So here’s a question: isn’t it incredibly inefficient to be constantly running ‘OpenDoor()’ when you’re on the pressure plate and the door is already open?
Ideally, I would think you would only call OpenDoor() once as soon as you overlap the pressure plate.
On the other hand, to accomplish this (at least the way I would do it) is to check whether the is already open before you call OpenDoor(), and this check may add as much overhead as constantly calling OpenDoor()…thoughts?
Bryan