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Posted (edited)
On 5/17/2026 at 7:31 AM, Bzhydack said:

Nectar and Brine has the same density too.

Just checked the table, turned out there's more liquids with the same density, and they all can stack on top of each other layers by layer.

  • Nectar and salt water (not brine, they are a little bit heavier)
  • Sucrose and sulfur
  • Lead and gold
  • Iron and aluminium
  • Copper and steel
  • Tungsten and Iridium (and uranium)

I would like to ultilize this to make a fish tank with least required liquid and a bit less efforts, instead of multiple liquids layer i can use phyto oil and biodiesel. But i don't like the idea that they will collapse in the next patch :D

Edited by MinhPham
Posted (edited)

This is cool. It's like a better version of how O2 and PO2 have the same density. It gets tedious hunting down every last gas bubble with deodorizers.

 

On 5/16/2026 at 10:59 PM, MinhPham said:

I would like to ultilize this to make a fish tank with least required liquid and a bit less efforts, instead of multiple liquids layer i can use phyto oil and biodiesel. But i don't like the idea that they will collapse in the next patch :D

If you want to make a super low effort aquarium at the cost of space you could just make 1 giant tank that takes up a good portion of the map. Given the math from the other post. The tank does need to be temp controlled at 10-25C like you said so the fish don't die off from morphs. Or you could go a different approach and make 3 giant tanks that account for morphs. All of them require that you sweep the eggs out anyways.

 

The 1-tank way

So at 9 pacu before overcrowding misery [accounting for -1 happiness to temps] in 8 tiles of full liquid, we can add 1 more for every 8 tiles. The equation for how many tiles of liquid you need per pacu is now [8 + ((X-9) * 8)] This is simplified to [8X-64]. Each dupe needs 25 pacu to sustain themselves on raw filet alone. That is 136 tiles for the first dupe and +200 tiles for every subsequent duplicant. [25,50,70 -> 136, 336, 536]

The 3-tank way

Now it is 10 pacu before overcrowding and better range of temp control so slightly more efficient, but you also need to use 3x the space to account for morphs whether or not they are born at all. You will need insulated tiles between each tank and put them in order of temps to limit conductivity. The equation is now [24 + ((X-30 ) * 24)] or [24X-696] tiles per pacu. The equilibrium point between the 1-tank and 3-tank system is [24X-696 = 8X - 64] or 39.5 pacu after which the 3-tank system requires more space per fish. That is 24 tiles for the first dupe and +600 tiles for every subsequent duplicant.

 

Pros of 1-tank and 3-tank

- Less sweeper/loader and power used for the bottom of each tank [Use a thin tall tank to maximize refined metal efficiency].

Cons

- Space inefficient

 

If you want to be more efficient with space then you would have to split them up into 8 tile segments.

 

Pros of 8 segment pacu patties

- Space efficient

Cons

- Expensive in refined metal as each patty requires its own sweeper/loader and power.

 

tldr;

I think the best way is to have 1 large tank that is temperature controlled. Should be easier to do this since the thermal mass is so low if you stack a few grams of liquid at a time.

 

in my case with 6 dupes, that's 1136 tiles of liquid 😵. I think I'll stick with the 160 cycles it takes to print hatches. At least I can limit them to one tile.

I might build an aquarium later using this.

Edited by BLACKBERREST3
Posted (edited)

Just thought of a way to bolster the 3-tank system. If you could intelligently transfer each egg to the correct tank with a critter sensor and buffer space to account for X amount of pacu that scales with the amount you currently have, you could make use of all of the tiles of space a three-tank system offers minus the buffer. ex. wrong pacu egg is laid in Tank A, transfer a pacu that would normally be in tank B or C back to tank A to make up for the loss. The math involved uses probability so you could use egg chance to figure out a reasonable buffer space here. I guess the logic would look something like this:

Tank A B C -> A is gulp ->  B is pacu -> C is tropical

6 Scenarios where X laid Y egg. 2 of those scenarios have 2 degrees of separation and the other 4 have 1 degree of separation.

A -> B, send 1 B egg back to A

A -> C, send 1 C egg to B AND 1 B egg to A

B -> A, send 1 A egg to B

B -> C, send 1 C egg to B

C -> A, send 1 A egg to B AND 1 B egg to C

C -> B, send 1 B egg to C

If you were only after 1 morph then the buffer would just be the sizes of the other two tanks.

 

I don't think liquids of same density are changing anytime soon btw.

Edited by BLACKBERREST3

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