在圆形雷达数学方法上表示点 [英] Representing Points on a Circular Radar Math approach
问题描述
我正在编写一个简单的应用程序,它可以向您展示您周围的朋友,但不是在法线地图中,而是在像 UI 一样的真正圆形雷达上:
(是构建 4x4 变换矩阵背后的数学运算
将所有点转换为 RADAR 坐标系
只需将所有点乘以RADAR变换矩阵M
Q(i) = P(i)*M
所以点 Q(i)
现在是 RADAR
(0,0,0)
表示雷达原点(中心)(1,0,0)
指向北(0,1,0)
指向东(0,0,1)
指向上
所以现在只需将所有坐标乘以雷达比例
scale = RADAR_radius/RADAR_range;
RADAR_radius
是您RADAR在屏幕上的大小,以像素或坐标单位为单位RADAR_range
是RADAR最大圆圈代表的最大距离[m]
在此之后只需将点绘制到 RADAR(交换 x,y
因为我使用 X
作为北而不是 Y
) 并且您也可以丢弃所有比范围更远的点.您也可以像旧的 Elite 一样添加 3D RADAR,方法是将 Z
坐标添加到垂直轴(或画一条 L 线)
希望它有点帮助,不要太混乱......
I am coding a simple app that can show you what friends are around you, but not in the normal map but on a really circular radar like UI:
(http://i.imgur.com/9Epw0Xh.png)
Like this, where i have every users latitude, longitude, and of course my own being the center.
I also measure the distance of every user to position them so the data I know is their lat, longitude and distance to me.
For mathematical reasons let's say the radar is 100 pixels radius, I can distance them by the distance from me using the left or right positioning, but in terms of top or bottom it gets a litte trickier, since i try to convert their latitude - my latitude into a percentual result and then put them on the radar... but I think there are maybe better ways with polar to cartesian coordinates, but im really kinda clueless.
Is there a best approach with these types of interfaces or anything implemented around there?
convert long,lat of all points to cartesian 3D space coordinates
it is conversion spherical -> cartesian 3D space. Math behind is here. After this all points
(long,lat,alt)
will became(x,y,z)
where(0,0,0)
is center of the EarthX
axis islat=0,long=0 [rad]
Y
axis islat=0,long=+PI/2 [rad]
Z
axis is NorthXY
plane is equator
If you want more precision handle Earth as ellipsoid instead of sphere
long = < 0 , +2*PI > [rad] lat = < -PI/2 , +PI/2 > [rad] alt = altitude above sea level [m] Re =6378141.4; [m] Rp =6356755.0; [m] R=alt+sqrt( (Re*cos(lat))^2 + (Rp*sin(lat))^2 ) x=R*cos(lat)*cos(long) y=R*cos(lat)*sin(long) z=R*sin(lat)
create RADAR local cartesian coordinate system
Basically you need to obtain 3D vectors for
X,Y,Z
axises. They must be perpendicular to each other and pointing to the right direction from RADAR origin point(P0)
.You can use vector multiplication for that because it creates perpendicular vector to its multiplicants. Direction is dependent on the order of multiplicants so experiment a little.
//altitude this one is easy Z = P0 //north (chose one that is non zero, resp. bigger to avoid accuracy problems) X = (1,0,0) x Z // old X axis * Altitude X = (0,1,0) x Z // old Y axis * Altitude //east is now also easy Y = X x Z // now normalize all of them to unit vectors X = X / |X| Y = Y / |Y| Z = Z / |Z| // and check if they are not negative (X,Y) // if they are then swap multiplicants or multiply by -1 // do not forget that X is computed by two methods so swap the correct one
convert all points to RADAR coordinate system
just multiply all points by RADAR transform matrix
M
Q(i) = P(i)*M
so the points
Q(i)
are now local to RADAR(0,0,0)
means radar origin (center)(1,0,0)
points to north(0,1,0)
points to east(0,0,1)
points up
so now just multiply all cordinates by RADAR scale
scale = RADAR_radius/RADAR_range;
RADAR_radius
is size of you RADAR on screen in pixels or units of coordinatesRADAR_range
is the max distance the RADAR biggest circle represents [m]
after this just draw the dot to RADAR (swap
x,y
because I useX
as North notY
) and also you can discard all points that are more distant then range. Also you can add 3D RADAR like in old Elite by addingZ
coordinate to vertical axis (or draw an L line)
Hope it helps a little and was not too much confusing...
这篇关于在圆形雷达数学方法上表示点的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!