Image projection and object resolution

Normally the vertical pixel pitch is equal to the horizontal pixel pitch: $LS_{pitch} = P_{pitchv}$. The achievable vertical and horizontal object resolution are dependent from depth positions of objects to be represented by the screen - according to the fan projection, shown in fig.2:

$$\Delta X_{obj}=\Delta Y_{obj}=\frac{\left( Z_{obj}-Z_{obs}\right) LS_{pitch}}{Z_{screen}-Z_{obs}}$$ (5)

The possible depth resolution of the objects depends in addition from the $\Delta$horizontal distance of the viewing centers, say distance of the two watching eyes: $D_{eye}$. The smallest sampling angle might be denoted by $\Delta\beta$: It is slightly dependent on the angle $\beta$ itself because $\Delta X_{obs}$ is constant.

$$\Delta \beta =\frac{\Delta X_{obs}\cdot \left[ \cos \left( \beta \right) \right] ^{2}}{Z_{screen}-Z_{obs}}$$ (6)

Let k be the number of views between the eyes. Then the eye distance corresponds to the sampling angles as follows:

$$D_{eye}=k\cdot \Delta X_{obs}$$ (7)

where a symmetrical eye position is assumed.

With respect to the two projections centers the following depth sampling planes $Z_{sp}\left( k,i\right)$ for objects are given:

$$\left[ Z_{obs}\left( i,k\right) -Z_{screen}\right] =\frac{i\... ...dot \left[ Z_{obj}\left( i,k\right) -Z_{obs}\right]} {D_{eye}}$$ (8)

Let be the sampling distances on the screen $\Delta X_{screen} = LS_{pitch}$ and those at the eye plane of the observer $\Delta X_{obs}$. Then $\left( \ref{Eq_Zobs1} \right)$ results into the following equation:

$$Z_{obj}\left( i,k\right) =\frac{Z_{screen}\cdot k\Delta X_{o... ... \cdot i\Delta X_{screen}}{k\Delta X_{obs}-i\Delta X_{screen}}$$ (9)