In college when I studied electrical engineering at Vanderbilt U., Nashville, TN, 1956-1960 all students regardless of engineering specialty were required to take elementary surveying and to do this we all went to the surveying camp atop the ridge east of Sparta, TN. Around the base of the rock clifts at the top there was what appeared to be a rudimentary road but was in fact the right of way of the NC&StL railroads branch to Ravenscroft & Clifty where there were coal mines. The line was abandoned in 1938 or so. The text for the course was "Elementary Surveying" by Rayner & Schmidt, published by D. Van Nostrand. My book is the 1955 printing. On page 169, there begins a chapter on laying out horizontal curves, either road or railroad, the only difference really being the degree of curve. Anything sharper than 17 degrees is rare on a standard gauge line, 24 degrees on narrow gauge. There is a constant, 5729.7, that is found to relate radius and degree. This is important for our discussion because default curve values are of radius, while M-track values are in degree of curve. We are interested in the degree of curve. This constant for metric material is 5729.7/3.28 = 1746.86 where 3.28 is the number of feet in a meter. The degree of the curve is also the lateral deviation at each 100 foot location, i.e. end of each 100 ft. chord of the curve. If I stand over one point and sight back to the previous point on the curve this is one chord. If I then look ahead and turn my head 5 degrees to align with the next point, I am on a 5 degree curve. Of course I don't turn my head, instead I use a instrument called a transit that is handy for measuring turn angles.
So you have carefully digitized the track center line from Google EArth or whatever source you prefer and you have a track chart of the line which shows that there is a 5 span timber trestle at a certain point. When I see this I try to determine the center of the trestle and mark that. But wait, this trestle is on a 5 degree curve, or when you lay the track it took a 300 meter curve piece to fit your digitizing points. Uh Oh, now what. Well I scratched dandruff on several routes over a 20 year period but then a couple months back, I woke up in the middle of a cool night with my oldest (of 5) cats snuggling up to me for warmth and then a thought occurred. So the next morning while working on the N. Montana line of the Milwaukee Road (which has A LOT of timber trestles) I tried the method and it works. I use TSM but whatever your preferred object creator program is, the process should be similiar.
First you need to have already created 3 parts for the trestle. First you need a bent, which is the supporting structure under the trestle. Typically these have 5 to 7 vertical members or "poles", a top cap, and diagonal braces. Most Milwaukee Road trestles are 6 pole bents, typically 12"x12" and 20 feet in length. The horizontal cap on top is typical 11-12 feet in length, also 12"x12" in size. You need a span, which is typically made with 12x12 boards and is typically 10 feet wide by 14 feet in length. Lastly you need an end abutment and berm with the abutment being either wood or concrete and the berm with a texture applied that closely matches that on the terrain at the location. This will be the "kit" from which you can easily create timber trestles. Likewise, if you created the same parts in concrete, where is the typical span is 10 feet wide and 17 feet in length you will have a "kit" for concrete trestles. Depending on what you used for the top of the span, i.e. the part immediately under the track, you can use either track pieces for a ballast deck, or bridge rail pieces for an open deck. Most Milwaukee Road trestles were ballast deck.
So we will lay out a trestle on a level curve of radius 300 meters. But we need to know the degree of curve, so 1746.86/300 = 5.82 degrees. I will expain this for TSM but other quasi-CAD programs should work the same. Five spans is a odd number so my method is to place the first span straight on an axis line, say the z axis. Now The span on one end will turn (14 (or 17 if concrete)/100) X 5.82 or .81 degrees, while the one on the other end will turn -.81 degrees. The next pair will turn plus or minus 1.62 degrees relative to the first span we placed, and the end berms will rotate Plus or minus 2.43 degrees relative to the first span. Now pretty it up so there are no additional offsets or gaps and then place the bents. Usually those on the ends of the center span can be perpindicular to it, while the other two will turn plus or minus about 2 degrees.
This works well on a level or nearly level curve and if it is not too long a trestle. If the trestle has 7 or 8 or more spans (or panels) or it is on a 1% or greater elevation then a level trestle will not look right. In that case I lay out the curve first and then place the trestle without the bents, but with the end abutments, in place. Then it is a matter of doing a series of tiny elevations of each piece (unlike for the lateral curve they will all be the same). Once satisfied, add the bents and place the complete trestle.
