The LM317 circuit provides constant current while using different "loads" (actually current limiting resistances) in series provides a varying current to the battery being charged.
When using the LM317 circuit to charge the battery, it looks like this:
Vin--[Lm317]--Vout-+ Adj | | [R1] | | +----------+ | [Battery] | Gnd----------------+
The battery _is_ the load. As the battery charges, the voltage across the battery increases. However, the current remains the same - it is whatever value R1 is divided into 1.25, regardless of the battery voltage. Let's say R1 is 10 ohms, and the battery voltage is 10.5 when you start charging. The current will be
1.25/10 or .125 amps. Now at the end of charge say the battery is charged to 13.8 volts. The current will _still_ be .125 amps. The LM317 circuit acts as a constant current source.With the alternative you mention (using only a resistor as the current limiter), the circuit looks like this:
Vin---[R1]---+ | [Battery] | Gnd----------+
The current depends on two variables, Vin and the battery voltage. The formula is I = (Vin - Vbattery)/R1. But the vbattery voltage is continually increasing, meaning the current is continually decreasing. If Vin is 15 volts and Vbattery is 10.5 at the start of charge, and R1 is the same 10 ohms we used in the constant current circuit, the initial charge current will be (15 - 10.5)/10 or .450 amps. When the battery voltage reaches 13.8 at the end of charge, the current will be (15 - 13.8)/10 or .120 amps. That is not a constant current - the LM317 circuit is.
You can choose whatever approach you want. With the LM317 approach, you can set the exact current you want mathematically. With the approach of putting various different things into the circuit to act as limiting resistors, it's hit or miss. It's also a kluge of jumpers and devices, whereas the LM317 is straightforward.
If you use the LM317 circuit I would recommend you add a diode rated at 3 amps, like a 1N5404:
Vin--[Lm317]--Vout-+ Adj | | [R1] | | D1 +----------+--->|---+ | [Battery] | Gnd-------------------------+
The diode prevents the battery from discharging through the LM317 in the event that Vin is not present (eg you turn off the Schumacher without disconnecting the battery.) The LM317 will not be "happy" with current going through it backwards, and will let the magic smoke out.
Also, pay attention to the wattage rating of R1. The minimum wattage is found by I^2*R, and it's best to use higher wattage. So, for example, say R1 is 1 ohm to give you 1.25 amps charging current. That resistor will need to dissipate
1.5625 watts - I'd use a 5 watt resistor.Ed