186 lines
6.0 KiB
Plaintext
186 lines
6.0 KiB
Plaintext
X-NEWS: cudnvr rec.pyrotechnics: 12454
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Path: carbon.denver.colorado.edu!copper!csn!ncar!gatech!europa.eng.gtefsd.com!paladin.american.edu!zombie.ncsc.mil!cs.umd.edu!newsfeed.gsfc.nasa.gov!jat741
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From: jat741@rs740.gsfc.nasa.gov (Jordan A. Truesdell)
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Newsgroups: rec.pyrotechnics
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Subject: Re: Trajectory of a shell
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Message-ID: <2mpmdi$4eg@paperboy.gsfc.nasa.gov>
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Date: 23 Mar 94 15:14:58 GMT
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References: <Cn0oIq.1B3@newcastle.ac.uk> <kwdCn42HL.KHF@netcom.com>
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Organization: NASA, Goddard Space Flight Center
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Lines: 170
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For your projectile fired without the hindrance of atmosphere,
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it's easiest to solve your problem in rectangular coordinates,
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such that the x axis is the distance from the mortar and the y
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axis is the height of the projectile.
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Let me use capital letters for variables (V), small letters
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_
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for subscripts (Vx) and, if necessary, vectors noted as V.
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Some definitions:
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A = acclereation G = acceleration of gravity =9.8 m/s^2
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V = velocity Vi = initial velocity =32.2 ft/s^2
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R = distance from origin
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S, X = horizintal Distance H,Y = vertical distance (height)
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T is time (from launch)
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dX (or dV...) is the differential of X (or V...)
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_ _
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I and J are unit vectors along the X and Y axes
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_ _ _ I'm gonna call
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A = AxI + AyJ, A^2 = Ax^2 + Ay^2 this "theta"
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_ _ _ /
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V = VxI + VyJ, V^2 = Vx^2 + Vy^2, tan @ = Vy/Vx
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_ _ _
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R = RxI + RyJ, Rx = X = S, Ry = Y = H, R^2 = S^2 + H^2
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| Y
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| _ _ ____ Hmax
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| - -
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| - -
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| / \
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| / \
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| / @ \
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_/____________________\_________ X
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|Smax
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Okay, employing a little calculus yields:
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dX = Vx dT (or "the change in X is equal to the velocity in the
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X direction times the change in time.")
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/T
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Integrating: X = | Vi cos@ dT => X = Vi T cos@
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/0
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This gives us the position as a function of time, initial
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velocity, and angle of fire. If we knew the time of flight
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(be patient... :-) we would then know the range.
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To find the maximum height, we need to solve for Y when Vy = 0.
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To find what Vy is:
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/Vy /T
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dVy = Ay dT => | dVy = | (-G) dT => Vy = Vi sin@ - GT
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/Vi sin@ /0
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/T
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and dY = Vy dT => Y = | (Vi sin@ - GT) dT => Y = ViTsin@ - .5GT^2
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/0
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Solving for Vy = 0, or when 0 = Vi sin@ - GT
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T = (Vi sin@) / G
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Substituting this into the expression for Y gives Hmax:
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2 2
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/Vi sin@\ 1 /Vi sin@\2 Vi sin @
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Hmax = Vi |-------| sin@ - - G |-------| = -----------
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\ G / 2 \ G / 2G
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and using T when y=0 (T>0) yields Smax:
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2
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/Vi sin@\ Vi sin2@
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Smax = Vi |-------| cos@ = ---------
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\ G / 2G
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which, it was pointed out, is at a maximum when @=45 degrees.
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If you wanted to know the equation of the path (height in terms
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of distance from the mortar), T can be eliminated to yield:
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2
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G X 2
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Y = X tan@ - ----- sec @
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2
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2Vi
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This is, of course, for and ideal situation where there is no
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atmosphere. So, if you use the proper values for the moon, this
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would be a very useful calculation for Lunar fireworks.
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Of course, we do have atmosphere to worry about (thank
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goodness), and there are ways to partially account for it,
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especially if you have "nice" geometery, like a sphere.
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The drag on a sphere is given by:
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2
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Drag = C q (pi d /4)
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D inf
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where
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C = Coefficient of drag
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D 2
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q = free stream dynamic pressure = 1/2 rho U
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inf inf inf
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pi = 3.1415926...
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d = diameter of the sphere
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rho = density (inf = "at infinity" or the in the free stream)
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U = speed of the incoming flow
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mu = viscosity; for air, we may use Sutherland's equation,
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1.5
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-6 T
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mu = 1.458 x 10 ----------- , T is in Kelvin
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T + 110.4
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-5 kg m
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the units are kg/s m (mu=1.7894x10 s at 15 C)
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1.5
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-8 T
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or mu = 2.27 x 10 ---------- , T is in degrees Rankine
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T + 198.6
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2 -7 lbf s
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the units are lbf s / ft (mu=3.7383x10 ft^2 at 59 F)
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The coefficient of drag is dependant on the Reynolds number of
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the flow. the Reynolds number may be calculated as:
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Re = rho U d/ mu
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inf inf inf
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The following graph is an aproximation of the Cd for a sphere
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in axisymmetric flow (no spin)
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2.0|
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1.0|
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0.5| ***
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_| ******** *
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_|************ *****
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| *
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_| *
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| * *
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| * **
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0.1|_______________________________*****____
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| 3 | 4 | 5 | 6
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4x10 10 10 10
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In order to account for drag, you "simply" need to add the drag
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term to the ideal case above for acceleration. Don't forget to
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remember that as your velocity changes, so will your Reynolds
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number, and your drag.
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Have fun!
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--
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+ Jordan Truesdell + +
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+ jat741@rs740.gsfc.nasa.gov + "It doesn't take a rocket scientist..." +
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+ NASA-Goddard Space Flight Center + -Unknown +
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