Question on ASA Obstruction Award Procedure

[HawkeyeCubP]

Good scenario, WMB. It hurts my brain, but here goes:

Originally Posted by WestMichBlue

At the end of the play he wants to send B-R to 2B - but R1 is there!

Part of me thinks that the only way this award can be made is if, in the BU's judgement, the BR would've reached 2B before R1 got back to it.....But then the rule simply says "reach," not "own" or "be protected on," so perhaps my first statement doesn't matter...

Originally Posted by WestMichBlue

We are supposed to protect the runner to the base they would have reached had obstruction not occured. This runner was physically capable of reaching 2B ahead of any throw. But even without obstruction, the B-R would not have gone to 2B because R2 was still there.

Originally Posted by WestMichBlue

So does it make sense to send B-R to 2B and force R1 to 3B?

I would use the term "advance" instead of "force" if explaining the awards to a player or coach, but I suppose yes (if we're following my logic from above), because of the following pertinent exerpt from RS 36:

Originally Posted by Rule Book

When an obstructed runner is awarded a base that they would have reached had obstruction not occurred and a preceding runner is on that base, the obstructed runner shall be awarded that base and the runner occupying it is entitled to the next base without liability to be put out.

Originally Posted by WestMichBlue

R1 chose not to advance, and possibly would have been thrown out at 3B because of the close proximity of the ball to 3B. How can we force a runner to advance because we artifically placed an obstructed runner on her base?

Because of RS 36, I guess.

Originally Posted by WestMichBlue

PS - just to make it a little more fun - assume that B-R did advance to 2B even while R1 was returning to 2B. Suppose F6 throws ball to F4 who tags both runners on the base and looks at you, the BU. R1 "owns" the base so it is the B-R that would be called out. But she cannot be out due to obstruction. Would you leave B-R at 2B and send R1 on to 3B?

Yes, I suppose, because BR "reached" 2B, which is all the rule and RS say has to be able to happen.

Other takes?