Uhlenbrock decoder 67800

Hej

Er der nogle af jer der kan forklare mig hvordan dette regnestykke ender med en rest på 208? Jeg vil benytte adressen på min decoder 2000 samt 2001. Men jeg kan ikke se hvordan tallet 208 fremkommer, en der kan hjælpe? Jeg er med på de 2000 skal divideres med 256 og det giver 7 som skal skrives i CV 120 og så 208 i CV121 men hvordan pokker fremkommer tallet 208??

Address 1 and 2 (CV 120 and 121, 160 and 161)The addresses for each servo output can be freely selected. The valid range of address is1-2048.The servo is brought to the stop by the Address 1 in accordance with CV122 (red) andCV123 (green).Address 2 brings the servo to stop positions in accordance with CV162 (red) and CV163(green).Note: Address 2 can be configured by CV programming.Addresses to 255 can be entered directly as values in the CV for the Low byte (e.g.CV121). CV for the High byte (e.g. CV 120) remains at value 0 (factory setting).Addresses from 256 the values for the High byte and the Low byte must be calculated. Forexample, programming of the address 2000 is as follows.• Divide the address value by 256 (2000/256 = 7 remainder of 208). Register the integer result (7) as value in CV for the High byte (e.g. CV 120).• Register the remainder (208) as value in CV for the Low byte (e.g. CV 121).Setting Stop positions (CV 122 and 123, 162 and 163)The setting of the stops is done by numerical values between 0 and 127.CV value = 0 maximum value for stops 1 and 3 (CV 122, 162)CV value = 127 maximum value for stops 2 and 4 (CV 123, 163)Setting time (CV 124)Time constant by which servo position is incremented or decremented in 1 ms steps.Process time = (difference between "red" and "green" values) * setting time * 1 msSetting time = (desired procedure time in seconds) * 1000 difference of the values forretaining position "red" and "green"