Its a pretty simple calculation but you have to tell us what measurement you are making and the details of everything between what you are measuring and the surface of the belt.
If you know the rpm of the drive wheel and the diameter of it you can calculate the speed. Maybe there are easier methods, but i never work with converyors.
First thing is to figure out what the actual rpm of the drive wheel would be. Then the circumference because that will determine how much the belt will move in 1 turn (providing optimal transfer). and then calculate
Example, using metric here.
driving wheel, 10rpm
driving wheel circ, 0,5m
0,5*10=5m/min
5/60=0.08333m/s
[driven pulley dia] / [drive pulley dia]
Divide motor RPM by above result
Divide result by 40 (gearbox), save result
[driven sprocket teeth] / [drive sprocket teeth]
Divide saved result by sprocket result
This gives you the RPM of the head pulley
[head pulley dia] * pi (3.1415926) = travel per revolution (inches)
[travel per revolution (inches)] / 12 = feet travelled per revolution
RPM * feet travelled per revolution = feet per minute of belt driven by head pulley
If I'm interpreting correctly, you have a motor at 1750 RPM driving the input of the gearbox through a 4.5"/22.5" belt reduction.
The output shaft of the 40:1 gearbox drives the 30" diameter conveyor pulley through a 15/60 chain or timing belt reduction.
The input speed of the gearbox is 1750 * 4.5 / 22.5 or 350 RPM
The output speed of the gearbox is 8.75 RPM
The conveyor pulley speed is 8.75 * 20 / 60 or 2.91 RPM
The conveyor pulley is 30" (2.5 feet) in diameter and 2.5 * PI or 7.85 feet in circumference.
The surface speed of the conveyor pulley is 2.91 (Revolutions per minute) times 7.85 (feet per revolution) or 22.9 feet per minute.