How many weighings will I need to find the heavy coin?

Weighing #1 - 16 coins on each side. If they are equal, the 27th coin is the odd weight.

Weighing #2 - Using the lightest stack, weigh 8 coins on each side, chose lightest stack --

Weighing #3 - Weigh four coins on each side, chose lightest stack

Weighing #4 - Weigh 2 coins on each side, chose lightest stack

Weighing #5 - Weigh 1 coin each side, find lightest coin

My guess is a maximum of 5 weighings, minimum of 1.

Ooops sorry - I was trying to find the lightest coin - replace light with heavy in each of the above weighings.....

and I'd suggest replacing 16 with 13 in the first line...

Duh! You're right!

Let's see if that makes it less weighings....

13/13 - chose heaviest or know heaviest one

6/6 from heaviest pile or known heaviest one

3/3 from heaviest pile

1/1 from heaviest pile or known heaviest one

That makes a total of only 4 weighings!

Weigh 9 against 9. If it balances, the heavy coin is in the remaining 9, otherwise it's on the heavy side. It's now down to 9 coins.

Weigh 3 against 3. If it balances, the heavy coin is in the remaining 3, otherwise it's on the heavy side. It's now down to 3 coins.

Weigh 1 against 1. You can now identify the coin using the same reasoning as in the previous steps

3 weighings

i agree with markr. since it takes a weighing for 3 coins and 2 for 9 coins, it will probably takes 3 weighings for 27 coins. in other words :
3^n=coins
where n=number of weighings