1

I am trying to both take items out of a chest and put the same type of item into the chest so that it will always have less than a full stack of the item but never run out of it either. For this I am using one slice of a standard item sorter with a chest on top keyed to one type of item rather than a hopper and a chest full of the item above the keyed one which feeds into the keyed chest through another hopper. I fill all but one slot with a filler item and some number of the relevant item between 0 and a full stack. When the item reaches 0 or some lower limit the comparator should drop one signal level, unpowering the hopper above the chest and filling it back up till the comparator gives the neutral output again. If the item reaches a full stack or some upper limit the signal should go up one, allowing me to activate the hopper below to funnel the excess into storage.

Is there some combination of filler items which will allow the upper and lower bounds to be within one stack of one another?

  • I didn't understand the explanation of your circuit, a screenshot would be helpful. The important part is: You want three different comparator output levels within one stack of items, correct? – Fabian Röling Jul 14 at 7:45
0

The power levels of a comparator are spread evenly over the space in a chest. Since there are much more than 15 slots in a chest, you can't have three different comparator outputs from changing just one slot, not even with items of stack size 16 or 1.

What you could do is using a hopper, hopper minecart or furnace (you can put any item in, just be careful not to shift-click something smeltable and also one of the surprisingly many fuels).

The Minecraft wiki has a good table for the amount of items in different containers and their comparator power levels: https://minecraft.gamepedia.com/Redstone_Comparator archive

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.