Now what if we have a 6 span, or panel, trestle. In that case, place two span pieces along one of the axis lines butted together at the origin point. Then for the mentioned 300 meter radius curve turn one piece +.405 degrees and the other -.405 degrees. The next pieces will respectively turn plus or minus 1.215 degrees, and the next two, plus or minus 2.035 degrees, and the end abutments plus or minus 2.855 degrees. All angles are relative to zero, So for an odd number of spans, start with one span centered along an axis line and for an even number, start with two pieces, the point between them being centered.
Jerry Sullivan, P.E. (retired) retired from CSX, Southern, and as docent - C&TSRR
So you have carefully digitized the track center line from Google EArth or whatever source you prefer and you have a track chart of the line which shows that there is a 5 span timber trestle at a certain point. When I see this I try to determine the center of the trestle and mark that. But wait, this trestle is on a 5 degree curve, or when you lay the track it took a 300 meter curve piece to fit your digitizing points. Uh Oh, now what. Well I scratched dandruff on several routes over a 20 year period but then a couple months back, I woke up in the middle of a cool night with my oldest (of 5) cats snuggling up to me for warmth and then a thought occurred. So the next morning while working on the N. Montana line of the Milwaukee Road (which has A LOT of timber trestles) I tried the method and it works. I use TSM but whatever your preferred object creator program is, the process should be similiar.
First you need to have already created 3 parts for the trestle. First you need a bent, which is the supporting structure under the trestle. Typically these have 5 to 7 vertical members or "poles", a top cap, and diagonal braces. Most Milwaukee Road trestles are 6 pole bents, typically 12"x12" and 20 feet in length. The horizontal cap on top is typical 11-12 feet in length, also 12"x12" in size. You need a span, which is typically made with 12x12 boards and is typically 10 feet wide by 14 feet in length. Lastly you need an end abutment and berm with the abutment being either wood or concrete and the berm with a texture applied that closely matches that on the terrain at the location. This will be the "kit" from which you can easily create timber trestles. Likewise, if you created the same parts in concrete, where is the typical span is 10 feet wide and 17 feet in length you will have a "kit" for concrete trestles. Depending on what you used for the top of the span, i.e. the part immediately under the track, you can use either track pieces for a ballast deck, or bridge rail pieces for an open deck. Most Milwaukee Road trestles were ballast deck.
So we will lay out a trestle on a level curve of radius 300 meters. But we need to know the degree of curve, so 1746.86/300 = 5.82 degrees. I will expain this for TSM but other quasi-CAD programs should work the same. Five spans is a odd number so my method is to place the first span straight on an axis line, say the z axis. Now The span on one end will turn (14 (or 17 if concrete)/100) X 5.82 or .81 degrees, while the one on the other end will turn -.81 degrees. The next pair will turn plus or minus 1.62 degrees relative to the first span we placed, and the end berms will rotate Plus or minus 2.43 degrees relative to the first span. Now pretty it up so there are no additional offsets or gaps and then place the bents. Usually those on the ends of the center span can be perpindicular to it, while the other two will turn plus or minus about 2 degrees.
This works well on a level or nearly level curve and if it is not too long a trestle. If the trestle has 7 or 8 or more spans (or panels) or it is on a 1% or greater elevation then a level trestle will not look right. In that case I lay out the curve first and then place the trestle without the bents, but with the end abutments, in place. Then it is a matter of doing a series of tiny elevations of each piece (unlike for the lateral curve they will all be the same). Once satisfied, add the bents and place the complete trestle.
Now what if we have a 6 span, or panel, trestle. In that case, place two span pieces along one of the axis lines butted together at the origin point. Then for the mentioned 300 meter radius curve turn one piece +.405 degrees and the other -.405 degrees. The next pieces will respectively turn plus or minus 1.215 degrees, and the next two, plus or minus 2.035 degrees, and the end abutments plus or minus 2.855 degrees. All angles are relative to zero, So for an odd number of spans, start with one span centered along an axis line and for an even number, start with two pieces, the point between them being centered.
Jerry Sullivan, P.E. (retired) retired from CSX, Southern, and as docent - C&TSRR