如何在 LAT/LON NMEA 坐标中获得 6 位小数

Question

I'm using this module:

https://github.com/inmcm/micropyGPS/blob/master/micropyGPS.py

Unfortunately, float in micropython are limited to 5 decimal places eg. 80.12345

How to get at least 6 eg 80.123456?

I know it is accurate to 1 meter, but I need more accuracy.

Answer 1

I had a look at the class MicropyGPS , and concluded that if you don't pass 'dms' or 'dd' as the location_formatting (3rd) parameter of the constructor, you will get the latitude and longitude properties of the class directly in their internal representation, which is a list of 3 items:

• int degrees
• float minutes
• str hemisphere ( 'N' , 'S' , 'E' or 'W' )

For this reason you have 2 or 3 more digits of precision than a float, in this representation. I don't know how you need latitude and longitude to be represented for your purposes. If you need them to be a single float each, you lost. If you can treat degrees and minutes separately, or if you need strings, you won.

I also had a look at MicroPython, which I didn't know. It seems that while floats have a low precision, integers have unlimited precision. So what you could do, if this is ok for your purposes, is convert degrees and minutes to an int, in units of 10**-6 degrees (microdegrees). Something like (untested):

DEGREES_FACTOR = 1000000
MINUTES_FACTOR = 1000000.0 / 60.0

degrees, minutes, hemisph = my_micropyGPS.longitude
microlongitude = degrees * DEGREES_FACTOR + round(minutes * MINUTES_FACTOR)
if hemisph == 'W':
    microlongitude = -microlongitude

degrees, minutes, hemisph = my_micropyGPS.latitude
microlatitude = degrees * DEGREES_FACTOR + round(minutes * MINUTES_FACTOR)
if hemisph == 'S':
    microlatitude = -microlatitude

The resulting integers would be in decimal fixed point representation , with the point 6 decimal positions from the right edge. This is handy for visualization, but you could also use different units, like minutes scaled up by a power of 2, which would eliminate any conversion error.

Please note also that a 5-digit precision would mean that the number in your example would better be represented in decimal as 80.123 (all digits count in floating point, not only those to the right of the point).

Answer 2

Thanks for the answer. The problem is that after receiving the result I have to calculate it further, eg the distance to another point or the direction between one and the other waypoint. If I understand correctly, would your examples be good only for presentation of the result and not for further calculation?

Examples code for my caluclates:

    def dist_waypoints(self, lat1, lon1, lat2, lon2):
        """
        Calculate distance betweeen two coordinates.
        """
        try:
            lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
            result = 6371 * (acos(sin(lat1) * sin(lat2) + cos(lat1) * cos(lat2) * cos(lon1 - lon2)))
            meters = int(result * 1000.0);
            return meters
        except Exception as e:
            self.exception_save(e)
            sys.exit()
            return 0

    def bearing_dest(self, lat1, lon1, lat2, lon2):
        """
        Calculate bearing destination betweeen two coordinates.
        """

        try:
            theta1 = radians(lat1)
            theta2 = radians(lat2)
            delta1 = radians(lat2-lat1)
            delta2 = radians(lon2-lon1)
            y = sin(delta2) * cos(theta2)
            x = cos(theta1)*sin(theta2) - sin(theta1)*cos(theta2)*cos(delta2)
            brng = atan2(y,x)
            bearing = degrees(brng)
            bearing = (bearing + 360) % 360
            return bearing
        except Exception as e:
            self.exception_save(e)
            return 0