毕业设计&课设-在MATLAB中执行SI系统.zip

[](http://oldhouse.ddns.net:7575/job/FEX%20-%20Physical%20Quantity/) [](http://oldhouse.ddns.net:7575/job/FEX%20-%20Physical%20Quantity/) [](https://www.mathworks.com/matlabcentral/fileexchange/69125-physicalquantity) [](https://www.paypal.com/cgi-bin/webscr?cmd=_s-xclick&hosted_button_id=4M7RMVNMKAXXQ&source=url) # PhysicalQuantity Consistent, fully automated, intuitive handling of physical units Can't remember how to convert feet to meters? Wish there was a way around the dreaded radians-versus-degrees problem? Ever stared yourself blind on a bug that was due to inconsistent units? Look no further. This set of tools will resolve all of those problems, once and for all. Preparations: - put the PhysicalQuantity and PhysicalVectorQuantity directories on the MATLAB search path Example session: >> L = Length(4, 'm') L = 4 meters >> R = Length(2, 'ft') R = 2 feet >> A = Area(L*R, 'm^2') A = 2.438400000000000e+00 square meters Wait...really? That can't be right. Let's see, 1 ft = 0.3048 meter, so 2 ft * 4 m = 2*0.3048 * 4 = 2.4384 ... square ... meters. Wow! What if I try to break it? >> A = Area(L*L*R, 'm^2') Error using Area (line 13) Can't create 'Area' (dimensions [L]²) from a quantity with dimensions [L]³. >> A = Area(L*R, 'm^3') Error using Area (line 13) Dimensions implied by the given string ([L]³) are incompatible with those of an Area ([L]²). >> tan( Area(L*R, 'm^2') ) Can't compute the tangent() of an Area. OK, well, that ... could actually be very useful! How's it handle angles? >> theta = Angle(-1.2, 'deg') theta = -1.200000000000000e+00 degrees >> phi = Angle(0.5, 'rad') phi = 500 milliradians >> cos(theta) ans = 9.997806834748455e-01 % <- yup, same as cosd(-1.2) # Overview In physics, a quantity's *unit* is intricately connected to the quantity itself. Typically, quantities with different units cannot be used in the same operation, without some sort of conversion between them - "you can't add apples to oranges". This is not unlike data types in most programming languages. What do you expect to get when you divide two integers? Add a boolean to a character? You can't just do that without some sort of conversion. Therefore, it makes sense to create a data type that takes physical units into consideration for all operations. Preferably, this data type is also super-easy to use, and produces intelligible error messages when it's used in a physically meaningless way. And that is exactly what this tool set aims to provide. # How to use PysicalQuantities are constructed from scratch like this: Q = <quantityname>(<value>, <units>) Take a look in the PhysicalQuantity directory for an overview of currently supported <quantities>. The <value> can be any numeric value (including sparse, complex, etc.). The <units> should be a string specifying any of the supported long/short unit names (see below), possibly combined by the following operations: '*': multiply '/': divide '^': exponentiate Example: F = Force(300, 'Newton') r = Density(2, 'kg/m^3') To get a list of available units: >> L = Length(); >> L.listUnits() Length supports the following units of measurement: - Chinese mile (li) - Parsec (pc) - astronomical unit (AU) - foot (ft) - furlong (fur) - inch (in) - lightyear (ly) - meter (m) - mile (mi) - nautical mile (n.mi) - smoot (smt) - statute mile (st.mi) - yard (yd) - Ångström (Å) Conversions between compatible units are seamless: >> L = Length(20, 'yards'); >> L('meters') ans = 1.828800000000000e+01 meters >> P = Length(3, 'inch'); >> Q = Length(18, 'foot'); >> Length( L*P/Q, 'meter' ) ans = 2.54e+02 millimeters Operations work as expected: >> L = Length(1, 'inch'); >> t = Duration(32, 'seconds'); >> M = Mass(18, 'lb'); >> F = Force( M*L/t/t, 'Newton' ) ans = 2.025219058242187e+02 milliNewtons If you find this work useful, please consider [a donation](https://www.paypal.com/cgi-bin/webscr?cmd=_s-xclick&hosted_button_id=4M7RMVNMKAXXQ&source=url).