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Old May 1st, 2004, 07:47 AM   #1
Thyrn
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help with a sorting problem

My problem is a have two sensors one high and one low the convylor keeps moving. and i have three plungers only using two high part add a low part need to be sorted having problems with the three loops a wrote keep runing at the same time.

the ladder i have done is this:

start/stop starts convayor

first loop
see if high sensor was hit latch a high bit
if high sensor bit is not latched and low sensors was hit latch low bit

high bit latched start rto timer

low bit latched start rto timer

On the DN bit of both rto start a ton timer for plumger

On the dn bit of the Rto for high bit latch the plunger A

On the dn bit of the rto for the low bit latch the plumger B

On the DN bit of the ton unlatch all and res all rto

this is one loop i need to make it work with three loops so the parts keep sorting with out stoping can i get some help please
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Old May 1st, 2004, 09:51 AM   #2
Terry Woods
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Thyrn...

It sounds like a rather simple problem, but it is really hard to develop the question, let alone develop the answer.

Please, take another shot at describing the purpose of this "system".

Then, describe the physical layout and relative placement of hardware.

Then, take another shot at describing the sequence... showing "cause" and "effect".

(026)
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Old May 1st, 2004, 10:06 AM   #3
Thyrn
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ok

The program is to sort high parts and low parts on a moving convayor
there are two sensors and three air hammers.

the parts are put on the convoyer and pass though two light sensors

one high and one low , the high part will hit both sensors the low will hit only one sensors

The hight sensor is first the second sensor is the low.

then the 2 air hammer are to hit the part off the convayor in the right bin one for high part one for the low part

Also the parts keep coming

having problems making keep sorting
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Old May 1st, 2004, 10:23 AM   #4
Terry Woods
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Are both sensors at the same horizontal position? That is, is the upper sensor physically located directly above the lower sensor?

It sounds like you simply need to indicate to the first hammer whether or not to "kick" the part into the bin. If not, then let the part pass the first hammer. Then, at the second hammer, if a part is seen at all, then it must belong in this bin because it doesn't belong in the first bin.

Surely there is more to this than you have said.

(040)
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Old May 1st, 2004, 10:26 AM   #5
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the first sensor is about 6 inches form the second sensor
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Old May 1st, 2004, 10:31 AM   #6
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now it look simpel

the first condition.The space between the parts and the conveyor speed must be enable the hammer to go in and out without hit another part.
Logic
1.High sensor ON and low sensor ON hammer 1 hit for x sec(hit and go back without to touch another part)
2.Low sensor ON hammer 2 hit for x sec

If you canot kick it out where the sensor located then we have to use shift register.
How it look like?
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Last edited by ArikBY; May 1st, 2004 at 10:33 AM.
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Old May 1st, 2004, 10:39 AM   #7
Terry Woods
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Does this line process only those two parts?
If so, could it be arranged for the conveyor to dump into the second bin (without needing a hammer) ?
How big are the parts?
What is the gap between parts?
If the parts are smaller than 6", is it possible that one part might be in front of the first sensor and another part in front of the second... at the same time?

(060)
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Old May 1st, 2004, 10:42 AM   #8
Thyrn
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I wish it was that simple

the convayor does not stop moving. the 2 air hammers are at the end of the convoyor

the parts are put on the convayor one after the other non-stop.

how do i get the ladder to keep sorting the parts.

when the first part passes the high senors than it hits the low sensor and continces to the air hammers, but in a second or two another part is coming and it hits the first senoser before the first part hits the second. Also a low part will never it the hight sensor.

Please help
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Old May 1st, 2004, 10:45 AM   #9
Doug-P
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Build a shift register

When a part is detected, latch your size bits and start a timer (RTO1) set to the time it takes to get from induction to pusher #1 (P1). When RTO1 is done, a. actuate P1 if needed, b. start RTO2, which is set to the time from P1 to P2, c. use size bits to condition a second set of bits, d. reset RTO1. When RTO2 is done repeat a. through c. to decide whether to actuate P2. Repeat for P3.

I gotta ask, school project?
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Old May 1st, 2004, 10:50 AM   #10
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Thyrn

It look like the part is too close one each other.That what you mean?
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Old May 1st, 2004, 10:56 AM   #11
Terry Woods
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Do you have sensors at the hammers?

If not, how can you tell when a part is in front of a hammer?

Does the conveyor "index" or does it run continuously?

Is it possible for the parts to move on the conveyor? That is, once a part is placed on the conveyor, is it possible that vibration might cause the part to slip or slide a little?

C'mon Thyrn... I'm just trying to define the problem. We can't see what you see. We don't know what you know. We can't help you develop an answer if we can't "see" the problem.


(077) Surely you must have it by now, Eric?

Last edited by Terry Woods; May 1st, 2004 at 11:09 AM.
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Old May 1st, 2004, 12:11 PM   #12
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I usually don't like to wrestle with undefined (or at least, not completely defined) problems. But... since Thyrn hasn't been able to explain why this particular problem is so difficult...

It appears to me that the simplest answer to the problem is to locate the sensors at the ejectors.

Place the High sensor at the first ejector and the Low sensor at the second ejector.

If the sensors can be placed like that then there would be no need for any kind of shift register management.

If the High sensor "sees" a High-Part then, eject the part into the High-Bin.

Low-parts will pass the High sensor without being sensed.

Any part that shows up at the Low sensor must be a Low-part and is therefore ejected into the Low-Bin.

Thyrn, is there anything to prevent the implimentation of this solution?

(096)
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Old May 1st, 2004, 12:52 PM   #13
Thyrn
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i will send you my ladder work i have mybe that will help
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Old May 1st, 2004, 12:56 PM   #14
Terry Woods
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I don't have a "Ladder Reader" at home. If I did, it probably wouldn't be for your type of PLC.

Can you please answer the questions I asked?

Showing the ladder code does not explain the physical constraints you are under.

(140)
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Old May 1st, 2004, 01:03 PM   #15
Thyrn
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the code is in a word doc i scan in.


the problem is this I have high part and low parts that have to be sorted by sizes.

the two sensors can not be moved.

the convoyor does not stop at all.

the parts are put on the belt one at a time every 2 secs.

the part will hit the high senors first the contunce down the converyer and hit the low sensor, by the time it its the low sensor the next part is hitting or passing the high sensor.
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