RE: Cheese factor

Sorry the company would be roughly equivilent to 50 soldiers. I forgot
to
double the PA cost.

Company Commander      5
3x Cmd Squad	       3
3x Plt Leaders	       9
21x Plt Troops	      21
1x PA Plt Leader       3 
5x PA Troops	      10
		      --
		      50 Rough Comparison Points

---
Brian Bell
bkb@beol.net
http://www.ftsr.org/
---
-----

> -----Original Message-----
> From: Bell, Brian K (Contractor) [SMTP:Brian.Bell@dscc.dla.mil]
> Sent: Monday, April 02, 2001 9:22 AM
> To:	'gzg-l@csua.berkeley.edu'
> Subject:	RE: Cheese factor
> 
> 
> [Bri] I have often seen PA selected as roughly equivilent to 2 normal
> soldiers. Has anyone toyed with the idea of making command worth more
in
> this rough firguring? 
> An untested example would be to make Leaders = 1 soldier x command
level.
> So
> it would be: Squad Leader x2 soldier, Platoon Leader x3 soldier,
Company
> Commander x5 soldier (then 10x, 20x, 40x...). 
> Or Leader = 1 + 0.10 (round up) of the forces he leads (PA count as
2). So
> a
> Squad Leader would be 1 + .10 * 3 soldiers = 2 soldiers. Platoon (8)
> Leader
> would be 1 + .10 * 7 soldiers = 3 soldiers. A PA Platoon (6) Leader
would
> be
> 2 + .10 * (2 * 5 soldiers) = 3 soldiers. A Company Leader (1 PA
Platoon, 3
> Rifle Platoon, 1 Command Squad (4)) would be 1 + ((.10 * 6 * 2) + (3 *
> (.10
> * 8)) + (.10 * 3)) = 1 + 1.2 + 2.4 + .3 = 4.9 = 5 soldiers
> So taking a Company of 1 Command Squad, 1 PA Platoons and 3 Rifle
Platoons
> (34 soldiers) would count as 44.
> 
> 
> > Allan Goodall		   awg@sympatico.ca
> > Goodall's Grotto:  http://www.vex.net/~agoodall
> --- End Original Message ---
> 
> My comments above marked by [Bri]
> 
> ---
> Brian Bell
> bkb@beol.net
> http://www.ftsr.org/
> ---