Benchwork for an Ho helix or "Doughnut Delema"


D

dmiller

Member
Happy new year!
Can someone tell me the best way to design benchwork to support an HO helix, which is basically a doughnut table!

I would like to be able to shimmy up inside the doughnut hole for repair work.
I am using L girder benchwork. The helix is for two levels about 18 inched separated.

So if I create an L girder frame as in the drawing, How do I support like 14" 1x4 cross beams to the L girders?? ,

Perhaps I need to cut 1" plywood in the doughnut shape and attach directly to the L girder frame (table) level and glue risers directly to it for the elevations??

attached pdf scan drawing

Eating to many Krumpy's Doughnuts in Hagerstown, Md while trying to plan this construction!!

Thanks Engineers!
-David Miller
 

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dmiller said:
Can someone tell me the best way to design benchwork to support an HO helix, which is basically a doughnut table!
Click to expand...
threaded rods. They can then be easily adjusted.
observation.JPG
 
Looks like it is best to abort using L girder method and go with an open grid and plywood top design, since my layout is mostly 2' deep shelf. Any suggestions on how to figure out the what angles to cut the 1x3s to form a hexagon or an octagon? or is it best to put one board over the other and draw the cut line on the board. Fortunately I bought a nice compound miter box saw. (never was good at geometry). And I am use an ASHLAND Helix kit also, so thank you 'fcwilt' for your drop box pictures.

dmiller
 
Well you have a miter saw so just set the angle and cut - being sure to cut the ends of each board to "mirror" one another.

As to the angles:

A circle has 360 degrees. An octagon 8 sides. So 360 / 8 = 45 degrees.

The angle of the cut on the end of each board therefore needs to be 45 degrees / 2 = 22.5 degrees

Your saw should already have a marking for this angle.
 
Iron Horseman said:
threaded rods. They can then be easily adjusted.
observation.JPG
Click to expand...

That looks like a great idea.
How far apart are the double tracks in order to safely accommodate the longer cars (such a the modern passenger cars) ??

What overall diameter does that make the helix ??

Brian
 
Nice Helix Horseman, I was going to suggest threaded rod as you have done, it only makes scence.
 
beiland said:
How far apart are the double tracks in order to safely accommodate the longer cars (such a the modern passenger cars) ??

What overall diameter does that make the helix ??
Click to expand...
This is on the Laramie River Route Layout. There are actually two helix like this on it. These tracks are 32" radius and 34 1/4" radius. Add 4" for the supports make the whole thing 72.5".

This configuration is so flexible and adjustable that this helix is has now been reversed. That is, in the photo notice the train is climbing clockwise. The 2nd level of this layout was extended out such that the camera would now be in the layout. To get the tracks onto the extension, the bolts of the helix were all adjusted. Today the track would run directly from under the camera to the tracks in front of the loco but would be headed down not up. Did not have to take up and re-lay the track, just crank on the bolts.
 
Building an octagon support for a Helix

How does one determine the outside length of wood if building a hexagon or octagon support for a helix.. ? Say 42" diameter
to fit under the helix

Free "pi" for the answer! is there a formula?


Thank you Engineers!




fcwilt said:
Well you have a miter saw so just set the angle and cut - being sure to cut the ends of each board to "mirror" one another.

As to the angles:

A circle has 360 degrees. An octagon 8 sides. So 360 / 8 = 45 degrees.

The angle of the cut on the end of each board therefore needs to be 45 degrees / 2 = 22.5 degrees

Your saw should already have a marking for this angle.
Click to expand...
 
dmiller said:
How does one determine the outside length of wood if building a hexagon or octagon support for a helix.. ? Say 42" diameter
to fit under the helix

Is there a formula?
Click to expand...

Have you forgotten all of your high school geometry?

OK, between any two sides of the octagon you have three pieces of wood, right?

Two pieces are at 45 degrees, one is at 90 degrees, with respect to the sides, right?

So what is the total distance between the two side pieces?

Hint: It involves the value for the sine of 45 degrees.
 
