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Does anyone have any clue on how to start this question off?

The highly scary helter skelter at the fair is a cylindrical tower of height 30 metres and circumference 8 metres. The slide is wound around the tower exactly 5 times. What’s the total length of the slide?

I think it's to do with Helixes or something and we have not covered any of that in class. I've thought about 'unravelling it' as in making it a rectangle with the width being 30 and the length being 8π but do not know what to do next! Any help is much appreciated as I have no clue on what to do!!

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    This question is part of the [Newcastle University Senior Challenge '17](http://www.ncl.ac.uk/maths/outreach/teachers/challenge/) (question 3 _Highly Scary Helter Skelter_ of [this pdf](http://www.ncl.ac.uk/media/wwwnclacuk/mathematicsandstatistics/files/mathschallenge2017/Newcastle%20Senior%20Maths%20Challenge%202017.pdf)). The stated due date for entries is 10 March 2017, and this question will remain locked with answers temporarily deleted until after this date. – 2017-02-25

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For a single winding the height would be $6$ metres.

The circumference of the cylinder is given as $8$ so no $\pi$ required.

Unfold the cylinder and the single winding unravels as a straight line corresponding to the diagonal of a right-angle triangle of side lengths $6,8$.

By Pythagoras' theorem, the length of the diagonal is $\cdots\bullet $.

Total spiral length is $5$ times this, which is $\cdots\bullet $.

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    By Pythagoras' Theorem would it be 10? Therefore: 5*10= 50?? – 2017-02-25
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    Yes, and so ... – 2017-02-25
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    Ermm, the slide is also 50..? – 2017-02-25
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    how does 5*10=30? – 2017-02-25
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    I fixed it. It was a typo sorry ;-;. – 2017-02-25
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    Yes, that's correct. Take a piece of A4 paper, draw a diagonal across it, and roll it up into a cylinder. That might be a good illustration. :) – 2017-02-25
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    Thank you very much! This question was bugging me for days!! :) – 2017-02-25
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    You're welcome! Glad you found it helpful :) – 2017-02-25
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If you know a bit of calculus then you could parameterize the helix as $$ x(t) = \frac{4}{\pi}\cos(10\pi t), \quad y(t) = \frac{4}{\pi}\sin(10\pi t), \quad z(t) = 30 t, \quad t \in [0,1] $$ The arc length is then \begin{align} A &= \int_0^1 \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}\,dt\\ &= \int_0^1 \sqrt{1600 + 900}\,dt\\ &= 50 \end{align}

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    I have no clue! But thanks for your input! – 2017-02-25