Finding the length of a diagonal

Discussion in 'Physics & Math' started by Beaconator, Feb 3, 2019.

  1. Beaconator Valued Senior Member

    Messages:
    1,486
    So I was playing with triangles and I was trying to find out the legnth for a diagonal 22" by 30" up so I took 30/22 and times it by pie. My theory was sine over cosine... So they guy at the store was sitting there about to cut the ten foot pipe... I told him four feet which fits perfectly for the railing straight across. Which is what I want to make now anyways.

    I know its completely the wrong math but the only thing I can think of is that's its some kind of vetruvian scalar.
     
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  3. James R Just this guy, you know? Staff Member

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    39,426
    What you should have done was used Pythagoras. The diagonal is:

    \(\sqrt(22^2 +30^2)=37.2" = 3' 1.2"\)

    Your calculation, on the other hand, was

    \(30/22\times 3.142 = 4.5\text{ (no units)}\)

    Given that your calculation doesn't even produce a length, let alone the correct length, it's not much use.

    Theory of what? Sine of what? Cosine of what? How does this relate to the calculation you actually performed?

    4 feet is about 11 inches too long. So, either your initial measurements were wrong, or the guy at the store cut the wrong length, or you did some trimming somewhere, or there was overhang, or something else.

    What's a vetruvian scalar? And what in this case do you think was a vetruvian scalar?

    At this point, my advice to you would be: get somebody else to do the maths if anything important ever depends on getting the right answer. A wild stab in the dark might be good enough for many purposes, but if you need precision you'll need to get it right.
     
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  5. Beaconator Valued Senior Member

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    1,486
    It is something... Ten based number system vs. 16 parts in a whole.
    Pretty useful when the frame is every two feet.
    my teacher always taught us purple elephant theory.
    Better longer than shorter. I can always trim but growing back a pipe is hard,
    English is a funny language. Many meanings for single words as opposed to many words for single things.

    Knowing the parameters helps.

    Not saying its something known just saying it worked. Vetruvian man there Leonardo.
     
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  7. James R Just this guy, you know? Staff Member

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    Beaconator:

    The price of cheese when the elephant explodes.

    Yes, and also when the hoop ball flounces.

    Yours too? Good.

    Well, that at least makes some sense. One out of four ain't bad.

    You didn't answer my question. Never mind. It sounds like you're in no fit state.

    Being able to string a coherent thought together helps.

    Serendipity then.

    Yes. And..?
     
  8. Beaconator Valued Senior Member

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    1,486
    It worked can you explain why?
     
  9. sculptor Valued Senior Member

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    8,476
    Never underestimate the power of dumb luck to see you through when planning and intellect fail.
     
  10. DaveC426913 Valued Senior Member

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    18,960
    JamesR did.

    • your initial measurements were wrong,
    • the guy at the store cut the wrong length
    • you did some trimming somewhere
    • there was overhang
    • something else
    However it worked, it wasn't because you calculated correctly.
     
  11. sculptor Valued Senior Member

    Messages:
    8,476
    Alternately:

    It seems that there has been an assumption that you were dealing with a right triangle.
    If your length measurements were accurate, then that is most likely not the case.
     
  12. James R Just this guy, you know? Staff Member

    Messages:
    39,426
    The word he used was a "diagonal" and I assumed a rectangular cross-section, especially since he gave the measurements of two sides.

    On the other hand, there's no way to know what he means by "diagonal". Most of the rest of what he has written hasn't made any sense.
     
  13. sculptor Valued Senior Member

    Messages:
    8,476
    yeh
    but, the length dimensions given work for an
    obtuse scalene triangle.
    with angles
    19°11'17"
    26°37'39″
    134°11'4″

    does "diagonal" necessitate a right triangle?

    ........................
    long ago
    I had a customer who wanted me to make cabinets which fit into recesses in his wall.
    The recesses were trapezoidal ----
    ok it is easy to accidentally make a cabinet with a trapezoidal footprint
    making one intentionally proved to be a tad more interesting
     
  14. DaveC426913 Valued Senior Member

    Messages:
    18,960
    • it was not a right triangle
     
  15. sideshowbob Sorry, wrong number. Valued Senior Member

    Messages:
    7,057
    That was my thought. A carpenter I once worked for used to joke, "I cut it twice and it's still too short."

    But it's the only thing in your posts that makes any sense.
     
  16. DaveC426913 Valued Senior Member

    Messages:
    18,960
    Excellent point sculptor.

    So, given that the OP specified everything except that it was a right triangle, we can safely conclude that the setup looked like this:

    Please Register or Log in to view the hidden image!

     
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  17. Beaconator Valued Senior Member

    Messages:
    1,486
    It was an adjacent triangle placed over the initial right triangle.

    So sin over cos gives you the tangent times pi gives you the area unsquared inscribed or length inscribed?
     

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