algebra in degree minutes seconds

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AbelM Kusemererwa
AbelM Kusemererwa 2015 年 6 月 19 日
回答済み: Mukul Rao 2015 年 6 月 23 日
The format of the variable 'a' are in degrees minutes but written as for example a= '4534.7899', where 45 is degrees and 34.7899 is minutes.
The vector cell 'c' is populated by data format such 'a' c= {a1; a2; a3}. 'c' is attached at lat How can I subtract a constant, angle 'a' from each cell in 'c' and obtain a column of cells 'd' which in seconds?
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Mukul Rao
Mukul Rao 2015 年 6 月 22 日
I do think I understand the format completely. How would this format represent say 45 degrees 34 minutes and 4.55 seconds?
AbelM Kusemererwa
AbelM Kusemererwa 2015 年 6 月 22 日
he attached is a degree minutes vector cell in the format 5257.1172944 where deg=52 and min=57.1172944 a=5257.1172000 , 'a' is a single input angle of the same format as the attached file.
I want to subtract a from each cell in the vector cells attached and multiple the difference by 60 to obtain an answer in seconds

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Mukul Rao
Mukul Rao 2015 年 6 月 23 日
You can make use of either the "arrayfun" or the "cellfun" to perform elementwise operations. Here are the links for these functions:
The commands would look like,
a = 5257.1172000;
load lat
%Convert entries in lat_cell to a number with "str2num"
output = cellfun(@(cellin) (str2num(cellin) - a)*60,lat_cell);
The variable "output" is a double array that contains the required differences.

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