# Figuring Scale distance



## GearDrivenSteam (Jul 3, 2008)

Can someone tell me how to figure scale distance? Particularly in 1:20.3? Thanks.


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## Guest (Aug 26, 2008)

an inch on your layout should represent about one foot and eight inches in reality.


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## Del Tapparo (Jan 4, 2008)

(Distance on your 1:20.3 model) X 20.3 = Distance in the 1:1 world


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## GearDrivenSteam (Jul 3, 2008)

So, the Durango and Silverton would be 2.2167487 miles in 1:20.3.


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## Guest (Aug 26, 2008)

forget it! 
not even in Z-gauge will one be able, to build the lengths of any railroad in th correct scale length. 

one has to compromise. 
if the space between two stations is two or three train-lengths, that should do. 

my house is 60'x 50' - i would need 3' x 3' just to make that one house in 1:20.3 ! 
one has to condense things. 
or do you own a national park?


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## GearDrivenSteam (Jul 3, 2008)

I understand about compromise. That's not what I asked. 

Posted By kormsen on 08/26/2008 6:51 PM
forget it! 
not even in Z-gauge will one be able, to build the lengths of any railroad in th correct scale length. 
one has to compromise. 
if the space between two stations is two or three train-lengths, that should do. 
my house is 60'x 50' - i would need 3' x 3' just to make that one house in 1:20.3 ! 
one has to condense things. 
or do you own a national park?


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## Dave F (Jan 2, 2008)

Posted By kormsen on 08/26/2008 6:51 PM

or do you own a national park?




For those of us who are U.S. citizens, yes, as a matter of fact, we do own the National Parks. Why do you ask?


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## East Broad Top (Dec 29, 2007)

not even in Z-gauge will one be able, to build the lengths of any railroad in th correct scale length. 

Z gauge--6.5mm gauge, which when equated to 2' gauge is 3.25mm/foot. So, if you're doing the Monson RR (6 miles long), you're looking at around 340' of track.  

Later, 

K


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## Richard Smith (Jan 2, 2008)

5280 feet divided by 20.3 = 260 feet = 1 scale mile.


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## Guest (Aug 27, 2008)

So, if you're doing the Monson RR (6 miles long), you're looking at around 340' of track. 


i stand corrected... 

wouldn't that be something? a Z-gauge gardenrailway?


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## chrisb (Jan 3, 2008)

This is the way I do it. Multiply prototype distance in feet 
to get prototype inches. Then divide prototype inches by 20.3 
I do this on a excell spread sheet. Column A is Proto Feet, B is proto inches and C is the dimension is 1:20.3 scale. 
Use a formula to do the math for you so you can go from 1 foot to whatever and fill in the column. 

Example a building is 24 feet by 36 feet. 
(24x12)/20.3= 14.187 inches or 14 3/16 inches 
(36x12)/20.3= 21.281 inches or 21 1/4 inches 

To build this building at 1:20.3 scale you would make it 
14 3/16 inches by 21 1/4 inches


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## GearDrivenSteam (Jul 3, 2008)

Thanks for the serious answers. ....and though not a National Park, yes, I have enough acerage to run almost 3 miles of trackage.....though as I said was not my question. Again, thanks for the serious answers. I was figuring it correctly all along. I just wanted to make sure.


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