Downstroke escape with light supination? Help!

I always thought that the downstroke escape needed to be performed with a pronated wrist. In the video titled “Downstroke escape,” the downstroke escape is described using a slightly supinated wrist, with the plane of motion being shown (from what I can tell) as a simple 9 to 3 wrist deviation.
But a supinated wrist with a strict 9 to 3 plane of motion would put the pick, at the end of a downstroke, trapped underneath the next string. Wouldn’t the plane of motion need to be something more like 8 to 2 in order to perform a downward pick escape with a lightly supinated wrist? I really want to get this. What am I missing here?


In the video, Troy specifically describes John McLaughlin’s downstroke escape as an 8 to 2 motion with a lightly supinated wrist. You’re correct, if it was a 9 to 3 motion, the wrist would have to be pronated in order to not end up trapped. Sounds like you’re already on to the answer. Best of luck!

Oh I must have missed that. Thanks for getting back to me!

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