Number of parts left on a reel? - Electronic Circuits Projects Diagrams Free

Richard Principal · 3rd October 2007, 11:29 AM

At the moment I operate a machine tha

Leftyretro · 3rd October 2007, 02:29 PM

Quote:

Originally Posted by Richard Principal

Hi here is a problem.
At the moment I operate a machine that puts surface mount components on a PCB, the boss asked me how many parts left on a used reel, It might of been a reel that normally had 1000 parts on it, I had a rough guess of say 600, and believe or not, I came out "not far off" (I wish I was this lucky with every guess)

But I was thinking one should be able to calculate it, if you measure the inside radius and had a good idea where the outside radius was, measure the radius of the parts left and apply this together with the number of parts that is normally on the reel, one should be able to work out from a formula how many parts on a real.

Tonight I have been thinking of it and think the "Triangular series" could be the answer to this problem. and I came up with the following table.

% of reel left....... % of components left
10.....................16.94
20.....................20.02
30.....................24.64
40.....................30.8
50.....................38.5
60.....................63.14
70.....................58.52
80.....................70.84
90.....................84.7
100.................100.

lets say a full reel holds 1000 parts and the reel is half gone, I think there should be if you look at the table about 385 parts left (well roughly)

Has anybody else had this problem? and can add something, it is almost 11:30pm in New Zealand now, I must go to bed and would be keen to see how my table fits in with the true case tomorrow when I go to work.

Interesting problem and a math solution would be cool. However I suspect that one needs to know how many parts per unit length are stored on the strip as well as the thickness of the component/strip, inside/outside diameter, etc.

I've never seen pro parts reels in operation but I would think that such a automated process would have to have a component counter that keeps track and is reset with each loading of a new reel and alarms or stops when reel is empty

Lefty

Pommie · 3rd October 2007, 02:52 PM

There cannot be a simple solution because as the hub gets bigger the relationship must become nearly linear. Imagine a real with a 10 inch hub and 1 inch of components. Compare that to a 1 inch hub with 10 inch of components.

Mike.

Richard Principal · 3rd October 2007, 05:03 PM

Just to answer Leftyretro first

Sceadwian · 4th October 2007, 05:44 AM

Aside from machine counting of deposited components shouldn't there be some kind of tick marks on the tape?

Richard Principal · 4th October 2007, 06:02 AM

N o

mvs sarma · 4th October 2007, 09:03 AM

It appears we can change approach,like this

we know how many jobs are completed after mounting a reel. we also know how many componenets are used per job.
we can start a counter at the time of loading a reel - for each component- acquire the data of jobs done from then till now use a formula and the balance is known-- Ofcourse there may be some dropouts or missed components- we can always give allowance for that.

Richard Principal · 4th October 2007, 01:29 PM

Now we seem to be getting into another problem,

**3v0** · 5th October 2007, 01:52 AM

You could easily setup a program to figure this out.

The program would calculate the carrier on the reel and use that to figure out how many parts remained.

The crude program (sort of). This is off the top of my head and could quite possibly have an error two. It is always a crapshoot when on posts untested code; even more so with this sort of thing.

Code:

// The inputs would be:
// ED: Empty hub diameter
// FD: Full hub diameter (not needed but interesting to know)
// PD: Partial diameter.
// DD: Delta diameter, diameter change per turn (carrier thickness)
// PPL: Part count per unit length of carrier

WD = ED;     // working diameter = empty diameter
LEN = 0;     // total length of carrier

while (WD < PD)  // ***
{
   LEN = LEN + pi*WD;  // circumference = pi * diameter
   WD = WD +  DD;      // adjust circumference 
}

return (WD * PPL)   // answer

*** This may end up being better expressed as (WD < (PD + (.5 DD)) or similar.

Measure the tape the same way every time. maybe measure out from the start of the tape on the inside hub.

As a check the program should be able to calculate the number of parts on a full reel. +- 1 turn

This all depends on the reels being uniformly tensioned.

j.p.bill · 5th October 2007, 02:29 AM

If you have a decent scale, weigh a full reel and an empty one. The difference in weight is equal to the components. Check with one component to insure you have adequate resolution.

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3rd October 2007, 11:29 AM	(permalink)
Richard Principal	Number of parts left on a reel? At the moment I operate a machine tha Last edited by Richard Principal; 31st October 2007 at 06:02 AM.

3rd October 2007, 02:52 PM	(permalink)
Pommie	There cannot be a simple solution because as the hub gets bigger the relationship must become nearly linear. Imagine a real with a 10 inch hub and 1 inch of components. Compare that to a 1 inch hub with 10 inch of components. Mike.

3rd October 2007, 05:03 PM	(permalink)
Richard Principal	Just to answer Leftyretro first Last edited by Richard Principal; 31st October 2007 at 06:03 AM.

4th October 2007, 05:44 AM	(permalink)
Sceadwian	Aside from machine counting of deposited components shouldn't there be some kind of tick marks on the tape? __________________ "Because I be what I be. I would tell you what you want to know if I could, mum, but I be a cat, and no cat anywhere ever gave anyone a straight answer, har har."

4th October 2007, 06:02 AM	(permalink)
Richard Principal	N o Last edited by Richard Principal; 31st October 2007 at 06:03 AM.

4th October 2007, 09:03 AM	(permalink)
mvs sarma	It appears we can change approach,like this we know how many jobs are completed after mounting a reel. we also know how many componenets are used per job. we can start a counter at the time of loading a reel - for each component- acquire the data of jobs done from then till now use a formula and the balance is known-- Ofcourse there may be some dropouts or missed components- we can always give allowance for that. __________________ Regards, Sarma.

4th October 2007, 01:29 PM	(permalink)
Richard Principal	Now we seem to be getting into another problem, Last edited by Richard Principal; 31st October 2007 at 06:04 AM.

5th October 2007, 01:52 AM	(permalink)
3v0	You could easily setup a program to figure this out. The program would calculate the carrier on the reel and use that to figure out how many parts remained. The crude program (sort of). This is off the top of my head and could quite possibly have an error two. It is always a crapshoot when on posts untested code; even more so with this sort of thing. Code: // The inputs would be: // ED: Empty hub diameter // FD: Full hub diameter (not needed but interesting to know) // PD: Partial diameter. // DD: Delta diameter, diameter change per turn (carrier thickness) // PPL: Part count per unit length of carrier WD = ED; // working diameter = empty diameter LEN = 0; // total length of carrier while (WD < PD) // *** { LEN = LEN + piWD; // circumference = pi diameter WD = WD + DD; // adjust circumference } return (WD * PPL) // answer *** This may end up being better expressed as (WD < (PD + (.5 DD)) or similar. Measure the tape the same way every time. maybe measure out from the start of the tape on the inside hub. As a check the program should be able to calculate the number of parts on a full reel. +- 1 turn This all depends on the reels being uniformly tensioned.

5th October 2007, 02:29 AM	(permalink)
j.p.bill	If you have a decent scale, weigh a full reel and an empty one. The difference in weight is equal to the components. Check with one component to insure you have adequate resolution.