使用 WhirlyGlobe 创建 CLLocation
我尝试使用tapmessage 的wheregeo 和worldloc 属性创建一个cllocation 对象。然而，经纬度和纬度我从它们获得的 x、y 值无法用于创建 cllocation 对象。也许我的格式不正确？
感谢提前
记录这
GeoCoord coord = msg.whereGeo;
Point3f point = msg.worldLoc;
NSLog(@"x point is %f",point.x());
NSLog(@"y point is %f",point.y());
NSLog(@"from wheregeo %f",coord.lat());
NSLog(@"from wheregeo %f",coord.lon());
给了我这个
2012-01-19 12:30:22.491 WhirlyGlobeApp[1068:4007] x 点是 -0.138646
2012-01-19 12:30:22.492 WhirlyGlobeApp[1068:4007] y 点是 0.554229
2012-01-19 12:30:22.493 WhirlyGlobeApp[1068:4007] 来自 wheregeo 0.962698
2012-01-19 12:30:22.494 WhirlyGlobeApp[1068:4007] 来自 wheregeo 1.815925
我正在使用 cllocation 默认值使用纬度和经度方法初始化
I have attempted creating a cllocation object using the tapmessage's wheregeo and worldloc properties. however the lat and lon & x, y values that i get from them are not able to be used to create a cllocation object. Maybe i am formatting incorrectly?
Thanks in advances
logging this
GeoCoord coord = msg.whereGeo;
Point3f point = msg.worldLoc;
NSLog(@"x point is %f",point.x());
NSLog(@"y point is %f",point.y());
NSLog(@"from wheregeo %f",coord.lat());
NSLog(@"from wheregeo %f",coord.lon());
gives me this
2012-01-19 12:30:22.491 WhirlyGlobeApp[1068:4007] x point is -0.138646
2012-01-19 12:30:22.492 WhirlyGlobeApp[1068:4007] y point is 0.554229
2012-01-19 12:30:22.493 WhirlyGlobeApp[1068:4007] from wheregeo 0.962698
2012-01-19 12:30:22.494 WhirlyGlobeApp[1068:4007] from wheregeo 1.815925
I am using the cllocation default init with latittude and longitude method
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
这就是弧度与度数。坐标以弧度为单位，但您可能需要 CLLocation 的度数。
只需使用
GlobeMath.h中的RadToDeg()模板即可。看起来很奇怪，但只要像函数一样调用它即可。That would be radians vs. degrees. The coordinates are in radians, but you probably need degrees for
CLLocation.Just use the
RadToDeg()template inGlobeMath.h. It looks weird, but just call it like a function.