I've got my release angle "method" posted on several other sites looking for discussion and feedback, and one response I got back threw me for a bit of a loop.  One of my examples uses a 20* release angle on the lane, and he said that's impossible because the lane isn't wide enough.  He said that 15 degrees is a difference of 5 feet from the foul line to the arrows.  To which I said it's not in relation from the foul line, it's from the pins, or in other words, from one side of the foul line to the other would be 90*, it's just like calculating angle of rotation.  But of course that got me second guessing myself.  I could swear I've always been told and have always understood that it's from the pins, not the foul line.  A 90* angle of rotation would be sideways or parallel to the foul line, right?  And a 0* angle of rotation would be end over end straight at the pins?  I can't imagine why it would be calculated the other way, doesn't make any sense.  It wouldn't make sense for a 10 degree angle of rotation to be steeper or more sideways than a 60 degree angle of rotation . . but then in the same sentence that he said a 15 degree release angle was impossible, he said "most release angles are only a couple degrees," which directly contradicts what he said in the first place.  If 15* is 5 feet from the line to the arrows, what would 5* be?  If he's using the foul line as 0 and the headpin as 90, wouldn't that make most release angles up in the 70s or 80s?