Assuming the rigidity and resistance to chatter is dependent on stiffness of the tool (a cylindrical beam).
After some digging, it seems deflection is proportional to the cube of length over the fourth power of diameter.
In practice I've never had problems with a 3/8" bar extended by well over 1 1/2" taking decent cuts on steel.
A 1" bar will be 50 times as stiff as a 3/8" bar of the same length (fourth power of the difference in diameter), and so can be 3.70 (cube root of 50) as long to be similarly stiff.
3.70 x 1.5 = 5.55"
The 1" boring bar should be fine, and you can always halve the depth of cut to halve the load.
Neil
<edit> I think we are all familiar from practice, if not calculation, that stiffness of a bar decreases rapidly with length, and many of us know the cube root relation (e.g. a 1" long bar is eight times stiffer than a 2" one). We also have an intuitive understanding that stiffness also increases with diameter, but is suspect that most of us are less aware it is a fourth power relationship (e.g. 2" diameter bar 16 times as stiff as a 1" diameter bar). I'd come across this before but it had fallen off then end of my memory until I looked it up. It's hard to find a simple answer on line as most websites use the "area moment of inertia" in the calculation so it applies to any shape. For a circle, this is proportional to radius to the power of four. (Yes I looked all this up)
Edited By Neil Wyatt on 03/11/2019 14:05:05