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Equations of motion and Navy deck guns

  1. Aug 21, 2006 #1
    Dear All,

    Currently I'm working out a formula for Destroyers in the online game BattleGround Europe to be able to blast inland towns with their deck guns. To do this I've had to use the equations of motion. I could use some help on dechipering the final part of the clue. Points to note are:

    1)Muzzle velocity is constant
    2)Air resistance does not factor in the game for shells
    3)I am not concerned with crosswind effects
    4)I have to be able to hit targets higher than the ship so curve is not a nice symmetrical parabola-type

    So I did some theory tests. I have started with a muzzle velocity of 50ms and have chosen a fictional target 165m away 45m above sea level. Here's how I went about solving the problem:

    I worked out that for both horizontal motion and vertical motion the time 't' would be the same at the instance where the horizontal distance and vertical distance were aligned (on the target).

    I worked out an equation for 't' using horizontal motion. Using a=0 it came back with a nice formula which I susbstituted into the equation for vertical motion. At this point all unknowns except the required firing angle disappeared and the equation worked through to the formula below which I have checked several times and I'm pretty certain is correct:

    (cos^2(x))(1100tanx - 300) = 363

    Where ^ denotes 'to the power of' and x is the angle. This just leaves me with the angle to equate but it's been a good while since I did any trig identities and I'm not even sure this can be solved? Can anyone help me out with this last little bit? I'd be most grateful :wink:
  2. jcsd
  3. Aug 21, 2006 #2


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    Multiply through by cos2(x). Then use the identities for sin(2x) and cos(2x). Finally, use the angle addition formula:

    [tex] \sin(a+b) = \sin(a) \cos(b) + \sin(b) \cos(a) [/tex]

    Can you figure out how to use this formula to turn any expression of the form A sin(x) + B cos(x) into C sin(x+a)?
  4. Aug 21, 2006 #3
    Bingo! Some help from a real boffin o:) I'll try what you have just suggested and will be back in touch later tonight. It's quite embarassing to think I could do this kind of stuff easily 8 years ago :rolleyes:
  5. Aug 21, 2006 #4


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    boffin? I don't know what that means, but it sounds like an insult. In which case: hey?!
  6. Aug 21, 2006 #5
    It's an English slang word for 'expert' or someone who is very knowledgable on their subject. It is not an insult, it is a big compliment.

    I ran the cos2 term through the equation and got this:

    1100sin(x)cos(x) - 300(cos2(x)) = 363

    Which using trig identities reduced to this:

    550sin2(x) - 300(cos2(x)) = 363

    Have tried using the cos2x trig identity but it just seems to cause more trouble. Have tried a few times but can't get any further. What should I do for the next step :confused:
  7. Aug 21, 2006 #6


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    Well, do use the cos (2x) identity, and then normalize it. If you have
    A sin x + B cos x
    then divide by C = sqrt(A^2 + B^2) so that (A/C)^2 + (B/C)^2 = 1
    then do what statusx said. If you let b = arctan (B/A), what is cos b and sin b?
    Last edited: Aug 21, 2006
  8. Aug 21, 2006 #7
    I've looked at both suggestions and I'm still clueless to be honest :rolleyes: I'll have another try tonight and get back to you in the morning. Thanks for the help so far guys :wink:
  9. Aug 22, 2006 #8
    Ok guys I finally got it down to this:

    550sin2(x) + 300sin2(x) = 663

    I put the equation into Excel and used the Goal Seek function to get an answer for 'x'. It came back with 0.69 radians or 39.6975 degrees. When I used the formula again using the muzzle velocity still at 50ms and this computed value for 'x' it came back with the right answer!! :bugeye:

    It stated that at the time 't' (4.298s) when the horizontal distance travelled was 165m the vertical distance was 45m which is spot on where I wanted it.

    So I'm happy with this but I still need an easier expression to solve which can be done quickly in the environment without the use of Excel and to be honest chaps I didn't really understand your suggestions, so I could do with a little more explanation :redface:
  10. Aug 22, 2006 #9


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    cos(2x)= cos2(x)-sin2(x)=2 cos2(x)-1. So you can substitute cos(2x) in, and then use the trick mentioned above to simplify the expression to sin(2x+a)+c
  11. Aug 22, 2006 #10


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    How quickly do you need to solve this sort of problem?

    In real life, people would pre-make charts for this sort of thing before there were computers. (Of course, in real life the earth is round, and spins.)
  12. Aug 22, 2006 #11
    Thanks again StatusX for the reply. I'll go try that out tonight mate :smile:


    Well myself and other members of the game don't really want guys to have to be carrying loads of charts around with them each time they play. I'd rather there were a formula they could use to get data quickly. Especially seeing as when you fire the Destroyer's deck guns the recoil is so powerful that it moves the entire ship. After a few salvos you'll need to recalculate. The game changes very quickly indeed as it's entirely a multiplayer environment so one may need to quickly change targets and not want to have to sift trhough a load of charts to do it.

    Plus, using a formula will give more accurate results:wink:
  13. Aug 22, 2006 #12
    Okay I have written down all your suggestions and collected them together. I have tried them all with no luck. I did however manage to successfully reduce the equation using the cos2(x) identity to this:

    11sin2(x) - 3cos2(x) = 10.26

    Now I'm sure once I've broken through the next barrier I'll be ok I just need to know what exactly should I do here?

    I'm most grateful for the help by the way and if I can get this to work I'll post up some screenshots of it in game being used :smile:
  14. Aug 22, 2006 #13


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    ok, now sqrt(11^2 + (-3)^2) = 11.4, so divide both sides by 11.4:
    .965 sin (2x) - .263 cos (2x) = 0.9
    b = arctan (-.263/.965) = -.266
    a = 2x
    .965 sin (2x) - .263 cos (2x) = sin(2x) cos(-.266) + cos(2x) sin(-.266) = sin (2x - .266) = 0.9
    x = .693
    Last edited: Aug 22, 2006
  15. Aug 22, 2006 #14
    That's awesome mate, thanks alot. Just one question though (and this is the bit I've been missing) where did you get the arctan part for b from? That's the bit I just wasn't getting. The rest is all substitutions and the use of the C part which I understood.

    That's wicked work, thanks very much indeed :smile: o:)
  16. Aug 22, 2006 #15


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    Well, you want to use the sine double angle identity on an expression of the form
    A sin a + B cos a
    This means that A has to be interpreted as the cosine of some angle b, and B has to be interpreted as the sine of the same angle, since the identity is sin(a + b) = sin a cos b + cos a sin b. Since sin^2 b + cos^2 b = 1, A^2 + B^2 must equal 1, so the first thing is to normalize the equation so that happens (assume that's already done here). Now, b is an angle such that cos b = A and sin b = B. Look at the diagram:
    http://img206.imageshack.us/img206/64/trianglexs2.png [Broken]
    You can see from this that b will be arctan (B/A).
    Last edited by a moderator: May 2, 2017
  17. Aug 23, 2006 #16
    That's great stuff. I'm most grateful for all your all help and that goes for everybody. If I ever get stuck with anything in the future I know where to come to get some advice:cool: :wink:
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