Winch

The steel rope has a diameter of 6mm and a length of 20m.
We are winding on drum width 60mm, starting diameter 50mm. What is the final diameter after winding?

Result

D2 =  138 mm

Solution:

D1=50 mm l1=60 mm d=6 mm l2=20 m=20⋅ 1000 mm=20000 mm  r1=D1/2=50/2=25 mm  V1=π⋅ r12⋅ l1=3.1416⋅ 252⋅ 60≐117809.7245 mm3 V2=d2⋅ l2+V1=62⋅ 20000+117809.7245≐837809.7245 mm3  V2=π⋅ r22⋅ l1  r2=V2π⋅ l1=837809.72453.1416⋅ 60≐66.6687 mm  D3=2⋅ r2=2⋅ 66.6687≐133.3374 mm D2=d⋅ ⌈D3/d⌉=6⋅ ⌈133.3374/6⌉=138 mmD_{1}=50 \ \text{mm} \ \\ l_{1}=60 \ \text{mm} \ \\ d=6 \ \text{mm} \ \\ l_{2}=20 \ m=20 \cdot \ 1000 \ mm=20000 \ mm \ \\ \ \\ r_{1}=D_{1}/2=50/2=25 \ \text{mm} \ \\ \ \\ V_{1}=\pi \cdot \ r_{1}^2 \cdot \ l_{1}=3.1416 \cdot \ 25^2 \cdot \ 60 \doteq 117809.7245 \ \text{mm}^3 \ \\ V_{2}=d^2 \cdot \ l_{2} + V_{1}=6^2 \cdot \ 20000 + 117809.7245 \doteq 837809.7245 \ \text{mm}^3 \ \\ \ \\ V_{2}=\pi \cdot \ r_{2}^2 \cdot \ l_{1} \ \\ \ \\ r_{2}=\sqrt{ \dfrac{ V_{2} }{ \pi \cdot \ l_{1} } }=\sqrt{ \dfrac{ 837809.7245 }{ 3.1416 \cdot \ 60 } } \doteq 66.6687 \ \text{mm} \ \\ \ \\ D_{3}=2 \cdot \ r_{2}=2 \cdot \ 66.6687 \doteq 133.3374 \ \text{mm} \ \\ D_{2}=d \cdot \ \lceil D_{3}/d \rceil=6 \cdot \ \lceil 133.3374/6 \rceil=138 \ \text{mm}

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