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Thread: What are the common triangles, OT, laying floors ??

  1. #11
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    It goes back to units. If you want the unit to be 1/2", just take any of the triples (which will all have two odd components), and divide each component by 2.

    Ed
    For just a little more, you can do it yourself!

  2. #12
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    Assuming you're working in feet and inches, there are several in the list that mklotz posted that could be useful if you simply take them as inches.

    For example, 65,72, 97. 48, 55, 73. 36, 77, 85.

    5'-5", 6"-0", 8"-1". 4"-0", 4"-7", 6"-1". 3"-0", 6"-5", 7"-1"

    If you want larger triangles, multiply by 1.25, 1.5, or 2. No fractions smaller than 1/4" if you multiply by 1.25.
    Last edited by cameron; 02-10-2019 at 01:46 PM.

  3. #13
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    If you do the C squared = A squared + B squared and come out with a long decimal it is easy to convert to feet and inches. Say your answer is 10.7125 feet, multiply .7125 X 12 inches = 8.55 inches, so a bit over 8 1/2 inches. If you multiply the .55 inches by 16 you will get the answer to the nearest 1/16 of an inch, .55 x 16 = 8.8 sixteenths or 9/16 inch. With a tape measure and all the variables of measuring you won't get closer than a 1/16 of an inch. Concrete calculators have this feature built in for squaring foundations.

  4. #14
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    If you're just doing a floor why on earth do you need anything more than the 3-4-5 combinations?
    Keith
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  5. #15
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    Quote Originally Posted by LKeithR View Post
    If you're just doing a floor why on earth do you need anything more than the 3-4-5 combinations?
    Because, for some floor layouts, there are better triplets available, which is what the OP was apparently aking for.

  6. #16
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    If you read the article I referenced you would know that there is a formula due to Euclid for generating Pythagorean triplets...

    Euclid's formula is a fundamental formula for generating Pythagorean triples given an arbitrary pair of integers m and n with m > n > 0. The formula states that the integers

    a = m^2 − n ^2 , b = 2 m n , c = m ^2 + n ^2

    form a Pythagorean triple.

    So, if you don't like any in the list I showed, just generate one that suits your requirements.
    Regards, Marv

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  7. #17
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    Quote Originally Posted by cameron View Post
    Because, for some floor layouts, there are better triplets available, which is what the OP was apparently aking for.
    yes, exactly. some of the 45-45-90 will fill the room better, and give me longer legs to go from.
    and, I got to layout wood planking, starting in the back of a room, and meet up with planking in the hallway, and the hallway goes around a corner heading into said room.

  8. #18
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    Quote Originally Posted by cameron View Post
    Because, for some floor layouts, there are better triplets available, which is what the OP was apparently aking for.
    Could you explain why that would matter? How would room dimension affect (strip) floor layout?

  9. #19
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    Sometimes the simple "triple" won't work, and you have to solve a right triangle for special occasions.

    This is when the wise old Indian SOHCAHTOA comes in.

    S=O/H
    C=A/H
    T=O/A

  10. #20
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    Quote Originally Posted by Ringo View Post
    yes, exactly. some of the 45-45-90 will fill the room better, and give me longer legs to go from.
    and, I got to layout wood planking, starting in the back of a room, and meet up with planking in the hallway, and the hallway goes around a corner heading into said room.
    But 45-45-90 is always going to end up with a fractional number as noted already. Either the sides or the hypotenuse is going to have a fraction in it. Now there might well be a particular set value that works out to where both the sides and the hypotenuse are integers for one and the other is close enough to an actual integer. Like up to perhaps .01 or .02 out. But that's just one particular size.

    A 3-4-5 is only "longer and skinnier" by 1/3 part for the longer side. So hardly "tall and skinny" as suggested. And as we're seeing it's so easy to arrive at a reasonably enlargement of this basic integer relationship that it seems like folly to try to keep in mind a single particular set of dimensions for a 45-45-90 that isn't even going to be a perfect simple integer relationship.

    I also cannot imagine a room squareness test where I'd need better than 3ft x 4ft to look for 5ft to confirm trueness our "outness". I'd only use 6ft x 8ft and then look for 10ft where I've got two folks with one in the corner holding the dumb end and the other with the tape measure body at the ends and they want room to get around the guy stuck in the corner. So 6-8-10 just to get some extra room for two bodies rather than it needs to be that much more accurate.

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