Hey guys, when you are talking about the cylinder diameter, are you talking about the rod or the entire diameter? If the entire diameter is 3", the rod could be 1.25". If this is the case, a 3" cylinder is not out of the question. My apologies if this is what is being discussed. I thought you were talking about a 3" rod on the cylinder!!
What I have learned here, or at least made to realize is that hydraulic cylinders can act as force multipliers in both directions. Either using a small amount of fluid at high pressure to generate large forces for work, or using large forces acting on the cylinder to generate high pressure in a small amount of fluid. The big difference being that one way you have a PRV to protect the components. With no PRV in the circuit beyond the valve, there is potential to generate pressure high enough to exceed the designed ratings of the components.
... a 1.5" or 2" cylinder simply isnt going to work. But hey, if you want to find out the hard way, go for it. But I am one that can learn from someone elses mistake and not repete their failure. Chilly807 in the link I gave already tried this with a 2" cylinder. He overextended the cylinder and bent the rod. And he only has a L3400 like myself. Which is on the smaller/lighter side of a CUT.
Not my area of expertise, but my RBT35 8' has hydraulics on all four functions (with cushion valve on power angle) and the specs are probably available or I could measure if req'd ...but, the reason for the post is to ask just what you want to be the weak link if you somehow manage to generate unacceptable forces? I would think the hydraulic system's relief valve would be a good choice. It is only the normal operating forces that you would normally need to contend with and, if the blade casts the material as it should then the forces should be manageable. If you snag an immovable object, then where do you want the give?