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Velocity triangles Mechanics watch

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    Hi guys, so where have I went wrong in my working out here?Can someone please help? For part i, it's meant to be 11.8 knots

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    (Original post by sienna2266)
    Hi guys, so where have I went wrong in my working out here?Can someone please help? For part i, it's meant to be 11.8 knots
    Everything looks fine, except you have the angle incorrect.

    The current is coming from 150 degrees, so it's going on a bearing of 150+180 = 330 degrees. So, the angle in the triangle is...?
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    (Original post by ghostwalker)
    Everything looks fine, except you have the angle incorrect.

    The current is coming from 150 degrees, so it's going on a bearing of 150+180 = 330 degrees. So, the angle in the triangle is...?
    Thanks, but when they say from 150degrees is it 150 degress away from the positive x axis or from negative x axis and anticlockwise or clockwise or from a bearing of 150 degrees? I am slightly confused here. In my diagram, I did 150 degrees anticlockwise from the positive x axis which is clearly wrong as you have kindly pointed out. So the current is coming from 150degrees from where?
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    (Original post by sienna2266)
    Thanks, but when they say from 150degrees is it 150 degress away from the positive x axis or from negative x axis and anticlockwise or clockwise or from a bearing of 150 degrees? I am slightly confused here. In my diagram, I did 150 degrees anticlockwise from the positive x axis which is clearly wrong as you have kindly pointed out. So the current is coming from 150degrees from where?
    The question is refering to a bearing. 150 is measured clockwise from due North.
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    (Original post by ghostwalker)
    The question is refering to a bearing. 150 is measured clockwise from due North.
    Thanks this makes sense but why do you add 180degrees to the 150?
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    (Original post by sienna2266)
    Thanks this makes sense but why do you add 180degrees to the 150?
    Because the bearing of 150 is the direction the current is coming from. We add 180 to get the bearing of the direction the current is going.
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    (Original post by ghostwalker)
    Because the bearing of 150 is the direction the current is coming from. We add 180 to get the bearing of the direction the current is going.
    Thanks so much so I finally got 11.8 knotts. What do you do for the course of the ship bit? So for the course set, I would say bearing of 270N but that is wrong. the answer is 283 degrees. Could you let me know your thoughts please?

    Cause in some questions you put the north in between the 2 components that make up the resultant velocity and find the bearing from there to the end point
    For example,in this question:
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    And in some questions, you put the north at where the resultant velocity starts and find the bearing from there to the end point
    For example, in this question:
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    (Original post by sienna2266)
    ...
    Going to take more time than I have available today, so will have to leave for someone else - can pick it up tomorrow if it's not been dealt with.
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    (Original post by ghostwalker)
    Going to take more time than I have available today, so will have to leave for someone else - can pick it up tomorrow if it's not been dealt with.
    Sure no worries! Thanks for the help so far! Would really appreciate it if you have the chance to look over it tomorrow
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    (Original post by sienna2266)
    Thanks so much so I finally got 11.8 knotts. What do you do for the course of the ship bit? So for the course set, I would say bearing of 270N but that is wrong. the answer is 283 degrees. Could you let me know your thoughts please?
    It cannot be steered at a bearing 270 because then the current from the water will offset it and hence the resultant motion will not be westward.

    Having said that, it goesn't make much sense for the answer to be more than 270 degrees as then the motion of the ship would have a +ve vertical component and the current already has +ve vertical component, so no way would they cancel to give a vertical velocity of 0, which is a necessary condition for the ship to travel west.
    Clearly, we must have the ship travelling at a bearing between 180 and 270.

    I say the bearing of the ship for it to travel west is 255 degrees.

    The idea is as follows:

    The ship goes at some bearing \theta with speed 10 knots.
    The water current flows from a bearing 150 with speed 3 knots.

    You are to find \theta so that the resultant velocity is due west.
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    (Original post by RDKGames)
    It cannot be steered at a bearing 270 because then the current from the water will offset it and hence the resultant motion will not be westward.

    Having said that, it goesn't make much sense for the answer to be more than 270 degrees as then the motion of the ship would have a +ve vertical component and the current already has +ve vertical component, so no way would they cancel to give a vertical velocity of 0, which is a necessary condition for the ship to travel west.
    Clearly, we must have the ship travelling at a bearing between 180 and 270.

    I say the bearing of the ship for it to travel west is 255 degrees.

    The idea is as follows:

    The ship goes at some bearing \theta with speed 10 knots.
    The water current flows from a bearing 150 with speed 3 knots.

    You are to find \theta so that the resultant velocity is due west.
    Thanks for your reply! Here is what I've got so far:
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    Could you please let me know where the theta angle that you are referring to is?
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    (Original post by sienna2266)
    Thanks for your reply! Here is what I've got so far:
    Could you please let me know where the theta angle that you are referring to is?
    Draw a new diagram. \theta would be this; the bearing of the 10 knot vector

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    (Original post by RDKGames)
    Draw a new diagram. \theta would be this; the bearing of the 10 knot vector

    But I thought bearing is from the north?
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    (Original post by sienna2266)
    But I thought bearing is from the north?
    Yes, got distracted there, you got it though.
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    (Original post by RDKGames)
    Yes, got distracted there, you got it though.
    But as shown in my diagram below, when you're finding the bearing of the direction of the course of the ship, you do 90 + 180 and you get 270degrees which is the wrong answer as you kindly pointed out.
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    (Original post by sienna2266)
    But as shown in my diagram below, when you're finding the bearing of the direction of the course of the ship, you do 90 + 180 and you get 270degrees which is the wrong answer as you kindly pointed out.
    OK well I don't understand why you're doing this procedure, and I explained why 270 is incorrect.

    In part (i) the ship is being steered to the west. In part (ii) it is not, so you need to find the bearing of the steering.
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    (Original post by RDKGames)
    OK well I don't understand why you're doing this procedure, and I explained why 270 is incorrect.

    In part (i) the ship is being steered to the west. In part (ii) it is not, so you need to find the bearing of the steering.
    So I am trying to do part i) the course set bit and I still don't know which angle I am finding.

    Could you please kindly show me?

    In your diagram, You advised that the angle theta is this:
    Attachment 726416
    But then you said that this is not the angle we are meant to find
    So what is the angle we are finding?
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    (Original post by sienna2266)
    So I am trying to do part i) the course set bit and I still don't know which angle I am finding.

    Could you please kindly show me?
    Ah OK, I was talking about part (ii) [oops, should've read previous replies before jumping in!]

    So, you are finding this angle in part (i):




    Which is the bearing of the resultant vector. To find that, you simply need to do 270+(angle between the 10 and R vectors)
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    (Original post by RDKGames)
    Ah OK, I was talking about part (ii) [oops, should've read previous replies before jumping in!]

    So, you are finding this angle in part (i):




    Which is the bearing of the resultant vector. To find that, you simply need to do 270+(angle between the 10 and R vectors)
    Thank you!! I got part i) and for part ii) here is the diagram I've got so far:
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    Could you kindly let me know what would be the next step? Would the resultant velocity be different from part i)?
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    (Original post by RDKGames)
    Ah OK, I was talking about part (ii) [oops, should've read previous replies before jumping in!]

    So, you are finding this angle in part (i):




    Which is the bearing of the resultant vector. To find that, you simply need to do 270+(angle between the 10 and R vectors)
    Actually here is a better diagram to my previous post for ii):
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    Could you kindly let me know what would be the next step? Would the resultant velocity be different from part i)? Many thanks
 
 
 
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