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Math Help - Length needed from a known length and angle

  1. #1
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    Length needed from a known length and angle

    Hi,

    I am not sure what the right title should be or even what area of maths this is (geometry?)
    It's only been 6 or 7 years since I did this at school... can't believe how easy it is to forget, if you don't do it everyday.

    I am currently designing a piece of metal work and need to work out the depth of a screw.

    The datasheet says ...



    However I'm completely stumped at what the formula for finding this out is! If someone could show me the formula, that would be great.

    Thanks

    Ps. apologies for the bad pic ! The screw is equal angles at both sides.
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  2. #2
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    Hello mrman

    Welcome to Math Help Forum!
    Quote Originally Posted by mrman View Post
    Hi,

    I am not sure what the right title should be or even what area of maths this is (geometry?)
    It's only been 6 or 7 years since I did this at school... can't believe how easy it is to forget, if you don't do it everyday.

    I am currently designing a piece of metal work and need to work out the depth of a screw.

    The datasheet says ...



    However I'm completely stumped at what the formula for finding this out is! If someone could show me the formula, that would be great.

    Thanks

    Ps. apologies for the bad pic ! The screw is equal angles at both sides.
    Subtract 3 from 5.05 and then divide by 2, to work out the amount by which the greater radius exceeds the smaller: answer: 1.025.

    Then, to work out x, use the fact that you have a right-angled triangle with an angle of 80^o (= 180 - 100), the adjacent side is 1.025 mm and you want the opposite: x.

    So x = 1.025 \tan 80^o = 5.813 mm (3 d.p.)

    Does that make sense?

    Grandad
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  3. #3
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    Hi,

    Thanks. That makes sense.

    Unfortunatly 5.813mm is too deep as I'm trying to go through a 1.2mm piece of metal so x has to be 1.2mm or less. 100degrees screw is out then.

    Is there a way of saying that I want X to be 1.2 and the thread to be 3 so therefore angle must be "Z"?

    If it's a lot of work then don't worry, I can just see what the biggest screw angle is and work try it out.
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  4. #4
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    Hello mrman
    Quote Originally Posted by mrman View Post
    Hi,

    Thanks. That makes sense.

    Unfortunatly 5.813mm is too deep as I'm trying to go through a 1.2mm piece of metal so x has to be 1.2mm or less. 100degrees screw is out then.

    Is there a way of saying that I want X to be 1.2 and the thread to be 3 so therefore angle must be "Z"?

    If it's a lot of work then don't worry, I can just see what the biggest screw angle is and work try it out.
    If the angle is \theta, then working as we did before:

    \tan(180 - \theta) = \frac{\text{Opp}}{\text{Adj}}=\frac{1.2}{1.025}= 1.171

    \Rightarrow 180-\theta = 49.5^o

    So \theta \approx 130.5^o

    Grandad
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  5. #5
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    Hi,

    Thanks so much. That's great.

    That formula will come in handy alot.

    I don't think they do 130degrees screws, but I have found some that may be suitable (Flat undercut).

    Thanks again for your help.
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