We proceed to take up the general condition of the second problem,
before left unexamined, namely, that in which the vertical distances
_BC′_ and _AC_ (Fig. 6. page 13), as well as the direct distances _TD_
and _TD′_ are unequal.
In Fig. 6., here repeated (Fig. 76.), produce _C′B_ downwards, and
make _C′E_ equal to _CA_.
[Illustration: Fig. 76.]
Join _AE_.
Then, by the second Corollary of Problem II., _AE_ is a horizontal
line.
Draw _TV_ parallel to _AE_, cutting the sight-line in _V_.
∴ _V_ is the vanishing-point of _AE_.
Complete the constructions of Problem II. and its second Corollary.
Then by Problem II. _ab_ is the line _AB_ drawn in perspective; and by
its Corollary _ae_ is the line _AE_ drawn in perspective.
From _V_ erect perpendicular _VP_, and produce _ab_ to cut it in _P_.
Join _TP_, and from _e_ draw _ef_ parallel to _AE_, and cutting _AT_
in _f_.
Now in triangles _EBT_ and _AET_, as _eb_ is parallel to _EB_ and _ef_
to _AE_;—_eb_ ∶ _ef_ ∷ _EB_ ∶ _AE_.
But _TV_ is also parallel to _AE_ and _PV_ to _eb_.
Therefore also in the triangles _aPV_ and _aVT_,
_eb_ ∶ _ef_ ∷ _PV_ ∶ _VT_.
Therefore _PV_ ∶ _VT_ ∷ _EB_ ∶ _AE_.
And, by construction, angle _TPV_ = ∠ _AEB_.
Therefore the triangles _TVP_, _AEB_, are similar; and _TP_ is
parallel to _AB_.
Now the construction in this problem is entirely general for any
inclined line _AB_, and a horizontal line _AE_ in the same vertical
plane with it.
So that if we find the vanishing-point of _AE_ in _V_, and from _V_
erect a vertical _VP_, and from _T_ draw _TP_ parallel to _AB_,
cutting _VP_ in _P_, _P_ will be the vanishing-point of _AB_, and (by
the same proof as that given at page 17) of all lines parallel to it.
[Illustration: Fig. 77.]
Next, to find the dividing-point of the inclined line.
I remove some unnecessary lines from the last figure and repeat it
here, Fig. 77., adding the measuring-line _aM_, that the student may
observe its position with respect to the other lines before I remove
any more of them.
Now if the line _AB_ in this diagram represented the length of the
line _AB_ in reality (as _AB_ _does_ in Figs. 10. and 11.), we should
only have to proceed to modify Corollary III. of Problem II. to this
new construction. We shall see presently that _AB_ does not represent
the actual length of the inclined line _AB_ in nature, nevertheless we
shall first proceed as if it did, and modify our result afterwards.
In Fig. 77. draw _ad_ parallel to _AB_, cutting _BT_ in _d_.
Therefore _ad_ is the sight-magnitude of _AB_, as _aR_ is of _AB_ in
Fig. 11.
[Illustration: Fig. 78.]
Remove again from the figure all lines except _PV_, _VT_, _PT_, _ab_,
_ad_, and the measuring-line.
Set off on the measuring-line _am_ equal to _ad_.
Draw _PQ_ parallel to _am_, and through _b_ draw _mQ_, cutting _PQ_ in
_Q_.
Then, by the proof already given in page 20, _PQ_ = _PT_.
Therefore if _P_ is the vanishing-point of an inclined line _AB_, and
_QP_ is a horizontal line drawn through it, make _PQ_ equal to _PT_,
and _am_ on the measuring-line equal to the sight-magnitude of the
line _AB_ _in the diagram_, and the line joining _mQ_ will cut _aP_ in
_b_.
We have now, therefore, to consider what relation the length of the
line _AB_ in this diagram, Fig. 77., has to the length of the line
_AB_ in reality.
Now the line _AE_ in Fig. 77. represents the length of _AE_ in
reality.
But the angle _AEB_, Fig. 77., and the corresponding angle in all the
constructions of the earlier problems, is in reality a right angle,
though in the diagram necessarily represented as obtuse.
[Illustration: Fig. 79.]
Therefore, if from _E_ we draw _EC_, as in Fig. 79., at right angles
to _AE_, make _EC_ = _EB_, and join _AC_, _AC_ will be the real length
of the line _AB_.
Now, therefore, if instead of _am_ in Fig. 78., we take the real
length of _AB_, that real length will be to _am_ as _AC_ to _AB_ in
Fig. 79.
And then, if the line drawn to the measuring-line _PQ_ is still to cut
_aP_ in _b_, it is evident that the line _PQ_ must be shortened in the
same ratio that _am_ was shortened; and the true dividing-point will
be _Q′_ in Fig. 80., fixed so that _Q′P′_ shall be to _QP_ as _am′_ is
to _am_; _am′_ representing the real length of _AB_.
But _am′_is therefore to _am_ as _AC_ is to _AB_ in Fig. 79.
Therefore _PQ′_ must be to _PQ_ as _AC_ is to _AB_.
But _PQ_ equals _PT_ (Fig. 78.); and _PV_ is to _VT_ (in Fig. 78.) as
_BE_ is to _AE_ (Fig. 79.).
Hence we have only to substitute _PV_ for _EC_, and _VT_ for _AE_, in
Fig. 79., and the resulting diagonal _AC_ will be the required length
of _PQ′_.
[Illustration: Fig. 80.]
It will be seen that the construction given in the text (Fig. 46.) is
the simplest means of obtaining this magnitude, for _VD_ in Fig. 46.
(or _VM_ in Fig. 15.) = _VT_ by construction in Problem IV. It should,
however, be observed, that the distance _PQ′_ or _PX_, in Fig. 46.,
may be laid on the sight-line of the inclined plane itself, if the
measuring-line be drawn parallel to that sight-line. And thus any form
may be drawn on an inclined plane as conveniently as on a horizontal
one, with the single exception of the radiation of the verticals,
which have a vanishing-point, as shown in Problem XX.