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Math Help - Analysis of crossbow action

  1. #1
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    Analysis of crossbow action

    This is a home-engineering problem, but there's not much mechanics involved. It's exclusively trig/calculus. I haven't done any of that since I left school more than 40 years ago, so I'm brushing away the cobwebs. I've spent two days trying to puzzle this out, but I just can't crack it.

    It's to do with the kind of Roman crossbow called an "inswinger". I'll point it to shoot skywards, so we can talk about "vertical" and "horizontal", to make things easier.

    You have two axles 2 feet apart. The bow arms attached to the axles are rigid levers driven by torsion springs. The arms are 1 foot in length, and the bowstring (2 ft long) is strung between the tips. In the rest position, with the bowstring taut, the arms point directly upwards at the 12 o'clock position. To cock the bow you pull the string down, forcing the levers inwards and downwards until they point directly at each other - a 90 deg turn. The tips can be regarded as touching at that point. And obviously then the angles subtended by the two halves of the string is zero. On release, the arms fly upwards and outwards, pulling the bowstring taut while propelling the missile up the arrow shute.

    Supposing a constant velocity from the arm tips (to simplify things), I need to calculate the missile velocity at any point during the release. And that is a function of the changing geometrical relationships in the configuration. I can see that the velocity ratio missile/armtip is 1:1 at the start, and something indeterminately large at the finish (infinite in theory). But finding a formula to fix the missile's position at any point is defeating me.

    I can simplify a little by ignoring everything to the right of the arrow chute, because it's just the mirror image of the left side. If I draw a line between the axles and call that the "baseline", I can be sure about just one position, apart from the starting and finishing positions. It's where the arm tip is at 60 deg to the baseline. Since the left half of the baseline is the same length as the arm, and the left half of the string is the same length, you have an equilateral triangle. So at that point the missile is positioned exactly on the baseline.

    But after that, I can't use the baseline as a reference, because the line between the missile and the axle is no longer horizontal, and no longer the same length. The only line that isn't moving is the chute, so it might be better to deal with a right-angle triangle involving that. The arm-tip is at A, missile at B, and the horizontal from A crosses the chute at C. The angle between the arm and the baseline is \theta. The angle of the string to the chute is \phi.

    The missile is getting its velocity ( \text{v}_m) from two sources. Call the tip's linear velocity \text{v}_t. The vertical component of that, \text{v}_v, does not cause the shape of the triangle ABC to change. It just raises it. So there is a 1:1 relationship. One inch of upward movement of the tip causes one inch of movement of the missile.

    So I've got the beginnings of an equation.

    \text{v}_m = \text{v}_v+...something.

    ...where \text{v}_v must be \text{v}_t \text{ cos }\theta, and "something" is the movement of the missile caused by the horizontal movement of the tip ( \text{v}_h), which changes the angle of the string, thus forcing the missile up the chute. And \text{v}_h=\text{v}_t\text{ sin } \theta. But I can't just plug it into the equation, the way I can with \text{v}_v. There is no 1:1 relationship. It depends on \phi. But how do you calculate that? What is the relationship between \phi and \theta? That's where I'm stuck.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Oh my... getting a diagram oh what you are trying to describe would be very helpful. To be honest, you lost me at "The bow arms attached to the axles are rigid levers driven by torsion springs"..
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  3. #3
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    I would include a diagram if I knew how to do it. I had enough trouble with the LATEX codes! But the torsion springs are twisted bundles of fibres wrapped around the axle. Does that clarify it?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Okay, I did a google search and the first image that I think is in your problem is that:



    Is that close?

    Except I'm not sure which part is which without any label... I could add some labels to clear it up, but without your help here, I don't think that it'll be possible,
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  5. #5
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    Yes, that's it, basically. But the arms in that diagram start below what I described as the "baseline" (the line drawn between the two axles). My model starts with the arm-tips exactly on the baseline, and they finish at the 12 o'clock position, unlike the one in the diagram, which stops 10 degrees or so before that. Also the arms on mine each occupy half the width between the axles, with their tips just touching in the cocked position, so the angle between the two halves of the string would be 0 at the starting point.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Lol, seems I'm more practising my image editing skills

    Okay, is that better? (and my labels good?)



    Assume the vertical arm is at 12 o'clock (it's 4 degrees too much to the left though)

    EDIT: Oh, I don't know whether there are 2 strings or not... and from your description, it appears to have only 1...?
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  7. #7
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    I think you might have misinterpreted the first diagram. No, there aren't two strings - or four arms. That diagram just shows the arms and string in both the starting and finishing positions. There are just two arms and one string. Each arm travels through 94 degrees on that model. On mine it travels through 90 deg. Your amended diagram is correct, except that you have written "baseline" in the wrong position. The "baseline" is a line drawn directly between the two axles. And the arm is also 1 ft long, so the tips are touching at the start position, and the missile would start 1ft below the baseline, since the string is 2ft long and is folded double at that point.
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Ok, I think that I understand now, and I see how it is. (they could've put arrows in the first diagram, I really thought there were 4 arms!)

    So, if I understand well, the relationship is 1:1 from the 'set' position until the missile reaches the baseline (ie, exactly between the axles), in terms of position. And the first part is trying to find the position from the 'set' position up to the baseline?

    EDIT: Going to bed now. A little past midnight.
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  9. #9
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    Well, no. The 1:1 velocity ratio between the missile and the arm tips only applies at the start position, when they're moving in the same direction. By the time the missile has reached the baseline, the relationship has changed. Looking only at the left half of the diagram, when the missile is directly between the axles, the arm is at 60 deg. And at that point the arm, half-string and half-baseline make the three equal sides (1 ft each) of an equilateral triangle. That's the only position I can be sure of in terms of a geometric relationship. What happens after that is something I need a formula for.
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  10. #10
    MHF Contributor Unknown008's Avatar
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    Okay, if I understood well now:



    At the start, the velocity v_m of the missile is equal to the velocity of the arm, v_a

    So, at time t = 0 and for very small values of t,

    v_m \approx v_a

    Assuming constant acceleration for the missile and constant velocity for the arm tip and considering, the arm tip moves \dfrac{\pi}{3}\ ft while the missile moves 1 ft and we have:

    \dfrac{\pi}{3} = v_a t

    1 = \dfrac12 at^2

    1 = \dfrac12 a\left(\dfrac{\pi}{3v_a}\right)^2

    a = \dfrac{18v_a\ ^2}{\pi^2}

    So, the function modelling the velocity at any particular time between the point where the missile is at rest and the point where it is between the axle is:

    v_m = \dfrac{18v_a\ ^2t}{\pi^2}

    EDIT: If I consider it all, from the start to the finish, the arm does a quarter of a turn (and 1 ft vertically) while the missile does 2 ft, seems very much like the position of the missile follows:

    s_m = 2 \sin(\theta) for 0 \leq \theta \leq \dfrac{\pi}{2})

    And to get the velocity at any point, we take the derivative:

    v_m = 2 \cos(\theta)

    Hm... not convincing.

    I'll try get where you were stuck, ie at phi and theta, notice that:

    \sin\phi + \cos\theta = 1

    Last edited by Unknown008; February 6th 2011 at 10:46 PM.
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  11. #11
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    Quote Originally Posted by Unknown008 View Post
    At the start, the velocity v_m of the missile is equal to the velocity of the arm, v_a

    So, at time t = 0 and for very small values of t,

    v_m \approx v_a

    Assuming constant acceleration for the missile and constant velocity for the arm tip and considering, the arm tip moves \dfrac{\pi}{3}\ ft while the missile moves 1 ft and we have:

    \dfrac{\pi}{3} = v_a t

    1 = \dfrac12 at^2

    1 = \dfrac12 a\left(\dfrac{\pi}{3v_a}\right)^2

    a = \dfrac{18v_a\ ^2}{\pi^2}

    So, the function modelling the velocity at any particular time between the point where the missile is at rest and the point where it is between the axle is:

    v_m = \dfrac{18v_a\ ^2t}{\pi^2}
    You can assume constant velocity of the arm-tip, because I stated that as a given. But it isn't, in fact, true. The arm starts from rest, so it obviously does accelerate. But at the moment all I'm interested in is the relative velocity of the missile to the arm, so we can work with a constant arm velocity without invalidating the logic. However, you cannot assume constant acceleration of the missile. That definitely isn't true. Most of its acceleration will occur in the last few degrees of movement of the arm. You could assume it over a very short distance, but then your \dfrac{\pi}{3} formula is not true over a short distance.



