That link doesn't open - at least not on my laptop.
It is easy as.
The 8" rim is 2" narrower than the 10" rim - that's 1" on each side - the outside of the rim is therefore 1" further inside. That applies if you have the same offset on both wheels.
In maths it is like this:
You have a 10" wide rim, thats 254mm. neg. 30 would mean that the mounting face is 30mm from the middle to the inside of the rim. So the outer shoulder of the rim is about 157mm away from the mounting face. (254/2=127+30=157)
If you have 8" wide rim, thats 203.2mm. neg. 30 would again mean that the mounting face is 30mm from the middle to the inside of the rim. So the outer shoulder of the rim is 131.6mm away from the mounting face. (203.2/2=101.6+30=131.6)
So the outside of the 8" rim is 25.4mm or 1" further in as opposed to the 10" rim.
If I understand your question correct you want to know the offset of a 8" rim which outer edge sits at the same spot than the 10" rim.
In that case you want a rim with 30+25.4mm offset in other words neg. 55.4mm offset.
You might then want to worry about your steering geometry, wheel bearings and general driving behavior. I believe such a rim is not available off the shelf anyway. If you want a neutral steering behavior you want the middle of the tyre contact patch where the extended line of the kingpin meets the ground level. (Front axle)
You just have to get your head around that zero offset is when the mounting face of the rim is right in the middle. negativ=wider track / positiv narrower track.
I wrote that out of the top of my head directly before bedtime - someone might correct me if I got the maths wrong.
That link doesn't open - at least not on my laptop.
It is easy as.
The 8" rim is 2" narrower than the 10" rim - that's 1" on each side - the outside of the rim is therefore 1" further inside. That applies if you have the same offset on both wheels.
In maths it is like this:
You have a 10" wide rim, thats 254mm. neg. 30 would mean that the mounting face is 30mm from the middle to the inside of the rim. So the outer shoulder of the rim is about 157mm away from the mounting face. (254/2=127+30=157)
If you have 8" wide rim, thats 203.2mm. neg. 30 would again mean that the mounting face is 30mm from the middle to the inside of the rim. So the outer shoulder of the rim is 131.6mm away from the mounting face. (203.2/2=101.6+30=131.6)
So the outside of the 8" rim is 25.4mm or 1" further in as opposed to the 10" rim.
If I understand your question correct you want to know the offset of a 8" rim which outer edge sits at the same spot than the 10" rim.
In that case you want a rim with 30+25.4mm offset in other words neg. 55.4mm offset.
You might then want to worry about your steering geometry, wheel bearings and general driving behavior. I believe such a rim is not available off the shelf anyway. If you want a neutral steering behavior you want the middle of the tyre contact patch where the extended line of the kingpin meets the ground level. (Front axle)
You just have to get your head around that zero offset is when the mounting face of the rim is right in the middle. negativ=wider track / positiv narrower track.
I wrote that out of the top of my head directly before bedtime - someone might correct me if I got the maths wrong.
