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From: jat741@rs740.gsfc.nasa.gov (Jordan A. Truesdell)
Newsgroups: rec.pyrotechnics
Subject: Re: Trajectory of a shell
MessageID: <2mpmdi$4eg@paperboy.gsfc.nasa.gov>
Date: 23 Mar 94 15:14:58 GMT
References:
Organization: NASA, Goddard Space Flight Center
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For your projectile fired without the hindrance of atmosphere,
it's easiest to solve your problem in rectangular coordinates,
such that the x axis is the distance from the mortar and the y
axis is the height of the projectile.
Let me use capital letters for variables (V), small letters
_
for subscripts (Vx) and, if necessary, vectors noted as V.
Some definitions:
A = acclereation G = acceleration of gravity =9.8 m/s^2
V = velocity Vi = initial velocity =32.2 ft/s^2
R = distance from origin
S, X = horizintal Distance H,Y = vertical distance (height)
T is time (from launch)
dX (or dV...) is the differential of X (or V...)
_ _
I and J are unit vectors along the X and Y axes
_ _ _ I'm gonna call
A = AxI + AyJ, A^2 = Ax^2 + Ay^2 this "theta"
_ _ _ /
V = VxI + VyJ, V^2 = Vx^2 + Vy^2, tan @ = Vy/Vx
_ _ _
R = RxI + RyJ, Rx = X = S, Ry = Y = H, R^2 = S^2 + H^2
 Y


 _ _ ____ Hmax
  
  
 / \
 / \
 / @ \
_/____________________\_________ X

Smax
Okay, employing a little calculus yields:
dX = Vx dT (or "the change in X is equal to the velocity in the
X direction times the change in time.")
/T
Integrating: X =  Vi cos@ dT => X = Vi T cos@
/0
This gives us the position as a function of time, initial
velocity, and angle of fire. If we knew the time of flight
(be patient... :) we would then know the range.
To find the maximum height, we need to solve for Y when Vy = 0.
To find what Vy is:
/Vy /T
dVy = Ay dT =>  dVy =  (G) dT => Vy = Vi sin@  GT
/Vi sin@ /0
/T
and dY = Vy dT => Y =  (Vi sin@  GT) dT => Y = ViTsin@  .5GT^2
/0
Solving for Vy = 0, or when 0 = Vi sin@  GT
T = (Vi sin@) / G
Substituting this into the expression for Y gives Hmax:
2 2
/Vi sin@\ 1 /Vi sin@\2 Vi sin @
Hmax = Vi  sin@   G  = 
\ G / 2 \ G / 2G
and using T when y=0 (T>0) yields Smax:
2
/Vi sin@\ Vi sin2@
Smax = Vi  cos@ = 
\ G / 2G
which, it was pointed out, is at a maximum when @=45 degrees.
If you wanted to know the equation of the path (height in terms
of distance from the mortar), T can be eliminated to yield:
2
G X 2
Y = X tan@   sec @
2
2Vi
This is, of course, for and ideal situation where there is no
atmosphere. So, if you use the proper values for the moon, this
would be a very useful calculation for Lunar fireworks.
Of course, we do have atmosphere to worry about (thank
goodness), and there are ways to partially account for it,
especially if you have "nice" geometery, like a sphere.
The drag on a sphere is given by:
2
Drag = C q (pi d /4)
D inf
where
C = Coefficient of drag
D 2
q = free stream dynamic pressure = 1/2 rho U
inf inf inf
pi = 3.1415926...
d = diameter of the sphere
rho = density (inf = "at infinity" or the in the free stream)
U = speed of the incoming flow
mu = viscosity; for air, we may use Sutherland's equation,
1.5
6 T
mu = 1.458 x 10  , T is in Kelvin
T + 110.4
5 kg m
the units are kg/s m (mu=1.7894x10 s at 15 C)
1.5
8 T
or mu = 2.27 x 10  , T is in degrees Rankine
T + 198.6
2 7 lbf s
the units are lbf s / ft (mu=3.7383x10 ft^2 at 59 F)
The coefficient of drag is dependant on the Reynolds number of
the flow. the Reynolds number may be calculated as:
Re = rho U d/ mu
inf inf inf
The following graph is an aproximation of the Cd for a sphere
in axisymmetric flow (no spin)
2.0


1.0



0.5 ***
_ ******** *
_************ *****
 *
_ *
 * *
 * **
0.1_______________________________*****____
 3  4  5  6
4x10 10 10 10
In order to account for drag, you "simply" need to add the drag
term to the ideal case above for acceleration. Don't forget to
remember that as your velocity changes, so will your Reynolds
number, and your drag.
Have fun!

+ Jordan Truesdell + +
+ jat741@rs740.gsfc.nasa.gov + "It doesn't take a rocket scientist..." +
+ NASAGoddard Space Flight Center + Unknown +