Translating Trig Expressions into Code
Mostrar comentarios más antiguos
Hello everyone! I am a beginner at matlab and am attempting to calculate the area of each individual grid cell of climate data. I know that the area can be calculated with ((2*pi*R)/360)^2)*cos(theta)*delta_latitude*delta*longitude but am lost as to how to translate this into code. Can anyone offer suggestions please?
2 comentarios
Joseph Cheng
el 22 de Ag. de 2014
Editada: Joseph Cheng
el 22 de Ag. de 2014
well unless there is something complicated what you typed in should work. if any of those are an array (more than 1x1) then you may want to look into the element by element multiplication '.*'
also did you make a typo? should it be delta_longitude?
((2*pi*R)/360)^2)*cos(theta)*delta_latitude*delta_longitude
and make sure theta is in radians or use cosd() for cosine with input of degrees.
Stephanie Goldstein
el 22 de Ag. de 2014
Respuestas (2)
Roger Stafford
el 22 de Ag. de 2014
0 votos
Your formula is only a valid approximation if the two delta's are small and are measured in degrees. Also theta must be the latitude. If it is measured in degrees, in matlab you would use either cosd(theta) or cos(theta*pi/180). Otherwise you can copy it directly into a matlab assignment.
3 comentarios
Stephanie Goldstein
el 22 de Ag. de 2014
Roger Stafford
el 22 de Ag. de 2014
Editada: Roger Stafford
el 22 de Ag. de 2014
Yes, that would produce a complaint from matlab. If you have a 224 x 1 size vector first multiplied by a 1 x 240 size vector second, it would give you a 224 x 240 matrix of products. Is that what you are after? You haven't adequately explained what you are really attempting to do.
Stephanie Goldstein
el 25 de Ag. de 2014
Image Analyst
el 22 de Ag. de 2014
Editada: Image Analyst
el 22 de Ag. de 2014
I think you both made a typo. There should be no ) after the R. You have three ( and four ).
degreeToRadiansConversionFactor = pi / 180; % Call it out so people know what it is.
area = (R * degreeToRadiansConversionFactor)^2 * delta_latitude * delta_longitude;
I also took out the cos(theta) since I don't know why it's there. Since the arc length on the surface of a circle is S=R*theta, if the patch is rectangular, you'd get an area of length * width = R^2 * delta_latitude * delta_longitude, where delta_latitude & delta_longitude are in radians, so where do you get the cos(theta) from?
6 comentarios
Stephanie Goldstein
el 22 de Ag. de 2014
Image Analyst
el 22 de Ag. de 2014
You must be using arrays. Just use a scalar for delta.
Stephanie Goldstein
el 22 de Ag. de 2014
Roger Stafford
el 22 de Ag. de 2014
No, the cos(theta) factor is needed! If you cut the earth with a plane parallel to the equator's plane but at a higher latitude, the circle you get has a smaller circumference than 2*pi*R, and that is the plane the longitude angle is measured in. The cos(theta) is the required correction.
Roger Stafford
el 22 de Ag. de 2014
@Image Analyst. If you have ever used USGS topographical maps, you will note that maps in the U.S. of the seven-and-a-half-minute series are not squares but are rectangles, even though the two angle delta's are the same. That illustrates the need for the cosine correction.
Stephanie Goldstein
el 25 de Ag. de 2014
Categorías
Más información sobre Lengths and Angles en Centro de ayuda y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!