Question :)

[Non-Mathematician]

Hi I am not a mathematician and haven't studied calculus ever. I am doing a small project and need to calculate something that looks simple but I feel it need calculus knowledge to get the answer. :shock:

A.What is the maximum number of revolutions that a tape can rotate around a shaft of diameter 6cm if the tape has a thickness of 4mm and length 120cm.

 

[tkhunny]

Just add it up.

6 cm = 60 mm

1st lap: pi * 60 mm
2nd lap: pi * (60 + 4) mm
3rd lap: pi * (60 + 2*4) mm

nth lap: pi * (60 + (n-1)*4)

All laps, from 1 to n: pi * {(n*60) + [1 + 2 + 3 + ... + (n-1)]*4)}

You may or may not know the sum of those values in the inner-most parentheses. It's (n(n-1))/2

This gives: pi * {(n*60) + [(n(n-1))/2]*4)} = pi * {(n*60) + [(n(n-1))]*2)}

Since we started with 1200 mm, this results in: 1200 = pi * {(n*60) + [(n(n-1))/2]*4)} How's your algebra? You're almsot done. Note: This will result in the total number o flaps, including the last partial lap. You will need to think about what you want to do with that last partial lap.

Also, this is not perfect. It takes a little bit of tape to get up to the next layer.

 

[mmm4444bot]

We can answer this question using algebra.

The maximum number of "whole" revolutions is 5, but there's just enough tape left over to make about 1/50th of a revolution.

So, I say 5.02 revolutions.

You'll have to let us know whether or not you want an explanation. For now, I'll just post a few comments.

The distance around a circle is called the circumference.

One formula for circumference is Pi*d, where Pi is the number 3.14159 (rounded) and the symbol d represents the diameter. (The asterisk denotes multiplication).

For the first revolution, the diameter is 6.

Once the tape wraps around the shaft once, 0.4 cm gets added to each end of the diameter (4mm = 0.4 cm).

In other words, the diameter of the circle around which the tape wraps increases by 0.8, with each whole revolution.

Therefore, the distance around follows this progression (per "wrap"):

6 Pi
6.8 Pi
7.6 Pi
8.4 Pi
9.2 Pi
et cetera

The sum of these first five lengths is 38 Pi, which totals 119.38 cm (rounded).

The remaining 0.62 cm of tape will only go about 1/50th the way around.

That progression of numbers is called an "arithmetic sequence". There are formulas to deal with generating these numbers and calculating sums of the "first n elements" in such a sequence.

I used those formulas to generate what's called a quadratic equation. The solution coming out of that equation is 5.02 revolutions.

Cheers ~ Mark

 

[mmm4444bot]

Ah, I see that tkhunny posted, while I was still typing.

I note that tk's last equation has solution n = 5.53 revolutions, versus my result of 5.02 revolutions.

I must have made a mistake, somewhere, setting up my equation.

tk's equation simplifies to this quadratic equation:

n^2 + 29n - 600/Pi = 0

My quadratic equation is:

n^2 + 14*n - 300/Pi = 0

But, if you're talking complete revolutions, then we both say, "5 complete revolutions is the maximum possible, using 120cm of tape".

 

[lookagain]

Just add it up.

6 cm = 60 mm

1st lap: pi * 60 mm
2nd lap: pi * (60 + 4) mm
3rd lap: pi * (60 + 2*4) mm

It is closer to this situation:

With the tape going around, it makes approximately a torus (doughnut shape.)
The circumference of the center circle of the torus is what will be used as the distance
around the shaft, not the inside circumference of the torus (inside of the curved tape).

The distance from the middle of the tape, through the shaft's diameter, to the middle of the tape
on the opposite side is 2 mm + 60 mm + 2 mm = 64 mm.

Then the next diameter is 2 mm + 64 mm + 2 mm = 68 mm, and so on with the successive
diameters (of the center of the tape on one side to the center of the tape on the direct opposite side)
increasing by 4 mm.

The sum of the corresponding circumferences is \(\displaystyle (64\pi + 68 \pi + 72 \pi + 76 \pi + 80 \pi + ... ) \ \ in \ mm\)

or \(\displaystyle \pi(64 + 68 + 72 + 76 + 80 + ...) \ \ in \ mm.\)

----------------------------------------------------------------------------------------------------------------------------

\(\displaystyle \pi(64 + 68 + 72 + 76 + 80) \ mm \approx 1,131 mm\)

\(\displaystyle \pi(64 + 68 + 72 + 76 + 80 + 84) \ mm \approx 1,395 mm\)

\(\displaystyle 120 \ cm = 1,200 \ mm\) falls between these two, so there are


\(\displaystyle \boxed{5 \text{ complete revolutions of the tape with some leftover tape}}\)

 

[mmm4444bot]

I studied the other posts in this thread, and now I know where I went wrong.

I took a simplistic view of one segment of tape wrapped exactly once around the shaft, forming a band that essentially expands the diameter of the "shaft" for the next band.

I mean, I assumed this was possible, and it's not. That was my error.

