Homing In

So here are the rules for the new homing challenge I'm going to try to run this weekend.  The goal of this encounter is to find a transmitter that the players know the frequency of.  The players, however, do not have access to a radio direction finder, so they are forced to play a game of warmer/colder across the search zone (in this case, California.)

For each "turn" of the air search, the party will attempt to ascertain the distance to the beacon (+1/50km, round down).  The check will be a Communications/Conventional check, with difficulty set by their actual distance from the beacon.  A success will give them a band that is (20 + 2D6)% of the distance to the beacon, divided by the Margin of Success (0 is consider 1 MoS for these purposes) wide, and the beacon will lie somewhere in that band.  Divide the band width by 10, and roll 1D10 to determine which sub-band of the return band the beacon is in, and give minimum and maximum ranges accordingly.

Failure is handled the same way, but with the final return values having 10% * 1D6 added to them.

I understand the above may be confusing, so let's walk through an example.

Randall is attempting to locate a beacon some 200 km distant.  He makes a +4 check (200 km/50 km) against Communications/Conventional (he has it at +3).  He rolls a 9, for a total result of 12.  The target is 11, so he succeeds by 1.  The GM rolls 2D6 for a result of 3, meaning the band Randall detects the beacon in is 23% of 200km wide, or about 46 km.  The GM then divides the 46km band into 10 sub-bands and rolls 1D6 with a result of 2.  That means the beacon is in the middle of the second 4.6-km-wide sub-band.  The GM reports that the beacon is between 193 km (200 km minus the 1.5 4.6-km sub-bands) and 239km (200km plus the 8.5 4.6-km sub-bands.)


Randall now moves closer (by accident, since he doesn't know where on the circle the beacon is), and his new position is only 100km out from the target (which Randall knows to be somewhere in the 46-km-wide band he saw before.)  Randall now rolls again, with a +2 modifier (100km/50km).  He rolls a 3, for a total  result of 6, failing the roll by three.  The GM again rolls 2D6, for a result of 10, which gives the band the beacon is in a 10-km width (20+10)/MoF.  The GM then divides the 30-km band by 10, and rolls 1D10 for a seven, meaning the "base" band is between 93 km (100 km - 7.5 * 1km) and 103 km (100 km + 2.5 * 1km).  The GM then rolls another D6 for a 1, representing the error.  Both the short and the long range have 10% added to them, giving a final band of 102km to 113km.  Note that the actual beacon does not lie in this band, despite the fact that the band is quite tight.


It occurs to me that if that example didn't intimidate you, you've probably played a lot of BattleTech.  One of the advantages of this system is that if you fail by a great deal the band you get back get more and more specific about being wrong.

Obviously, the math here is a bit dense, but not overly so.   Finding distance on a sphere is about as complicated, and a quick bit of code will go a long way.  The real trick here is that the GM can't tell the characters their MoS or MoF, because knowing the modifier will give them a hard bracket to range the beacon and help eliminate areas picked up by close failures.

So that's the bulk of Sunday's session, if the players do what I expect them to do.  Now I need a map of California and a compass.