Calculating Angles

remember all saws show zero degrees as the start point but thats actually 90 degrees so further cause confusion:D(y)
 
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You need to define vectors normal to two adjacent pieces and find their dot product; from that you can get the angle.

Define x and y in the horizontal plane and z vertically.
For one piece let x = 0; y and z are defined by the pitch.
If I understand correctly, the base is 290mm and the height is 500mm, so one piece leans inwards by half of 290 = 145mm. So a vector normal to that would be p = (0,500,145).
The adjacent piece is rotated by 45 degrees around z, so its x and y values are equal and their magnitude is 500, i.e. q = (500/sqrt2, 500/sqrt2, 145)
The dot product is 0 + 500x500/sqrt2 + 145x145 = 197802.

Now to get the angle, cos T = p . q / |p||q|
|p| = |q| = sqrt( 500x500 + 145x145 ) so |p||q| = 500x500 + 145x145 = 271025
so cos T = 197802 / 271025 = 0.72983
so T = 43.128 degrees.

But the angle you want is half that, 21.56 degrees, because you remove half of that from each piece.

Disclaimer: worked out late at night, might be wrong!

(Would love to know what you’re actually making!)

P.S. If the pieces were vertical the answer would be 22.5 degrees, i.e. half of 45; it doesn’t reduce by much for a relatively pointy “spire” like this.

Edit: fix typos!

Edit: and in relation to the angle shown in the picture in the original post, that is of course 90 - 21.56 = 68.44 degrees.
 
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Now to get the angle, cos T = p . q / |p||q|
|p| = |q| = sqrt( 500x500 + 145x145 ) so |p||q| = 500x500 + 145x145 = 271025
so cos T = 197802 / 271025 = 0.72983
so T = 43.128 degrees.

This sort of maths makes my brain ache now, but yes, that makes sense. In another life 40 years ago, in the merchant navy (before satnav) I used to be able to work out sun and star sights, great circle sailings and many other things long hand from books of (nories) tables. That part of my brain seems to have disappeared!
 
Nice! Easier in many ways than all those trig formulas.

You have it right I think for the saw tilt angle, but how would you calculate the mitre cut angle? I can visualise the vector space you describe, but cannot see how to get the mitre angle (about 6.51deg) using this method.

Some typos:
.....and find their cross product......
The cross product is .....
isn't that the dot product ? https://en.wikipedia.org/wiki/Dot_product

....x and y in the horizontal plane and y vertically.
z vertically ?

(Would love to know what you’re actually making!)
A hollow octagonal pyramid is my guess, but it is going to look like a spike, I think. Presumably the OP is going to stick it on top of some more 'roof' (unless it's part of a model).
 
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....I need a diagram ...

from the line drawing at https://woodgears.ca/miter/
splay_angles.png
 
Mitre angle: I don’t think I’d resort to vectors for that.
Find the dimensions of each triangular piece:
Height by pythagoras = sqrt( 500x500 + 145x145 ) = 520.6
Base b: b/sqrt(2) + b + b/sqrt(2) = 290, so b = 290/(sqrt(2)+1) = 120.12
tan M/2= 120.12/2 / 520.6 = 0.1154
so M/2 = 6.58 degrees, M = 13.16 degrees.
 
(Would love to know what you’re actually making!)

.
I am making a fancy nesting box for an Owl

It will be a nine sided polygon with a witches hat type roof

this is the roof -
pol_1120.JPG

this is bigger than it may look - 433 high and about 480 in diameter.

I will be biscuit joining it, but what sort of glue? (it will often get wet)
 
we really need to stop the input off too many people this is a void that needs to be owl freindly rather than interesting ??
holes in trees and walls are what they are a void that will attract a nesting creature or not if its wrong
yes if its for your pleasure with a nest box being a possible option if your lucky but in general easy to observe will often mean if you open up it will not give security
 

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