Expert: Sherry Wallin - 8/21/2009

Question
Hi Sherry,

Thanks for offering your time.

I work in electronics manufacturing and we have a huge inventory of electronic components.  Every few months we have to perform an inventory count.  We have an automated counter, but it takes a long time and I'm interested in speeding up the process with some math.
The components are packaged in "Tape & Reel". Most of the parts we have are all packaged in a paper tape, and wrapped around a plastic reel. The thickness of the tape is 2mm, and the spacing between every part on the tape is also 2mm. The tape is wrapped around a reel at a starting diameter of 50mm.  
If I measured the distance that the tape traveled up the side wall of the reel, I think I'd have enough info to calculate the total length of the tape.  If I knew this, I could use some math to tell me how many parts are on the reel.  The problem is, I'm not too good at math.  
Do you think you can help me with this problem?

Thanks!
-Dax

Answer
Hi Dax~
    I can help you think through this problem. If the empty reel is 50mm and you add one winding then you have 54mm which is a circumference of 54*pi since the circumference is pi*diameter. The next winding is 58*pi, and the 3rd winding is 62*pi and so on. Each windings circumference is increased by 4mm. Suppose you were to measure the diameter and it was 70mm. Take the sum of the circumferences: [54*pi + 58*pi + 62*pi + 66*pi + 70*pi]/2. You divide by 2 since that is the spacing between parts and we want to know how many 2's are in the sum of the circumferences. Notice there is a pi in each term so you can factor that out and get (54+58+62+66+70)*pi/2 = 310*pi/2 = 155*pi ~=486.7 so that is 486 parts. Thus the formula is to add
[54 + 58 + 62 + ...+ (50 + 4n)]*pi/2 where n = 1,2,3,... and n is the number of windings. n is also the [final diameter - 50mm]/4. I hope this makes sense.

Math Prof