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### Foil Wrapped Around a Spindle

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Date: 12/17/97 at 17:18:00
From: Liz Brown
Subject: Integration?

I have a problem in an engineering work environment. I have a tightly
rolled sheet of foil wrapped around a spindle. I know the diameter of
the spindle (Do) and the external diameter of the combined spindle and
surrounding foil (Di) and the thickness of a sheet of foil (t). I need
to know the length of foil (L) needed to generate the external
diameter.

I have been given a formula by word of mouth

L = pi (square(Do) - square(Di) ) / 4t

I need to know if this is an approximation of a more accurate formula
as I am dealing with a thickness of 0.01 mm and the length calculation
needs to be accurate for production purposes. Is it possible to prove
from first principles?

I would be grateful of any help you can give me. I did A level Maths
but that was at least 15 years ago. Thank you.
```

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Date: 12/22/97 at 15:53:20
From: Doctor Mark
Subject: Re: Integration?

Hi Liz,

If I understand your problem correctly, (Do) should be the diameter of
the *spindle and surrounding foil* (so that it is Do(uter)), while
(Di) should be the diameter of the *spindle alone* (so that it is
Di(nner) - not that I am expecting you to eat it!). With that change
in the definitions, then the formula you were given seems correct to
me, though it is 1) an approximation of a sort and 2) more complicated
than is necessary.

I expect that the diameter of the spindle is much larger than the
thickness of the foil. Is that right?  That would be the case if the
0.01 mm represents the thickness of the foil, and if Di is of the
order of a few mm or larger. If so, then the formula you were given is
correct to a *very* good approximation.

However, it is should also be the case that the thickness t of the
foil should be the difference between Do and Di,

t = Do - Di

and so your formula actually reduces (by factoring the polynomial as:
a^2 - b^2 = (a+b)(a-b), with a = Do, and b = Di) to

L = pi ( [Di + Do]/2 )

That formula could more easily be derived by drawing a picture (which
we can use to get the original formula as well). Draw two circles with
the same center, one of radius Di (the i[nner] circle--the spindle),
one of radius Do (the o[uter] circle - the spindle plus foil). The
foil is represented by the region outside the inner circle and inside
the outer circle.

If you now draw a circle, with the same center as the first two, with
its boundary (the circumference) halfway in between the two circles,
then the diameter of this circle should be the average of the inner
and outer diameters, i.e., of Di and Do. The circumference of this new
circle will be pi times that average, and this circumference is (to a
very good approximation, if t is small compared to Di) just L, given
by the formula above this paragraph.

Your original formula comes from finding the area of that region in
between the two circles, which is:

area between the two circles = area of outer circle -
area of inner circle

= pi [ (radius of outer circle)^2 - (radius of inner circle)^2 ]

= pi [ (Do/2)^2 - (Di/2)^2  ]  =  pi [ {(Do)^2}/4 - {(Di)^2}/4 ]

= pi [ (Do)^2 - (Di)^2 ]/4

If t is very small compared to Di and Do, then the region between
the circles looks sort of like a very thin rectangle of length L
and width t, wrapped around the circle of radius Di. Equating the
area between the two circles to the the area of the rectangle, which
is (length) x (width) = Lt, we get:

Lt = pi [ (Do)^2 - (Di)^2 ]/4

Divide both sides by t, and this is your original formula.

Personally, I would use the L = pi*(Do + Di)/2 formula, since it is
less susceptible to small errors in the measurement of Do and Di. In
fact, since Do and Di are almost the same, I would probably use the
even simpler formula

L = pi(Di).

That is, the length of the foil is just the circumference of the
spindle. I suspect that (again, if the thickness of the foil is a lot
less than the diameter of the spindle) this would be good enough for
most practical purposes.

None of what I said is going to be true if, in fact, the foil is *not*
really thin compared to the diameter of the spindle. Write back if
that is what you need.

Hope this has been of some help.

-Doctor Mark,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Practical Geometry

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