Railway Engineering Curve Design Through Hallade Method

  • Thread starter tomtomtom1
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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.
 

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  • #2
I like Serena
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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]
 
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  • #3
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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.
 
  • #4
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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?
Yes.


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).
Yes. R = Radius.

But when you calculate with angles, you should use radians instead of degrees.
So you should use Alpha = 0.099668652.
 
  • #5
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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.
 
  • #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.
 
  • #7
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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.
 
  • #8
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You mentioned the following:-

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

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 [itex]2 * {4v \over C} * \cos (\alpha)[/itex] ?
The 2 that was already there (before the equal sign).
I only replaced sin(alpha).

Where does the *1 come from, from [itex]{8v \over C} * 1[/itex]?
It's an approximation.
The cosine of a small angle is approximately 1.
For reference: cos(10°)=0.985.

And finally why cant we use cos instead of sin?
Huh? :confused:
I don't understand your question.
 
  • #9
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Thanks serena i finally understand thanks to you.
 
  • #10
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You're welcome! :smile:
 

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