Free Pi for this answer...
How would one determine would be the outside length of a board to fit an octagon or hexagon? (before the angle is cut)??


would an easier way to build the bench work be to place a square with in a square. I.E. Build a square frame that fits the four corners and add 45 degree cross braces to add more support ?


trying to save trees...



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Well you have a miter saw so just set the angle and cut - being sure to cut the ends of each board to "mirror" one another.

As to the angles:

A circle has 360 degrees. An octagon 8 sides. So 360 / 8 = 45 degrees.

The angle of the cut on the end of each board therefore needs to be 45 degrees / 2 = 22.5 degrees

Your saw should already have a marking for this angle.
 
dmiller,

The outside length of the board used doesn't alter or change the angle of the cut, all it does is alter the physical size of the end result.

I think, however, what you are asking is how to determine the size of the hexagon/octagon for your layout. Probably the quickest and easiest way to do that is pretty much what you have said. Make a square that conforms to the maximum size you can use, then use a 45 degree square to ascertain the length of the individual 'boards' to be cut.

While this will not be 100% accurate, it will give you the basis for the length of each board for the hexagon/octagon. The shorter the board, the smaller the area the end product will fit into.

If you want to save trees, try this. Determine the maximum width for your hexagon then cut a board, with 90 degree cuts, to match that length. Then make 45 degree cuts to either end of that board, about an inch in from either end, and use that to make a template of the hexagon as it will be IF you were to use the length of the cut board. Gradually make additional 45 degree cuts until the board is the proper length for creating an accurate hexagon that fits into your space. Once that length is established, use that piece of board as your template for cutting the remaining 5 boards to form the hexagon.

Alternatively, you could find a picture of a hexagon then change that pictures size to reflect the external width of the hexagon that will fit in your defined space and then measure the outside length of the boards.

The important thing here is to make sure each board is the same length. If going with an octagon though, you can have 4 or the boards one length and the other 4 a different length; although, that will give you a distorted octagon but would also work if your space does ot have the same east/west dimensions as your north/south dimensions.

PS: Out of curiosity, you give a free 'Pi' for information? That sounds great but what in heck is a Pi? :)
 
wombat457 said:
That sounds great but what in heck is a Pi?
Click to expand...
A pun. Pie the desert vs Pi the Greek letter. Used for the irrational number representing the ratio of the circumference of a circle to its diameter. 3.14159...

dmiller said:
How does one determine the outside length of wood if building a hexagon or octagon support for a helix.. ? Say 42" diameter
to fit under the helix

Free "pi" for the answer! is there a formula?
Click to expand...
No formula that I know of, As it would depend how you want the frame to fit. I would probably start with a box 45" wide and 8" thick or so, and then make a diagonal brace 45 degrees through the corners.
 
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fcwilt said:
Have you forgotten all of your high school geometry?

Hint: It involves the value for the sine of 45 degrees.
Click to expand...
Sines are trig not geometry! Trig was not required in my high school, and I don't believe it is today either. :) For a geometric solution one would have to apply the Pythagorean theorem. Or in the case of 45 degrees the square root of 2 over 2. 45 degrees is easy since in the unit circle it is a isosceles triangle.
 
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Iron Horseman said:
Sines are trig not geometry! Trig was not required in my high school, and I don't believe it is today either. :) For a geometric solution one would have to apply the Pythagorean theorem. Or in the case of 45 degrees the square root of 2 over 2. 45 degrees is easy since in the unit circle it is a isosceles triangle.
Click to expand...

When I went to school in the 60s, we studied, among other things, Geometry, Trigonometry and Calculus.

Perhaps that is not common.

In any case the values are the same.

But strictly speaking you are absolutely correct.
 
it was a bad pun on pie and Pi Nut then I was always lousy at math

Pi (π) is one of the most important and fascinating numbers in mathematics. Roughly 3.14, it is a constant that is used to calculate the circumference of a circle from that circle's radius or diameter. It is also an irrational number, which means that it can be calculated to an infinite number of decimal places without ever slipping into a repeating pattern. This makes it difficult, but not impossible, to calculate precisely.


 
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