    EDIT: If I consider it all, from the start to the finish, the arm does a quarter of a turn (and 1 ft vertically) while the missile does 2 ft, seems very much like the position of the missile follows:

    s_m = 2 \sin(\theta) for 0 \leq \theta \leq \dfrac{\pi}{2})

    And to get the velocity at any point, we take the derivative:

    v_m = 2 \cos(\theta)

    Hm... not convincing.

    I'll try get where you were stuck, ie at phi and theta, notice that:

    \sin\phi + \cos\theta = 1



    Bingo! Well spotted! I was looking for a relationship between \phi  and \theta, and I believe this is it. I have to go out now, but I'll back back online later. In the meantime I'll be thinking about the implications....
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  12. #12
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    I've had to spend some time getting re-aquainted with trigonometry. After all these years most of it has fallen through a hole in my memory.

    Now that I have a connection between \theta and \phi - i.e, \cos \theta+\sin \phi=1 (thanks for that!), I can proceed with the velocity analysis.

    V_a = velocity of arm tip.
    V_v = vertical component of V_a.
    V_h = horizontal component of V_a.

    V_m=velocity of missile
    V_m_1=velocity of missile due to V_v
    V_m_2=velocity of missile due to V_h

    V_m=V_m_1+V_m_2

    - or, to put it in words, the velocity of the missile due to the vertical movement of the arm must be added to the velocity of the missile due to the horizontal movement of the arm.

    The vertical part is easily solved. Since the vertical movement doesn't change the angle of the string, the relationship is 1:1.

    V_m_1=V_v

    So, V_m=V_v+V_m_2

    And, V_v=V_a \cos \theta,

    giving, V_m=V_a \cos \theta + V_m_2

    V_m_2, the vertical movement of the missile produced by the horizontal movement of the arm, V_h, must be governed by the ratio of the vertical and horizontal sides of the ABC triangle. And since it increases as the angle increases, it must be horizontal/vertical. That resolves as,

    V_m_2 = V_h \tan \phi

    So our main equation becomes,

    V_m=V_a \cos \theta + V_h \tan \phi

    And V_h=V_a \sin \theta

    giving V_m=V_a \cos \theta + V_a \sin \theta .\tan \phi

    I have to get rid of \phi and have everything expressed in terms of \theta, using \cos \theta + \sin \phi=1. That will involve some conversions. I've had to look up trigonometric identities on the internet, because I've forgotten them all. This one looks promising:

    \tan \phi = \frac{\sin \phi}{\sqrt{1-\sin^2 \phi}}

    It will allow me to convert my \tan \phi to an expression involving \cos \theta

    \sin \phi = 1-\cos \theta

    Giving,

    \tan \phi = \frac{1-\cos \theta}{\sqrt {1-(1-\cos \theta)^2}}

    = \frac{1-\cos \theta}{\sqrt{1-(1-2\cos \theta+\cos^2 \theta)}}

    = \frac{1-\cos \theta}{\sqrt {2\cos \theta-\cos^2 \theta}}


    So our main equation for missile velocity against arm velocity is,

    V_m=V_a \cos \theta + V_a \sin \theta .\frac{1-\cos \theta}{\sqrt {2\cos \theta-\cos^2 \theta}}

    That's where I'll stop for the moment, because the next step is calculus. I have a book on calculus, which has my name on the flyleaf, and a Malawi address where I haven't lived since 1965. I don't believe I've opened it with a purpose since the late '60s. So it will take me some time to go through it and learn all over again how to differentiate an expression like the one I have to deal with. It'll probably take me a few days, or even a week. But I don't want to embark on it with the wrong equation. I'd appreciate a check on what I've done so far, because even my basic algebra is pretty rusty, and I could have made mistakes.
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  13. #13
    MHF Contributor Unknown008's Avatar
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    Okay, this is become complicated for me...

    Calculus? If it's about the derivative or integration, wolfram does it for you

    http://www.wolframalpha.com/input/?i=\frac{1-cos\theta}{\sqrt{2cos\theta+-+cos^2\theta}}
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  14. #14
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    Okay, thanks unknown008, I'll take it from here.
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