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Insights Make Units Work for You - Comments

  1. Dec 1, 2017 #1

    scottdave

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  2. jcsd
  3. Dec 1, 2017 #2

    berkeman

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    Such a fundamental and useful trick. I still remember when I learned it in one of my first university basic engineering classes. What an epiphany! :smile:
     
  4. Dec 1, 2017 #3
    Great first Insight @scottdave!
     
  5. Dec 1, 2017 #4

    scottdave

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    Thanks,
    I had fun writing it, and learned a little more about LateX, as well.
     
  6. Dec 1, 2017 #5
    Thanks @scottdave for a great Insight. Dimensional analysis is in my opinion a underrated method/trick to check complicated expressions.
     
  7. Dec 1, 2017 #6

    Charles Link

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    I have an additional comment or two. I like the way @scottdave shows the conversion by multiplying by 1, as in the example with ## 1=\frac{2.54 \, cm}{1 \, inch} ##. There is one other thing that I find useful in doing some of the more difficult conversions, and that is to write the quantity such as length ## \vec{L} ## as a vector. In that case ## \vec{L}=L_{cm} \hat{cm}=L_m \hat{meter} ##. We can solve for ## L_{cm} ## in terms of ## L_m ## if we know that ## 1 \hat{meter}=100 \hat{cm} ##. In that case ## L_{cm}=100 \, L_m ##. This kind of thing comes in handy, e.g., in converting E&M (electricity and magnetism) quantities in c.g.s. to M.KS. and visa versa. This is a more advanced application, but on occasion, it is useful also on simpler conversions.
     
  8. Dec 1, 2017 #7

    scottdave

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    These techniques can definitely be applied to more complex calculations. It is probably more critical to use these methods in those situations.

    I chose to keep my examples fairly simple, so that the article might appeal to a broader readership.
    I can't tell how many questions I have seen on PhysicsForums and other sites, where these techniques would have solved the problem for the OP.
     
  9. Dec 3, 2017 #8

    Mister T

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    I like everything about your article, but there is one point of contention I would like to make. Setting these ratios equal to ##1## is a matter of preference, and I prefer to follow the advice of Arnold Arons and :sedulously avoid" it. I understand the motivation for using it, which is to imply that it's okay to use it as a multiplicative factor in the same way that it's okay to use ##1## as a multiplicative factor. But expressions like $$frac{2.54 mathrm{cm}{1 mathrm{m}}=1$$ can be a source of confusion for learners. I think it better to simply state that multiplying by the factor $$frac{2.54 mathrm{cm}{1 mathrm{m}}$$ is a perfectly acceptable practice. After all, it is not the same thing as multiplying by ##1## because it has the effect of changing the units.

    But I realize that this is entirely a matter of preference. I have a colleague who disagrees, stating that students are likely to encounter it presented that way. But I disagree.

    Reference https://www.physicsforums.com/insights/make-units-work/
     
  10. Dec 3, 2017 #9

    Mister T

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    Let's try getting the LaTeX coding right one more time ...

    I meant to say that I prefer not using expressions like $$frac{2.54 mathrm{cm}}{1 mathrm{m}}=1$$

    But instead prefer $$frac{2.54 mathrm{cm}}{1 mathrm{m}}$$
     
  11. Dec 4, 2017 at 7:59 PM #10

    scottdave

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    Interesting point, @Mister T. I see where you are coming from. The point that I intended was it is easier to figure out if the factor by going through and setting 2.54 cm equal to 1 inch. Then, you can have either (2.54 cm) / (1 in) or (1 in) / (2.54 cm), depending on which way you need to go. I believe this is easier than trying to remember "If I want to convert from centimeters to inches, then divide by 2.54".

    It seems that LaTeX works differently in the comment section than in regular posts? At least on this browser.
     
  12. Dec 4, 2017 at 8:17 PM #11

    Mark44

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    What you wrote would be a source of confusion for anyone.
    I think you meant ##\frac{2.54 \text{ cm}}{1 \text{ inch}} = 1##, which to me is perfectly fine. Since 2.54 cm = 1 in, divide both sides by "inch" to produce the equation I wrote, or even ##2.54~\frac{\text{ cm}}{\text{ inch}} = 1##.
    Except that 2.54 cm is NOT equal to 1 m.
    If you accept, say, that 2.54 cm represents the same lengh as 1 inch, then we can write 2.54 cm = 1 in, recognizing that we're talking about the same length measure, but with different units. It follows that the ratio ##\frac{2.54 \text{ cm}}{1 \text{ inch}}## is dimensionless and equal to 1. Being 1, we can multiply whatever we want by it without changing the underlying value of what it is multiplying.
     
    Last edited: Dec 4, 2017 at 11:28 PM
  13. Dec 4, 2017 at 8:52 PM #12

    DS2C

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    This is so useful for me. Ive watched a bunch of videos on it and this explains it much better. So are you treating the units like variables essentially?
     
  14. Dec 4, 2017 at 10:17 PM #13

    Mister T

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    Yes, I agree completely. Teaching students to multiply by the appropriate ratio is the way to go. My point is that it's not necessary, and adds to student confusion, when you set those ratios equal to 1. While I agree that it's valid in the sense that it's being used, its only utility is to justify the practice of multiplying by ratios. There's no need for a student to ever see an expression like 2.54 cm = 1 in. Some people like to instead write 2.54 cm ##\hat{=}## 1 in. to make the distinction clear that the two quantities represent the same length and in that sense are equivalent rather than being equal.

    Well, I buggered that up pretty good, didn't I? Especially when I wrote m instead of in.
     
  15. Dec 5, 2017 at 6:05 PM #14

    scottdave

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    That would be a way to think about it, or as vectors of certain lengths, as @Charles Link stated in his post.
     
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