Basic calculations!

[JohnB2]

I'm planning a bird feeder with a

Gable Roof

in effect a 4 sided prism. As I vary the pitch, and thus the height, the angle at which each ridge meets will vary; but how do I calculate the cutting angle to ensure butt joints?

I'm thinking of a 15 degree slope for each face.

JohnB2

 

[matt]

90 degrees - 15 degrees = 75 degree cut.

And, to cut the horizontal plane to butt 4 pieces to make a basic structure you would produce a 90 degree point to each piece by cutting two 45 degree angles to meet in the centre. If you wanted eight pieces to make the sructure then you'd cut 45 degree points by cutting two 22.5 degree angles (to meet in the centre etc).

For such a small project you could draw each piece full size.

 

[Chris Knight]

John,
Are you seeking a compound mitre calculation?

http://www.csgnetwork.com/sawmitercalc.html

 

[Steve Maskery]

the angle at which each ridge meets will vary

John, I may well be wrong, but you say EACH ridge. Unless the bird-feeder is L-shaped, there can be only one. Do you mean each HIP? If so, that is a different calculation.

I used to have a spreadsheet which did that for me, but there was a error somewhere and I don't have the trig skills to find out exactly where.

Is this thing square? If so, I'll see about modelling it up in SU and take the angles from there.

S

 

[JohnB2]

Thanks Chris / Matt

I should perhaps have said Mitre not Butt joints for where the four pyramid sides come together.

JohnB

 

[JohnB2]

Steve

My inexperience shows clearly!

The roof consists of four equal sized triangular pieces in the shape of a pyramid. Each face will slope at 15 degrees. If the apex is H cm above the base then a mitre is needed along each edge ~ which I thought would depend upon the value of H.

I can't do the trig either!

 

[Steve Maskery]

OK
You have 4 hips and no ridge at all. If it were rectangular, you would have a ridge (if all sides sloped at 15 deg). I'll see what I can do.
S

 

[Steve Maskery]

According to my SU model;

Twist your mitre gauge by 44.008 deg, leaving a 45.992 included angle

Tilt your blade to 79.454 deg.

Anybody get anything different?

S

 

[mr grimsdale]

It's the same problem as in this thread but perhaps not to read it or you could get seriously confused!
What you are after is the dihedral angle at the corners, which will be greater than 90º. Trigonometry, graphical projection or computer program are the ways to do it.

 

[Chris Knight]

Steve,
The mitre calculator I pointed to has a different result. Because, I think I have taken a 15 degree slope from horizontal and yours is from vertical.

 

[Steve Maskery]

No, Chris, I have taken 15 deg from horizontal too.

Perhaps I should have said tilt blade 10.5 deg off vertical. Or are the numbers still different? If so, I'll take another look at my model.

S

 

[Steve Maskery]

This is what I have drawn:

The two orange faces are parallel, the cube above is taken as a lump out of the roof to get the axes ref'd off the hip edge.

S

 

[Sgian Dubh]

According to my SU model;

Twist your mitre gauge by 44.008 deg, leaving a 45.992 included angle

Tilt your blade to 79.454 deg.

Anybody get anything different?

S

Steve, your numbers are spot on, assuming John's roof rises at 15º from the base line and he's after a mitred joint. The dihedral angle is 158.91º which divided by 2 = 79.5º for the bevelled edge-- the blade tilt setting. The mitre angle is 46º as near as we woodworkers can set it. As you say, set the mitre gauge to 44º.

All this assumes John has a table saw or perhaps a chop saw to do the job. Slainte.

 

[thomvic]

There seems to be some confusion as to which angle is the dihedral angle. Is it the angle between the base and one of the sides or the angle between two triangular sides? I reckon the latter.

This site seems helpful as it specifically for woodworking.
http://www.josephfusco.org/Articles/Dih ... edral.html

 

[Sgian Dubh]

The dihedral angle is the angle at the intersection of two planes. So, for example, if you look at a plan view of a rectangular box the dihedral angle at the corner is 90º. Slainte.

 

[mr grimsdale]

Wrong - that's the plan angle (90º in this case). The dihedral is the angle square on between the two planes and will be more than 90º i.e the angle you would see if you look along the ridge from the apex, which is what Richard is (too hastily!) describing but he's put the marks in the wrong place.
It's Steve's 79.454 times 2 = 160º approx. in that example.
The shallower the pitch the bigger the angle until at 0º pitch the dihedral angle reaches 180º, i.e. flat.

 

[MickCheese]

I'm lost :?

Mick

 

[Steve Maskery]

Well what Richard says is true, but what he has drawn is not quite right.

Sorry Richard

I think that the image is supposed to indicate that the dihedral angle is what you see if you look from the apex down to one of the corners (i.e. straight along a hip). This is correct.

However, the angle he has labelled as the dihedral is not quite right. It does not go along one side of the base, through the bottom corner and along the other side of the base (as drawn) but it starts at some point along the base edge THEN MOVES TO A POINT SOME WAY UP THE HIP, then back down the other face tot he bottom edge again. If you think of the angle making a triangle, you have to position that triangle square on to the line of sight to read the true dihedral angle. In Richard's drawing it is parallel with the base, not square on to the line of sight.

Ooh, that sounds complex. It's not really.

Just a slip of the pencil, I think.
S

 

[xy mosian]

So.. the dihedral angle at the intersection of two planes is the angle which lies on a plane which itself is at right angles to both?

Taking the pyramid example. If one of the corners is attacked with a saw in a manner such that the cut is square, in both directions, to the edge. The exposed angle is the dihedral angle?

?????
xy

 

[Sgian Dubh]

Steve, and Mr Grimsdale have another look at the drawing because I believe it's correct, and as you are both trying to describe.

The drawing indicates that you should look from the apex and along the ridge at the intersection of the two planes to see the angle between those planes.

I do admit that I simply drew the angle you would see on the base plane, and I should perhaps have inclined the indicated angle perpendicular to the angle the ridge rises from the base plane. Slainte.