Kanluwen said:
macd21 said:
Again, it depends on the realities on the ground. However if a SM refuses a reasonable demand from an Inquisitor then he better have some Inquisitorial allies in his back pocket or else his fellow marines will be ordered to execute him. A smart Inquisitor makes sure that his demands are reasonable.
Of course, a really smart Inquisitor asks nicely. A reallly really smart Inquisitor retains a lot of his fellow Inquisitors as allies, so that if the SM complains to the Lord Inquisitor the response will be "do as your told and stop whining". The point is that there is nothing unusual or unreasonable about an individual Inquisitor having a couple of marines to serve him.
Ehhhhhhhhhhhh. The Astartes don't complain to "a Lord Inquisitor".
They petition the High Lords of Terra directly, which includes an Inquisitorial representative whose power is solely based on the good will of the Astartes.
Why would the Inquisitor's power be based on the good will of the Astartes?
Also: realistically, no. Unless the issue is a serious one, the Astartes are not going to be bothering the High Lords with this one. The Chapter Master does not go running to the High Lords every time an Inquisitor demands a few marines to accompany him on a mission. He'll go to the Lord Inq first. If the LI thinks the local Inquisitor is right, then odds are there's no point sending a message to Terra. They'll compare the CM's report with that of the LI and probably side with the LI. That makes the CM look stupid, cost him political capital and possibly gets him punished for failing to obey the =I=.
The only time the astartes can get away with refusing an =I= is when he's an outsider without allies in the Inquisition. That kind of Inquisitor will know his limits (otherwise he'd be dead already) and won't be the one asking for Marines.
). The Sisterhood in particular would be very hard to convince to leave any threat to His Divine Majesty's Imperium alone unless there's an Inquisitor around ordering them to do so.
That;s the beauty of it.