# Length needed from a known length and angle

• July 21st 2009, 04:15 AM
mrman
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 ...

http://i27.tinypic.com/24pw08g.jpg

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.
• July 21st 2009, 05:20 AM
Hello mrman

Welcome to Math Help Forum!
Quote:

Originally Posted by mrman
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 ...

http://i27.tinypic.com/24pw08g.jpg

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?

• July 21st 2009, 05:48 AM
mrman
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.
• July 21st 2009, 06:03 AM
Hello mrman
Quote:

Originally Posted by mrman
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 $\theta \approx 130.5^o$