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Death Zone of an Artillery Cannon.

  1. Dec 2, 2005 #1
    nomenclature: vi = initial speed, z = height x,y are the 'ground', φ = elevation angle of the barrel, α. = α 'dot' = dα/dt, α: = double dot

    show that an artillery cannon can hit a target anywhere in the surface given by:
    g²r²=(vi²)²-2gz(vi²)
    ignore air resistance. no wind.
    note: this is not intended to be done as a surface integral.

    so what i did was decided that it can shoot something directly above it:
    zmax = vi²/2g, zD = 0
    then considering the 2-d case (y=0)
    and then as φ decreases, so rotates the point of max height (up to φ = ╥/2, but then the max height moves back inside this trajectory). This is going to make a paraboloid from (0,zmax) to (?,0). The equation of tha paraboloid is z = Ax + Bx² + C
    at zmax x=0 so zmax = C = vi²/2g
    z. = Ax. + Bx.²
    z: = Ax: + Bx:² = -g, but x: = 0 so the equation for the paraboloid is wrong???

    help please.
     
  2. jcsd
  3. Dec 2, 2005 #2
    nomenclature: vi = initial speed, z = height x,y are the 'ground', φ = elevation angle of the barrel, α. = α 'dot' = dα/dt, α: = double dot

    show that an artillery cannon can hit a target anywhere in the surface given by:
    g²r²=(vi²)²-2gz(vi²)
    ignore air resistance. no wind.
    note: this is not intended to be done as a surface integral.

    so what i did was decided that it can shoot something directly above it:
    zmax = vi²/2g, zD = 0
    then considering the 2-d case (y=0)
    and then as φ decreases, so rotates the point of max height (up to φ = ╥/2, but then the max height moves back inside this trajectory). This is going to make a paraboloid from (0,zmax) to (?,0). The equation of tha paraboloid is z = Ax + Bx² + C
    at zmax x=0 so zmax = C = vi²/2g
    z. = Ax. + Bx.²
    z: = Ax: + Bx:² = -g, but x: = 0 so the equation for the paraboloid is wrong???

    help please.
     
  4. Dec 2, 2005 #3

    siddharth

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    First of all you need to find the equation of the trajectory for a fixed value of [tex] \phi [/tex]. (Where [itex] \phi [/itex] is the elevation angle)

    Then as you change the value of [tex] \phi [/tex], the trajectory changes.

    That is, for each value of [tex] \phi [/tex], you will have a different parabola. So, you need to find the envelope of this family of parabolas.
    Parabola because if you work with [itex] r [/itex] being the displacement along the x-y plane and [itex] z [/itex] being the hieght above this plane, you will get [itex] z=f(r, \phi) [/tex] which will be the equation of the parabola. So the envelope of this family of parabolas will also be a parabola.

    As this does not depend on which direction you shoot the cannon in the x-y plane, The same equation will give you the required surface.

    Have you been taught how to find the envelope of a given family of curves?
     
    Last edited: Dec 2, 2005
  5. Dec 2, 2005 #4

    siddharth

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    emptymaximum, I have replied to your thread in the Introductory Physics sub
    forum. In the future, don't double post the same question.
     
  6. Dec 2, 2005 #5
    no.
    that's my problem.
    finding the enveloping function.

    the trajectory equation for fixed φ is: z= Ax +bx² where:
    A = z.i/x.i and B = -g/2(x.i)²

    how to get enveloping function:

    it is going to be a paraboloid of the form z = Dx + Ex² + C, where C is the maximum height for φ = 90. C = vi²/2g
    x is going to be 1/2 xmax, bacause the max height ouccurs at halfway to the max range.
    what's D and E?
    how do i figure them out?
     
  7. Dec 3, 2005 #6
    YES!!!!! I Got it!

    When you come online Ill help you solve it.

    Stop complicating things with a dot a double dot and so on....not necessary...

