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Having trouble in designing logic gate


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Hello, I was designing my Puft farm, and hit the wall in logic gate design.

Please refer to attached image. I want my puft to stand by in the left room until right room is empty, and open the door when right room is emptied, and then keep the door open until right room is occupied by puft just moved.

To get it clear, I want following logic results when critter sensor is set to send ON when critter number is above 0

A   T  T  F  F

B   T  F  T  F

C   F  T  F  T

Could anyone help me with earning this logic result using ONI automation system? 

 

 

P.S. I used to connect AND gate result of A and NOT-B into C, but I found puft sometimes stuck in the door time to time, so I want the door keep open until puft completely arrive to next room.

gate.jpg

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1 hour ago, GnyoR said:

Hello, I was designing my Puft farm, and hit the wall in logic gate design.

Please refer to attached image. I want my puft to stand by in the left room until right room is empty, and open the door when right room is emptied, and then keep the door open until right room is occupied by puft just moved.

I do with Hatches via a Buffer and Not gate. 

B (set to above 0) - > Buffer (set for however long you want/need to delay reopening the door) - > Not - > door. (Going Sensor to Not to a Filter results in the same effect) 

Once the Puft arrives in the room with sensor B, the door will close and remain closed for however long the Buffer is set to after the Puft leaves that room. In my experience (with Hatches) sensor A is useless for this specific door travel. 

Here's an older version (I use weight plates now) of my Hatch ranch showing 3 of these door setups in a row to serve as a Pez dispenser for dropping Hatches into the ranches I need them in. I do 3 cause I've had problems in the pat of 2 (or more) Hatches dropping. 3 doors seems to prevent this. Considering how slow Pufts move, this might be a concern for you if you ever get multiple Pufts on the A side. 

Spoiler

5db6481a9ad88_PezMultiLevel-Overview-Automation.thumb.jpg.cb7d85ae8b0d47d7504a3b172661d12b.jpg

 

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3 hours ago, GnyoR said:

To get it clear, I want following logic results when critter sensor is set to send ON when critter number is above 0

A   T  T  F  F

B   T  F  T  F

C   F  T  F  T

With the complete truth table you can get the reducted form of your equation and will give you what operation (gates) you need to use here link

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4 hours ago, beowulf2010 said:

Considering how slow Pufts move, this might be a concern for you if you ever get multiple Pufts on the A side. 

@beowulf2010 Thank you for sharing your design. I love it's simplicity and it seems perfectly working as I intended. It seems that I had some kind of compulsion that closed door should be a default state. And it turned out to be groundless. For pufts, even filter gate in your design doesn't need to exist. I'm quite happy with the simple solution.

 As you mentioned, 2 or more pufts in same room makes some malfunctioning, but it seems to be ignorable problem if hatching eggs have some time interval with each other.

1 hour ago, Edoc_ said:

With the complete truth table you can get the reducted form of your equation and will give you what operation (gates) you need to use here link

@Edoc_ This is exactly what I wanted to get initially. Thank you very much for the help. Very happy to find a strong tool for designing automation.

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Np @GnyoR. I'm not great with the complex automation, but I tend to notice when a "simpler" setup would work just as well. 

And yes, if you cam guarantee a 1 at a time feed to the door (incubator hooked up to sensor A maybe), you will only need 1 door. 

I need the 3 since I just dump all Stone Hatch eggs in a room just outside the dispenser so I can't guarantee one at a time hatching. 

 

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I’m sure this doesn’t answer your question but according to the logic table all you need it the not of b as the changing a doesn’t seem to give you a different result at c 

I might be reading it wrong but it looks like you are saying c is only true when b is false regardless of the position of a 

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On 5/5/2020 at 10:58 PM, peachkillu said:

I’m sure this doesn’t answer your question but according to the logic table all you need it the not of b as the changing a doesn’t seem to give you a different result at c 

I might be reading it wrong but it looks like you are saying c is only true when b is false regardless of the position of a 

Oh, you're absolutely right about that. I can't believe that I couldn't realize that. And that very well explains why @beowulf2010's simple solution just fitted my intention.

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