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  1. MHB Apprentice

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    #1
    What is the correct method to solve this type of problem?

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    #2
    Quote Originally Posted by frankie View Post
    What is the correct method to solve this type of problem?
    Sorry, I don't understand the language.
    I can understand: $ \displaystyle \text{component velocities }A: \begin {Bmatrix}x = 8.5\text{ m/s} \\ y = -6\text{ m/s} \\ z = 10 \text{ m/s} \end{Bmatrix}$
    But why do we have: $ \displaystyle \vec{A}: (2,8,8) \;x,y,z\text{ pos.}$

    And what is a "closing velocity"?



  4. MHB Apprentice

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    #3 Thread Author
    Quote Originally Posted by soroban View Post
    Sorry, I don't understand the language.
    I can understand: $ \displaystyle \text{component velocities }A: \begin {Bmatrix}x = 8.5\text{ m/s} \\ y = -6\text{ m/s} \\ z = 10 \text{ m/s} \end{Bmatrix}$
    But why do we have: $ \displaystyle \vec{A}: (2,8,8) \;x,y,z\text{ pos.}$


    And what is a "closing velocity"?



    But why do we have: $ \displaystyle \vec{A}: (2,8,8) \;x,y,z\text{ pos.}$
    The positions of the vectors may aid to solve the problem geometrically, but I'm
    not sure if they are strictly needed to solve the problem. I think it depends
    what method is used.

    The closing velocity is how fast A is going towards B plus how fast B is going towards A combined.

    Thanks

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    #4
    Quote Originally Posted by frankie View Post
    But why do we have: $ \displaystyle \vec{A}: (2,8,8) \;x,y,z\text{ pos.}$
    The positions of the vectors may aid to solve the problem geometrically, but I'm
    not sure if they are strictly needed to solve the problem. I think it depends
    what method is used.

    The closing velocity is how fast A is going towards B plus how fast B is going towards A combined.

    Thanks
    Hi frankie! Welcome to MHB!

    If I'm not mistaken, the closing velocity is the velocity that the planes get closer to each other.
    And the closing speed is its magnitude, that is, the rate of change of the distance between the airplanes.

    If the planes are at positions $\mathbf p_1$ and $\mathbf p_2$ and have velocities $\mathbf v_1$ and $\mathbf v_2$, then the closing velocity is:
    $$\mathbf v_{closing} = (\mathbf v_2 - \mathbf v_1) \cdot (\mathbf p_2 - \mathbf p_1) \frac{\mathbf p_2 - \mathbf p_1}{\|\mathbf p_2 - \mathbf p_1\|^2}$$
    where the center dot $\cdot$ denotes the dot product of vectors.
    And the closing speed is:
    $$v_{closing} = \frac{(\mathbf v_2 - \mathbf v_1) \cdot (\mathbf p_2 - \mathbf p_1)}{\|\mathbf p_2 - \mathbf p_1\|}$$


    Since you posted this in Calculus, did you intend to derive these formulas?
    Otherwise we can just fill in the numbers.

  6. MHB Apprentice

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    #5 Thread Author
    Quote Originally Posted by I like Serena View Post
    Hi frankie! Welcome to MHB!

    If I'm not mistaken, the closing velocity is the velocity that the planes get closer to each other.
    And the closing speed is its magnitude, that is, the rate of change of the distance between the airplanes.

    If the planes are at positions $\mathbf p_1$ and $\mathbf p_2$ and have velocities $\mathbf v_1$ and $\mathbf v_2$, then the closing velocity is:
    $$\mathbf v_{closing} = (\mathbf v_2 - \mathbf v_1) \cdot (\mathbf p_2 - \mathbf p_1) \frac{\mathbf p_2 - \mathbf p_1}{\|\mathbf p_2 - \mathbf p_1\|^2}$$
    where the center dot $\cdot$ denotes the dot product of vectors.
    And the closing speed is:
    $$v_{closing} = \frac{(\mathbf v_2 - \mathbf v_1) \cdot (\mathbf p_2 - \mathbf p_1)}{\|\mathbf p_2 - \mathbf p_1\|}$$


    Since you posted this in Calculus, did you intend to derive these formulas?
    Otherwise we can just fill in the numbers.

    Thanks for your post!

    My first thought was to do the dot product of the 2 velocity vectors V1 \cdot V2 and that's it.
    The numbers didn't make any sense so I knew it was wrong.

    Yes it would be very helpful to me if I understood the formulas and how they are derived.

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    #6
    Quote Originally Posted by frankie View Post
    Thanks for your post!

    My first thought was to do the dot product of the 2 velocity vectors V1 \cdot V2 and that's it.
    The numbers didn't make any sense so I knew it was wrong.

    Yes it would be very helpful to me if I understood the formulas and how they are derived.
    Well, the Calculus approach is to model the movement of the planes with:
    \begin{cases}
    \mathbf p_1(t) = \mathbf p_1(0) + \mathbf v_1 \cdot t\\
    \mathbf p_2(t) = \mathbf p_2(0) + \mathbf v_2 \cdot t\\
    \end{cases}
    So the distance $d$ between the planes is:
    $$d(t) = \|(\mathbf p_2(0) + \mathbf v_2 t) - (\mathbf p_1(0) + \mathbf v_1 \cdot t)\|
    = \|\mathbf p_2(0) - \mathbf p_1(0) + (\mathbf v_2 - \mathbf v_1) t\|$$

    The closing speed is the derivative of $d(t)$...

