Rope

How many meters of rope 10 mm thick will fit on the bobbin diameter of 200 mm and length 350 mm (central mandrel have a diameter 50 mm)?

Result

l =  313.4 m

Solution:

a=10 D1=50 D2=200 x=350 n=(D2−D1)/a=(200−50)/10=15 o=π⋅ D x1=D1=50 d=2⋅ a=2⋅ 10=20 x2=x1+(n−1)⋅ d=50+(15−1)⋅ 20=330 s=(x1+x2)/2⋅ n=(50+330)/2⋅ 15=2850 l1=π⋅ s=3.1416⋅ 2850≐8953.5391 l2=l1/1000=8953.5391/1000≐8.9535 m=x/a=350/10=35 l=m⋅ l2=35⋅ 8.9535≐313.3739=313.4  m a = 10 \ \\ D_{ 1 } = 50 \ \\ D_{ 2 } = 200 \ \\ x = 350 \ \\ n = (D_{ 2 }-D_{ 1 })/a = (200-50)/10 = 15 \ \\ o = \pi \cdot \ D \ \\ x_{ 1 } = D_{ 1 } = 50 \ \\ d = 2 \cdot \ a = 2 \cdot \ 10 = 20 \ \\ x_{ 2 } = x_{ 1 }+(n-1) \cdot \ d = 50+(15-1) \cdot \ 20 = 330 \ \\ s = (x_{ 1 }+x_{ 2 })/2 \cdot \ n = (50+330)/2 \cdot \ 15 = 2850 \ \\ l_{ 1 } = \pi \cdot \ s = 3.1416 \cdot \ 2850 \doteq 8953.5391 \ \\ l_{ 2 } = l_{ 1 } / 1000 = 8953.5391 / 1000 \doteq 8.9535 \ \\ m = x / a = 350 / 10 = 35 \ \\ l = m \cdot \ l_{ 2 } = 35 \cdot \ 8.9535 \doteq 313.3739 = 313.4 \ \text{ m }

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