I kinda need to know how to do the opposite of that...I have an x,y,z coordinate and I need to convert it to a lon/lat positon.
Anyone know how to reverse this formula: X = R * sin(LAT) * sin(LNG); Y = R * cos(LAT); Z = R * sin(LAT) * cos(LNG); Thats pretty much what I need to do. =) Thanks so much for all the help so far guys! On Mar 24, 5:35 am, Plopsi <[email protected]> wrote: > hi, imho you justed swapped some axis, here is a snippet from my lat/ > long stuff: > > m_PositionBubble[i].x = 200 * Math.sin(Deg2Rad(90-m_Bubble[i]._lat)) * > Math.sin(Deg2Rad(m_Bubble[i]._lng+90)); > m_PositionBubble[i].y = 200 * Math.cos(Deg2Rad(90-m_Bubble[i]._lat)); > m_PositionBubble[i].z = 200 * Math.sin(Deg2Rad(90-m_Bubble[i]._lat)) * > Math.cos(Deg2Rad(m_Bubble[i]._lng+90)); > > so > > X = R * sin(LAT) * sin(LNG); > Y = R * cos(LAT); > Z = R * sin(LAT) * cos(LNG); > > look if you have to do this 90-LAT and LNG+90 thing, i'm not sure > about that at the moment if it is only a fix for my angles. > > Plopsi > > On 12 Mrz., 07:15, dyc <[email protected]> wrote: > > > Hey guys, > > > I am currently trying to go from an XYZ coordinate to a lon/lat > > coordinate. > > > The formula to go from lon/lat to XYZ is as follows: > > > LAT = latitude * pi/180 > > LON = longitude * pi/180 > > x = -R * cos(LAT) * cos(LON) > > y = R * sin(LAT) > > z = R * cos(LAT) * sin(LON) > > > Can anyone help me figure out how to reverse that formula so I can get > > lon/lat from xyz? > > > Thanks so much! > > > `Scott To unsubscribe from this group, send email to away3d-dev+unsubscribegooglegroups.com or reply to this email with the words "REMOVE ME" as the subject.
