Powering 1.8v GPS with Arduino
That module requires 65 mA max.  A linear regulator from the 5V supply would only dissipate 210 mW.  That's low enough that no heat sink is required.  While there are more efficient and fancy ways, if you have to ask here this is probably the better solution since it's simple.

If the Arduino has a 3.3V supply that can spare a extra 65 mA, even better.  That would mean the 1.8V regulator would only drop 1.5V, for a dissipation of 100 mW.  Make sure you put a 1 &micro;F or so ceramic cap immediately in front of and after the regulator.

If you are stuck with 5V in, then you can use a dropping resistor on the input of the regulator to split up the dissipation between the regulator and the resistor.  Let's say whatever LDO you get should have at least 1V headroom.  That means you need 1.8V + 1.0V = 2.8V after the resistor, and the resistor can drop up to 5.0V - 2.8V = 2.2V.  From Ohm's law, 2.2V / 65mA = 34&Omega;, so the standard value of 33&Omega; in series with the regulator will work fine.  Again, don't forget the ceramic caps on the input and output of the regulator.  The input cap is especially important with the dropping resistor.

Added:

Here is the circuit I was talking about:



At the maximum current of 65 mA, the resistor drops 33&Omega; x 65mA = 2.1V and will dissipate 140 mW.  The regulator will drop the remaining voltage of 1.1V and will dissipate 70 mW.  140 mW dissipation in the resistor is enough that you need at least a 1206, although check the specs.  A "1/4 W" thru hole resistor will work fine for this.

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