Vertical Curves on Railways

[TheoBald]

I hope I'm using the correct Forum. Reading a HS2 thread I learnt something of the limits on permitted horizontal curvature. Eager to learn more I searched for further detail which led me to a couple of brief mentions on vertical curves. I'm keen to understand more about these but despite searching the internet I've found little of help re our railways. I've found some detail of American Railways (both old and new methods!), Indian, Irish, Japanese and Israeli but the only British reference of any real value has been an old Track Standards Manual Section 8 - https://www.rssb.co.uk/rgs/standards/GCRT5017 Iss 2.pdf .

I'd be grateful for any help members might be able to offer that will further my understanding. In particular I have the following queries:

(1) Can anyone reference any helpful online documents re theory/examples/regulations on the subject?

(2) The above document uses a centripetal acceleration of 0.06g in calculations whilst a HS2 one used 0.07g. Is there a consistent standard?

I've been able to derive the equation R=V^2/7628 contained in Appendix Section A6 (taking g=9.81). Relating to this:

(3) In Section 6 (Normal Limiting Values) 6.8 states "vertical radii 1km". Substituting in the above, this suggests to me that no normal vertical curve needs to have a restriction of less than 87 km/h (54 mph) due to its vertical curvature. Is my understanding correct?

(4) Section 7 (Exceptional Design Values) 7.6 allows radii down to 500m on passenger lines (62 k/h?). How often is an EDV used? Are there any well known examples?

(5) Do the same regulations/limits apply to the Tube? My eyes suggest not!

(6) Is there any requirement for a "transition curve" as there is with horizontal ones? I'm guessing not since vertical curves don't involve cant. If there is, then how is this calculated? If not, is the instantaneous acceleration noticeable?

(7) Finally (you'll be relieved to read) is anyone able to check my understanding/calculations for a chord taking a level track to a gradient of 1 in 30 (suitable for 50 mph, so radius 1 km)? I estimate it would be 33.3 m long, rising/falling a measly 0.555 m.

 

[John Webb]

It may be worthwhile looking at https://www.thepwi.org/ - website of the Permanent Way Institution. Possible though in part it may be only accessible to members?

 

[edwin_m]

I think you're right on the last one. The angle subtended by the transition curve at the centre of curvature must be the same as the difference between the gradients, so the length of the curve is 1000m/30 (give or take the odd small angle approximation).

 

[TheoBald]

Thanks John - I will follow your lead when time permits.

Edwin I used your approximate method (giving 33.333333 initially) then more precise trigonometry to get 33.32099. I mention this because elsewhere in the article I referenced there is a mention of the Versine of an angle (no, me neither, so I looked it up - the second article I found "The Versine formulae / A railway track blog"). Lo and behold, the Versine can be used in the calculation of height gained/lost!

 

[edwin_m]

I always chuckle when I see profiles and radii with the positions and radii of H and V curves set out to the nearest milimetre...

Incidentally the thing a transition curve avoids is not acceleration, it is change in acceleration. I've never seen a vertical transition curve, and I think there are two reasons:

• The vertical acceleration due to a vertical curve is only a small part of that being experienced anyway due to gravity, so an instantaneous change (in fact it will be spread over the time the length of the coach between bogie centres takes to pass over the change of curvature) isn't really perceptible.
• A sudden sideways acceleration when standing passengers were previously experiencing none could cause people to fall over. Hence there are limits on rate of change of lateral acceleration and traction/braking systems also limit rate of change of longitudinal acceleration.