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Railway Engineering Curve Design Through Hallade Method

  1. Nov 22, 2011 #1
    Hello all

    Please note: This is not a homework question. I work in Engineering.

    Firstly Thank you for reading my post.

    I am trying to learn a technique called Hallade which is used on the Railway to design curves.

    I have some notes (attached).

    The bit I fail to understand is on page 8.

    Three equations are presented:-

    1) Sin (Alpha) = 4V/C
    2) Sin (Beta) = Sin(2Alpha)
    3) Sin (Beta) = 8V/C

    I understand how equation 1 was derived but I do not understand how equation 2 & 3 was derived???

    I was wondering if anyone could help shed some light into how equation 2 & 3 was derived, please do not miss any steps.

    I have been working on this for the last few weeks and got nowhere so I really appreciate any help you can give me.

    Thank you for your help.

    Please note: i have been informed by a few people that equation 2 could actually be wrong and should be Sin(Beta) = 2Sin(Alpha) but either way i still dont undertsand how its dervied.

    Attached Files:

  2. jcsd
  3. Nov 22, 2011 #2

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    Welcome to PF, tomtomtom1! :smile:

    Your text says: "the versine is small in relation to the chord, therefore there is no practical difference between the arc and chord lenghts."

    It also says this was previously stated, but I don't see that, and anyway, it is only true for angles alpha that are "small enough".

    The chord length is [itex]2 \cdot R \sin \alpha[/itex].
    The corresponding arc length is [itex]R \cdot 2 \alpha[/itex].
    So what they are saying is that [itex]2 \cdot R \sin \alpha \approx R \cdot 2 \alpha[/itex].

    From this follows the approximate equality that you asked about.

    Alternatively you can see it this way:
    [tex]\sin \beta = \sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \cdot {4v \over C} \cdot \cos \alpha \approx {8v \over C} \cdot 1[/tex]
    Last edited: Nov 22, 2011
  4. Nov 24, 2011 #3
    You were kind enough to reply to my post on Railway Engineering Curve Design Through Hallade Method, thank you for this.

    I was hoping you could explain your response a little as I am struggling to understand it.

    You said “The chord length is. “ but you have not gone any further. Am I correct in thinking that the chord length is C?

    Secondly you went onto say that “Arc Length = R * 2 * alpha”
    Am i correct in thinking that R = Radius?
    If that is the case then using an example where:-
    R = 100000
    Alpha = 5.739170477
    Then Arc Length = 100000 * 2 * 5.739170477 = 1147834.095

    Is this correct?

    (I got 5.739170477 from a right triangle where the opposite = 10,000 and hypotenuse = 100,000)

    You then said that “ 2 * R * Sine Alpha approx equal R * 2 * Alpha”

    Using the same R and Alpha above, 2 * R * Sine Alpha = 20000

    So 20000 does not approx equal 1147834.095

    Could you shed some light on this?

    If you can reply then I would be very grateful.

    Thank you for you time and support.
  5. Nov 24, 2011 #4

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    Yes. R = Radius.

    But when you calculate with angles, you should use radians instead of degrees.
    So you should use Alpha = 0.099668652.
  6. Nov 26, 2011 #5
    Thank you for your reply.

    You said that “ Sin Beta = sin 2 Alpha “

    Where does Sin 2*Alpha come from, how do you derive it???

    I must admit i am struggling a little with understanding the intuition, so I was wondering if we could use an example.

    If we consider an isosceles triangle with the longest sides being 5 and the base being 8, the height being 3.

    The angle that is made by the two longest intersecting lines (Beta) is:-

    Beta = 1.854590436 (In radians)

    What would Sin (1.854590436) = ?

    Thank you for you help.
  7. Nov 26, 2011 #6

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    Let's take your isosceles triangle with the definitions for alpha and beta as given in your pdf file (that I can't seem to open any more btw).

    Beta is the angle between the long lines of length 5.
    Alpha is defined as half of that, the angle between the length 5 line and the length 3 line.
    So beta = 2 alpha, and so sin(beta)=sin(2 alpha).

    In your example, from the definition of the sine (sine = opposing / hypotenuse), you get: sin(alpha) = 4/5

    Note that your example won't work very well to verify your equation (3), since it's only approximately true for reasonable small angles of beta.
  8. Dec 5, 2011 #7
    Thank you for your reply it is starting to make senseand yes i am very slow (lol).

    You mentioned the following:-

    sin(beta) = sin (2alpha) = 2 sin(alpha) * cos(alpha) = 2 * {4v \over C} * cos (alpha)
    approx {8v \over C} * 1

    I know that 2 sin(alpha) * cos(alpha) was dervied from the double angle forumla but could you explain the rest of it:-

    Where does the 2 come from, from 2 * {4v \over C} * cos (alpha) ?

    Where does the *1 come from, from {8v \over C} * 1?

    And finally why cant we use cos instead of sin?

    BTW i have attached the PDF notes again, hopefully you should be able to open it.

    Thank you.
  9. Dec 5, 2011 #8

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    The 2 that was already there (before the equal sign).
    I only replaced sin(alpha).

    It's an approximation.
    The cosine of a small angle is approximately 1.
    For reference: cos(10°)=0.985.

    Huh? :confused:
    I don't understand your question.
  10. Dec 11, 2011 #9
    Thanks serena i finally understand thanks to you.
  11. Dec 11, 2011 #10

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    You're welcome! :smile:
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