To recap, this is roughly the layout plan for Battledown Flyover (top half of photo). Note that the flyover west and east tracks in red, are dead ends. Therefore the bridge scene is almost a static diorama. Almost because a train can run across the flyover from west to east (up line) but, it would need to be a short train.
Since the track does not need to be the steep gradient (1:30) of a model railway in order to join other tracks at ground level I can make the gradient prototypical. But, what is the gradient of the prototype?
Foolishly I did not initially search the web for the answer. Probably because as the bridge is within walking distance of my home I can go and look where the gradient starts, relate this to google maps to measure the horizontal distance and from this calculate the gradient.
Problem is I have not used trigonometry since school days!
Using Goggle maps I measured the horizontal distance of the gradient to the bridge to be approximately 7376 mm (4 mm scale). I new the height of the model to be 80 mm. Now I needed the angle. I remembered from school the acronym SOHCAHTOA to help remember when to use sine, cos or tan but, I choose the easy path of using the web to do the calculations for me!
An angle of 0.621 degrees was calculated here.
and the gradient of 1:92 here.
Only then did I search the web and found in this forum posting stating the gradient to be 1:90.
I was impressed that my calculation was pretty close to that!
Now I had to work out the rail height at the extremities of the gradients on the model. I measured the track lengths to be 1300 mm on the west end and 1400 mm on the east end.
Recalculating using the same angle and subtracting 1300 from 7376 etc. gave heights of 65.8 mm and 68.5 mm respectively.
Now I can proceed to build the bridge embankments.
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