Wikipedia:Reference desk/Archives/Mathematics/2018 October 21
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October 21
[edit]Helix rod geometry
[edit]I think this is the first time I've asked a question on the mathematics desk. I normally can't understand the answers here (or even the questions) so I don't hang around as much as I do at the other desks.
I am building a model railway that involves a helix of track connecting two scenic levels built on a plywood track-bed that supports spiraling 'laps' of track with semi-circular curves joined by short straight sections. A large rectangle is cut out of the centre of each piece of plywood, leaving a band of plywood around the edges to support the track and a cut is made perpendicular to and bisecting one of the straight sections so multiple laps can be joined to make the helix.
The laps are supported by adjustable bolts on four vertical rods that pass through holes in my plywood. The rods are suspended from a frame in a rectangular arrangement that intersects the plywood just inside the route of the track. Initially I positioned and drilled the holes at the same horizontal coordinates as the rods but as I began to assemble the helix I noticed and then realized that inevitably the sloping plywood tries to bend the rods so their horizontal coordinates converge as the helix adapts to the vertical pitch. I visualize that if the pitch were expanded ridiculously rods would be made to meet.
In practice I have no problem - firstly my holes are about three times the diameter of the rods, and secondly I have been attacking them in situ with a drill to make inelegant but functional eccentricities that allow the rods to remain vertical. (The convergence only seems to occur in the direction of the shorter dimension of the system, perpendicular to the sections of straight track.)
For unrelated reasons I need to start again. (The unrelated reason is that I want to install metal strips under the track that will be attracted by magnets under the locomotives to increase traction and hauling ability. Currently the locomotives spin their wheels instead of pulling all the desired coaches.) How could I calculate the correct location of the holes - drilled when the lap is flat - so that they align with the free hanging rods?
The dimensions that I think are relevant are:
Rod arrangement: corners of a rectangle 95 cm long and 27.4 cm wide. (Is this difference the reason the convergence only seems to be in one direction?)
Vertical pitch: 6 cm
The dimensions that I don't think matter are:
Radius of track curves: 39.25 cm
Length of straight track connecting curves: 33.5 cm
Outer length of plywood: 116 cm
Outer width of plywood: 82.5 cm
Thickness of plywood: 1 cm
Track gauge: OO9 (9 mm) - similar to N gauge
Number of laps: 7
Can anybody tell me the correct offset for the holes? (Should I provide any more information?) Hayttom (talk) 06:27, 21 October 2018 (UTC)
- I have tried reading what you’ve written a couple of times but I find it impossible to picture what you are trying to do. Mathematically as a problem it’s lacking rigour and formalism. A diagram would be helpful, or a more precise description in terms of mathematical objects.--JohnBlackburnewordsdeeds 00:03, 22 October 2018 (UTC)
- Sorry I'm not a mathematician. A flat ring of negligible thickness (perhaps, to a non-mathematician, a 'washer') sits in an x/y/z coordinate system with the ring's axis on z-axis. The ring is cut radially along the x-axis, let us say in the x>0 direction, and the resultant and initially adjacent ends are displaced variably but symmetrically along the z direction in opposite directions to form a helix. How are the x and y values of arbitrary points on the disk altered? They will converge, but along what vector?Hayttom (talk) 05:26, 22 October 2018 (UTC)
- Well this version is more jargony anyway. It took a number of readings but I think I've got the gist of the problem. First, the track is irrelevant; presumably its weight is negligible so unless it's exerting some other force on the plywood there it won't make a difference in the shape. What may be relevant but which I didn't see in the description are the dimensions of the hole cut in the middle of the plywood. In any case, what you have is a stiff but elastic object which is held in place at certain boundary points but is otherwise allowed to assume the shape which minimizes stress, and this is the subject of elastica theory. It's not something I know much about but I do know the math gets difficult very quickly and even the solution to the simplest case of a thin flexible rod in 2 dimensions is known not to be expressible in elementary functions. Not only do you have surface instead of curve, but you're bending it in 3 dimensions so there's twisting as well. Fortunately though, it looks like you don't need an exact solution since you're making the holes large enough to allow for a certain amount of error. So at the risk of provoking the ire of the spherical cow, I think the best approach is to make some drastic simplifying assumptions. As a first approximation, I'm going to take the