Good afternoon and tag der deutschen einheit.

I'm using a LOGO in conjuction with a Vision System that is controlling the production of a machine.

When the camera gets a trigger, it takes a picture and if the part that it's being produced it's not ok, then it sends a signal to a LOGO's input.

If the part is ok, then it sends a signal to another LOGO's input.

Working this way, I can setup a bad-in-a-row counter that increases by 1 when there's a not ok part and reset this counter when a good part is produced.

If we do have like 3 or 5 bad parts in a row we can just stop the machine with one of the LOGO's relay outputs.

The thing is that I would like to also control if there has been a lot of bad parts during the last 100 parts produced (or 50).

It can happen that the machine is doing a ok-nok-ok-ok-nok-nok-ok-nok-nok-ok series... and then my "Stop when there are 5 bad parts in a row" will never work.

Is there any way of storing the last signals from the camera and stop if there has been more than x parts nok?

Like having an array that dismisses the first value and stores the last 100 each time a signal from the camera is received?

Thanks in advance for any help that you can provide to a completely newbie.

Greets from Barcelona

Hello AdolfoMallenco,

as I understand it, you only want to evaluate the last 100 (50) reviews of a property. The 101 (51) would fall out of consideration, wouldn't it ?

In my opinion, the counter modules would not lead to the desired goal. This can be done with a shift register. But then you have to say goodbye to the idea of the last hundred entries.

If you use the shift registers of the Logo, only a maximum of 32 acquisitions can be evaluated (4x 8-bit shift registers).

If you build this up conventionally (see program), you will exceed the maximum number of markers in a logo with 100 evaluations. But then there is no more space for the rest of the program. Even with only 50 evaluations, there is very little space left.

Ella_68

 Hi, Thanks a lot for your answer.

ixo65

Why so complicated?

Just make another counter that only counts the bad parts.

 The goal is to control the production and stop the machine in case anything goes wrong.

Doing the 5-in-a-row bad parts is fine if there's a machine's operator next to it:

Example of 5 in a row bad production

But during the night, I need to check that the production is not doing something like this:

Example of Never-5-in-a-row bad scenario

I want to stop as soon as there are more than x bad parts during the last 32 (maybe 10 bad parts of the last 32).

Ella_68

Hello AdolfoMallenco,

as I understand it, you only want to evaluate the last 100 (50) reviews of a property. The 101 (51) would fall out of consideration, wouldn't it ?

In my opinion, the counter modules would not lead to the desired goal. This can be done with a shift register. But then you have to say goodbye to the idea of the last hundred entries.

If you use the shift registers of the Logo, only a maximum of 32 acquisitions can be evaluated (4x 8-bit shift registers).

If you build this up conventionally (see program), you will exceed the maximum number of markers in a logo with 100 evaluations. But then there is no more space for the rest of the program. Even with only 50 evaluations, there is very little space left.

Ella_68

Hey, thanks a lot for your assistance.

I'm still having a headache when I look into your example 

It's really impressive, but even clicking on I1 8 times in a row (or I2 8 times in a row) it always ends triggering M1.

I'm going to check information about the shift registers of the LOGO! to see if I can use it in my aplication.

Greets from Barcelona

I have tried the shift register and it looks like just what I was looking for.

Once again, thanks for your tips and assistance.

I have two more questions:

How can I increase the number of registers from 8 to 32? Using 4 registers?

And how can I count the number of 1 or 0 bits of a register?

If I've 8 bits, 4 of them are 1's, is there any function to get how many of them are 1s or zeros?

Any help would be truly appreciated.

Greets from Barcelona

That is absolutely correct. The counter is reset with marker M1.

With B003 the count is converted into an analog value. The differential threshold switches (B004, B007, B011 ...) then react to this analogue value. These only react to a specific counter value and enable the setting or resetting of the RS elements (B006, B008, B012, ...). The rest of the program adds up all the RS elements set by I1. Subtracting these sums from the total gives the value of I2.

Eso es absolutamente correcto.El contador se pone a cero con el marcador M1.

Con B003 el contaje se convierte en un valor analógico.Los interruptores de umbral diferencial (B004, B007, B011...) reaccionan entonces a este valor analógico. Estos solo reaccionan a un valor de contador específico y permiten configurar o restablecer los elementos RS (B006, B008, B012, ...). El resto del programa suma todos los elementos RS establecidos por I1. Restar estas sumas del total da el valor de I2.

Ella_68

(translated with Google)

Take a look at the attached program.This is built with shift registers.

Ella_68

 Hello Ella_68

Ella_68

Take a look at the attached program.This is built with shift registers.

Ella_68

Thanks for the example.

It's massive! I'm going to study it.

Thanks again

Hi AdolfoMallenco,

please attach your files as an attachment and don't use ibb as a filesharing hoster. I can't open and check it, because the website is blocked. 

If you want to attach more than one file, you can create a zip file to attach more than one file. 

Best regards,
Jen_Moderator

Ella_68

Take a look at the attached program.This is built with shift registers.

Ella_68

 Hello Ella_68

Sorry to bother you, but there's something I haven't been able to figure out by myself: Once there are x bad pieces, I can attach a coil to AQ1 in your example to stop the machine.

Once the operator fixes the problem, is there a way to reset all shift registers to zero? It looks like the only way would be to turn off the LOGO.

Thanks again for any hint or tip that you could provide.

Greets from Barcelona