Gentlemen,
I think that using power calculations in this example is just plain wrong... Your contact is not dissipating 120 Watts. It has a rated resistance (we will neglect impedance and capacitance for sake of simplicity), and that means that the power "seen" by the contacts is just P = R * I^2 . In your case, the manufacturer has determined that if the current is greater than 4A, then the dissipated power becomes enough to start to heat up contacts and lead to eventual failure or loss of reliability. When you contact is closed, the relay doesn't "see" the applied voltage, only the difference across it's contacts!
In your case, that means that current-wise you would be fine, because you are below the rated 4A.
But Richard is right, the rated voltage is there to avoid your contacts from arching between themselves when they are open. I wouldn't go past the rated voltage of 30 VDC either.
Hope this helped!