import math


def latlon2equirectangular(lat, lon, phi_er=0, lambda_er=0):
    """Naive equirectangular projection.  This is the same as considering (lat,lon) == (y,x).
    This is a lot faster but only works if you are far enough from the poles and the dateline.

    :param lat:
    :param lon:
    :param phi_er: The standard parallels (north and south of the equator) where the scale of the projection is true
    :param lambda_er: The central meridian of the map
    """
    x = (lon - lambda_er) * math.cos(phi_er)
    y = lat - phi_er
    return y, x