ronsal I am having a little trouble getting from 1:1 to 1:160. How do I convert full size measurement to a working useable measurement in n scale, such as .030"?
I am having a little trouble getting from 1:1 to 1:160. How do I convert full size measurement to a working useable measurement in n scale, such as .030"?
Well, you apparently have a lot of calculating to do. But what real (non)railroad engineers do once they get out of school is use a table or a conversion program. Here is one that might help:
http://urbaneagle.com/data/RRconvcharts.html
MILW-RODRhe tightest clearance I can think of off hand is bearing clearance on Chevy V8's which is .0018"-.0024".
Wow, that's the tightest clearance? I used to work on steam turbine mechanical controls and the pilot valve to bushing clearances on some of them were in the order of two to three mils (0.002-0.003 inch).
R. T. POTEET ronsal I am having a little trouble getting from 1:1 to 1:160. How do I convert full size measurement to a working useable measurement in n scale, such as .030"? I'm just al little confused, my friend: .030 what? If you are talking about a prototype measurement of ".030" and you wan't to convert that to N-Scale please call me from your local asylum to let me know how you are doing. .030"/160 equals .0001875" or 1/5333.333333". To the best of my knowledge only engineers and machinists work with figures like that . If, on the other hand you have an N-Scale measurement of .030" you multiply that out by 160 and it renders to 4.8"--.030" X 160.
I'm just al little confused, my friend: .030 what? If you are talking about a prototype measurement of ".030" and you wan't to convert that to N-Scale please call me from your local asylum to let me know how you are doing. .030"/160 equals .0001875" or 1/5333.333333". To the best of my knowledge only engineers and machinists work with figures like that .
If, on the other hand you have an N-Scale measurement of .030" you multiply that out by 160 and it renders to 4.8"--.030" X 160.
Converting scale measure to prototype has already been given, just multiply by 160. To convert prototype measurement to scale just the opposite, divide by 160.
Ex.= .030 x 160= 4.8" (as mentioned)
624 / 160= 3.9"
So a loco prototype 52' in length bewteen pilot faces (early 'Geeps' and RS's) would scale out to about 4 inches, not including couplers. I still haven't tackled accuratly measuring scale coupler length on a model locomotive.
wm3798 Drop a few bucks and get a scale ruler... Sheesh.
Drop a few bucks and get a scale ruler... Sheesh.
LOL! I fully agree..
Of course,I simply double the size when I design my General Rubber..One Pikestuff warehouse is 30' x 80' scale feet..I am combining two kits so its comes out to 30' x 160'..To keep things simple I used my N Scale ruler and made a foot print by simply measuring the feet.
No mind boggling math for me.
Larry
Conductor.
Summerset Ry.
"Stay Alert, Don't get hurt Safety First!"
Route of the Alpha Jets www.wmrywesternlines.net
.030" as in strip styrene.
andrechapelonBesides, true HO gauge is 16.47646383467279 mm, not 16.5 mm. Andre
Andre
Well, actually its 16,49425287356322 mm
/stefan
From the far, far reaches of the wild, wild west I am: rtpoteet
Quick and dirty - convert the full-scale dimension to inches and decimal fractions (4' 8 1/2" to 56.5") and divide by 160 (0.353125.)
In metric, the same dimension (1435mm) divided by 160 equals 8.96875mm.
Yeah, but you have to multiply the English equivalent ((.353125 inches) by 25.4 to get the metric equivalent which works out to 8.969375. Besides 1435/25.4 = 56.496", which is not 56.5" if you want to get picky about it
Regardless of which metric value you use, the variance from true 9 mm is only about .031 mm, or 0.00122 inches. Close enough to call it 9 mm.
Besides, true HO gauge is 16.47646383467279 mm, not 16.5 mm.
wm3798 Or if you're a math moron like me, do this. ...Get a scale ruler. OR You have a 50' boxcar. Multiply 50 x 12 (to get the number of inches) then divide that number by 160. Now you know how many inches your N scale car should be (more or less) 50 x12 600 600 ./. 160 = 3.75" Tah Dah!
Or if you're a math moron like me, do this.
...Get a scale ruler.
OR
You have a 50' boxcar. Multiply 50 x 12 (to get the number of inches) then divide that number by 160. Now you know how many inches your N scale car should be (more or less)
50
x12
600
600 ./. 160 = 3.75"
Tah Dah!
Forgetting for a moment that that length is the INSIDE length, and that it's probably more like 50'6" or 606" so your starting measurement isn't going to close to accurate, the method is of course sound, but you better make sure that you're prototype dimension is accurate. :-)
Chris van der Heide
My Algoma Central Railway Modeling Blog
And there you probably thought that 9mm was an accurate scale reduction of standard gauge. Not!
(It's even less accurate as HOn30 - but still a lot closer than the 2 foot and 3 foot gauge prototype models being sold as On30)
Chuck (Modeling Central Japan in September, 1964 - in twice-N, 1:80, AKA HOj)
This is the type of situation, where the metric system really comes in handy...
In N scale 1 (prototype) foot equals .075" and an inch is .00625". An object being 5´ 2 3/8" thus computes to
5 x .075 = .375" plus
2x .00625 = .01250 plus
3 x .00625/8 = .002344, the total being .3898" or roughly .40!
In the metric world, the same object is 1.584 m or 1584 mm - in N scale that´ll be 1584/160 = 9.9 mm or roughly 10 mm.