Where:$p$ = principal point
$i$ = isocenter
$n$ = nadir point
$O$ = perspective center (lens)
$f$ = focal length of photograph
$\alpha$ = $\angle pOa$ = angle at the perspective center $O$ measured counter-clockwise from the photograph perpendicular to any point $a$ on the upper side of the photograph
$\beta$ = $\angle pOb$ = angle at the perspective center $O$ measured clockwise from the photograph perpendicular to any point $b$ on the lower side of the photograph
$a, b$ = images of ground points in the principal plane of the photograph
$a’, b’$ = corresponding points on the equivalent vertical photograph
$+r$ = distance $ia$ = distance $ic$
$-r$ = distance $ib$ = distance $ie$
$d$ = the displacement for an image, parallel to the principal line and due to tilt, on the upper side of a photograph
$d’$ = the displacement for an image, parallel to the principal line and due to tilt, on the lower side of a photograph
Formula: \[ d = \frac{r^2}{\frac{f}{\sin t}-r} \] Derivation: \[ \angle Oip = 90^{\circ} – \frac{t}{2} \] In the isosceles triangle iac,
\[ \angle iac = 90^{\circ} – \frac{t}{2} \] Therefore lines $Oi$ are parallel (corresponding opposite interior angles) and,
$\Delta Oia’$ is therefore similar to $\Delta aca’$
By proportion,
\[ \frac{d}{r+d} = \frac{ac}{Oi} \] However,
\[ ac = 2r\sin\frac{t}{2} \] \[ Oi = \frac{f}{\cos\frac{t}{2}} \] Substituting,
\[ \frac{d}{r+d} = \frac{2r\sin\frac{t}{2}}{\frac{f}{\cos\frac{t}{2}}} \] \[ d = \frac{2r\sin\frac{t}{2}\cos\frac{t}{2}}{f}(r+d) \] By trigonometry,
\[ 2\sin\frac{t}{2}\cos\frac{t}{2} = \sin t \] Therefore,
\[ d = \frac{r\sin t (r+d)}{f} \] \[ df = r^2\sin t + dr\sin t \] \[ d(f-r\sin t) = r^2\sin t \] \[ d = \frac{r^2\sin t}{f-r\sin t} \] Dividing the numerator and the denominator by $\sin t$, \[ d = \frac{r^2}{\frac{f}{\sin t}-r} \] This form of the equation applies to both sides of the photograph when the algebraic sign of $r$ is observed.
