I am going to draw out my layout on some graph paper and dont know how to figure in the curve radius. Could someone help me on this.
Its the distance from the center of the curve to the curve itself.
Dave H.
The radius of a curve is 1/2 its diameter. If you use a 1/2" to the foot scale to draw your layout plan, a 36" radius curve would be represented by circle 3" in diameter.
Which leads me to the question. Don’t they teach geometry in high school anymore?
Andre
I suggest purchasing some Bachmann EZ track that is your minimum radius and use it as a template. From there you can scale up or down by measuring from the template.
David B
ON PAPER you need a Compass and cross-hatch pad.
ON A LAYOUT you need a pencil with string length tied around it to act as a ‘Trammel’; I use a yard stick with holes drilles through it (pencil point sized through ‘1’) and ‘nail sized’ for radii - (minus 1’') - for larger curves.
I think KALMBACH has a ‘How to’ book out out that shows you .
THe only thing I have is where on the track is where they measure it from? Center of the track, inside, inside rail, outside, ouside rail?
Maybe I’m thinkining too much…
I understand how to do it on the table but how do you scale it down to a piece of paper? I am using N scale and am look to have 30" radius curves.
Center of track.
Andre
I use graph paper at a scale where the grid represents 3 inches. and 18 inch radius takes up 6 squares for the radius.
If you used my scale, you’d end up making a circle 10 squares for the radius.
Most graph paper has the squres arranged on a 1/4" grid. I’ve seen some that go down to 1/8 inch grid.
Can someone tell me how to do the math to scale it down onto a piece of paper. I am doing N scale and am going to have 21" all the way up to 30" radius.
What scale are you using for the drawing? IIRC, standard graph paper uses 1/4" grids. If you use 1 grid per foot, the diameter of a 30" curve would be 5 grids. You’d put your compass in the middle of grid #3.
However, I’d suggest you use 2 grids/foot (1/2" to a a foot). That makes the diameter of your drawn circle 10 grids and the center of the circle 5 grids. It’s easier to draw if you’re not trying to make the scale of the drawing too small.
Andre
The scale drawn on the paper is your decision! If you are using graph paper with 1/4" squares, you decide what that 1/4" represents. For example, when drawing floorplans of a house, the scale is typically 1/4" = 1’. In your case, I would suggest using 1/4" = 3". A 36" radius would then consume 12 squares from the centerpoint of the radius, or 3" of graph paper.
Do you understand this explanation?
Don Z.
Yes I am useing 1/4" graph paper so I just need to figure out what a 20" curve and a 30" curve would be on the paper. I am going to have 1 square equal 3" for 4 squares are 1’
Geometry ??? … Apparently simple math isn’t taught either !!!
If your squares are 3" and you want a 30" radius … 30 divided by 3 = 10 (squares) !!!
Mark.
This would make the diameter of your circle 20 squares (or 5 actual inches).
Quick. What’s the area of a circle 5 inches in diameter?
Extra credit: Find the volume of a cone 12 inches high whose base is 5 inches in diameter.
Extra extra credit: convert the result to cubic centimeters.
Oh wait a minute. Model railroading is supposed to be fun. Therefore, it cannot include math problems.
Never mind.
Andre
Your Graph paper has small squares. Each square can represent X inches - say 3 inches, and a 30"r. curve will take up say 10 squares.
Stretch your compass the appropriate # of squares,and scribe you circle. To find the CENTER take the elonagted compass and lightly draw two segments away from the starting points of the curve. WHERE these intersect is the CENTER of your curve.
SIMPLER is to use KATO’s ‘Unitrack’ in their largest radius curve (in millimeters) and Plug together. Remember, 30" radius in HO is equivalent to 60" in 'N’gage.
Hey, its why I went to school for political science. Math just gets in our way! Unless you’re in to quantitative research but why would anyone want to do that?
Ok serious post: I know there has to be a way to do this but my math simple brain doesn’t know where the knowledge is stored. Suppose you have a curve, a piece of flex track as it were, that you laid out by eye at a curve that “looked good.” Perhaps its for a siding that goes off into the woods or something. Obviously there has to be a way to determine the radius of that curve, so that you can determine the longest possible cars that can navigate it. I mean, plus it might just come in handy later for some reason, even if it does work out to 24.5" or something and 18" would get the job done. I know that if its not a perfect arc it doesnt have the same radius the entire way through. And would have multiple center points. But…how would I determine, I guess, the minimum radius of that arc?
Right? Does that make a lick of sense?
The very first thing that streaked through my mind … does it look good ? … do your cars track through it ok ? … then does it really matter ??? [swg]
But, then I realized you were serious ! [;)] - Years ago, I made abunch of templates from plain thin card board, each about three feet wide. One edge had the cut curvature of every radius from 18 up to 30 inches. I used these to the point of wearing out the edge drawing lines on my early stages of benchwork. You could use these to determine the smallest radius within a given french curve. But like my first thought … if it looks good and works fine - does it really matter ?
Mark.
Because it’s not on the government test to pass to the next grade, there for no need to teach it. It’s just useless knowledge and a waste of time [or so I was told by a teacher], like history and ect.
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