Sorry for the incomplete information and thanks for the input!
The frame is a FKL Tomcat from I don't know when, I would guess it weighs 150 lbs, 300 with driver. Tires are 4.10 / 3.50 - 5 with a diameter of 10.75" jack shaft has a 10 teeth #41 sprocket, axle has 40 teeth. Speed at 3600 rpm should be 32mph at the max ratio of the torque converter.
In the 2nd pic of the first post I've put a plate on the flat edge of the flange toward the engine of the driven to rest the ruler on and project the position of this plane in relation to the same edge on the drive.
With the spacer provided in the kit the distance you see in the pic is 0.174" but the flanges have different width. The drive is 0.321" at the top and the driven is 0.247" at the top. So in order to put the inner sheaves at the same position, my 0.174" should need to be 0.074" which would means to add a 0.100" spacer to move the drive further away from the engine and when I do so the belt looks crooked in the pulleys.
But since the pulleys are different diameters and because the belt moves toward the engine when it climbs in the drive and in the opposite direction in the driven, I still don't know what my 0.174" should be.
Anyways your input is appreciated!
Here's the requested pic:
Here's what the belt looks like: