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Posers and Puzzles

Posers and Puzzles

  1. 20 Nov '13 10:16
    The next project that I am doing with my 13 year old daughter is to build a catapult.

    The experiment is to find the angle of release that lets the projectile travel furthest. Also it is an exercise where I can teach her to use hand tools to build the machine.

    The idea is to construct it so the 'firing' arm will begin in the horizontal position and travel through 90 degrees to the vertical, hit a stop and release the projectile. With it mounted on a timber base we can lift one end to alter the elevation in say 5 degree increments to carry out the experiment.

    My question to you all is can the answer be found by maths calculation and if so how?

    Thanks for reading this
  2. Subscriber AThousandYoung
    Just another day
    20 Nov '13 19:03
    To calculate the answer requires some simple trigonometry. It's a classic mechanics (physics) problem.

    http://www.wired.com/wiredscience/2010/09/maximum-range-in-projectile-motion/
  3. 20 Nov '13 19:32
    Thanks for the link.
    Read it through twice.
    Brilliant.
    Understand it perfectly but never would have worked it out myself.

    Cheers
  4. Standard member forkedknight
    Defend the Universe
    20 Nov '13 20:15 / 2 edits
    And in order to understand the derivation for why a projectile takes the (vertical) path

    y = y0 + Vy0 * t - 1/2 * g * t^2

    requires basic calculus:

    position = y
    velocity = y' = Vy0 - g * t
    acceleration = y'' = -g

    Although that's pretty advanced stuff for a 13-year-old. I was still doing this stuff in first-year physics as an engineering student.
  5. 20 Nov '13 21:27
    Thanks to both of you for your comments.

    Yes Mr Knight, agreed. The maths is for me for now. My daughter will be making the machine and conducting the practical side of things. She is quite keen to chop up pieces of wood, drill holes and make something.

    She has already asked about the strength of the spring which of course lets me introduce her to Hooks law.
  6. Subscriber joe shmo
    Strange Egg
    20 Nov '13 23:30 / 1 edit
    Originally posted by AThousandYoung
    To calculate the answer requires some simple trigonometry. It's a classic mechanics (physics) problem.

    http://www.wired.com/wiredscience/2010/09/maximum-range-in-projectile-motion/
    Actually, that's not quite the proper scenario. The angle which maximizes the distance traveled is 45 deg only when the initial height of the object is zero.

    Since you have catapult, you should take into account the initial height of the object when it is released.

    However, analytically that is a tricky task

    From (1)

    x = V*cos(A)*t

    Where;
    x = distance
    V = magnitude of initial velocity
    A = angular direction of initial velocity
    t = time

    and (2)

    y = h + V*sin(A)*t -1/2*g*t^2

    where;
    y = height at any time "t"
    h = initial height
    g = accel. due to gravity
    V = magnitude of initial velocity
    A = angular direction of initial velocity
    t - time

    After Sub(1) into (2) and simplifiy to eq (3)

    y = h + tan(A)*x -1/2*g*x^2/cos^2(A)

    the total distance traveled occurs when y = 0 So,

    0 = h + tan(A)*x -1/2*g*x^2/cos^2(A) eq(4)

    Then you would have to solve for x(A) (distance "x" as a function of the angle "A" )

    Take the first derivative of that x(A) and set it equal to zero and solve

    dx/dA = 0

    I should note that the last couple steps are a disaster if you are attempting to arrive at an analytic result.

    However, you could always make a plot by plugging in various values of "A" between 0 and 90 deg into eq(4), solve the quadratic for x, then graph x vs. A. and locate the maximum "x" and the corresponding "A" from the chart.
  7. 21 Nov '13 00:29
    Thanks for your input Jo.
    Missed that point altogether i.e an offset of the release height above the landing point.
    I think the old saying goes 'If it was easy then everyone would do it'
    And of course the classic 'A problem is the gift wrapping of a solution'

    Next time there is athletics on the TV I will be watching the throwing events with a keen eye.

    Cheers
  8. Standard member forkedknight
    Defend the Universe
    21 Nov '13 01:35 / 1 edit
    Originally posted by eltricky

    Next time there is athletics on the TV I will be watching the throwing events with a keen eye.
    One problem with classic projectile physics is that it totally neglects to account for air resistance of any kind.

    Many of the Olympic style throwing events involve a significant amount of aerodynamic flight, which changes things considerably as far as optimal trajectories.
  9. Subscriber joe shmo
    Strange Egg
    22 Nov '13 02:35
    Originally posted by forkedknight
    One problem with classic projectile physics is that it totally neglects to account for air resistance of any kind.

    Many of the Olympic style throwing events involve a significant amount of aerodynamic flight, which changes things considerably as far as optimal trajectories.
    Thats not so much a probem for classical mechanics, as it is a problem of mathematical complexity.

    You have to solve for x(t) from

    m*d²x/dt² = - k*(dx/dt)²

    and y(t) from

    m*d²y/dt² = k*(dy/dt)² -mg

    So what to do (unless you can derive the solutions from above) is to say that its not going to go as high, and not as far as what our theory predicts.
  10. 29 Nov '13 21:56 / 1 edit
    Let me try this:
    Say we have the Vx and Vy (speeds) of the projectile:
    The time in flight merely depends on Vy and initial height H
    First work out time to zero Y speed (T_0)
    Speed at any time = Initial_velocity + acceleration*time

    When the projectile has gone up and back down to its initial height then
    it must be moving at the opposite y-speed to its initial speed

    since change in speed = acceleration*time
    2Vy = Gxt
    t = 2Vy/G

    (which is why starting and ending at the same height is simple)

    However in our case it must still get to the ground, lets say this time is T_1
    H = T_1 + Vy.T_1 + G.T_1**2/2

    T_1^2 + (2/G)(1+Vy).T_1 = 2H/G
    T_1^2 + T1.2(1+Vy)/G + ((1+Vy)/G))^2 = 2H/G + ((1+Vy)/G)^2
    ( T_1 + (1+Vy)/G ) ^2 = 2H/G + ((1+Vy)/G)^2
    T_1 + (1 + Vy)/G = sqrt( 2H/G + ((1+Vy)/G)^2)
    T_1 = sqrt( 2H/G + ((1+Vy)/G)^2) - (1 + Vy)/G

    so total time in flight = T_1 + t =
    sqrt( 2H/G + ((1+Vy)/G)^2) - (1 + Vy)/G + 2Vy/G

    and total distance = Vx times that

    so
    Distance = Vx.( sqrt( 2H/G + ((1+Vy)/G)^2) - (1 + Vy)/G + 2Vy/G )

    we know speed_squared (SS) = Vx^2 + Vy^2
    so Vx^2 = SS - Vy^2

    hmm, gets a bit tricky
  11. Subscriber joe shmo
    Strange Egg
    07 Dec '13 02:30 / 1 edit
    Originally posted by joe shmo
    Actually, that's not quite the proper scenario. The angle which maximizes the distance traveled is 45 deg only when the initial height of the object is zero.

    Since you have catapult, you should take into account the initial height of the object when it is released.

    However, analytically that is a tricky task

    From (1)

    x = V*cos(A)*t

    Where;
    ...[text shortened]... for x, then graph x vs. A. and locate the maximum "x" and the corresponding "A" from the chart.
    I should note that eq(4) (which neglects friction as F.K. pointed out) that I derived can be explicitly solved for x(A) from the quadratic formula.

    0 = c + b*x + a*x²; where

    c = h
    b = tan(A)
    a = 1/2*g*sec²(A)

    so,

    x = -tan(A) ± Sqrt(tan²(A) + 2*g*h*sec²(A))/(-g*sec²(A) )

    from which you can find dx/dA = 0, and theoretically solve it ( however I suspect a method of approximation would need to be employed).

    Anyhow, I pretty sure the topic is dead, but I just wanted to clarify.
  12. Subscriber joe shmo
    Strange Egg
    07 Dec '13 02:41
    Originally posted by iamatiger
    Let me try this:
    Say we have the Vx and Vy (speeds) of the projectile:
    The time in flight merely depends on Vy and initial height H
    First work out time to zero Y speed (T_0)
    Speed at any time = Initial_velocity + acceleration*time

    When the projectile has gone up and back down to its initial height then
    it must be moving at the opposite y-speed to i ...[text shortened]... y/G )

    we know speed_squared (SS) = Vx^2 + Vy^2
    so Vx^2 = SS - Vy^2

    hmm, gets a bit tricky
    Something isn't quite right here I believe. You seem to be adding time and distance with this equation.

    "H = T_1 + Vy.T_1 + G.T_1**2/2"

    H = time + distance + distance ??
  13. 08 Dec '13 12:39
    Originally posted by joe shmo
    Something isn't quite right here I believe. You seem to be adding time and distance with this equation.

