the shape of things to come

OmniVehicles: The type of armor, its weight, number (and
allocation) of points, and any reduction of weapon and equipment
slots required must be established when designing an
OmniVehicle’s base confi guration, and may not be altered in its
completed primary or alternate confi gurations.
Michael’s Hephaestus has 16 tons
remaining. Given its weight, he computes
that the maximum armor he can place
on the vehicle is 145 points ([3.5 x 30 tons]
+ 40 = 145 points). Mounting that many
points—even using Clan ferro-fi brous armor—would cost
him 8 tons, however (145 points ÷ [16 Base Points per Ton
x 1.2 Clan Ferro-Fibrous Multiplier] = 7.552, rounded up to
8). Unwilling to go quite that far, Mike decides to install
only 5 tons of Clan ferro-fi brous armor, which gives him 96
armor points to allocate instead (5 tons of armor x 16 Base
Armor Points per Ton x 1.2 Clan Ferro-Fibrous Multiplier =
96 points).
Michael decides to distribute most of this armor toward the
front and sides of the vehicle, where he feels it is most likely
going to be hit, and so he places 20 points of armor each in
the Front, Left Side and Right Side locations. The remaining 36
points he divides evenly among the Rear and Turret locations
at 18 points apiece. Verifying that all points are accounted
for, Mike adds all the armor values together: 20 [Front] + (20
+ 20 [Left Side, Right Side]) + 18 [Rear] + 18 [Turret] = 96. In the
appropriate non-shaded locations of the Hephaestus’ Armor
Diagram, Michael blacks out all extra circles, leaving only the
assigned amount for each location.
With 5 tons spent on armor now, the Hephaestus has only
11 tons remaining for weapons and other equipment (16 – 5
= 11). Michael also notes he has 1 fewer item slots remaining
in which to assign equipment, thanks to his use of Clan ferrofibrous
armor, meaning that his vehicle can only mount 10
items now (11 slots – 1 item = 10 slots).
Still looking to keep costs low and
preserve as much weight for weaponry
as possible, Nathaniel chooses to mount
3.5 tons’ worth of standard armor on his
Yellow Jacket. Like Michael’s Hephaestus,
the 30-ton VTOL has a maximum armor potential of 145
points ([3.5 x 30 tons] + 40 = 145 points), but using standard
armor would cost more than 9 tons to accomplish
(145 points ÷ [16 Base Points per Ton x 1 Standard Armor
Multiplier] = 9.06, rounded up to 9.5 tons).
The 3.5 tons Nathaniel chooses provides 56 points of
armor (3.5 tons x 16 Base Points per Ton x 1.0 Standard
Armor Multiplier = 56 points). He chooses to mount the
majority of this armor on the Yellow Jacket’s nose, with 20
points to that location. As the rotor location cannot receive
more than 2 points of armor, Nathaniel places the maximum
of 2 points there, before mounting 13 armor points
on each of the VTOL’s sides and 8 points on the rear. Verify-
ing his numbers, he adds up the armor points: 20 [Nose] +
(13 + 13 [Left and Right Sides]) + 8 [Rear] + 2 [Rotor] = 56
points. Nathaniel blocks out all the excess armor points in
each location on the Yellow Jacket’s Record Sheet.
Because he used standard armor rather than ferrofi
brous, the Yellow Jacket sees no loss in item space on its
inventory. It now has 17 tons of unspent weight remaining
(20.5 tons – 3.5 tons = 17 tons).
At 75 tons, Patrick’s Zhukov could
mount as much as 302 points of armor
([3.5 x 75 tons] + 40 = 302.5 points, round
down to 302), but doing so would cut
deeply into the 43.5 tons remaining to be spent. Pat thus
decides to devote 11 tons to armor, and chooses standard
armor to keep costs low. At 11 tons, the Zhukov will receive
176 armor points (11 armor tons x 16 Base Points per Ton x
1.0 Standard Armor Multiplier = 176 points).
Patrick distributes his armor points with a heavy
emphasis on the Front and Turret locations, where he
places 44 points each. On the sides, he assigns 32 points
each, before allocating the remainder—24 points—to
the rear. Verifying his numbers, he adds up the assigned
armor points: 44 [Front] + (32 + 32 [Left and Right Sides]) +
24 [Rear] + 44 [Turret] = 176 points. On his Vehicle Record
Sheet, Pat blacks out all the excess points for each of the
Zhukov’s non-shaded armor locations, based on these
fi g u r e s .
With 11 tons spent on armor, the Zhukov now has 32.5
unspent tons remaining (43.5 – 11 = 32.5). Because Pat
chose standard armor over ferro-fi brous, the vehicle also
retains all 20 of its original item slots.
Rick decides that his 100-ton Neptune
can aff ord to spend 14.5 tons on standard
armor, which will yield a total of 232
points (14.5 armor tons x 16 Base Points
per Ton x 1.0 Standard Armor Multiplier = 232). Just to
be sure, however, he verifi es that he is not exceeding the
vehicle’s maximum armor potential, and fi nds that the
submarine could take on as much as 390 points in all ([3.5
x 100 tons] + 40 = 390 points).
With no turret to place armor on, Rick has only 4 facings
among which to distribute his 232 armor points. He assigns
78 points to the front, while the sides receive 58 points
each and the rear receives 38. Summing up, he verifi es the
armor is fully assigned: 78 [Front] + (58 + 58 [Left and Right
Sides]) + 38 [Rear] = 232 points. On the Neptune’s Record
Sheet, he then blacks out all excess armor circles in each of
the aff ected locations, making sure to black out all armor
circles for the turret.
Because Rick used standard armor rather than ferrofi
brous, the Neptune does not lose any item spaces from its
inventory. It now has 23.5 tons of unspent weight remaining
(38 tons – 14.5 tons = 23.5 tons).