hi transistor,
Just seen you last post, I had prepared some notes for you which I will post.
I'll look at the spec you have posted.OK
This is how the designer intended the charger to work.
The voltage drop across R4 due to the charge current controls the conduction state of T3.
When the charge current is high, the voltage drop across R4 is sufficient to make T3 conduct.
The collector of voltage of T3 is close to the +V supply rail, so T2 is not conducting, so T1 base current
flows from the 0V rail via R2.
This makes T1 conduct and the relay is energised.
The closed contact of the relay keeps the mains supply connected to the transformer input.
When the battery charge current falls to a value such that the voltage drop across R4 is too low to keep T3 conducting, it turns off.
This causes T2 to conduct as current now flows thru R3 causing T2 collector voltage to be close to the +V rail.
So T1 stops conducting and the relay de-energises and the mains supply is switched off to the transformer input.
Your modification.
You have removed the relay, T1 and T2 so that there is no auto shut off when the battery charge current falls.
The effect of that is that the voltage drop across D6 falls due to the reduction in charge current, the voltage on the battery slowly rises.
You will have initially set the Vcharge voltage across the battery at 6.3V when the initial charge current was high.
Measure the Vout from the LM317 when the charge current is high with the battery voltage set at 6.3V.
NOTE:
If the voltage drop across D6 is 0.9V at 1 amp it will be close to 0.7V at very low charge currents.
Subtract 0.7V from Vout and that will be the voltage the battery will try to charge to,
it will be 6.3+ [0.9-0.7]=
approx 6.5V..
Which is close what you have actually measured.