Re: Why are ibanez guitars easier to string bend than washburns?
The amount of pressure it takes to get a string to go from one note to one whole step higher is constant no matter what. (assuming scale length is the same) It's the distance that changes. If you have a jazzbox with several inches of string behind the bridge and behind the nut, bending the string stretches those parts too. If you have a locking bridge (locked at the nut and bridge) the 25.5" length is all you're bending/stretching. If that locking bridge is a tremolo, then you're stretching the springs and bending the bridge forward. This is where I believe you are experiencing the difference, assuming you're lock nuts are locked. If they are unlocked, the reverse headstock completely reverses the behind the nut lenght and therefore the distance required for each string to reach pitch. The whole feel is inverted. So perform these tests with the nut locked! If it's still the case, the difference is in the balance of the bridge. The Ibanez bridges have (generally) a shorter block, and more mass than "OFR or Shaller-style" bridges. More importantly, the mass is distributed differently in relation to the block and studs. So when you bend, as others have said, the Washburn bridges are pulling flat to a different degree than the Ibanez bridges.
But remember, the word "tension" is often misapplied. At 25.5", it takes the same "tension" to reach any particular pitch regardless of all other factors. You just might have to push the string FARTHER to get to pitch because other things are giving way.