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[-] Someonelol@lemmy.dbzer0.com 5 points 3 hours ago

The lack of labeling each variable (with units!) in equations really boils my piss. Yes the author knows them by heart, but even peers in the same field could struggle to understand what they mean. If introductory chemistry and physics instructors beat the practice into their students I see no excuse for authors to leave them out in a thesis.

[-] Buddahriffic@lemmy.world 2 points 46 minutes ago* (last edited 28 minutes ago)

Variables can be in arbitrary units. If you put the units through the same steps as the values, you'll end up with the right unit, plus you need the values to be in compatible units for the operations to even make sense. At least as long as any constants are also given the correct units. This can also help discover cases where you accidentally mix similar but different units (like litres and gallons).

I also find it very satisfying to do those equations on the units, adding ones that get multiplied in and cancelling ones that get divided out, and then ending up with a unit that makes sense, given what the equation was supposed to express.

Though we might be saying the same thing, because while I don't think variables need units, expanding that variable with a value absolutely should involve a unit, as should any examples of that equation's use.

Edit: an example to show what I mean:

If you have a table that is 1.5m long and know that you want at least 2 feet between the wall and the table on each side so that people can get in and out if chairs comfortably, you'd have 1.5m + 2*2ft as the equation, but you can't add an m to a ', so you need to multiply one of them by a conversion factor:

1.5m + 2 * 0.3048m/ft * 2ft

The ft terms cancel out and you're left with

1.5m + 2 * 0.3048m * 2

= 1.5m + 1.2192m

~= 2.72m

So your space must be at least 2.72m to fit that table and chairs comfortably (by the above definition of comfort, at least).

Funny enough, this use of units helped me avoid a mistake in this example because I had the ft/m conversion initially but saw that would result in trying to add m to ft² so it was obvious that I had made a mistake. Otherwise I might have looked like an idiot trying to say that a 1.5m table with 4 feet of buffer space requires a room over 12m long.

this post was submitted on 16 Oct 2024
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