The Hawker rigging upgrade reads as follows:

Hawker Rigging: One carried item is concealed and has no load. For example, you could carry a load of drugs or a weapon, perfectly concealed, for zero load.

Does this apply to items that require multiple load count to carry? For example armor requires 2 slots of total load. So in this case, could I use the Hawker rigging bonus to wear concealed armor for 0 load?


1 Answer 1


Any carried item, yes.

In contrast to most other crews, which apply their benefits to multiple items or multiple load, the Hawker crew lets its scoundrels both carry and conceal any one item for free when you get the rigging upgrade.

This could theoretically apply to any amount of Load, but most standard items don't go above 2 or so; at that point your GM's just inventing heavy things for you to sneak around.

Speaking of things GMs do - as a GM, that "carried item" stipulation looks important, though it doesn't have explicit rules text behind it. It also shows up in Smuggler rigging, which can conceal two carried items but doesn't impact Load. Most of the items in the game can be considered carried items, hanging out in your nebulous inventory until you're ready to use them, but there are a few that the "carried" distinction doesn't apply to.

You need to wear armor, for one. You can't use +heavy armor (chain and metal plates, a metal helm) to protect yourself in such a way that a search won't find it. The Slide's fine clothes and jewelry also need to be worn to be effective, though in their case they have a noted separate Load cost for treating them as a carried item (to change into and out of) rather than being worn, so Hawker Rigging can still apply to them.

  • \$\begingroup\$ Are there actual rules to support the carried vs worn distinction? Or is this more of a GM ruling in your eyes? \$\endgroup\$
    – Destruktor
    Nov 9, 2020 at 15:21
  • \$\begingroup\$ Have clarified - the carried vs. worn distinction is my own call, based on the assumption that "carried" means something, rather than nothing. \$\endgroup\$
    – Glazius
    Nov 9, 2020 at 16:47

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