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I own MathCad 14 and in general love it to death, but I don't like the way it handles units using Solver, as the final answer in Solver never has any units. But if you solve the equation for the unknown and type that in, the answer will have the correct units.
This is inconsistent. I have attached a MathCad 14 worksheet that illustrates what I mean.
My question is: Are there any later versions of MathCad where Solver solves the equation and attaches the correct units? If so, I will put in on my wish list. If not, I will wait until it does appear.
(It has become clear MathCad was written by a bunch of pure mathematicians who don't know anything at all about these flaky things called "units" and really don't want to know. Should they be flying along in an airplane and the wings suddenly fall off, as they plunge to a fiery doom the thought might wander through their fuzzy-academic brains perhaps that attitude was a mistake.)
Jeff Corkern
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Consider the following as a statement of logic, and rank it as "True" or "False":
"If people possess immortal souls, it should be possible to logically deduce this by objective analysis of their actions."
I'm not sure how 14 handles the example you gave but here's an output pdf from the current mathcad 15.
Can you print what 14 does for comparison?
I've added the ',x' after the solve to show explicitly the calculation that is to be performed but it made no difference to the result.
Regards
Andy
You have an extra 'liter' in the solve routine when compared to the division. Just take it out, so that it is 2x, not 2xliter, and 'liter' will appear in the answer!
Hope this helps.
Bill
I'm aware leaving a unit out of the calculation will make MathCad put the non-canceled-out unit in the answer.
Basically, you're violating the rules in order to make it look right.
This is not what I want. What I want is for MathCad to do it right in the first place---all units properly labeled, including your unknown, and Solver puts the correct units next to the answer.
Just like it does when you enter an explicit calculation for x.
Jeff Corkern
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Consider the following as a statement of logic, and rank it as "True" or "False":
"If people possess immortal souls, it should be possible to logically deduce this by objective analysis of their actions."
Let's be precisely consistent. Solve the top equation for x, and get the true equality.
It's not inconsistent, and there are no rules being violated to fix it. If you put x*liter on the left then x is 0.025 (if you were to substitute 0.025 for x then the original equality is correct). If you put only x on the left, then for the equality to be correct x is 0.025*liter.