Program to calculate size of map and point to point distance.

[Elros]

Most fantasy maps that I have seen don't specify a projection because it usually isn't important to the story. Equirectangular is a horrible projection because its distortion is 1/cos(latitude), meaning that it's literally infinite distortion at the poles. If you're not drawing a latitude/longitude graticule on your map, you can assume that it's in a projection that's flat enough for storytelling purposes.

The simplest way to give users a sense of scale is to use a scale bar on the map, even if it's not precisely correct. The simplest way to ensure scale transfers from FT to something line Photoshop is to draw a rectangular island or two of known size off of one or more coasts. Then mask out those islands in your final map using a legend or title layer.

So i dont have to worry about projection, right?
Ok i will try to do what you say on FT demo to export the scale in photoshop.
I dont have any question, you solved me all my problems .


Thanks again for everything, really thank you.
If i have problem or something to ask about FT, can i ask to you?

[waldronate]

I'm always happy to answer questions.

The only reason that you would want to worry about projection is if your area of interest is large and you want to use it's default measuring tools. If you are willing to declare that your world is planar rather than spherical and want to use ft, set ft to planar mode and never worry about projection issues again.

[Elros]

I'm always happy to answer questions.

The only reason that you would want to worry about projection is if your area of interest is large and you want to use it's default measuring tools. If you are willing to declare that your world is planar rather than spherical and want to use ft, set ft to planar mode and never worry about projection issues again.

I cant declare my world is flat, i designed a story for a spherical world.
Can i make difficult measurements on FT?

I don't think i need long measurements, but if i need them, how could i solve the problem?
I have an idea, can i approximate long distances?
Writing something like "they traveled in that sailer for about two months".
Imagining 6-8 knots of speed, a 15-20miles at day, will be about 600miles in two months (I am approaching the miles at day of a boat at random).
I can?

In this case, how could i miss the measurement? Of months or days?

[waldronate]

FT's measurement tool provides distances in miles or kilometers (plus net change in degrees latitude and longitude, which isn't particularly useful for anyone). You can chain together multiple measurement segments by holding down the shift key to add a new segment. To get distance in units other than miles or kilometers, you'll need to do the math yourself to convert from the provided distance units to your preferred units. Similarly, if you want a non-constant travel cost (e.g. 5km/day in mountains and 20 km/day on plains), you'll need to measure each segment separately and apply appropriate cost per segment via an external tool such as a calculator.

Why doesn't FT do this sort of thing? Nobody asked for it (or not recently). The UI is also a bit complex because users would need to be able to specify what they're trying to measure and have some way to set up the various costs. Some users would undoubtedly ask for a way to store and recall paths to measure distances in different seasons (seasons are another thing that FT doesn't directly support). There are many possibilities for this sort of tool, but FT only provides simple distance measurement at the moment.

[Elros]

FT's measurement tool provides distances in miles or kilometers (plus net change in degrees latitude and longitude, which isn't particularly useful for anyone). You can chain together multiple measurement segments by holding down the shift key to add a new segment. To get distance in units other than miles or kilometers, you'll need to do the math yourself to convert from the provided distance units to your preferred units. Similarly, if you want a non-constant travel cost (e.g. 5km/day in mountains and 20 km/day on plains), you'll need to measure each segment separately and apply appropriate cost per segment via an external tool such as a calculator.

Why doesn't FT do this sort of thing? Nobody asked for it (or not recently). The UI is also a bit complex because users would need to be able to specify what they're trying to measure and have some way to set up the various costs. Some users would undoubtedly ask for a way to store and recall paths to measure distances in different seasons (seasons are another thing that FT doesn't directly support). There are many possibilities for this sort of tool, but FT only provides simple distance measurement at the moment.

It's not a problem to make those that measurements by hand.

But if i have to travel in all the world in a great distance as in the example above. How wrong would it be?

Example:
Suppose i have to travel north starting from the equator, and i say that a month of travel by ship is 600 miles, from point A to point B, is it possible that B is a month farther or closer than i think?
If yes, how much am i doing wrong? An acceptable term?

