cable

$$\text{Diameter} = \sqrt{\frac{4 \times 2.5}{3.14}} = \sqrt{3.18} = 1.78\,\text{mm}$$Diameter=3.144×2.5​​=3.18​=1.78mm

But this is just copper ❗
Actual cable outer diameter ≠ this number
Because we must add:

✔ Insulation thickness
✔ Inner/Outer sheath (for multi-core)

So outer diameter is usually:

(Different brands ⇒ slightly different sizes)

🌟 4️⃣ Cable Tray Filling

Tray filling rule (common):

✔ Only 40% of tray area can be filled with cables

Steps:
1️⃣ Get cable outer diameter
2️⃣ Area of one cable:$$A = \pi \left(\frac{D}{2}\right)^2$$A=π(2D​)2

3️⃣ Tray usable area:$$A_{\text{usable}} = \text{Tray width} \times \text{height} \times 0.4$$Ausable​=Tray width×height×0.4

4️⃣ Total cables allowed:$$N = \frac{A_{\text{usable}}}{A_{\text{cable}}}$$N=Acable​Ausable​​

🔹 Small Example:

Tray: 200 mm width x 50 mm height
Usable area = 200 × 50 × 0.4 = 4000 mm²

Cable OD: 8 mm → area:$$A = 3.14 \times 4^2 = 50.24\,\text{mm}^2$$A=3.14×42=50.24mm2

Number of cables:$$4000 / 50.24 ≈ 79 \text{ cables}$$4000/50.24≈79 cables

✅ Summary You Can Remember

If you want — I can create an Excel sheet for you:
✔ You enter tray size + cable OD → It calculates number of cables

Would you like the Excel?
And tell me your commonly used cable sizes! 😊

sample calculation

Sure! Here is one complete sample calculation step-by-step 👇
(You can save this as your reference)

📌 Sample Cable Tray Fill Calculation

🔹 Given:

✏ Step-1: Find Tray Usable Area

Tray Total Area$$A_{tray} = 300 \times 50 = 15000\ mm^2$$Atray​=300×50=15000 mm2

Usable Area (40%)$$A_{usable} = 15000 \times 0.4 = 6000\ mm^2$$Ausable​=15000×0.4=6000 mm2

✏ Step-2: Area of One Cable

Outer Diameter = 6 mm
Radius = D/2 = 3 mm$$A_{cable} = \pi \times r^2 = 3.14 \times 3^2$$Acable​=π×r2=3.14×32$$A_{cable} = 3.14 \times 9 = 28.26\ mm^2$$Acable​=3.14×9=28.26 mm2

✏ Step-3: Number of Cables Allowed

$$N = \frac{A_{usable}}{A_{cable}} = \frac{6000}{28.26} = 212.3$$N=Acable​Ausable​​=28.266000​=212.3

🎯 Final Answer

✔ Maximum cables allowed in tray:$$\approx \boxed{212\ cables}$$≈212 cables​

🔍 Quick Summary Table