Map Distortion - Worldographer

[MorelockTheWarlock]

Hi guys,

I recently picked up Worldographer/Hexographer II and it's going pretty well, for the most part...

I'm using the trace underlay function for an older map that was penned and scanned. I can import the image just fine however I'm having issues with fitting the map to grid. As you may know, you unfortunately can't import an image in it's native resolution (At least I don't believe you can) but rather have to set it's dimensions in hexes.

The image itself is 4422 x 3120 p with the hexes the standard 46.18 x 40 p, columns line up. The map is 4 pixels to a mile and I'd like to do 10 mile hexes. I'm not sure really sure what exactly is going on, mathematically the lengths and widths of the hex grid and image match up but visually it's looking rather distorted, more like a square than the rectangle it should be, and leaving my continent looking rather smushed. Hopefully the attached images of Worldographer and the original map help. Any advice?

Thanks.

[Redrobes]

I am not familiar with the software but it looks to me that if you have to specify the dimensions in hexes then you have the situation where the height of a hex is less than its width.

If the width of each of the 6 sides of the hex is 'h' then each hex width is 2xh. But the height is 2x( h cos 30 ) which is 2x( h x 0.866 ). So for every width you go along you have 86.6% of the height.

So if you have to specify the size in hexes then I would imagine that you either tell it how many hexes you have in width and then do the height but x 1.154 or you specify the height and set the width in hexes as you would but x 0.866.

So lets say you map is 2000 x 1000 pixels and its 4 pixels to the mile then that is 500 x 250 miles. If each hex is 10 miles then that is 50 x 25 hexes. But the with of the hex is larger than the height so to fix that up put in either 43.3 x 25 or 50 x 28.8.

It depends on how they measure the 10 miles to the hex. If 10 miles is the distance from the top of a hex to the bottom or 10 miles is the left to the right. If the 10 miles represents the distance across one bar of any hex side then you need to put in half the number in width and 0.577 x the height. In our example it would be 25 x 14.42.

Below is your image stretched by this factor. If it looks correct is aspect ratio once more and now the hexes are squished then I think that is what is going on.

[MorelockTheWarlock]

I am not familiar with the software but it looks to me that if you have to specify the dimensions in hexes then you have the situation where the height of a hex is less than its width.

If the width of each of the 6 sides of the hex is 'h' then each hex width is 2xh. But the height is 2x( h cos 30 ) which is 2x( h x 0.866 ). So for every width you go along you have 86.6% of the height.

So if you have to specify the size in hexes then I would imagine that you either tell it how many hexes you have in width and then do the height but x 1.154 or you specify the height and set the width in hexes as you would but x 0.866.

So lets say you map is 2000 x 1000 pixels and its 4 pixels to the mile then that is 500 x 250 miles. If each hex is 10 miles then that is 50 x 25 hexes. But the with of the hex is larger than the height so to fix that up put in either 43.3 x 25 or 50 x 28.8.

It depends on how they measure the 10 miles to the hex. If 10 miles is the distance from the top of a hex to the bottom or 10 miles is the left to the right. If the 10 miles represents the distance across one bar of any hex side then you need to put in half the number in width and 0.577 x the height. In our example it would be 25 x 14.42.

Below is your image stretched by this factor. If it looks correct is aspect ratio once more and now the hexes are squished then I think that is what is going on.

Thanks so much!