((1 - _a_)/_n_)_f_,
in the Line of Projection, and thence (at the Angle given)
((_nt_ - 1 + _a_)/_nn_)_g_,
in the Line of Descent; the Point in the Curve answering to these, is
the Place of the Project answering to that Moment.
46. I am aware of some Objections to be made, whether to some Points of
the Process, or to some of the Suppositions. But I saw not well how to
wave it, without making the Computation much more perplex'd. And in a
Matter so nice, and which must depend upon Physical Observations, 'twill
be hard to attain such Accuracy, as not to stand in need of some
Allowances.
47. Somewhat might have been farther added to direct the Experiments
suggested at ¶ 21, and 31. But that may be done at leisure, after
deliberation had, which way to attempt the Experiment.
48. The like is to be said of the different resistance which different
Bodies may meet with in the same _Medium_, according to their different
Gravities (extensively or intensively consider'd) and their different
Figures and Positions in Motion. Whereof we have hitherto taken no
account; but supposed them, as to all these, to be alike and equal.
_POSTSCRIPT._
49. The Computation in ¶ 41, 42, 43, may (if that be also desired) be
thus represented by Lines and Spaces. The Ablatives _a_, _ma_, _mma_,
&c. (being the same with the first Column taken backward) are fitly
represented by the Segments of _NF_ (beginning at _N_) in Figure 5 and
6, and therefore by Parallelograms on these Bases, assuming the common
height of _Fh_, or _NQ_; the Aggregate of which is _Nh_, or _FQ_. And,
so many times 1, by so many equal Spaces, on the same Bases, between the
same Parallels, terminated at the Hyperbola: The Aggregate of which is
_hFNQn_. From whence if we subduct the Aggregate of Ablatives _FY_; the
remaining Trilinear _hQn_, represents the Descent.
50. If to this of Gravity, be joined a projecting Force; which is to the
Impulse of Gravity as _hK_ to _hF_ (be it greater, less, or equal) taken
in the same Line; the same Parallels determine proportional
Parallelograms, whose Aggregate is _KQ_.
51. And therefore if this be a perpendicular Projection downwards; then
_hKkn_ (the Sum of this with the former) represents the Descent.
52. If it be a Perpendicular upwards; then the difference of these two
represents the Motion; which so long as _KQ_ is the greater, is
Ascendent; but Descendent, when _hQn_ becomes greater; and it is then at
the highest when they be equal.
53. If the Projection be not in the same Perpendicular, (but Horizontal,
or Oblique) then _KQ_ represents the Tangent of the Curve; and _hQn_ the
Ordinates to that Tangent, at the given Angle.
54. But the Computation before given, I take to be of better use than
this Representation in Figure. Because in such Mathematical Enquiries, I
choose to separate (as much as may be) what purely concerns Proportions;
and consider it abstractly from Lines, or other Matter wherewith it is
incumbred.
As to the Question proposed; whether the resistance of the _Medium_ do
not always take off such a proportional part of the Force moving through
it, as is the specifick Gravity of the _Medium_ to that of the Body
moved in it: (For, if so, it will save us the trouble of Observation.)
I think this can by no means be admitted. For there be many other things
of Consideration herein, beside the intensive Gravity (or, as some call
it, the specifick Gravity) of the _Medium_.
A viscous _Medium_ shall more resist, than one more fluid, though of
like intensive Gravity.
And a sharp Arrow shall bore his way more easily through the _Medium_,
than a blunt-headed Bolt, though of equal Weight, and like intensive
Gravity.
And the same Pyramid with the Point, than with the Base forward.
And many other like Varieties, intended in my ¶ 48.
But this I think may be admitted, namely, That different _Mediums_,
equally liquid, (and other Circumstances alike,) do in such proportion
resist, as is their intensive Gravity. Because there is, in such
proportion, a heavier Object to be removed, by the same Force. Which is
one of the things to which ¶ 33 refers.
And again: The heavier Project once in Motion, (being equally swift, and
all other Circumstances alike) moves through the same _Medium_ in such
proportion more strongly, as is its intensive Gravity. For now the Force
is in such proportion greater, for the removal of the same resistance.
And this Part of what my ¶ 32, insinuates.
But where there is a Complication of these Considerations one with
another, and with many other Circumstances, whereof each is severally to
be considered; there must be respect had to all of them.
[14]
R - 1_V_) _R_ (_V_ + _V/R_, _V/RR_, &c.
_VR_ - V
--------
+ V
+ V/VR
-------
+ V/R
+ VV/RRR
--------
+ V/RR
_&c_.
[15]
_m_ - 1) 1 (1/_m_ + 1/_mm_ + 1/_m_³ + &c.
