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ochoin
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by ochoin » Mon Oct 26, 2020 3:52 am
Partly because I need to consolidate my Polybian Roman army (mostly HaT & Zvesda figures) and partly because there was a Newline sale on, I bought another Roman legion : which make a fourth & final one.
The figures painted so far are Hastati. I've been thinking of giving them a Syracusan Greek ally as I have some units of Greek spear & slingers painted from years ago. I'll need to re-visit Newline to bring the slingers up to strength & buy some thureophori & a general.
The Hastati:
8 Triari, 8 Velites, 16 Principes (including command) to go.
donald
grizzlymc
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by grizzlymc » Mon Oct 26, 2020 10:07 am
Excellent. Now you can play with yourself and learn how to defeat them.
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by Essex Boy » Mon Oct 26, 2020 10:29 am
Very impressive, Donald.
The bases are a little cornery for my taste, but the whole is a splendid sight.
Iain
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by grizzlymc » Mon Oct 26, 2020 11:08 am
Ocho, you know the kewl kidz are using dodecagonal baser.
ochoin
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by ochoin » Mon Oct 26, 2020 11:35 am
I appreciate the feedback. The FoG rules are quite hidebound on mathematical precision.
Indeed, in the Index pages, on basing, the rule book states:
Let f(x)=log(1+|x|) for x∈R and g(x)=log(1+x) for x>−1
Then:
limx→0+f(x)−f(0)x−0=limx→0+g(x)−g(0)x−0=g′(0)=1
and
limx→0−f(x)−f(0)x−0=limx→0−g(−x)−g(0)x−0=−limx→0−g(−x)−g(0)−x−0=−g′(0)=−1
Can't argue with that.
donald
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by grizzlymc » Mon Oct 26, 2020 11:40 am
Tell me that you are joking, please!
BaronVonWreckedoften
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by BaronVonWreckedoften » Mon Oct 26, 2020 3:21 pm
Don't forget the optional ursine intercession rule - if all else fails, shoot the bear.
Kein Plan überlebt den ersten Kontakt mit den Würfeln. (No plan survives the first contact with the dice.)
Baron Mannshed von Wreckedoften , First Sea Lord of the Bavarian Admiralty.
Essex Boy
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by Essex Boy » Mon Oct 26, 2020 8:35 pm
ochoin wrote: ↑ Mon Oct 26, 2020 11:35 am
I appreciate the feedback. The FoG rules are quite hidebound on mathematical precision.
Indeed, in the Index pages, on basing, the rule book states:
Let f(x)=log(1+|x|) for x∈R and g(x)=log(1+x) for x>−1
Then:
limx→0+f(x)−f(0)x−0=limx→0+g(x)−g(0)x−0=g′(0)=1
and
limx→0−f(x)−f(0)x−0=limx→0−g(−x)−g(0)x−0=−limx→0−g(−x)−g(0)−x−0=−g′(0)=−1
Can't argue with that.
donald
I bet I could argue with it. I have a gift in that respect.
ochoin
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by ochoin » Mon Oct 26, 2020 11:30 pm
You do calculus?
I actually got that off the boy-genius who uses it in his job.
Something to do with corners, he tells me.
donald
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by Essex Boy » Mon Oct 26, 2020 11:49 pm
ochoin wrote: ↑ Mon Oct 26, 2020 11:30 pm
You do calculus?
I actually got that off the boy-genius who uses it in his job.
Something to do with corners, he tells me.
donald
Nope. I do arguments.
The boy-genius is a credit to his parents.