I understand about compromise. That's not what I asked.
Posted By kormsen on 08/26/2008 6:51 PM
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?
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
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.