The INNER surface of such a band of tape would meet, when joining the ends of the segment together, but there will be a gap at the OUTER surface of the band. This is because the end surfaces of the tape segment are perpendicular to the arc of the shaft. (Think of a V.)

Since the outer surface of one revolution becomes the outer diameter of the "shaft", the circumference around that circle is bigger than I thought because the next revolution of tape needs to bridge that gap.

In my mathematical model (versus the models used by tkhunny and lookagain), the error grows significantly, with each revolution.

By sheer luck, the original poster's scenario involves only a few revolutions, so, despite falling into the drink, I walked out dry!

 

[BigGlenntheHeavy]

A.What is the maximum number of revolutions that a tape can rotate around a shaft of diameter 6cm if the tape has a thickness of 4mm and length 120cm.

\(\displaystyle C \ = \ 2\pi r, \ let \ 2\pi \ = \ \theta \ (a \ complete \ cycle).\)

\(\displaystyle Then \ 2\pi r \ = \ 4\theta \ \implies \ r \ = \ \frac{\theta}{.5\pi}\)

\(\displaystyle Now \ since \ the \ radius \ of \ the \ shaft \ is \ 30mm, \ 15\pi \ < \ \theta \ < \ b\)

\(\displaystyle x \ = \ rcos(\theta) \ = \ \frac{\theta cos(\theta)}{,5\pi} \ and \ y \ = \ rsin(\theta) \ = \ \frac{\theta sin(\theta)}{.5\pi}\)

\(\displaystyle Now \ by \ arc \ length: \ 1200 \ = \ (2/\pi)\int_{15\pi}^{b}\sqrt{\theta^2+1}d\theta, \ b \ = \ 77.4 \ (trusty \ TI-89).\)

\(\displaystyle 77.4-15\pi \ = \ 30.276 \ and \ 30.276/2\pi \ = \ about \ 4.81 \ revolutions.\)

\(\displaystyle Ergo \ 4.81 \ revolutions \ of \ 1200mm \ around \ the \ shaft \ will \ use \ it \ up.\)

\(\displaystyle REMARK: \ The \ graph \ is \ called \ the \ Spiral \ of \ Archimedes.\)

 

[lookagain]

Again, the almost concentric rings of almost circular tape sections have diameters of about
64 mm, 68 mm, 72 mm, 76 mm, 80 mm, and so on. (See my earlier post.) Multiply each
of these by \(\displaystyle \pi\) to get their circumferences and add them up.

\(\displaystyle \pi(64 + 68 + 72 + 76 + 80) \approx 1,131 \ mm,\) which would mean there will be at \(\displaystyle 5\) revolutions,
plus leftover tape.

Even if each layer's distance around were instead exaggerated to 68 mm, 72 mm, 76 mm, 80 mm, and 84 mm,

respectively, then the total distance would be \(\displaystyle \pi(68 + 72 + 76 + 80 + 84) \approx 1,194 \ mm.\)


This is still \(\displaystyle less\) than the total length of the tape of \(\displaystyle 1,200 \ mm,\) so there would \(\displaystyle still\) be a fraction \(\displaystyle more\) than
\(\displaystyle 5\) revolutions.

There must be at least one wrong assumption in the calculus method from the above post,
because its number of revolutions works out to \(\displaystyle less\) than five.

 

[BigGlenntheHeavy]

\(\displaystyle C \ = \ \pi d \ = \ \pi(68+76+84+92+100) \ = \ 1319 \ and \ change.\)

\(\displaystyle Hence, \ number \ of \ revolutions \ of \ 1200 \ will \ equal \ 4 \ + \ change.\)

 

[lookagain]

The length of a torus (discussed in part here):

http://cc.bingj.com/cache.aspx?q=comput ... 4,cf87b84a

is \(\displaystyle 2\pi R,\) where \(\displaystyle R\) is the distance from the center of the torus to the center of a circle being rotated
around to make the torus.

Then, for each successive diameter in the approximation, it gains \(\displaystyle 4 \ mm,\) not \(\displaystyle 8 \ mm.\)

Then the following set-up does not apply:

\(\displaystyle "C \ = \ \pi d \ = \ \pi(68+76+84+92+100) \ = \ 1319 \ and \ change."\)

 

[BigGlenntheHeavy]

\(\displaystyle Look, \ if \ you \ have \ a \ circle \ of \ radius \ 30mm \ and \you \ wrap \ a \ string \ of \ thickness \ of \ 4mm\)

\(\displaystyle then \ the \ circumference \ of \ the \ circle \ with \ string \ will \ equal \ 2\pi(34mm) \ = \ 68\pi mm, \ not \ 64\pi mm.\)

 

[lookagain]

\(\displaystyle Look, \ if \ you \ have \ a \ circle \ of \ radius \ 30mm \ and \you \ wrap \ a \ string \ of \ thickness \ of \ 4mm\)

\(\displaystyle then \ the \ circumference \ of \ the \ circle \ with \ string \ will \ equal \ 2\pi(34mm) \ = \ 68\pi mm, \ not \ 64\pi mm.\)

