# Railway Engineering Curve Design Through Hallade Method

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.

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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I like Serena
Homework Helper
Welcome to PF, tomtomtom1! 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 $2 \cdot R \sin \alpha$.
The corresponding arc length is $R \cdot 2 \alpha$.
So what they are saying is that $2 \cdot R \sin \alpha \approx R \cdot 2 \alpha$.

Alternatively you can see it this way:
$$\sin \beta = \sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \cdot {4v \over C} \cdot \cos \alpha \approx {8v \over C} \cdot 1$$

Last edited:
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.

I like Serena
Homework Helper
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).

But when you calculate with angles, you should use radians instead of degrees.
So you should use Alpha = 0.099668652.

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:-

What would Sin (1.854590436) = ?

Thank you for you help.

I like Serena
Homework Helper
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.

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.

I like Serena
Homework Helper
You mentioned the following:-

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

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

Where does the *1 come from, from ${8v \over C} * 1$?
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? You're welcome! 