Here's a quote for you: "Most hard problems an alchemist may face can be better solved by a trip to bed and the bath than by spending the whole night staring at the alchemy bench." I found this to be true, as after I spent several hours on a timing issue with no real progress made, merely days before the deadline, I went to sleep. The next day, a simple idea presented itself to me in the bathtub... and it worked.

But first, let's talk about rate math. Rate math is all about ratios. *Normally*, you want nice ratios. Two iron makes a copper. Earth, wind, fire, and water make 1 chunky mors and some vitae. Figuring out what the min rate is requires figuring out what whole number of reagents makes what whole number of outputs, because the number of inputs you can take is capped at 1 per 2 cycles.

So, the critellium in the room: Critellium. I assign "points" to each metal: Lead is 1, Tin 2, Iron 4, Copper 8, Silver 16, and Gold 32. Purifying a metal thus doesn't change the total amount of points. For projection, we can think of quicksilver as an equivalent amount of points of the metal being projected—this makes it easy to see that it's better to project after purifying. 1 Critellium is therefore 144 points (plus 2 quicksilver), which is a nice round number that only has prime factors of 2 and 3.

Metal slag provides 7 points and 1 quicksilver. So for a single Critellium, we set aside 2 quicksilver and the rest can be used for projecting. Only 3 can be used for projecting silver to gold, and the rest would have to be for projecting copper to silver. So 9 slag would be 80 points of quicksilver alone... but alas, that's 143 points! 10 slag is 158 points, which is 14 extra. So we can count a whole number of outputs from a whole number of inputs as 10 slag plus 14*9 slag, i.e. 15 outputs and 136 inputs. So, min rate is 272/15.

You might think that's a particularly cursed ratio, but no. Critellium's madness involves even worse ratios to come. 15 critellia means 30 quicksilver, 45 gold, and 45 silver. So we split our 136 input quicksilver into 30 kept as quicksilver, 45 projecting silver to gold, and 61 projecting copper to silver. All of the metal we start with has to be purified, and the total copper produced from 136 each of lead, tin, and iron is 119 copper. That means, yes, 58 copper is purified, and 61 is projected. 58:61 is absolutely the worst ratio I have ever seen in this game. Thanks zorflax. You're the best.

Here's how this solution breaks down: 1 pipeline takes all the lead and turns it into copper. 136 lead becomes 17 copper. 2 pipelines separately turn 68 iron into 34 copper, and 2 pipelines separately turn 68 tin into 17 copper. One of those iron pipelines can use up all of its 68 neighboring quicksilver projecting the copper it makes all the way into gold, which accomplishes 34/61 Copper projections and 34/45 Silver projections, leaving 27 and 11. The lead pipeline and one tin pipeline produce copper at the same rate, so we can purify those together to get silver, and the remaining two pipelines produce 3 copper every 4 periods.

Oh, I forgot to mention periods. You see, it made sense to me to restrict to a low period to aim for a good instructions secondary, because I get to avoid using repeat instructions. I have no idea yet whether it worked, but it sure was a challenge to make!

We bond these metals together into a chain of triangles, each with 1 silver, 2 gold, and 3 copper. From our 136 inputs we make 17 triangles. A single triangle like this can use six quicksilver for 4 projections and 2 leftover, i.e. 1 critellium. But our math indicates we can only do that for 9 triangles of 17, and that uses up almost all our remaining projections. What's left for our 8 remaining triangles is 2 silver projections, making 6 critellia total.

9:8 is a peculiar challenge, especially given we have enough quicksilver to attach 6 to 2 of every 3 triangles. 3 and 17 are relatively prime, so it's better to do math on 51 triangles.  I was able to configure the machine to attach the 6 quicksilver to the first and third triangle in each loop, and now the mess begins.

Construction pipeline 1 (lower right) takes all the "#1" triangles and projects them, and to satisfy the math it needs 10 of the "#3" triangles. It has to move things through quickly because it could operate twice in a row, so it has to have a "period" of 4 periods. Construction pipeline 2 takes all the "#2" triangles and 7 of the "#3" triangles. Each triangle in a set position is easy to pick out, as it's 12 periods between each loop. 7:10 for the other is not a great ratio. Fortunately, there's enough delay before triangle #3 that I can delay pipeline 1 to grab only if pipeline 2 hasn't. So really we only need to select something 7 in 17 times, with 12 periods in between.

Piece of cake. Just make a track loop of length 204 and put 7 arms on it in specific places.

Construction pipeline 1 then became a simpler problem with a fixed input (the winged metal triangle) and a critellium output. Pipeline 2, however, wound up with all the rest of the complexity. The 24 triangles it's responsible for have to be turned into 18 outputs, and 6 silver has to be projected. This is a little more tractable considering 4 triangles makes 3 outputs with one projection, and from there it's not too too bad to make a brick with 4 triangles and 7 quicksilver.

I made this pipeline with a "period" of 6 periods because I had the time between bricks and it saves on instructions, but an interesting thing happened just when the pipeline was about to repeat: the last quicksilver 7-stick collided with the metal. This was the problem I solved by taking a long break from it.

Essentially, every fourth triangle causes the metal extrusion to be broken off into a brick. If it can be either of triangle #2 or #3, which occur 4 periods apart, then on some loops the brick doesn't advance fast enough or it arrives with the later half of the pipeline in the wrong state. My semi-arbitrary choice of which of the 17 #3 triangles to take needed a revision. I moved two arms... and that's it. No additional arms or translations or rotations to add to my existing design. Now it's always triangle #2, so it's guaranteed to happen at the same pipeline state.

Finally. Time for some good rest.