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I have a partial reel of tape with components. The label says 2500 pieces. Counting the outermost layer (60 pieces) and multiplying it with the number of turns (15) gives 900 pieces. But the inner turns will have fewer components.

The spool has an indicator like a ruler which goes from 1 to 11. The belt ends at ~3.5. But I have no idea how full it may have been originally.

As a hobbyist I don't have a counting machine or similar. Is there a known approach to get a better estimate except counting manually? If you do math, please mention the formula, since I have many other reels like this with different number of turns, diameters etc.

Here’s how my reel(s) look like:

Image of the belt

The component is ST72F324BK6T3.

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    \$\begingroup\$ At work in the past we would measure how many components were in 6in of reel and estimate the number remaining based on the length of tape \$\endgroup\$ Jul 13, 2022 at 21:33
  • \$\begingroup\$ How many on the outer ring? How many turns? It can be estimated from these two values using geometry. \$\endgroup\$
    – Mattman944
    Jul 13, 2022 at 21:35
  • \$\begingroup\$ @Mattman944: Updated. It's 60 outermost and 15 turns, which multiplies to 900. \$\endgroup\$ Jul 13, 2022 at 21:39
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    \$\begingroup\$ Some reels will have small numbers on the side representing the percentage of the length of the spiral, which is a very quick way of giving a ballpark. \$\endgroup\$
    – Wesley Lee
    Jul 13, 2022 at 22:51
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    \$\begingroup\$ I’m voting to close this question because it does not concern electronics \$\endgroup\$ Jul 14, 2022 at 19:24

6 Answers 6

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An interesting geometry problem. It is easiest to understand in Excel. Doing it in calculus would be cool, but I am rusty.

The radius measurements are in arbitrary units, it doesn't matter (I measured with a ruler in Photoshop). The bold numbers are the inputs.

enter image description here enter image description here

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CompuPhase has a reel quantity estimate calculator available online.

It works by the basic geometry calculations and gives results in number of components remaining:

Screenshot of calculator

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    \$\begingroup\$ This online calculator does exactly the same as the other solutions, it just combines all the steps into one. It claims to be "exact" but it is no better as it uses exactly the same maths. Some of the other methods don't even require you to know the component pitch. \$\endgroup\$ Jul 15, 2022 at 22:15
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The length will be

$$L = \frac {A_{spiral}} t $$ where \$ A \$ is the area of the spriral and \$t\$ the thickness of a layer.

$$L = \frac {A_{spiral}} t = \frac {\pi R^2 - \pi r^2} t$$

where \$R\$ is the outer radius and \$r\$ is the inner radius.

The number of pieces is given by $$ \frac L p $$ where \$p\$ is the pitch of the components.

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  • \$\begingroup\$ And t is the (R-r)/n where n is the number of turns. \$\endgroup\$
    – jcaron
    Jul 15, 2022 at 22:41
  • \$\begingroup\$ Then factoring the equation for L and substituting t gives you L=pin(R+r)(R-r)/(R-r). Which is simplifies to L=pin(R+r). Multiply top and bottom by 2 and we are back to the equations given on the length of a spiral calculator site. L=pi*n(D+d)/2. The average of the inner and outer circumferences X n. \$\endgroup\$ Jul 16, 2022 at 0:27
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You need to measure the inside and outside diameters of the spiral. From the circumference of the outermost turn you know the components per cm (or inch). There is an online spiral length calculator here.
The calculation is then trivial multiplication.

Using Mattman944's numbers (thanks Mattman944) the length of the spiral is 74.4 units and the count per length is 9.5 giving a total count of 707. Which is in complete agreement with Mattman944's answer, but doesn't require Excel.

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    \$\begingroup\$ Cool, there is always an easier way if you know where to look. \$\endgroup\$
    – Mattman944
    Jul 13, 2022 at 22:36
  • \$\begingroup\$ @Mattman944 No, all credit to you for spotting that the units (inch or cm) don't matter. And I wasted a lot of time doing the maths and working out the general case formula for the length of the spiral from scratch before googling "length of spiral". \$\endgroup\$ Jul 13, 2022 at 22:44
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You can get the average length of all loops by taking the average of the outermost and innermost loops. Then you can multiply the average you get with the number of loops you can see, and you get the total length, which you can use to calculate the number of components in the reel.

You can do the simplification I mentioned (outside length + inside length)/2 because the relationship between the length of the loop and the radius is linear.

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  • \$\begingroup\$ This was what came to mind when I saw the question. It is simple and as elegant (or not :-) ) as the other methods. \$\endgroup\$
    – Russell McMahon
    Jul 20, 2022 at 9:11
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Another way of calculating it is just from the 3 diameters

  • The spool's OD (when full and containing 2500 units) D
  • The diameter of the remaining units spiral S
  • The inner diameter of the spool, the hub. d The number of items is proportional to the area.
    Then \$2500 ∝ D^2 - d^2\$ (the area of the full spool omitting pi)
    remaining units \$∝ S^2 - d^2\$
    thus remaining units = \$2500 * (S^2 - d^2) / (D^2 - d^2)\$
    No need to know the units per length.
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    \$\begingroup\$ Or you could just temporarily unspool the reel, measure its length, count 10cm worth, do the sums and wind them back on. \$\endgroup\$ Jul 14, 2022 at 1:38
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    \$\begingroup\$ OP stated "I have no idea how full it may have been originally" so the value of D is unknown in this case. \$\endgroup\$
    – jcaron
    Jul 14, 2022 at 13:34
  • \$\begingroup\$ @jcaron Agreed, but any calculation other than actually counting the parts, is going to have a degree of approximation. In this case it is the assumption that the original manufacturers were unlikely to have used a spool with much spare space as their customers wouldn't want to be changing them more often than necessary. Hence assuming the spool was full, or nearly so, shouldn't result in too big an error. \$\endgroup\$ Jul 15, 2022 at 22:30

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