Here's what I did.
First, try assigning it to projection epsg:3857 - this is what Google Maps uses, and is in metres. So I did:
Now I can transform that to lat-long and see where on the earth it thinks it is:
x 34.12644 54.06393
y 28.78893 36.04800
Okay, that's the wrong place. Kentucky is 89W to 82W, a span of 7 degrees, but this was making it about 20 degrees. A factor of three. Which is about the same factor as metres to feet. So I figured it might be one of those American state coordinate systems that uses feet rather than metres.
A little searching led me to EPSG:3091...
So let's reassign and see where that projects to:
x -89.57120 -81.96479
y 36.49706 39.14773
Winner winner chicken dinner, as we say over here.
I can plot this on the Kentucky state map from the
> map("state","kentucky", col="red",lwd=4)
To use it in
leaflet you probably need to transform it, so:
add2 = spTransform(add, "+init=epsg:4326")
add2 in lat-long coordinates.
If you want to create a .prj file so that reading the shapefile results in the correct projection, download it from the ".PRJ file" link on the spatial reference web site: http://spatialreference.org/ref/epsg/3091/prj/
It should look something like this:
PROJCS["NAD83(HARN) / Kentucky Single Zone (ftUS)",GEOGCS["NAD83(HARN)",DATUM["D_North_American_1983_HARN",SPHEROID["GRS_1980",6378137,298.257222101]],P
plot(add)and see anything? That should draw a basic plot on the R graphics window. Shapefiles should have more than just a .shp file, there should be a .shx and a .dbf and a .prj file all together. Have you got any of them?
bbox(add)tell you? That should give the coordinates of the bounds of the data, and if it looks like degrees lat-long then I can tell you what to do next. If its big numbers (like metres or feet) then some thought required...
x 3798938 6018369
y 3348809 4307228