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    #7 Thread Author
    Quote Originally Posted by I like Serena View Post
    Well, the Calculus approach is to model the movement of the planes with:
    \begin{cases}
    \mathbf p_1(t) = \mathbf p_1(0) + \mathbf v_1 \cdot t\\
    \mathbf p_2(t) = \mathbf p_2(0) + \mathbf v_2 \cdot t\\
    \end{cases}
    So the distance $d$ between the planes is:
    $$d(t) = \|(\mathbf p_2(0) + \mathbf v_2 t) - (\mathbf p_1(0) + \mathbf v_1 \cdot t)\|
    = \|\mathbf p_2(0) - \mathbf p_1(0) + (\mathbf v_2 - \mathbf v_1) t\|$$

    The closing speed is the derivative of $d(t)$...
    I followed the initial steps you did, but setting up the equation to take the derivative is what I don't understand.
    I found a very similar answer to where you were headed, where they did show a few extra steps.
    However I don't understand how they did the steps. Maybe this will make sense to you and you can explain it?

    a = a0 + t * v1
    b = b0 + t * v2
    where a0, b0,v1 and v2 are vectors

    d = |a - b| = |a0 - b0 + t * (v1 - v2)|

    That part is exactly what you were providing, then this was given for the differentiation
    d' = (v1 - v2) dot ( (v1 - v2) * t + a0 - b0) / |(v1 - v2) * t + a0 - b0|

    and then it states to get the closing speed at time t = 0 the formula reduces to
    (v1 - v2) dot (a0 - b0) / |a0 - b0|

    which is exactly the formula you provided as the solution

    I'm really confused on the d' step and how the equation is setup, why the dot product is needed
    I could use a description in words what the equation is saying.

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    #8
    Quote Originally Posted by frankie View Post
    I followed the initial steps you did, but setting up the equation to take the derivative is what I don't understand.
    I found a very similar answer to where you were headed, where they did show a few extra steps.
    However I don't understand how they did the steps. Maybe this will make sense to you and you can explain it?

    a = a0 + t * v1
    b = b0 + t * v2
    where a0, b0,v1 and v2 are vectors

    d = |a - b| = |a0 - b0 + t * (v1 - v2)|

    That part is exactly what you were providing, then this was given for the differentiation
    d' = (v1 - v2) dot ( (v1 - v2) * t + a0 - b0) / |(v1 - v2) * t + a0 - b0|

    and then it states to get the closing speed at time t = 0 the formula reduces to
    (v1 - v2) dot (a0 - b0) / |a0 - b0|

    which is exactly the formula you provided as the solution

    I'm really confused on the d' step and how the equation is setup, why the dot product is needed
    I could use a description in words what the equation is saying.
    Wel... in words, the closing velocity is the projection of the relative velocities on the connecting positional vector.
    That's how we can "see" it without getting into calculus.
    But I guess something like a is not immediately obvious.

    So let's head down the calculus road.
    We have:
    $$d(t) = \|\mathbf a - \mathbf b\| = \|\mathbf a_0 - \mathbf b_0 + t (\mathbf v_1 - \mathbf v_2)\| \\
    = \sqrt{(a_{0,x}-b_{0,x} + t(v_{1,x}-v_{2,x}))^2+(a_{0,y}-b_{0,y} + t(v_{1,y}-v_{2,y}))^2+(a_{0,z}-b_{0,z} + t(v_{1,z}-v_{2,z}))^2}
    $$

    The derivative with respect to $t$ (applying the chain rule twice) is:
    $$d'(t) = \frac{2(a_{0,x}-b_{0,x} + t(v_{1,x}-v_{2,x}))(v_{1,x}-v_{2,x})
    + 2(a_{0,y}-b_{0,y} + t(v_{1,y}-v_{2,y}))(v_{1,y}-v_{2,y})
    + 2(a_{0,z}-b_{0,z} + t(v_{1,z}-v_{2,z}))(v_{1,z}-v_{2,z})}
    {2\sqrt{(a_{0,x}-b_{0,x} + t(v_{1,x}-v_{2,x}))^2+(a_{0,y}-b_{0,y} + t(v_{1,y}-v_{2,y}))^2+(a_{0,z}-b_{0,z} + t(v_{1,z}-v_{2,z}))^2}} \\
    =\frac{2(\mathbf a_0 - \mathbf b_0 + t (\mathbf v_1 - \mathbf v_2)) \cdot (\mathbf v_1 - \mathbf v_2)}{2\|\mathbf a_0 - \mathbf b_0 + t (\mathbf v_1 - \mathbf v_2)\|} \\
    = \frac{(\mathbf a - \mathbf b)\cdot (\mathbf v_1 - \mathbf v_2)}{\|\mathbf a - \mathbf b\|}
    $$

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    #9
    When using this formula, why do I get a closing speed of 5.14 instead of 6 when I have 2 planes, located at (-1, -2) and (2, 3)
    with respective velocities (0, 3) and (0, -3)? Aka, the planes are parallel to each other, moving in opposite directions with the same velocity. Intuitively, I would expect the closing speed to be 6 (the same as the relative speed of B according to A) but instead using this formula I got 5.14ish (30 divided by the square root of 34)...

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    #10
    Quote Originally Posted by catmitt98 View Post
    When using this formula, why do I get a closing speed of 5.14 instead of 6 when I have 2 planes, located at (-1, -2) and (2, 3)
    with respective velocities (0, 3) and (0, -3)? Aka, the planes are parallel to each other, moving in opposite directions with the same velocity. Intuitively, I would expect the closing speed to be 6 (the same as the relative speed of B according to A) but instead using this formula I got 5.14ish (30 divided by the square root of 34)...
    Hi catmitt98, welcome to MHB!

    You would be right if the planes were flying on the same line in opposite directions.

    However, consider the situation that the planes are passing each other at some distance.
    That is, consider the time they are closest to each other.
    Isn't the speed at which they approach each other then zero?

    We are not at that point yet, so the result is somewhere between 0 and 6.
    I find 5.14 as well.

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