Hi all,
I've had the following problem and I haven't come up with a reasonable way to solve it. I'll give my proposed solution at the end.
Enemy A is standing 100 feet from Player B. Enemy A runs away, at X2, 30' = 60'. Player B chases him, at 40', X2 = 80'. At the end of round 1, the distance separating them is 80'. This distance continues to decrease each round (60,40,20,0). The strange part is once you are able to "catch" the player/ Each of you takes a double move, and end up in adjacent squares.
Now player B, moving after Enemy A, wants to be able to take an action at the end of the round, but since he took a double move, he can't. The next round A moves another 60' and if B moves 60' again he is in the same situation that he was just in. However, he can't take a single move then attack since 40' is his limit and that puts him 20' away. So he has to take a double move to get by the enemy (which might or might not provoke an attack of opportunity for the enemy!), and then mow him down as he comes by, or something similar.
Of course, this makes no sense. I have looked this up in the PHB and DMG and didn't find anything satisfying - but this might be because I missed something. I see where it says about players with the same movement - use an opposed dex check, and if it goes long, an opposed con check. That doesn't satisfy what I need.
My thought was to run it as follows:
Depending on how long (initiative wise) the enemy takes off before the player chases, I calculate the difference in movement (in this case, 20' on a double move). If the enemy was 100' away, and A and B leave at about the same time, it will take B 5 rounds to catch A. This is the same number of rounds as the above situation. I then allow an attack from A on B, even though this means he gets an attack on a double move. I'm not too happy with this.
My second thought is to give half the difference (10'), and then have it takes 10 rounds to be able to catch A and get an attack in.
Any idea on how to do this?
Thanks a lot,
zyzzyr
I've had the following problem and I haven't come up with a reasonable way to solve it. I'll give my proposed solution at the end.
Enemy A is standing 100 feet from Player B. Enemy A runs away, at X2, 30' = 60'. Player B chases him, at 40', X2 = 80'. At the end of round 1, the distance separating them is 80'. This distance continues to decrease each round (60,40,20,0). The strange part is once you are able to "catch" the player/ Each of you takes a double move, and end up in adjacent squares.
Now player B, moving after Enemy A, wants to be able to take an action at the end of the round, but since he took a double move, he can't. The next round A moves another 60' and if B moves 60' again he is in the same situation that he was just in. However, he can't take a single move then attack since 40' is his limit and that puts him 20' away. So he has to take a double move to get by the enemy (which might or might not provoke an attack of opportunity for the enemy!), and then mow him down as he comes by, or something similar.
Of course, this makes no sense. I have looked this up in the PHB and DMG and didn't find anything satisfying - but this might be because I missed something. I see where it says about players with the same movement - use an opposed dex check, and if it goes long, an opposed con check. That doesn't satisfy what I need.
My thought was to run it as follows:
Depending on how long (initiative wise) the enemy takes off before the player chases, I calculate the difference in movement (in this case, 20' on a double move). If the enemy was 100' away, and A and B leave at about the same time, it will take B 5 rounds to catch A. This is the same number of rounds as the above situation. I then allow an attack from A on B, even though this means he gets an attack on a double move. I'm not too happy with this.
My second thought is to give half the difference (10'), and then have it takes 10 rounds to be able to catch A and get an attack in.
Any idea on how to do this?
Thanks a lot,
zyzzyr