1 - 1/_m_
-------
+ 1/_m_
+ 1/_m_ - 1/_mm_
--------------
+ 1/_mm_
+ 1/_mm_ - 1/_mmm_
----------------
+ 1/_mmm_
&c.
[16]
1
1 1
1 1 1
1 1 1 1
_&c._
[17]
1
---
_m_
1 1
----- ---
_m_² _m_
1 1 1
----- ----- ---
_m_³ _m_² _m_
1 1 1 1
----- ----- ----- ---
_m_⁴ _m_³ _m_² _m_
_An Instance of the Excellence of the _Modern Algebra_, in the
Resolution of the Problem of finding the _Foci_ of Optick Glasses
Universally. By _E. Halley_, S. R. S._
The Excellence of the _Modern Geometry_ is in nothing more evident, than
in those full and adequate Solutions it gives to Problems; representing
all the possible Cases at one view, and in one general Theorem, many
times comprehending whole Sciences; which deduced at length into
Propositions, and demonstrated after the manner of the _Ancients_, might
well become the Subjects of large Treatises: For whatsoever Theorem
solves the most complicated Problem of the kind, does with a due
Reduction reach all the subordinate Cases. Of this I now design to give
a notable Instance in the Doctrine of _Dioptricks_.
This Dioptrick Problem is that of finding the _Focus_ of any sort of
_Lens_, exposed either to converging, diverging, or parallel Rays of
Light, proceeding from, or tending to a given Point in the _Axis_ of the
_Lens_, be the _Ratio_ of _Refraction_ what it will, according to the
Nature of the transparent Material whereof the _Lens_ is formed, and
also with allowance for the thickness of the _Lens_ between the
_Vertices_ of the two Spherical Segments. This Problem being solved in
one Case, _mutatis mutandis_, will exhibit Theorems for all the possible
Cases, whether the _Lens_ be _Double-Convex_ or _Double-Concave_,
_Plano-Convex_, or _Plano-Concave_, or _Convexo-Concave_, which sort are
usually call'd _Menisci_. But this only to be understood of those Beams
which are nearest to the _Axis_ of the _Lens_, so as to occasion no
sensible difference by their Inclination thereto; and the _Focus_ here
formed, is by _Dioptrick Writers_ commonly call'd the principal _Focus_,
being that of use in _Telescopes_ and _Microscopes_.
Let then (in _Fig. 7. Tab. 5._) BEβ be a double Convex _Lens_, C the
Center of the Segment EB, and K the Center of the Segment Eβ, Bβ the
thickness of the _Lens_, D a Point in the _Axis_ of the _Lens_; and it
is required to find the Point F, at which the Beams proceeding from the
Point D, are collected therein, the _Ratio_ of Refraction being as _m_
to _n_. Let the distance of the Object DB = DA = _d_ (the Point A being
supposed the same with B, but taken at a distance therefrom, to prevent
the coincidence of so many Lines) the _Radius_ of the Segment towards
the Object CB or CA = _r_, and the _Radius_ of the Segment from the
Object Kβ or K = ρ; and let Bβ the thickness of the _Lens_ be =
_t_, and then let the Sine of the Angle of Incidence DAG be to the Sine
of the refracted Angle HAG or CAφ as _m_ to _n_: And in very small
Angles, the Angles themselves will be in the same proportion; whence it
will follow that,
As _d_ to _r_, so the Angle at C to the Angle at D, and _d + r_ will be
as the Angle of Incidence GAD; and again as _m_ to _n_, so _d + r_ to
(_dn + rn_)/_m_, which will be as the Angle GAH = CAφ; This being
taken from ACD which is as _d_, will leave (_m - nd - nr_)/_m_ analogous
to the Angle AφD; and the Sides being in this Case proportional to
the Angles they subtend, it will follow, that as the Angle AφD is to
the Angle ADφ, so is the Side AD or BD to Aφ or Bφ: That is,
Bφ will be = _mdr_/(_m - nd - nr_), which shews in what Point the
Beams proceeding from D, would be collected by means of the first
Refraction; but if _nr_ cannot be subtracted from _m - nd_, it follows
that the Beams after Refraction do still pass on diverging, and the
Point φ is on the same side of the _Lens_ beyond D. But if _nr_ be
equal to _m - nd_, then they proceed parallel to the _Axis_, and the
Point φ is infinitely distant.
The Point φ being found as before, and Bφ - Bβ being given,
which we will call δ, it follows by a Process like the former, that
βF, or the focal Distance sought, is equal to
_δρn_/(_m - δ + mρ_) = _f_.