    "H = T_1 + Vy.T_1 + G.T_1**2/2"

    H = time + distance + distance ??
    Aha! Thank you, yes, you are correct
    I am always too careless at maths.
    So, corrected:

    H = Vy.T_1 + G.T_1**2/2
    2H/G = (2Vy/G).T_1 + T_1^2
    sqrt(2H/G+Vy^2/G^2) = T_1 + Vy/G
    T_1 = sqrt(2H/G+Vy^2/G^2) - Vy/G {only want positive root}

    giving total time in flight = Vy/G + sqrt(2H/G+Vy^2/G^2)

    Distance = Vx(Vy/G + sqrt(2H/G+Vy^2/G^2))

    using
    Vx^2 = SS - Vy^2

    Distance = sqrt(SS - Vy^2 )(Vy/G + sqrt(2H/G+Vy^2/G^2))

    so differentiating

    d(distance)/d(Vy) =
    sqrt(SS - Vy^2 )(1/G + d/dy{ sqrt(2H/G+Vy^2/G^2))}
    + (Vy/G + sqrt(2H/G+Vy^2/G^2))(d/dy{sqrt(SS - Vy^2 )})

    from the chain rule
    d/dx f(g(x)) = f'(g(x)).g'(x)
    so
    d/dx sqrt(g(x)) = g'(x)/(g(x)

    d(distance)/d(Vy) =
    sqrt(SS - Vy^2 )(1/G + (2Vy/G^2)/(2H/G+Vy^2/G^2)
    + (Vy/G + sqrt(2H/G+Vy^2/G^2))(-2Vy/(SS - Vy^2 ))

    so we need to solve for
    0 =
    sqrt(SS - Vy^2 )(1/G + (2Vy/G^2)/(2H/G+Vy^2/G^2)
    + (Vy/G + sqrt(2H/G+Vy^2/G^2))(-2Vy/(SS - Vy^2 ))
  14. Subscriber joe shmo
    Strange Egg
    10 Dec '13 00:08
    Originally posted by iamatiger
    Aha! Thank you, yes, you are correct
    I am always too careless at maths.
    So, corrected:

    H = Vy.T_1 + G.T_1**2/2
    2H/G = (2Vy/G).T_1 + T_1^2
    sqrt(2H/G+Vy^2/G^2) = T_1 + Vy/G
    T_1 = sqrt(2H/G+Vy^2/G^2) - Vy/G {only want positive root}

    giving total time in flight = Vy/G + sqrt(2H/G+Vy^2/G^2)

    Distance = Vx(Vy/G + sqrt(2H/G+Vy^2/G^2))

    using
    Vx^2 ...[text shortened]... t(SS - Vy^2 )(1/G + (2Vy/G^2)/(2H/G+Vy^2/G^2)
    + (Vy/G + sqrt(2H/G+Vy^2/G^2))(-2Vy/(SS - Vy^2 ))
    I suppose that if all the math is done correctly it would be a valid result by first solving for V_y, then V_x from

    V_x = Sqrt(SS - V_y^2)

    and finding the corresponding angle from

    tan(Angle) = V_y/V_x

    But that final equation would be a absolute nightmare for sure! So I think I'm going to have to stick with my method. Who knows though, seems to be a difficult solution in either case.
  15. 11 Dec '13 00:46
    ok , thanks Jo

    Now we need to simplify:
    sqrt(SS - Vy^2 )(1/G + (2Vy/G^2)/(2H/G+Vy^2/G^2))
    + (Vy/G + sqrt(2H/G+Vy^2/G^2))(-2Vy/(SS - Vy^2 ))
    = 0

    {I replaced an errant bracket}

    so multiply through by sqrt(SS - Vy^2 )

    (SS - Vy^2 )(1/G + (2Vy/G^2)/(2H/G+Vy^2/G^2))
    + (Vy/G + sqrt(2H/G+Vy^2/G^2))(-2Vy/sqrt(SS - Vy^2 )) = 0

    Do some substitutions:
    (SS - Vy^2 ) = A
    (2H/G+Vy^2/G^2) = B

    so

    A(1/G + (2Vy/G^2)/B)
    + (Vy/G + sqrt(B))(-2Vy/sqrt(A) = 0

    multiply out the rightthand part and move the lefthand part to the other side

    2Vy^2/(G.sqrt(A)) + 2Vy.sqrt(B/A) = A(1/G + (2Vy/G^2)/B)

    square both sides:
    4Vy^4/(G^2.A)+4Vy^2.(B/A) + 8Vy^3.sqrt(B)/G = A^2(1/G + (2Vy/G^2)/B)^2

    now we can isolate sqrt(B) on its own, square both sides again, and we at least we have a polynomial without square roots in, but it still looks complex!