# How to find length of this tape ?

## outer diameter 5 cm inner diameter 3 cm thickness of tape 0.01 cm I used MATLAB d=5; % diameter x=0; % length of the tape while d > 3 x = x + pi * d; %circumference d = d - 0.010; end x / 100 % answer in meter I got 25.258404934862064 actually the right answer should be 50.26 and it looks like I got 25 half of 50. I can't find where the fail is.

##### 3 Answers
May 26, 2018

$50.51680 \ldots \ldots \text{metres}$
Had someone else look at this and they confirm my 12.7. They think that instead of diameter you should have written radius.
My programming works out the equivalent of radius as I only used one thickness when determining cycle count.

#### Explanation:

For some reason I could not fathom my why my calculation by hand gave an unrealistic answer. So I switched to writing a quick bit of code in VBA and found the answer to be

$\textcolor{g r e e n}{50.51680 \ldots \ldots m e t r e s}$

I ran it from the centre outwards. Subtracted length of 0 up to 3cm from length of 0 up to 5cm

$\textcolor{red}{\text{However, this is wrong!}}$

$\textcolor{red}{\text{I calculated the count of cycles by ONLY HAVING 1}}$color(red)("THICKNESS in a diameter. "ul("THERE ARE 2"))$\textcolor{red}{\text{So this is the equivalent of having radius of 5 and 3}}$
$\textcolor{red}{\text{However, at line 19 you will observe I used 2 thicknesses}}$$\textcolor{red}{\text{to the increment diameter which is correct. A happy accident!}}$

$\textcolor{red}{\text{Original output at 5/0.01 cycles for outer and 3/0.01 for inner}}$

length_outer = 5051.68098697234 $\approx 50.5 \text{ metres}$
length_total = 7869.68959724238
length_inner = 2818.00861027004

$\textcolor{g r e e n}{\text{New output at 5/0.02 cycles for outer and 3/0.02 for inner}}$
length_outer = 1269.20343205028 $\approx 12.7 \text{ metres}$
length_total = 1971.34939012759 $\textcolor{w h i t e}{\text{dddd}} \uparrow$
length_inner = 702.145958077319 color(white)("ddd")" |"
color(white)("dddddddddddddddddddddddddddd.d")" |"
color(white)("dddddddddddddddddddddddddd")"Not logical"

May 28, 2018

Manual calculation method

#### Explanation:

$\textcolor{red}{\text{There is a problem with my solution}}$

$\textcolor{red}{\text{I had not taken into account the accumulating hypotenuse}}$$\textcolor{red}{\text{length at the overlap step}}$

and the diagram

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{My solution without the hypotenuse taken into account}}$

Look at the comments at the bottom of my other solution. One of them explains how the system behaves and thus indicates the solution method.

Assuming given diameters should be radius (makes the numbers work).

5 cm gives 500 cycles of rotation from radius = 0 to 5 cm
3 cm gives 300 cycles of rotation from radius = 0 to 3 cm

Known Circumference $\to 2 \pi r \to \pi D$

Given: tape thickness $\to$ 0.01 cm so diameter incremented by 0.02 at each cycle

1st rotation cycle $\to 0.02 \pi \textcolor{w h i t e}{\text{d}}$ cycle circumference
2nd rotation cycle $\to 0.04 \pi \textcolor{w h i t e}{\text{d}}$ cycle circumference
3rd rotation cycle $\to 0.06 \pi \textcolor{w h i t e}{\text{d}}$ cycle circumference

Observing the pattern tells us that for ${n}^{\text{th}}$ rotations the circumference is $0.02 n \pi$

So the summation of circumferences is:

$0.02 \pi + 0.04 \pi + 0.06 \pi + \ldots + 0.02 n \pi$

$1 \left(0.02 \pi\right) + 2 \left(0.02 \pi\right) + 3 \left(0.02 \pi\right) + \ldots + n \left(0.02 \pi\right)$

$0.02 \pi \left(1 + 2 + 3 + \ldots + n\right)$

$0.02 \pi \left(\text{mean value"xx"count}\right)$

For the 5 cm radius we have:

$0.02 \pi \left(\frac{1 + 500}{2} \times 500\right)$

For the 3 cm radius we have:

$0.02 \pi \left(\frac{1 + 300}{2} \times 300\right)$

The length of tape between and including both 3 cm and 5 cm is:

$0.02 \pi \left(\frac{1 + 500}{2} \times 500\right) - 0.02 \pi \left(\frac{1 + 300}{2} \times 300\right)$

$= 5032.8314$cm to 4 decimal places

$\textcolor{g r e e n}{= 50.3283 \text{ metres to 4 decimal places}}$

May 28, 2018

$L = 12.5663 \text{m}$

Or if diameter was meant as radius

$L = 50.2655 \text{m}$

#### Explanation:

Here is another approach

Consider the following sketch

The area of tape (green area)

A_("Tape")=pi(5/2"cm")^2-pi(3/2"cm")^2=4pi"cm"^2

Imagine to role out the tape, we will have a rectangle (although a very thin one)

The area of this thin rectangle, must be the same as the area of the tape, and the small side of the rectangle must be the thickness of tape $T = 0.01 \text{cm}$

Thus

${A}_{\text{Tape")=L*T=>L=A_("Tape}} / T$

Using the values

L=(4pi "cm"^2)/(0.01"cm")=400pi"cm"~~1256.63"cm"=12.5663"m"

If diameter was meant as radius

${A}_{\text{Tape")=pi(5"cm")^2-pi(3"cm}}^{2} = 16 \pi$

Thus

L=(16pi"cm"^2)/(0.01"cm")=1600pi"cm"~~5026.55"cm"=50.2655"m"