Instrument tray
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May I ask: how many 1×3×1.5 shielded cables can fit in a 400×200 size cable tray?
The number of strands obtained by calculating (Length × Width) ÷ (3.14 × R2) × 0.6 is generally sufficient! Length × Width = area of the cable tray. 3.14 is the value of pi. 0.6 is the filling factor (it indicates that only 60% of the cable volume is accounted for; if it were 100%, there wouldn’t be any space left: lol). 3.14 × R² = area of a single cable. In reality, the number of cable cores varies, and there are also gaps between cables, so this can only be used as an estimate. This post was last edited by chmyh on 2009-3-15 16:56]
Cables can be laid without spacing on the cable tray; the filling rate of the cables within the cross-section of the tray should not exceed 40% for power cables； The control cable should not be more than 50%.
The answer is upstairs. . Instrument cable 50% cross-sectional fill rate