That is hard to beat. Maybe you can do something with this H to 3G that produces a
forward glider:
Code: Select all
x = 142, y = 93, rule = B3/S23
82b2o$82b2o7$61b2o$60bo2bo$61b2o7$65b3o$66bo$66b3o12$112bo$111bobo$
111bobo$112bo12$121b2o$121b2o2$103b2o$102bobo$102bo$11bo89b2o$11b3o
118b2o$14bo116bo2bo2b2o$13b2o116bobo4bo$112b2o18bo5bob2o$111bobo21b2ob
obo$111bo23bo2bo2bo$110b2o20bo4bo2b2o$132b5o$4bo$4b3o23b2o102b2obo$7bo
21bo2bo101bob2o$6b2o22b2o8$16b2o$16b2o9$8b2o$3b2o2bo2bo$3bo4bobo$2obo
5bo$bobob2o$o2bo2bo$2o2bo4bo$5b5o2$4bob2o$4b2obo!
Almost stable.
The beehive needs to be rebuilt with the two backward gliders...
Edit:
I just found this way of regenerating the beehive with a "reset" Herschel:
Code: Select all
x = 31, y = 20, rule = B3/S23
29b2o$29b2o3$b3o$bo$2o2$8b2o$7bo2bo$8b2o3$5b2o$5b2o3$12b3o$13bo$13b3o!
Amazingly, the reset circuitry (just one added block) does not interfere with the H to 3G.
The reset is actually from your H to G minus one block, it creates a beehive.
The second (ghosted) Herschel needs to appear 132 tics or more after the first.