    Also, your equations for those parablas.....another overcomplication not necessary...

    finding the enveloping function.....not necessary...
     
    Last edited: Dec 3, 2005
  8. Dec 3, 2005 #7

    siddharth

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    Perhaps, but it's not complicated either :smile: . Once you finish this, I'll post and show what I mean.
     
  9. Dec 3, 2005 #8
    finding the enveloping function makes it easier doesn't it?
    i know that that is the 'recommended' way to do it.
    also, i would like to learn how to evelope functions anyways.
    i like new skills.
     
  10. Dec 3, 2005 #9
    First thing you should do is start looking at simple planar projectile motion cases and look for a trend. Drawing pictures will be vital.

    Id be really interested to see what you mean, sid.
     
    Last edited: Dec 3, 2005
  11. Dec 3, 2005 #10

    siddharth

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    Here is the result

    http://mathworld.wolfram.com/Envelope.html

    I guess you could look up the proof in a Calculus text book or search for it on the net.

    As I said, first find the equation of the trajectory. (For a fixed value of [tex] \phi [/tex])

    It will be of the form
    [tex] z = kr - ar^2(1+k^2) [/tex]

    where a is a constant and k is some function of [tex] \phi [/tex]

    So, to determine the envelope we use the functions
    [tex] z = kr - ar^2(1+k^2) [/tex]
    and
    [tex] r-2akr^2=0[/tex]
    Eliminating k will give you the answer.
     
    Last edited: Dec 3, 2005
  12. Dec 3, 2005 #11
    why is y a function of the radius? z is a function of y or x for a fixed theta (along either the x or y directions). I have never seen this before, interesting, please tell me more. It was not in the calc I learned.

    I think you will find my way much simpler though :-)
     
    Last edited: Dec 3, 2005
  13. Dec 3, 2005 #12

    siddharth

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    Sorry, I should have mentioned that y is the height above the ground. For consistency, I should have used z instead of y.
    I have edited it now
     
    Last edited: Dec 3, 2005
  14. Dec 4, 2005 #13
    i got it i think. the equation of the path of the the line through the max heights is:
    z = Ax²+Bx +C
    my equation in the first post is wrong because the z-intercept isn't zero, it's max height.

    C is the max haight, when x = 0 = vi²/2g
    to find B:
    d(zmaz)/dx = 2Ax + B, but at zmax x = 0 so B = 0.

    A is found by setting z = 0, then x = xmax = vi²/g so:
    0 = A(vi²/g)² + vi²/2g
    0 = A(vi²/g +1/2)vi²/g
    0 = A(vi²/g +1/2)
    A =-g/2vi² so then:

    z = -gx²/2vi² + vi²/2g
    gx²/2 = (vi²)²/2g -zvi²
    g²x² = (vi²)² -2gzvi²

    making it 3-d is putting x = r so

    g²r² = (vi²)² -2gzvi²

    is that good?

    i looked up enveloping functions and i think it's more for Lagrangian mechanics isn't it? variational calulus? that's too advanced for where we are in the book.
    and i have no idea how you got
    [tex] z = kr - ar^2(1+k^2) [/tex]]
    i know the second expression is by d/dk but the first one is like...
    please can you help me out with that? that is something i would very much like to learn.

    cyrus what's your easy way? i'd like to see that too.
    thank you guys
     
    Last edited: Dec 4, 2005
  15. Dec 4, 2005 #14
    Ok, so here is what I did,

    The basics, which I wont bother proving but should be already commonly known:

    The max. Height is given by:

    [tex] z = \frac{v_0^2}{2g} [/tex] (1)

    The max. Range at 45 degrees is given by:

    [tex] R = \frac{v_0^2}{g} [/tex] (2)


