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what is "the formula" for waypoint projection


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Just out of curiosity. Apparently it is easy to type coordinates into a field, choose distance and bearing, press a button and then get the result for a waypoint projection.

 

But how is that mathematically done? What is that probably super long formula? It must be more than rather simple geometry for several reasons ?

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For short distances at a given latitude, you can get really close if you know the change in N coordinates for a given number of feet in the north direction and change in W coordinates for a given number of feet in the west direction.  Then you can use a couple very simple trigonometry formulas to convert a distance and heading to the projected coordinates.  This makes the assumption that the earth is flat and on a rectangular grid which is close to true for a small area.

Edited by medoug
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43 minutes ago, medoug said:

For short distances at a given latitude, you can get really close if you know the change in N coordinates for a given number of feet in the north direction and change in W coordinates for a given number of feet in the west direction.  Then you can use a single very simple trigonometry formula to convert a distance and heading to the projected coordinates.  This makes the assumption that the earth is flat and on a rectangular grid which is close to true for a small area.

 

Even easier if you use UTMs and trig because you don't need to find the ft/degree factors.  But most cachers don't have a clue as to what UTM coordinates are.

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2 hours ago, icezebra11 said:

 

Even easier if you use UTMs and trig because you don't need to find the ft/degree factors.  But most cachers don't have a clue as to what UTM coordinates are.

 

Including you, apparently.  UTM northing is not the same as north, for example.  And projection using azimuths in UTM does not work, in general.  In short:  UTM is not suitable for projections.

 

Vincenty's formula is good to a few centimeters over hundreds of kilometers. I recommend it.  I have a Windows app that you can use for it called FizzyCalc.

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9 hours ago, fizzymagic said:

 

Including you, apparently.  UTM northing is not the same as north, for example.  And projection using azimuths in UTM does not work, in general.  In short:  UTM is not suitable for projections.

 

Vincenty's formula is good to a few centimeters over hundreds of kilometers. I recommend it.  I have a Windows app that you can use for it called FizzyCalc.

FizzyCalc is a wonderful app on Windows. On Android, I highly recommend 'GCC - Geocache Calculator' for when you need to do projection (and many other) calculations in the field. 

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18 hours ago, fizzymagic said:

 

Including you, apparently.  UTM northing is not the same as north, for example.  And projection using azimuths in UTM does not work, in general.  In short:  UTM is not suitable for projections.

 

Vincenty's formula is good to a few centimeters over hundreds of kilometers. I recommend it.  I have a Windows app that you can use for it called FizzyCalc.

 

I am well aware of UTMs and that their axes are not parallel with Lat/Long but I was responding to medoug's post that specifically said "short distances" and "small area."  So let's do the math.  

 

How about a 1/4 mile projection (I'd call that a short distance) at 45 degrees from the coordinates N40 00.000, W105 00.000 or 500000mE, 4427757mN.  

 

A quarter mile equals 402.34 meters.  402.34sin45 = 284.5 meters therefore the projected coordinates are:

 

500284.5mE, 4428041.5mN

 

A 402.34 meter projection at 45 degrees, done on my Oregon 600 set to UTMs and from a starting point of 500000mE, 4427757mN results in coordinates:

 

500284mE, 4428041mN

 

I'd call that close enough for caching purposes.

 

 

 

Edited by icezebra11
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