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triganometry help!

locoboy - 14/1/10 at 09:22 PM

Guys i need to work out the 2 unidentified angles in the drawing below.

The angle that forms the kink in the vertical leg and the angle that takes it to horizontal.

Can anyone help given the dimensions already on the drawing?

bar

Cheers
Colin

MakeEverything - 14/1/10 at 09:30 PM

What size pipe?

omega0684 - 14/1/10 at 09:31 PM

168.7 degrees, if my trig serves me right

HTH's

[Edited on 14/1/10 by omega0684]

locoboy - 14/1/10 at 09:33 PM

57.15mm
Cheers

omega 24 v6 - 14/1/10 at 09:34 PM

quote:

168.46 degrees, if my trig serves me right

Not quite as you need the pipe diameter first.

deezee - 14/1/10 at 09:38 PM

I worked it out in AutoCAD to 166.1 degrees. I guessed the object as being 25mm box section. So I'm probably miles out

omega 24 v6 - 14/1/10 at 09:40 PM

171.82 degrees is my answer

locoboy - 14/1/10 at 09:42 PM

There are 2 angles though...............

omega0684 - 14/1/10 at 09:44 PM

quote:
Originally posted by omega 24 v6

quote:

168.46 degrees, if my trig serves me right

Not quite as you need the pipe diameter first.

no you dont, all you need to do is work out tan^-1of 0.2 and then minus that from 180. from his drawing it takes the pipe diameter into account from the bottom measurement. (as he has put his marker in the centre of the tubes)

all i did was 710-590 = 120 then do 120/2 = 60

430-130 = 300 then its just simple trig, tan (Alpha) = opp/adj = 0.2

tan^-1 0.2 = 11.31

180-11.3 = 168.7

i might be a little rusty, haven't done trig for 7 years

[Edited on 14/1/10 by omega0684]

omega 24 v6 - 14/1/10 at 09:47 PM

OK so overall bottom width is 710 plus 57.15
equals
767.15
Minus the top size of 590
equals 177.15
divide by 2 to give base of triangle
equals 88.575

Tan of angle = opp/adj

so 300 ( 430 -130) divided by 88.575
equals 3.387
so inv tan equals 81.72 and then add the 90 degrees
equals 171.723

omega 24 v6 - 14/1/10 at 09:50 PM

quote:

no you dont, all you need to do is work out tan^-1of 0.2 and then minus that from 180. from his drawing it takes the pipe diameter into account from the bottom measurement. (as he has put his marker in the centre of the tubes)

Was the way I did it at first but then wondered why someone asked for the diameter of the pipe.
So when you do it your way you are forming the triangle hypontenuse from the outside of the top measurement to the middle of the pipe in the bottom measurement. Not entirely the correct line IMHO

omega 24 v6 - 14/1/10 at 09:54 PM

quote:

There are 2 angles though...............

If you mean the top one then all angles of a rightt angle triangle add up to 180 degrees so 90 plus 81.72 equals 171.72 away from 180 leaves 8.28 plus the 90 degree so a total of 98.28

mattf - 14/1/10 at 10:19 PM

I made it 163.55 degrees. I make arctan 3.3869 to be 73.55

Toltec - 14/1/10 at 10:23 PM

quote:
Originally posted by mattf
I made it 163.55 degrees. I make arctan 3.3869 to be 73.55

I agree with that (checked on Excel, and a Casio) as well as the calculation to find lengths of the triangles sides.

omega 24 v6 - 14/1/10 at 10:25 PM

quote:

I made it 163.55 degrees. I make arctan 3.3869 to be 73.55

I concur I had my calc set on Grads instead of degrees DOH

[Edited on 14/1/10 by omega 24 v6]

locoboy - 14/1/10 at 10:47 PM

Your way above me guys!

so the 2 angles will be???????

omega 24 v6 - 14/1/10 at 11:10 PM

163.55 for the vertical angle ( the one you marked as unknown). And 106.45 for the other one

loggyboy - 14/1/10 at 11:26 PM

OOOPS didnt see your post giving the Dia so see below:
Rescued attachment AngleQuery.gif

02GF74 - 15/1/10 at 07:19 AM

quote:
Originally posted by mattf
I made it 163.55 degrees. I make arctan 3.3869 to be 73.55

that is what I make it, see below.

Rescued attachment angle.JPG

02GF74 - 15/1/10 at 07:21 AM

beaten to it ..... should be tan c = ... in the above ... oops!!!

.... and then 300/88.75

[Edited on 15/1/10 by 02GF74]

locoboy - 15/1/10 at 07:25 AM

Awsome,

Thanks a lot guys, It has helped me out of a bit of a sticky situation.

The power of LCB comes to the rescue again

Richard Quinn - 15/1/10 at 11:59 AM

Don't forget that if this is being done on a "programmable" bender, you will have to include an element of over bending to allow for spring back.