Electronics > Beginners
How can i put "gravity" to (Lat, Long to X, Y, Z) ?
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ammjy:
Hello, EEVBlog users !
Always thank you for your kind help and time.
I want make acceleration data through convert (Lat, Long to X, Y, Z).
but, i can't figure out how i can put "gravity" to X, Y, Z equations
to make acceleration data.
if latitude and longitude is,
Lat = [-30:2:30];
Long = [-90:1:90];
i can make latitude and longitude as follows,
lat = lat * pi/180;
long = long * pi/180;
phi = pi/2 - lat;
theta = long;
than, XYZ is
x = sin(phi) * cos(theta);
y = sin(phi) * sin(theta);
z = cos(phi);
Can I add g like this?
g = [ 0 0 9.8 ];
Acc_x = -g * sin(phi) * cos(theta);
Acc_y = g * sin(phi) * sin(theta);
Acc_z = -g * cos(phi);
I am so sorry, my question is so vague.
If you don't mind, could you give to me advise ?
Thanks you for reading, !
Please, let me know !
agehall:
If you just want a vector that represents the direction of g at every point of the sphere, that would essentially be the negative normal at each point.
So just use g=-9.81 (note: not a vector!) and then calculate the vector at each point by doing g * [x y z] where x, y and z are calculated as you suggested and you should end up with a vector that describes g as a directional vector at each point [x y z].
ammjy:
Thanks you so much, also Thanks you for your time, Ms. Agehall.
I understood your kind answer.
Have a good day !
Acc_x = -g * sin(phi) * cos(theta);
Acc_y = -g * sin(phi) * sin(theta);
Acc_z = -g * cos(phi);
soldar:
--- Quote from: agehall on April 18, 2019, 08:20:13 am ---If you just want a vector that represents the direction of g at every point of the sphere, that would essentially be the negative normal at each point.
So just use g=-9.81 (note: not a vector!) and then calculate the vector at each point by doing g * [x y z] where x, y and z are calculated as you suggested and you should end up with a vector that describes g as a directional vector at each point [x y z].
--- End quote ---
That is a simplification that might be sufficient or not depending on what the objective is. It assumes the world is a sphere and with uniform acceleration of gravity.
For more precise calculations search for "convert wgs84 to polar"
https://www.pgc.umn.edu/apps/convert/
ammjy:
Thanks you for your time, and your kind attention !
i understood very well your point :)
i totally agree.
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