Richard Principal
New Member
At the moment I operate a machine tha
Last edited:
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Richard Principal said: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.
// 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