435/14 watts (max dissipation)= 32ma
That should be 14w/435v=32ma
Your idea is sound but you should be aware that when you decrease the current flowing thru the tube, the plate voltage will rise. This is normal but it means that everytime you adjust the current to a new value, you need to remeasure the plate voltage and do the math again.
For that particular tube with a max dissipation of 14 watts, 70% dissipation would be 9.8 watts. So, adjust the current to where you think it should be (25ma is a good starting point), measure your plate voltage, and do the math. If the result is not 9.8 watts, readjust the current, remeasure the plate voltage, and recalculate. You may have to repeat a couple more times to get the current exactly where you want it, ie, providing 70% max dissipation.
Another thing to remember if you are measuring current using the 1 ohm cathode resistors method, cathode current does not equal plate current. Cathode current equals plate current plus screen current. I usually ignore the screen current for a 6V6, but this causes my calculations to be off a bit. That's OK with me because the error is on the safe side, ie, if my calculations say 9.8 watts, the actual dissipation will be a little less.
If you just gotta be exact you will need to know what the exact plate current is. You can measure it directly (dangerous or more trouble than it's worth IMO) or you can measure the voltage drop ACROSS the screen resistor and calculate the screen current using Ohm's Law. Then subtract the screen current from the cathode current.
Lord Valve wrote a decent procedure...
http://www.duncanamps.com/technical/lvbias.html