Corollary I:
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Rr = (DR * AB - RDGT) / N
Draw the line PY from P parallel to DC to the extended line of
AB, i.e. the point Y is the intersection of PY and AY.
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Draw the line DZ, which connects the points D and Y, mees the extended
line of CP, i.e. the point Z is the intersection of DZ and CZ.
And extend the perpendicular VX from the line RV, such that X is
on DZ.
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Now,
RX = (DR * AB) / N
and Xr = RX - Rr = RDGT / N
So, Xr is proportional to time. |
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Corollary II:
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Since we can take any point R on
DC to get various Rr, and the points D, a, and F are fixed, we can
find the projectile Draf easily by making a table with various Rr. |
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Corollary III: |
If we construct a parabola to D and make it satisfies teh following
conditions:
1) diameter DG produces downwards
2) latus rectum : 2DP
= resistance at the beginning of the motion : gravity
3) velocity = DP
It will just like constructing the projectile Draf in a nonresisting
medium, such that teh projectile will be a parabola.
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At the very beginning of the motion,
the latus rectum of this parabola = (DV)^2 / Vr
where Vr = tGT / N = (DR * Tt) / 2N |
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If we draw a tangent line to the
hyperbola GTS in G, it will be parallel to the line DK: |
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From the above diagram, we get
Tt / DR = CK / DC =>
Tt = (CK * DR) / DC
and
N : QB = DC : CP => N = (QB * DC) / CP
Therefore,
Vr = (DR * Tt) / 2N
= ((DR)^2 * CK * CP) / ((2DC)^2 * QB)
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Since the triangle DRV is similar
to the triangle DCP,
we have
DR / DV = DC / DP |
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Therefore,
Vr = ((DV)^2 * CK * CP) / ((2DP)^2 * QB) |
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So,
the latus rectum of the parabola
= (DV)^2 / Vr
= ((2DP)^2 * QB) / (CK * CP) |
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Since
QB / DA = CK / AC,
(DV)^2 / Vr = ((2DP)^2 * DA) / (AC * CP)
and DA : AC = DP : CP |
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As a result,
the latus rectum of the parabola
= (DV)^2 / Vr
= (2DP) * (DA / AC) * (DP / CP)
= resistance / gravity |
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Corollary IV:
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By corollary III, if we have the
initial velocity of a body which we projected, and the resistance
of the medium at the beginning of hte motion, then we can find the
projectile Draf. |
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If velocity DP is given, we can
find the latus rectum of the parabola. |
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Next, we can find DP by
2DP / latus rectum = gravity / resistance
i.e. DP = (1 / 2) * latus rectum * (gravity / resistance)
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Now, we can find A by
1) cutting DC in A, and
2) the fact that
(CP * AC) / (DP * DA) = gravity / resistance
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Corollary V: |
Converse to Corollary IV,
if the projectile DraF is given,
we can also find:
1) velocity of boby at r
2) resistance of the medium at r |
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If (CP * AC) / (DP * DA) [= gravity / resistance] = # is given,
we can get
1) the resistance of the medium at the beginning of the motion
= gravity * #
2) the latus rectum of the parabola
= 2DP
= (CP * AC) / (DA * #)
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Therefore, the velocity at the
beginning of hte motion = DP. |
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Then, from the tangent line rL, we can find
1) the velocity at r = rL
2) the resistance which is proportional to rL at r.
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Corollary
VI: |
The
length 2DP is proportional to the velocity only. It will not be augmented
or diminished by the change of the angle CDP. Only the change of velocity
will augment or diminish it. |
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Since
2DP : latus rectum of the parabola
= gravity in D : resistance in D
and from the velocity augmented, the resistance is augmented in the
same ratio, this property holds. |
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Corollary VII:
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Suppose there are two equal bodies in a homogeneous medium. Project
them with teh same velocity from D in different angles CDP and CDp.
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On one hand, we assume that these two bodies land at F and f on
the horizontal DC.
Let the length DP
and resistance / gravity = SM
By all these assumptions, find DF, Df and Ff / Df
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On the other hand, we actually
do the experiment and get the data F'f' / Df' |
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Now, find
Ff / Df - F'f' / Df' = MN
such that |
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if MN is negative, we get |
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Repeat the above step for two three
times, get |
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Mark the point the NNN cuts the
line S as X |
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So, we get the actual ratio:
resistance / gravity = SX
and find DF'
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Now, we can get the true DP':
DP' / DP = DF' / DF
i.e. DP' = (DF' / DF) * DP |
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Same as before, we can now get
1) the projectile DraF
2) the velocity of the body at r
3) the resistance of the body ar r.
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