In general the point is to filter your water. If you want to be sure about it, check the volume of your pool, and the flow rate of your pump. Run it long enough to go through that much volume 1-2 times a day. In practice, that's probably around 6-8 hours.
Make sure your chemistry is good, and then try it. If it looks nice for a few weeks, try lowering your operating time and see what happens.
As a side note, 1.5 HP is 1120W. That's probably peak power not normal consumption, but you should still assume it consumes somewhere around $0.10/hr or so.
Comparing those tables: Note that the speed switch in the circuit you show isn't using L.
A: L+2+3
B: L+1+3
C: L+1 (Maybe this is L+1+2 ???)
D: L+1+2+3
0: No connection (or no connection to anything but L)
1: 2+1 (possibly plus a connection to L)
2: 1+2+3 (possibly plus a connection to L)
3: 2+3 (possibly plus a connection to L)
Making them correspond with each other...
C is obviously equivalent to 0.
D is obviously equivalent to 2.
That leaves us with
A: L+2+3
B: L+1+3
1: 2+1 (possibly plus a connection to L)
3: 2+3 (possibly plus a connection to L)
We can make those match if we relabel the connections. If we just swap the labels on your terminals 3 and 2, then
A is equivalent to 3
B is equivalent to 1
If we renumbered them all (your 2 is their 3, your 3 is their 1, your 1 is their 2), then
A is equivalent to 1
B is equivalent to 3
Pick whichever you prefer; one will switch off-high-medium-low-off, and the other will switch off-low-medium-high-off.
As far as theory goes: I'm not sure either, but let's see what I can do with it.
3 (2->3) appears to be "slow" because power flows through the right half of the bottom coil, and then through the side coil, in series. More resistance, less current flow, less power.
1 (2->1) appears to be "fast" because the left side of the bottom coil, and the side coil, are powered in parallel. Both get the full house-current voltage applied across them rather than the reduced amount of power they got in series.
2 (2->1 and 3) is the tricky one. I am far from certain, so DON'T take my word for it. But I think what's happening here is that, since the middle and right sides of the bottom coil (1 and 3) are now connected to each other, that loop has a current induced in it by the motor's moving magnets, which creates a countering magnetic field, which acts as a magnetic brake to slow the motor... so fast with a bit of braking equals medium. Seems like an odd solution, but if I'm remembering my freshman Physics at all correctly it might actually be a reasonably efficient solution.
You might want to run this by the physics discussion, to get someone with more recent memory of electrodynamics to check and/or correct that last paragraph.
Gopher baroque...
Best Answer
I don't know who is saying those comments or in what context they are saying, but I don't agree. If the motor is actually designed to be jumpered between 240V and 120V, then it should perform the same in either configuration.
Induction motors of this type have an entirely passive rotor, so no brushes. The only windings are in the non-moving fields, so it is practicable to switch them. How it's done is to wind the fields with 2 wires half the cross-section. For 120V, those wires are connected in parallel. For 240V, in series. The same amount of current flows through each wire in either configuration, so motor performance is the same, and temperature will be the same.
If you are abusing a motor that is not manufactured to be switchable, then all bets are off. Or if you are dealing with a brushed or shaded-pole motor on a smaller appliance, the rules can change, but that is not OP's question.