Autor Wątek: Info: Question about the catenary and curves  (Przeczytany 785 razy)

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Offline jakhajay

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Question about the catenary and curves
« dnia: 25 Czerwiec 2016, 05:34:30 »
Hi guys, I'm trying to write a proper user manual for the editor, and I need to run something past you.

I know the posts for the catenary are approximately 70m apart, except for on curves, where they bunch up a bit closer. Marcin has given me an equation which he thinks might be right for calculating the distance between the posts on the curves. I just want to check with you guys, so we know for sure :) Anyway the equation is:


Where r is the radius of the curve, and p is the tolerance of lateral movement of the wires across the pantograph. (For non-programmers sqrt is the square root)

Does this look ok to you guys?
« Ostatnia zmiana: 28 Wrzesień 2016, 21:06:56 wysłana przez matek__9292 »

Offline Kacper9

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Re: Question about the catenary and curves
« Odpowiedź #1 dnia: 25 Czerwiec 2016, 05:44:07 »
Tworze pulpity do stacji dobrze wykonanych. Nie polecam swoich pulpity. Wypadkowa TD2.

Offline jakubg1

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Re: Question about the catenary and curves
« Odpowiedź #2 dnia: 25 Czerwiec 2016, 09:20:54 »
I asked this question 4 months ago. Here's the equation that I received from uetam:
a = 1.8 sqrt (R)
a = delay between two poles;
R = radius of curve.
« Ostatnia zmiana: 25 Czerwiec 2016, 09:23:58 wysłana przez jakubg1 »

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Od 24.05.17 rano do 26.05.17 popołudnie (wycieczka) oraz od 27.05.17 rano do 29.05.17 wieczór. Wtedy mnie nie ma!

Offline jakhajay

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Re: Question about the catenary and curves
« Odpowiedź #3 dnia: 28 Czerwiec 2016, 02:01:21 »
Great, thanks guys!

Offline Tolek

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Re: Question about the catenary and curves
« Odpowiedź #4 dnia: 29 Czerwiec 2016, 20:59:51 »
Well, not really... ::)

<tl;dr>
The catenary in a curve should form an open polygonal chain tangent to the curve (the straight line segments must be tangent to the curve). Let me rephrase: the polygonal chain should not intersect the curve. Let me rephrase again: the zigzag offset works only to the outside of the curve.
Knowing that, we can construct each catenary segment as a base of an isosceles triangle such that its height is equal to the curve radius and its arms’ length is equal to the sum of curve radius and zigzag offset and its tip is at the centre of the curve. We can then divide this triangle into two right triangles, such that each arm turns into the hypotenuse and the height becomes the adjacent side. Having established that, we can now find the ratio of the adjacent side to the hypotenuse which is equal to the cosine of the central angle (based at the centre of the circle). We can reverse this to get the magnitude of the angle given values for parameters r and o:

halfAlfa = acos(r / (r+o));
and hence get the angular distance between poles (points where the catenary is attached):
</tl;dr>

alfa = 2 * acos(r / (r+o));
Now, to get the actual length, we need to decide whether we want the length of the curve segment (length of the curved track segment between the points where catenary is attached) or the length of the actual catenary segment.
For the first, we need to multiply the angle by the

lengthOfCurveSegment = alfa * r;
lengthOfCurveSegment = 2r * acos(r/(r+o));

and for the other we need to find out the length of the base of the isosceles triangle which is equal to double the length of the opposite side of the right angle. Hence

lengthOfCatenarySegment = 2 * sqrt( (r+o)^2 - r^2);
after optimizing the equation:

lengthOfCatenarySegment = 2 * sqrt(o * (2r+o));
Not sure how Marcin arrived at his result. :<
Besides his equation should optimize into:

d = 2sqrt(2*p^3*r)

which makes no sense whatsoever (why the power 3?).
« Ostatnia zmiana: 29 Czerwiec 2016, 21:10:38 wysłana przez Tolek »