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Projectile motion with drag trouble

  1. Sep 5, 2014 #1
    Basically..I have a canon that shoots a pingpong ball at 30m/s from any angle between 0-90

    I've derived all my equations as can be seen below:

    So the velocity equations are:

    Vx(t) = (Vxo)e^(-kt/m)
    Vy(t) = (Vyo+gm/k)e^(-kt/m) − gm/k

    Integrate these to get X and Y as functions of time:

    X(t) = ∫(Vxo)e^(-kt/m)dt = (-m/k)(Vxo)e^(-kt/m) + C3
    Y(t) = ∫((Vyo+gm/k)e^(-kt/m) − gm/k)dt = (-m/k)(Vyo+gm/k)e^(-kt/m) − (gm/k)t + C4

    Define the initial position as (0,0), Figure out C3 and C4:

    X(0) = 0 = (-m/k)(Vxo)e^(-k(0)/m) + C3 = (-m/k)(Vxo) + C3; C3 = (m/k)(Vxo)
    Y(0) = 0 = (-m/k)(Vyo+gm/k)e^(-k(0)/m) − (gm/k)(0) + C4 = (-m/k)(Vyo+gm/k) + C4; C4 = (m/k)(Vyo+gm/k)

    So finally:

    X(t) = (m/k)(Vxo)(1 − e^(-kt/m))
    Y(t) = (m/k)(Vyo+gm/k)(1 − e^(-kt/m)) − (gm/k)t

    I'm using this formula for k=CAρ/2m (Not 100% about it)

    Problem is, where ever I look. The only solution to plotting this is via a computer program or excel. I can't code and the last time I used excel was in high school. If I could get a bump in the right direction as to how to go about this. It would be greatly appreciated :)
     
  2. jcsd
  3. Sep 5, 2014 #2
    Your formulas for the position seems correct. I've made a little program using C++ and RooT (surely not the easiest and best way) to plot the motion of your projectile. Unfortunately I can't understand exactly what kind of help you are looking for but if you'll give me a value of the angle and of the ratio [itex] \frac{k}{m} [/itex] I could start showing you the plot for those values. I hope this will be helpful for you.
     
    Last edited: Sep 5, 2014
  4. Sep 5, 2014 #3

    SteamKing

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    Homework Helper

    Well, a calculator and a piece of paper still works, as far as I can tell. It is a lot more effort, though.
     
  5. Sep 5, 2014 #4

    NascentOxygen

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    Staff: Mentor

    Hi Luchekv! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    I daresay you may avail yourself of Wolfram alpha's plotting facilities for this task. http://www.wolframalpha.com/examples/PlottingAndGraphics.html

    Type your one-line instruction into that little box at the top.

    Both plot and parametric plot should be useful. Start with some simple graphs to discover how it works.
     
    Last edited by a moderator: May 6, 2017
  6. Sep 6, 2014 #5
    Apologies gentleman I should have included more information. The specifications of the ping pong ball are as follows:

    Mass = 0.0027 kg
    Cross sectional area: 0.001256
    Drag Coefficient: 0.5
    Air density: 1.2041

    Vxo = 30 Cos 45
    Vyo = 30 Sin 45

    So using the formula provided for K I get: 0.14
    The other thing that confuses me is that the Drag formula is slightly different to what K is so I'm not sure which one I'm meant to use.

    Launch angle is 45° and I know roughly that should be ~9.1m in distance. Except when I use this data in the equations I get numbers either in negatives or 100+

    So just to clarify I need to be able to plot the distances with varying launch angles. I'm just not quite sure how to use the equations I've derived.

    As for Wolfram plotting. I've been using a graphics calculator to try and plot a very similar problem (A bullet through water) and even when I use the same numbers as in the textbook..the plot looks nothing like it. So I'm guessing I'm going about this the wrong way.

    Valerioperi those plots would be very helpful, is there anyway I could get a hold of that program? Apologies if that's asking too much.
     
  7. Sep 6, 2014 #6
    I'll show you soon the graph for the data you gave me. I would give you also the source code even if I have to figure out how since I'm new in the forum. Nevertheless to understand and use it you'll need:
    1. A little skills in C/C++ programming ( trust me, mine are very very poor)
    2. A C++ compiler
    3. RooT which is a free data analysis tool from CERN
     
  8. Sep 6, 2014 #7
    I'm not really good at fluid dynamics but the formula you used for k seems a bit wired. I don't get why it is already divided by m. More over, why you use [itex] k=\frac{1}{2}CA\rho [/itex] instead of [itex] k=-6\pi\eta R [/itex] where R is the radius of the ball and [itex] \eta [/itex] is the viscosity of air? I think that the first one has a dipendence of the square of v while the second has a linear dependance from v. It seems like you mixed them but in order to choose you need Reynolds number. I don't know if what I'm saying makes any sense, sorry :smile:
     
  9. Sep 6, 2014 #8
    Thanks valerioperi. I have a little experience in Java haha. I used k=CAρ/2m because that's what they used on another website with a similar problem (Measuring the distance of a cannonball) So I thought it would be the same (I agree though, bit odd)...As for Reynolds number. The situation is where no breeze is present. Only the drag induced by air. I thought it would be negligible.
     
  10. Sep 6, 2014 #9
    This graph (graph) shows the result using your formula and your data using the value 0.14 for k/m and not only k. Still there are big dimensional probems which make me believe that this is not the correct solution.
    [tex]
    F=ma=-kv
    [/tex]
    I think that's what you use according to your formulas.
    However, if we use the formula you suggested for k, we get:
    [tex]
    [kg\frac{m}{s^2}]=[\frac{kg}{s}]
    [/tex]
    This is obviously an absurd that shows how the force depends on the square of the velocity using this formula for k. This would mean that all the derivation and integration should be changed. I'm sorry, maybe it would be useful if you could link your source.



    Using instead the other formula for k that I suggested for which your formulas seems correct this would be the result (graph2). (using [itex]\eta=1.81*10^{-5})[/itex])
     

    Attached Files:

    Last edited: Sep 6, 2014
  11. Sep 6, 2014 #10
  12. Sep 6, 2014 #11
    I definitely think that's the best formula (consider the high velocity of the ball). Unfortunately, as I said, this formula says that [itex] F \propto v^2 [/itex] and then you have to calculate again from the beginning all your formulas. I promise I'll produce a new graph with your new formulas once you finished :smile:
     
  13. Sep 6, 2014 #12
    The drag coefficient given above is good only for the drag that is quadratic in speed. There is no closed-form solution of the problem with this drag, so one will have to integrate equations numerically to obtain the trajectory for any given initial conditions.
     
  14. Sep 7, 2014 #13
  15. Sep 7, 2014 #14
    Yes, that would be a simple method of doing it. Technically it uses Euler's method of numerical integration. This is probably the oldest integration method known to humanity, and so quite venerable, but it is not not very accurate.
     
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