Thanks for the great explanation, Ed! One question, though:
How is this circuit better than merely sticking different loads
in series with the battery being charged to adjust current?
It seems somewhat the same, just that there's an LM317 present.
What does the LM317 do that the load by itself wouldn't?
Kanon
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