Headphone King of Soundstage?

I own and listen to a lot of headphones.  I have to be honest, "soundstage" is not something I've ever bothered to think about when listening to headphones. 

Listening to headphones is a completely different experience than listening to speakers.  Not only are you surrounded by sound, but you often have a visceral sensation of the sound waves, even at lower volumes.  With headphones you have two little speakers right next to your ears, so you can't have that same immersive experience. 

There is some trickery that can be done, for example one of my headphone amps has a "crossfeed" circuit to send some of the right signal to the left side and vice versa to help simulate "spacialization" and reduce listening fatigue.  I've never used it. 

I have the HD800 and they are fine headphones, and from what I've read, HD800S may be even better.  I bought my HD800 second hand, so I can't speak to break in.  

If you want the sound of speakers, listen to speakers.

The only phones I've heard that produce "out-of-head" sound are the old Stax Lambda Pros, whose drivers were situated slightly in front of and aimed at the ears.

@ big_greg

Chord Dave has "crossfeed" function in it with 0 to 3.

0 means no crossfeed with 3 strongest.

I had done experiment to find some subtle differences.

Now I use 1 setting.

Of course headphone sounds different from speaker with dynamic limitation but more intimacy.

But I noticed some depth even with Stax 009S.

I'll have to try that cross feed thing sometime.

"Cross Feed". Isn’t this what the new Polk "Legend" speaker does? Also, there is (was?) a technology called MARS (McShane Ambience Retrieval System) that did the same thing. As for soundstage, sound goes into those holes on the sides of your head and your brain decodes them to reconstruct the original event. Why does the source of the sound make a difference? Speaker or headphones the result should be the same. I would recon the headphones should have a better soundstage as it’s removing the room effects from the equation.