[waldronate]

If head precisely north and south from the equator, scale doesn't change on a cylindrical projection like Equirectangular, so you're good. As you start to move east and west, though, the effects of projections start to come into play. The angular shrinkage east and west compared to north/south is proportional to cos(latitude), so at 25 degrees latitude (=111km/deg*25deg=2775km, the same number of degrees east and west is 90% of the size that it would be at the equator (94% at 20 degrees=2220km, 97% at 15 degrees=1665km). For a 600 mile trip near the equator, a straight line is probably a good enough approximation.

There are a number of useful calculation that relate distances on a sphere. Ed Williams' Aviation Formulary ( http://edwilliams.org/avform.htm ) lists many common formulae that might be of use to you. From that page: The great circle distance d between two points with coordinates {lat1,lon1} and {lat2,lon2} is given by: d=acos(sin(lat1)*sin(lat2)+cos(lat1)*cos(lat2)*cos (lon1-lon2)).

[Elros]

If head precisely north and south from the equator, scale doesn't change on a cylindrical projection like Equirectangular, so you're good. As you start to move east and west, though, the effects of projections start to come into play. The angular shrinkage east and west compared to north/south is proportional to cos(latitude), so at 25 degrees latitude (=111km/deg*25deg=2775km, the same number of degrees east and west is 90% of the size that it would be at the equator (94% at 20 degrees=2220km, 97% at 15 degrees=1665km). For a 600 mile trip near the equator, a straight line is probably a good enough approximation.

There are a number of useful calculation that relate distances on a sphere. Ed Williams' Aviation Formulary ( http://edwilliams.org/avform.htm ) lists many common formulae that might be of use to you. From that page: The great circle distance d between two points with coordinates {lat1,lon1} and {lat2,lon2} is given by: d=acos(sin(lat1)*sin(lat2)+cos(lat1)*cos(lat2)*cos (lon1-lon2)).

Is a bit difficult to undstand the advanced mathematics of which you speak.
Can you explain it to me a little easier?
So ifi move diagonally, can i run into a serious measurement error?

[waldronate]

You can largely ignore distortion for areas within 2000 km or so of the equator on an equirectangular projection. That information is important because it's image import needs its inputs in the equirectangular projection.

[Elros]

You can largely ignore distortion for areas within 2000 km or so of the equator on an equirectangular projection. That information is important because it's image import needs its inputs in the equirectangular projection.

I understand, but there is no way to remedy this problem?
I had decided to set my story in an area far enough north or south of the equator to have a cold and cold climate.
How can i make measurements in such an area?
I think the only solution is to ignore the rules of reality.

[waldronate]

If you're only interested in distances and your area of interest is less than 4000km across, importing that are into FT along the equator will let FT treat it as if your flat map is in a form that FT can understand (i.e. the Equirectangular projection). You're not saying that it's not cold. It's just that for convenience of measurement, you're treating your map in a way that makes it easy to import and measure in a specific tool (FT in this case). Using FT isn't the important part here: putting your map at the equator lets you conveniently ignore the projection when importing the image into FT.

If you have drawn a map that's just a few thousand km across, you can declare that it's in a projection that is "flat enough" to use a regular ruler on it and you'll likely get less than 10% measurement error (10% measurement error is probably less than the error in the original cartographer's survey for most medieval-level fantasy settings). If you later want to paste that map onto a globe, then you'd need to declare what that projection and center of projection are. If you want to use FT as a tool for measuring distances on such a map with that map positioned on its correct place on the globe, you'd need to reproject the base map to an Equirectangular map so that FT can properly measure distances. It's not hard to do, but might well be beyond what you need at this time ( http://fracterra.com/ReprojectImage.zip is a program that will let you graphically reproject an image in a number of projections back onto the Equirectangular image that FT wants ).

I hope that make sense. My earlier recommendation to put your map at the equator was purely to make it easy to import into FT. Being more correct with measurements FT requires more complexity in putting the map into FT.