For thin walled cylinders, hoop stress is calculated by multiplying the radius by the pressure in psi divided by the wall thickness.
If pressure and wall thickness remain constant, a larger radius causes a larger numerator which would be divided by a constant demominator; hence, a larger hoop stress. Based on this, if you have a smaller radius (12g. down to 20g for example)you can have higher pressure with the same wall thickness and still maintain the same hoop stress.
Well, lets just work the numbers. We do need to use mean radius (mean radius is half ID + half wall thickness, say 0.730/2 + 0.100/2. For 12 bore = 0.365" +0.050" = 0.415", assuming wall thickness at 0.100", Assume pressure at 5,000 psi. Then 12 bore hoop stress equal (5,000 X 0.415)/0.100 = 20,750 psi stress (pressure and stress have the same measurement units). For 20 bore, r = 0.615"/2 + 0.100/2 = 0.3575"; hoop stress = (5,000 X 0.3575)/ 0.100 = 17875 psi or, r for = (5,000 X 0.3505)/ 20,377 = 0.086". Yes, you are right!!This from Wikipedia.
Also from Wikipedia: Fracture is governed by the hoop stress in the absence of other external loads since it is the largest principal stress. Note that since the hoop stress is largest when r is smallest
(nope, when r is largest), cracks in pipes should theoretically start from inside the pipe
(nope, the outside should fail first).
Seems contadictory. Am I misunderstanding this?
Yes, I think so - as above. 
You were right to start with and then went astray. 
OK?.
