What's probably known is the layout of the shortest-period output-capable minecart repeater:
z+1 z+0 z+0, track
#▼▼# #▲▲# #══#
Cycle time is 18 steps. It can only "output" its period to another cart standing on level floor just outside the pit. I was curios, so tried my hand at building a counting circuit for such a fast oscillator. It requires halving the frequency first, which is a bit fiddly in and of itself, but it can be done.
..▼
..▼ <- 19-step oscillator
.#═╗
╔═▼║ <- track in pit: ╗
▼+▼╝# ╔ ╚
▼ <- track:║
═╗
xx
The "feeding" oscillator is to the north, the main pulse splitter is the central part, to the southwest the connection to the next counting circuit. The splitter contains two minecarts. Everytime the oscillator sends a pulse, a cart stands on the floor/track tile just south of the oscillator pit and is pushed into the angled-track pit from the north. It accelerates (further) on the SW track ramp, then tries to climb the NE ramp to the east. If the second cart of the circuit is not currently in the exit tile, the cart climbs out and finishes the circuit, returning to the oscillator's output tile, which it reaches after about 13 steps, in time to take the next pulse.
If the second cart of the splitter circuit is in the angled-track exit tile, the moving cart passes its impulse on to that cart (which cycles over to the oscillator output), accelerates to the north and leaves the pit to the northwest, turns around to the south and enters the pit which feeds the _next_ counter circuit (xx). It pushes that circuit's awaiting cart standing at the southern exit, then accelerates back north, takes the ramp corner to the east and climbs to the _smooth floor_ tile next to the splitter circuit's pit. Since it comes from normal floor, the cart doesn't enter the pit, but jumps over it and remains in flight until it hits the wall behind the track corner, coming to a stop on the splitter pit's exit (instead of following the track corner north). This entire process takes about 30 steps, significantly longer than one period of the oscillator; the remaining cart will go one full cycle in the meantime and has just started on its second round when the "transmitter" cart arrives in its resting tile, triggering the next transmission.
Result: on every second pulse of the oscillator, a signal is passed on to the counter to the southwest. Timing is pretty tight: the normal "push and return" routine of the splitter cart takes 14 steps, it barely clears the area before the next pulse from the oscillator arrives. Startup must be done carefully, the first cycle in the splitter can end up too slow to catch the second pulse. The receiving counting circuit still must be very tightly constructed, not every design is capable of handling an input with a 36-step period. The reaction periods in the circuit i built weren't all 18, they ended up 17,18,18,19 (repeat). That must have been an effect of build order discrepancies between carts.
There are faster oscillators possible (fastest non-powered that i know of is 4,833 steps), but this should be the fastest one that can actually generate a "push" output without power.
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Topic: What's going on in your fort? (Read 6106420 times)