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In first-person-shooters like Call of Duty, the hologram only appears when it is behind the glass portion of the sight.

In OpenGL, would it be implemented by rendering the glass to the stencil buffer and then rendering the hologram, but masked by the stencil buffer?

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    \$\begingroup\$ I doubt it is something that complicated. It is very likely just a 2D sprite. img1.wikia.nocookie.net/__cb20120119074042/callofduty/images/2/… \$\endgroup\$ – glampert Jul 3 '14 at 20:16
  • \$\begingroup\$ continuing on @glampert: .. where the sight only appears (2D) after the zoom-animation has finished. \$\endgroup\$ – Jean-Paul Jul 3 '14 at 20:29
  • \$\begingroup\$ @Jean-Paul, based on my memory of COD4, if you partially aim, you can still see the sprite, but only the part that's "inside" the glass. In MW2 and MW3, they definitely only show the part "inside" the glass. \$\endgroup\$ – Victor Jul 5 '14 at 3:05
  • \$\begingroup\$ Very similar to a Portal portal... \$\endgroup\$ – david van brink Sep 4 '15 at 19:32
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i think you mean the 3 dimensional offset of the holographic sight that is positioned relative to the actual frame. i think the best way to simulate the reflected projection of the dot / cross is to offset a sprite from the glass a little bit to the front of the gun ( but still ceeping it inside the frame / housing of the sight ). Also render it from the viewers side only. that way the holograph appears to be projected and positioned depending on the view angle. I hope that makes sense.

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Calculating UV in vertex shader:
uv.xy = mul(world2object, cameraWorldPos-mul(object2world, (0,0,0,1)))+(0.5,0.5)

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    \$\begingroup\$ Maybe you can expand on this answer to describe what you're doing and why it solves the problem. \$\endgroup\$ – MichaelHouse Jul 11 '14 at 4:21
  • \$\begingroup\$ It's project camera position onto sight XY plane. So if we sample texture with this coordinates and draw surface with alpha blend or add mode is will look exactly like reflector sight. \$\endgroup\$ – mouurusai Jul 11 '14 at 5:46
  • \$\begingroup\$ While it's an understandable answer, a much more elaborate explanation is probably needed. I believe that's also HLSL code, not GLSL. \$\endgroup\$ – Wolfgang Skyler Sep 4 '15 at 22:01

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