distance along the plywood from one hole to the next to be the straight line distance. The total horizontal distance to be covered is 2(95 + 27.4) cm, or about 245cm. The vertical distance to be covered, is 6cm. Applying Pythagoras, the total distance is 245.07cm or an increase of about .03%. This means you need to increase distances between the holes by the same about, in other words you need to drill the holes in the corners of a rectangle 95.03 cm long and 27.41 cm wide. Unless you're using sub-millimeter precision in your measurements the difference is negligible. If the vertical distance (pitch) is small then the increase in distance is approximately proportional the square of the pitch, so as long as the vertical pitch is only a few percent of total distance around the difference in hole position is probably too small to measure. The difference gets larger more and more rapidly as the pitch increases, so the rods would meet if you bent the plywood to the vertical, but in the range you're working with it looks like it doesn't make much difference. One thing to note here is that the size of the hole cut from the middle of the rectangles does not matter for this approximation. --RDBury (talk) 09:55, 22 October 2018 (UTC)
- Thank you very much RDBury for all your re-reading and solving effort! And your calculated 3 mm offset fits well with my observations. But yes, I not only don't need an exact solution - I don't need any mathematical solution at all. I'm really relying on a carpenters' solution, and carpentry is nearly as old a profession as mathematics. (Regarding the issue of the affect of the weight, I realize now that I should have mentioned that I'm making the structure quite rigid, both with wood and bolts that clamp each lap together and nuts and washers on the rods above and below the plywood.) I may post a new question about dynamic friction and my locomotives' pulling power but I'll head over to the science desk for that. Thanks again!
Resolved- Hayttom (talk) 10:58, 22 October 2018 (UTC)
- Well this version is more jargony anyway. It took a number of readings but I think I've got the gist of the problem. First, the track is irrelevant; presumably its weight is negligible so unless it's exerting some other force on the plywood there it won't make a difference in the shape. What may be relevant but which I didn't see in the description are the dimensions of the hole cut in the middle of the plywood. In any case, what you have is a stiff but elastic object which is held in place at certain boundary points but is otherwise allowed to assume the shape which minimizes stress, and this is the subject of elastica theory. It's not something I know much about but I do know the math gets difficult very quickly and even the solution to the simplest case of a thin flexible rod in 2 dimensions is known not to be expressible in elementary functions. Not only do you have surface instead of curve, but you're bending it in 3 dimensions so there's twisting as well. Fortunately though, it looks like you don't need an exact solution since you're making the holes large enough to allow for a certain amount of error. So at the risk of provoking the ire of the spherical cow, I think the best approach is to make some drastic simplifying assumptions. As a first approximation, I'm going to take the distance along the plywood from one hole to the next to be the straight line distance. The total horizontal distance to be covered is 2(95 + 27.4) cm, or about 245cm. The vertical distance to be covered, is 6cm. Applying Pythagoras, the total distance is 245.07cm or an increase of about .03%. This means you need to increase distances between the holes by the same about, in other words you need to drill the holes in the corners of a rectangle 95.03 cm long and 27.41 cm wide. Unless you're using sub-millimeter precision in your measurements the difference is negligible. If the vertical distance (pitch) is small then the increase in distance is approximately proportional the square of the pitch, so as long as the vertical pitch is only a few percent of total distance around the difference in hole position is probably too small to measure. The difference gets larger more and more rapidly as the pitch increases, so the rods would meet if you bent the plywood to the vertical, but in the range you're working with it looks like it doesn't make much difference. One thing to note here is that the size of the hole cut from the middle of the rectangles does not matter for this approximation. --RDBury (talk) 09:55, 22 October 2018 (UTC)
- Sorry I'm not a mathematician. A flat ring of negligible thickness (perhaps, to a non-mathematician, a 'washer') sits in an x/y/z coordinate system with the ring's axis on z-axis. The ring is cut radially along the x-axis, let us say in the x>0 direction, and the resultant and initially adjacent ends are displaced variably but symmetrically along the z direction in opposite directions to form a helix. How are the x and y values of arbitrary points on the disk altered? They will converge, but along what vector?Hayttom (talk) 05:26, 22 October 2018 (UTC)
