Hastati

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ochoin
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Hastati

Post by ochoin »

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:
Image

Image

Image

8 Triari, 8 Velites, 16 Principes (including command) to go.

donald
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grizzlymc
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Re: Hastati

Post by grizzlymc »

Excellent. Now you can play with yourself and learn how to defeat them.
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Essex Boy
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Re: Hastati

Post by Essex Boy »

Very impressive, Donald.

The bases are a little cornery for my taste, but the whole is a splendid sight.

Iain
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Re: Hastati

Post by grizzlymc »

Ocho, you know the kewl kidz are using dodecagonal baser.
ochoin
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Re: Hastati

Post by ochoin »

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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grizzlymc
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Re: Hastati

Post by grizzlymc »

Tell me that you are joking, please!
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Re: Hastati

Post by BaronVonWreckedoften »

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.
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Essex Boy
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Re: Hastati

Post by Essex Boy »

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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Re: Hastati

Post by ochoin »

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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Re: Hastati

Post by Essex Boy »

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.
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