What gauge of wire do I need for direct burial for a 300 ft run for a 70 amp subpanel? I am going to install a 50 amp plug for a 5th wheel and eventually add some lights to a green house about 70 feet away.
If you were simply concerned with ampacity of the conductors, you would be able to use 4 AWG copper or 3 AWG aluminum. Unfortunately, in your situation you'll have to concerned with the power quality, so you'll also have to take into account the voltage drop along the conductors.
Calculating voltage drop uses the simple formula
Voltage Drop = Resistance (R) * Current (I). However, figuring out resistance and current can be a bit tricky.
To determine the resistance of a conductor, you can use Table 8 from Chapter 9 of the National Electrical Code (which you should be able to find online somewhere). This table lists the ohms per thousand feet, of various types and sizes of conductors.
Once you have this value, you can divide it by 1000 to get the ohms per foot. You'll then multiply this value by 2 times the length of the conductor. So the formula for resistance will look something like this.
R = 2 * Length (L) * ohms/foot
The current is a bit trickier, since it will change depending on what is drawing power on the circuit. A lot of folks will say things like "It' highly unlikely you'll ever use the circuit at full load.". They'll then use a fraction of the circuit capacity to calculate the voltage drop. However, this can easily lead to the situation where a high load item turns on, and the lights dim. When calculating the voltage drop for feeders, I like to use the full capacity of the circuit to do the calculations. So in your case, I'd use 70 amperes.
It's recommended to only allow a 5% total voltage drop, from service to consumer. Because of this, you'll want the voltage drop along the feeder to be around 2 - 3%. For a 240 volt circuit, that means 4.8 - 7.2 volts (120 volt circuit = 2.4 - 3.6 Volts).
With all that said, let's look at some actual numbers. I'll be using values from the uncoated copper column of Table 8 for these examples.
4 AWG Uncoated Copper
VD = 2 * 300 * 0.000308 * 70 = 12.936 Volts
12.936 Volts / 240 Volts = 5.4%
3 AWG Uncoated Copper
VD = 2 * 300 * 0.000245 * 70 = 10.29 Volts
10.29 Volts / 240 Volts = 4.3%
2 AWG Uncoated Copper
VD = 2 * 300 * 0.000194 * 70 = 8.148 Volts
8.148 Volts / 240 Volts = 3.4%
1 AWG Uncoated Copper
VD = 2 * 300 * 0.000154 * 70 = 6.468 Volts
6.468 Volts / 240 Volts = 2.7%
1/0 Uncoated Copper
VD = 2 * 300 * 0.000122 * 70 = 5.124 Volts
5.124 Volts / 240 Volts = 2.1%
Depending on how much you actually care about power quality, you'll want to use either 2 AWG copper or larger. I'd personally probably use at least 1 AWG, because I know how power demand tends to grow over time. Once you have power in a location, you'll find all sorts of things to plug in.
I come up with 1/0AL or #2cu using a 50A load on this feeder.
Obviously, using a higher load will call for larger conductors. For instance, using the full 70A feeder size calls for 2/0AL or #1cu.
It is highly unlikely a 50A-120/240V travel trailer/motor home feeder will EVER see a full 50A. And "some lights to a green house" is a very vague potential load.