How to substitute expressions containing units when using the ezunits package? - maxima

Question

Without specifying units, I can express area and volume and have Maxima show the relationship:
(%i1) areaNoUnits: area = width * length$
(%i2) volumeNoUnits: volume = area * height$
(%i3) volumeNoUnits, areaNoUnits;
(%o3) volume = height length width
(%i4) subst(areaNoUnits, volumeNoUnits);
(%o4) volume = height length width
Now I want to specify units so I will use the ezunits package.
The ` (backtick) operator is the building block of ezunits:
An expression a ` b represents a dimensional quantity, with a indicating a nondimensional quantity and b indicating the dimensional units.
When I add units to the area and volume expressions, evaluation and substitution do not work:
(%i1) load ("ezunits")$
(%i2) areaWithUnits: area ` m^2 = (width ` m) * (length ` m);
2 2
(%o2) area ` m = length width ` m
(%i3) volumeWithUnits: volume ` m^3 = (area ` m^2) * (height ` m);
3 3
(%o3) volume ` m = area height ` m
(%i4) volumeWithUnits, areaWithUnits;
3 3
(%o4) volume ` m = area height ` m
(%i5) subst(areaWithUnits, volumeWithUnits);
3 3
(%o5) volume ` m = area height ` m
The expected output is:
volumeWithUnits, areaWithUnits;
3 3
volume ` m = height length width ` m
I do not see a function in the ezunits package to do evaluation or substitution. What is the right way to do this?

Answer 1

I would phrase it like this:
(%i2) load (ezunits) $
(%i3) width: W ` m;
(%o3) W ` m
(%i4) length: L ` m;
(%o4) L ` m
(%i5) area: width * length;
2
(%o5) L W ` m
(%i6) height: H ` m;
(%o6) H ` m
(%i7) volume: area * height;
3
(%o7) H L W ` m
I wrote each part as conceptualname: symbolforquantity ` unit and then wrote just conceptualname in further calculations, instead of conceptualname ` unit.
The substitution you tried in %i5 didn't work because subst is a purely formal substitution -- if there isn't a literal subexpression which is the same as the substituted-for expression, it doesn't match; subst doesn't look for rearrangements or factorizations which could help make a match. There are ways to work around that, so it might be possible to make your original formulation work, but I think it's better overall to sidestep the problem and work with conceptualname and symbolforquantity ` unit.
To say a little about what more one could do with expressions like %o7 above. There are at least two ways to replace symbols H, L, and W with specific values. One is to call subst:
(%i2) load (ezunits) $
(%i3) volume: H*L*W ` m^3;
3
(%o3) H L W ` m
(%i4) subst ([L = 20, W = %pi], volume);
3
(%o4) 20 %pi H ` m
Another is to make use of ev.
(%i5) ev (volume, L = 20, W = %pi);
3
(%o5) 20 %pi H ` m
Note that at the input prompt, something, someflags, somevalues is equivalent to ev(something, someflags, somevalues).
(%i6) volume, L = 20, W = %pi;
3
(%o6) 20 %pi H ` m
This is just a convenience. Within a function, one has to say ev(...); the shorter syntax isn't understood there.
ev is often convenient, but it's generally simpler to predict what the result is going to be with subst instead.