Foil Wrapped Around a Spindle

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.

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/