    Ok now for the real work. I started off by drawing a simple picture. When I shoot it straight up, it reaches its max height, z, given by equation (1). When I shoot it at its max range, the distance is given by equation (2). Now, as I vary [tex] \phi [/tex], I will sweep out a parabola, with a vertex at (x=0,y=0,z=H), down to (x=R,,y=0,z=0). (But I dont care what the equation of it is). All I know is that it makes some parabolic shape, its clearly not circular, as [tex] H not = R [/tex]. Similarly, its not an ellipse, because if it were, the slope at the endpoint, x=R would be infinite, so that the curvature would make the ellipse smooth. So the only shape left to conclude is a parabola. But this is just the profile, rotate this around 360 degrees and you get the solution for any direction, both phi and theta, where those are your standard spherical coordinate system angles. So we have a downward opening parabolid! That is given by the general equation:

    [tex] \frac{z}{c} = \frac {x^2}{a^2} + \frac{y^2}{b^2} [/tex]

    But we can solve this by using some simple geometric observations!
    We know that the horizontal traces are circles! Therfore, a=b=k, and we can just factor that out:

    [tex] \frac{z}{c} = \frac{x^2+y^2}{2k^2} [/tex]

    But we also know it is centered at (x,y,z)=(0,0,H),

    therefore the general equation of our parabaloid is given by:

    [tex] z-H = - \frac{1}{2k^2} (x^2+y^2) [/tex]

    But writting [tex] \frac{1}{2k^2} [/tex] is alot of work, so lets just make it a new constant p.

    [tex] z-H = - p [x^2+y^2] [/tex]

    Now, we know that (x,y,z)=(0,0,H), so we can check this equation for consistency:

    [tex] z-H = 0 [/tex]

    or,

    [tex] z=H [/tex]

    Ok! we know were on the right track.

    Now, we check at say, (x,y,z) =(R,0,0).

    [tex] -H = -p(R^2) [/tex]

    so that implies that p is:

    [tex] \frac{H}{R^2} = p [/tex]

    But we know from equations (1), (2), that this ratio is equal to:

    [tex] \frac{g} {2(v_0^2)} [/tex]

    So lets plug this back in for p,

    [tex] z-H = - \frac{g} {2(v_0^2)} [x^2+y^2] [/tex]

    Now, we make the observation that:

    [tex] r^2 = x^2+y^2 [/tex]

    and we-write the equation once more:

    [tex] z-H = - \frac{g} {2(v_0^2)} r^2 [/tex]

    Now, we replace H with its value from equation (1):

    [tex] z- \frac{v_0^2}{2g} = - \frac{g} {2(v_0^2)} r^2 [/tex]

    Isolate [tex] r^2 [/tex]

    [tex] - \frac{2(v_0^2)} {g}[z- \frac{v_0^2}{2g}] = r^2 [/tex]

    and now distribute the minus sign,

    [tex] \frac{2(v_0^2)} {g}[ \frac{v_0^2}{2g} -z] = r^2 [/tex]

    Now multiply both sides by [tex] g^2 [/tex] and simplify the equation and you get:

    [tex] (v_0^2)^2 - 2gzv_0^2 = g^2r^2 [/tex]

    And that completes that. :tongue:
    This problem was alot of fun, thanks for posting it!
     
    Last edited: Dec 4, 2005
  16. Dec 5, 2005 #15

    siddharth

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    Here is my way
    If the cannon is shot in the x-y plane at an angle [tex] \phi [/tex],

    [tex] r = v_0 t \cos\phi [/tex]

    and

    [tex] z = v_0 t \sin\phi - \frac{gt^2}{2} [/tex]
    where r is the distance along the plane from the point of firing and z is the height.
    Eliminating t from the above equations
    [tex] z = r \tan\phi - \frac{gr^2}{2v_0^2 \cos^2\phi} [/tex]
    Put [tex] \tan\phi = k[/tex]
    and
    [tex] a = \frac{g}{2v_0^2} [/tex]

    So the equation reduces to
    [tex] z = kr - ar^2(1+k^2) [/tex]
    The rest of it is as I showed in my previous post.

    That was a fun problem!
     
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