I have just purchased the Eagle Pass Truss Bridge from Midwest. I need to build this to fit a 24" radius curve on my HO layout.The Bridge is 86' Long. I need to know how wide I need to make it or the formula to calculate how wide to make it. The bridge will be going above the middle track of the Link below. Thanks in advance for your answers.
http://i220.photobucket.com/albums/dd51/Hwolf_photos/Best%20of%20Wolfcreek/IMG_2505.jpg
W = Width of the bridge
R = radius of the track
L = Length of the bridge
C = clearance width for your equipment
sq = squared
W = R- square root((Rsq)-(Lsq/4)) +C
The width is the Radius plus the clearance width for your equipment less the square root of the radius squared minus the length squared divided by 4.
Dave H. Painted side goes up. My website : wnbranch.com
I sent you an E-Mail for a little more clarification. Thanks
dehusman The width is the Radius plus the clearance width for your equipment less the square root of the radius squared minus the length squared divided by 4.
You've got to be kidding!
Mark
markpierce dehusman The width is the Radius plus the clearance width for your equipment less the square root of the radius squared minus the length squared divided by 4. You've got to be kidding! Mark
Why? He asked for the formula, I gave him the formula. If he wants to use it fine. If he wants figure it out graphically, fine. Since he has the track in place he could just measure it from the track. Put a mark on the track centers the length of the bridge apart. Put a ruler on those two points. At the midpont between the bridge ends measure the distance from the ruler to the center of track. Ad that distance to whatever clearance width you want. But he wanted a formula to calculate it. Its basic high school or jr. high school math.
Dave H.
Dave,
1. I asked if you were kidding because I didn't know if you were serious about the equation. Now I know you are.
2. I'm not a mathematician, so I can't prove or disprove the equation. But even if it is a correct equation, it can't be solved readily by itself. While the radius of curve and the length of the bridge (curved track centerline?) can readily be determined, the clearance width of the equipment cannot. That leaves two unknowns: the desired solution (bridge width) and the clearance width. With two unknowns, the solution cannot be solved with a single equation.
3. Clearance width depends on the width of the equipment, the length of the equipment, and something I'll call the "rotation factor." By rotation factor, I mean how far does the equipment swing outward and inward because of its center of rotation. For example, take two locomotives of the same overall dimensions but one is an 2-10-2 and the other is a 2-8-4. Since the center of rotation of the 2-10-2 is near the loco's center while the rotation center of the 2-8-4 is further forward, the clearance width of the 2-8-4 will be wider because its cab will swing out further from the track center for a given curve. And then there are (normal) articulated locomotives where the smokeboxes swing very wide because the center of rotation is so far to the rear. All equipment will swing parts of their bodies outward and parts inward on a curve, depending on the sharpness of the curve and the centers of rotations (their trucks). I'd hate to see the formula(s) for determining all of that.
4. The straight-forward solution that is nearly fool-proof is to test various equipment suspected to have the biggest clearance requirements and place them on the curve, meanwhile moving sticks on either side of the curve to see how far apart the interior part of the bridge above the tracks needs to be.
Sounds like a good plan guys. By the way what did you think about the scratch built trestle?
Whoa! Hold your horses. I hadn't looked at your photo and I was thinking of a single-span through-truss or through-girder bridge. That won't work because the other tracks and the trestle are too close. What you need is about a 3-span deck girder bridge (each span about 30') supported steel towers such as made by Micro-Engineering Company. On deck girder bridges the girders are placed not far from the rails. I believe three short spans will keep the rails not too far from the girders.
Yup, nice wooden trestle.
That is what I am going to use on the back track. The bridges will be underneath. I did not want to use this on the center track because I did not want to lose any of the effect of the trestle. This way the truss bridge can be above.
markpierce 2. I'm not a mathematician, so I can't prove or disprove the equation. But even if it is a correct equation, it can't be solved readily by itself. While the radius of curve and the length of the bridge (curved track centerline?) can readily be determined, the clearance width of the equipment cannot. That leaves two unknowns: the desired solution (bridge width) and the clearance width. With two unknowns, the solution cannot be solved with a single equation. 3. Clearance width depends on the width of the equipment, the length of the equipment, and something I'll call the "rotation factor." By rotation factor, I mean how far does the equipment swing outward and inward because of its center of rotation. For example, take two locomotives of the same overall dimensions but one is an 2-10-2 and the other is a 2-8-4. Since the center of rotation of the 2-10-2 is near the loco's center while the rotation center of the 2-8-4 is further forward, the clearance width of the 2-8-4 will be wider because its cab will swing out further from the track center for a given curve. And then there are (normal) articulated locomotives where the smokeboxes swing very wide because the center of rotation is so far to the rear. All equipment will swing parts of their bodies outward and parts inward on a curve, depending on the sharpness of the curve and the centers of rotations (their trucks). I'd hate to see the formula(s) for determining all of that.
The clearance width is the easiest part. the NMRA has done all the work, its available on line or you can just measure a clearance gauge. OR if you want something special like a walkway or anything like that you can create your own. Very simple.
Since you are determining the clearance width before hand, it ought be a "standard" width for your layout so you know how far away to put buildings, posts, scenery, signal masts, doorways for industries and engine shops, etc, etc, etc.
dehusman W = Width of the bridge R = radius of the track L = Length of the bridge C = clearance width for your equipment sq = squared W = R- square root((Rsq)-(Lsq/4)) +C
The length of the bridge is approximately 12 inches. Lsq/4 = 144/4 = 36.
The radius is 24 inches. 24sq is 576. 576-36 =540. The square root of 540 is about 23.25.
The radius minus 23.25 is about .75. That means that he needs to increase the width about 3/4 of an inch. If he is building a bridge from a kit, or plans there probably is a "clearance' already established, the width of the bridge for straight track. All he has to do is make the bridge at least 3/4 of an inch wider than stock and he'll be OK. make it an even inch wider and it should be more than ample.
That took all of three minutes to figure up, including the time it took to type the this post.