I don't need to "look" at all. These are the facts, and please don't ignore and/or try to refute the information given
in the link I already gave a couple of posts ago (as well as what is available at other sites):

The shaft has a radius of \(\displaystyle 30 \ mm,\) and the new/radius extends out \(\displaystyle 2 mm\) beyond the first/original radius,
out to the center of a cross-sectional circle of the torus (made up of the tape) \(\displaystyle 4 \ mm\) diameter tape, giving \(\displaystyle 32 \ mm.\)
The length (circumference) of the length of the torus-shaped tape lies in the middle of the inner circumference and the
outer circumference of the torus shape of the tape, which, the tape, has a thickness (diameter) of \(\displaystyle 4 \ mm.\)

Then, for the diameter, it is \(\displaystyle 2(32 \ mm) = 64 \ mm.\)

So the circumference of the first revolution of tape is closest \(\displaystyle \ ** \\) to \(\displaystyle 64 \pi \ mm.\)


\(\displaystyle **\) For the next (second) layer, the tape has to move up higher from the bare side of the shaft up over
onto itself, after completing one revolution to start making the second layer, so the initial circumference
is not exactly circular, but it is relatively slightly longer than if it were forming only a total of
one layer (revolution).





Signature:
"The hardest part of an argument is realizing that you are wrong."

 

[mmm4444bot]

"The hardest part of an argument is realizing that you are wrong."

Sometimes, the realization is easy, but the acceptance is hard. :wink:

 

[BigGlenntheHeavy]

\(\displaystyle You're \ splitting \ hairs \ lookagain, \ it'll \ wrap \ around \ the \ shaft \ a \ little \ under \ 5 \ revolutions.\)

\(\displaystyle Also \ note: \ The \ contention \ here \ is \ between \ I \ and \ lookagain, \ so \ if \ we \ want \ any\)

\(\displaystyle **** \ out \ of \ you \ mmm444bot, \ we'll \ squeeze \ your \ head.\)

 

[mmm4444bot]

we'll squeeze your head

I was actually thinking of myself, when I posted that comment on lookagain's signature, but your joke is a good one!

While I typed my original reply, I basically considered the spiral as a set of concentric circles. :roll: Heh, heh, heh.

lookagain's original reply prompted me to digest both it and tk's post, also.

The realization that I was wrong came quick and easy.

Yet, my acceptance of my goof is hindered because my error in logic gnaws at me still.

I feel no ill will toward you, Glenn.

I'll consider some experiments with wood doweling and cord, however, if my curiosity peaks.

Cheers ~ Mark

 

[lookagain]

\(\displaystyle Look, \ if \ you \ have \ a \ circle \ of \ radius \ 30mm \ and \you \ wrap \ a \ string \ of \ thickness \ of \ 4mm\)

\(\displaystyle then \ the \ circumference \ of \ the \ circle \ with \ string \ will \ equal \ 2\pi(34mm) \ = \ 68\pi mm, \ not \ 64\pi mm.\)

I'm **not** "splitting hairs," because it is cut and dried for the whole number part of the revolutions.
It is not an issue of rounding or something similar to that (my phrase) giving two different
whole number of revolutions. You're doing the part of the wrong process wrong with part of a wrong concept.

The outer circumference of the first layer is **not used,** because it is *not* the length of the (first layer of) tape.
Again (for at least the third time), it all comes back to the length of a torus, of which the ring of tape is an example.
Here, the length of the first layer of the tape is the average of the inner circumference (which is \(\displaystyle 60 \pi \ mm\) being circumference of the round edge of the shaft) and the outer circumference (which is the outer edge of the
tape and is \(\displaystyle 68 \pi \ mm).\) Therefore, the length of the first layer is \(\displaystyle (60 \pi \ mm \ + \ 68 \pi\ mm)/2 = 64 \pi \ mm.\)

Example:

What is the length of tape of \(\displaystyle 4 \ mm\) in width making exactly one revolution around a shaft
having a radius of \(\displaystyle 30 \ mm\)?

Looking at the circles formed by the shaft and the outer circumference of the tape, we can
calculate the area of the tape makes between these circles.

Area = \(\displaystyle \pi(34^2) \ mm^2 - \pi(30^2) \ mm^2 = 256 \pi \ mm^2\)

Then the area of the tape, when seen as a circle, also equals the cross-section (width) of the tape
times the length of the tape = \(\displaystyle (4 \ mm)(2 \pi R) = 256 \pi \ mm^2. \ \\)
Which radius, \(\displaystyle R,\) do we use?

Solve for \(\displaystyle R:\)

\(\displaystyle 8 \pi R \ mm = 256 \pi \ mm^2\)

Divide each side by \(\displaystyle 8 \pi \ mm:\)

\(\displaystyle R = 32 \ mm\)

So the length of the tape \(\displaystyle = 2 \pi R = 2 \pi (32 \ mm) = 64 \pi \ mm.\)