PROPOSTION IV PROBLEM II (In BOOK II):
 
Supposing the force of gravity in any homogeneous medium to be uniform, and to tend perpendicularly to the plane of the horizon: to define the motion of a projectile therein, which suffers resistance propotional to its velocity.
   
1 Draw the projectile with the initial velocity DP from D to F:
 

  Notice: Since there is the force of gravity in the medium, the projectile will not be in the shape of a parabola.
   
2 From P, fall the perpendicular PC on the horizontal line DC:
 
   
3 Let a be the maximum point of projectile DF. From a, draw aA perpendicular to DC cuts DC and DP:
 
 

Since the projectile before a is not affect by the force of gravity, i.e. only affected by the resistance, we can have the following ratio:
DA : AC = resistance : gravity

Moreover, since DP is the initial velocity, it is just affected by the resistance, and CP can represent the force of gravity, we can have the following ratio:
DP : CP = resistance : gravity
i.e. DP : CP = DA : AC

   
4 Use DC and CP as the asymptotes, draw any hyperbola GBS inside the triangle DPC, such that the hyperbola GBS will cut the projectile in two different points, with DG and AB as the perpendiculars.
 
  Notice: By Proposition II (in Book II), the area under the hyperbola represents the time.
   
5 Complete the rectangle DGKC and let the point where GK cuts AB be Q:
 
   
6 Complete the rectangle GKHE with EH perpendicular to AB and cuts AB at B:
 
  Notice: Since EH is perpendicular to AB, and GKHE is a rectangle, if we extend EG and HK to DC, they are perpendicular to DC. Also, ED and HC is a straight line.
   
7

Take a line N such that
N : QB = DC : CP

Take any point R on DC and erect RV perpendicular to DC with V on DP. And RV intersects the projectile DaF, EH, the hyperbola GBS, and GK at r, I, T, and t.

 
   
8

In this perpendicular RV,
take Vr = tGT / N.

In time RDGT, the projectile DraF will arrive at the point r.

Draw a tangent line rL at r.

 
  Therefore, velocity at r = rL
   
9

Since the triangle DRV is similar to the triangle DCP
and N : QB = DC : CP

We have
N : QB = DC : CP = DR : RV

Therefore,
RV = (DR * QB) / N

Moreover,
Rr = RV - Vr
= (DR * QB - tGT) / N
= (DR * AB - RDGT) / N

   
10

Now, take the time be represented by RDGT, and by Laws Corollary II (in Book I), we can distinguish the motion of the body into
1) lateral motion
2) ascent motion

For 1),
length = DR (by Proposition II in Book II)
height = (DR * AB - RDGT) / N = Rr (by Proposition III in Book II)

But, focus on the left bottom of the diagram:

 
 

At the beginning of the time RDGT, i.e. the very left of the above diagram,
RDGT = DR * AQ

Therefore, height = (DR * AB - RDGT) / N
i.e. height = Rr = (DR * AB - RDGT) / N , cancel out the length DR
= QB / N
= DC / CP
= resistance / gravity

   
11 DR = length, Rr = height
As a result, the body will move in the projectile Draf, which is the locus of the point r.

@

Summary:

 
 

DA : AC = DP : CP = N : QB = resistance : gravity
Vr = tGT / N
RV = (DR * QB) / N
Rr = RV - Rr = (DR * AB - RDGT) / N = DC / CP

There are seven corollaries followed Proposition IV Problem II:

 

 

Corollary I:

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.

 
 

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.

 
  Now,
RX = (DR * AB) / N
and Xr = RX - Rr = RDGT / N
So, Xr is proportional to time.
   

Corollary II:

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.
   
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.

   
  At the very beginning of the motion,
the latus rectum of this parabola = (DV)^2 / Vr
where Vr = tGT / N = (DR * Tt) / 2N
   
  If we draw a tangent line to the hyperbola GTS in G, it will be parallel to the line DK:
 
 

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)

   
  Since the triangle DRV is similar to the triangle DCP,
we have
DR / DV = DC / DP
   
  Therefore,
Vr = ((DV)^2 * CK * CP) / ((2DP)^2 * QB)
   
  So,
the latus rectum of the parabola
= (DV)^2 / Vr
= ((2DP)^2 * QB) / (CK * CP)
   
  Since
QB / DA = CK / AC,
(DV)^2 / Vr = ((2DP)^2 * DA) / (AC * CP)
and DA : AC = DP : CP
   
  As a result,
the latus rectum of the parabola
= (DV)^2 / Vr
= (2DP) * (DA / AC) * (DP / CP)
= resistance / gravity
   

Corollary IV:

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.
   
  If velocity DP is given, we can find the latus rectum of the parabola.
   
 

Next, we can find DP by
2DP / latus rectum = gravity / resistance
i.e. DP = (1 / 2) * latus rectum * (gravity / resistance)

   
 

Now, we can find A by
1) cutting DC in A, and
2) the fact that
(CP * AC) / (DP * DA) = gravity / resistance

   
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
   
 

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 * #)

   
  Therefore, the velocity at the beginning of hte motion = DP.
   
 

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.

   
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.
   
  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.
   
Corollary VII:

Suppose there are two equal bodies in a homogeneous medium. Project them with teh same velocity from D in different angles CDP and CDp.

 
 

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

   
  On the other hand, we actually do the experiment and get the data F'f' / Df'
   
  Now, find
Ff / Df - F'f' / Df' = MN
such that
 
  if MN is negative, we get
 
  Repeat the above step for two three times, get
 
  Mark the point the NNN cuts the line S as X
   
 

So, we get the actual ratio:
resistance / gravity = SX
and find DF'

   
  Now, we can get the true DP':
DP' / DP = DF' / DF
i.e. DP' = (DF' / DF) * DP
   
 

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.