Compound mitre angles

[MorrisWoodman12]

I am making a new hopper for a vertical mill we are renovating at Nutley windmill in East Sussex. The hopper is an inverted truncated square cone. Does anyone recommend or have a calculator to give the two angles that I have to cut the boards in order to mitre the sides together. The sides of the hopper slope at 50 deg to the horizontal.
Thanks.
Martin

 

[Adam W.]

Draw it full size on a piece of MDF and take off the angles with a sliding bevel or framing square.

That way it will work proper like.

Like so, but you should draw it yourself to get it right. (Ignore the 370 bit, as that's for something else.)

Use geometry, not maths!

 

[Inspector]

Search compound mitre angle calculator and you'll get lots. Here are a couple.

https://jansson.us/jcompound.html
http://www.pdxtex.com/canoe/compound.htm
Pete

 

[Sgian Dubh]

Assuming you're essentially describing a four sided pyramid, inverted in this case, and the sides 'rise' from the base line (which is at the top) at 50º from the horizontal you have something that looks like the sketch below (the other way up).

Looking down the sight line from the top of the pyramid gives you the dihedral angle which, in this case, is 114.4º. Divide the dihedral angle by 2 to establish the bevel cut required on each meeting edge i.e., the angle described by the edge to the panel's flat face. So,114.4/2 = 57.2º, the bevel angle of the meeting corners of the panels or parts. You may find you need to use the complementary angle of 32.8º (90º - 57.2º).

This leaves establishing the mitre gauge setting angle, which is the angle the panel has to be cut at the rising edges of each side from the base line (in the example below). In your case this angle is 57.3º. Basically, that's the angle in the sketch from the base line up each rising corner. As a bit of a footnote, the fact that the required bevel angle and the mitre angle are almost the same, i.e., ~57 degrees is coincidental. Slainte.

PS. I suggest you ignore any number after the decimal point in all the angles given above - spreadsheets can be overly precise, ha, ha. In fact, I suggest you simply try and get within a degree or three of any of the angles I've provided, because I suspect that will be more than close enough for what you're doing. I'm guessing your hopper doesn't need to emulate a finely wrought and delicate piece of furniture, art or sculpture, and near enough will be good enough. This last bit written with humour intended.

 

[Adam W.]

Hoppers are most likely butt jointed and fixed with nails, not mitred. With the outside cap cut off once in place, so you'll just have one angle to deal with, which is cut on two ends of the butt.

A templet would be good for this.

 

[MorrisWoodman12]

Search compound mitre angle calculator and you'll get lots. Here are a couple.

https://jansson.us/jcompound.html
http://www.pdxtex.com/canoe/compound.htm
Pete

Pete. Thanks for those. I'll store them away for another day should I need them again.
Martin

 

[MorrisWoodman12]

Assuming you're essentially describing a four sided pyramid, inverted in this case, and the sides 'rise' from the base line (which is at the top) at 50º from the horizontal you have something that looks like the sketch below (the other way up).

Looking down the sight line from the top of the pyramid gives you the dihedral angle which, in this case, is 114.4º. Divide the dihedral angle by 2 to establish the bevel cut required on each meeting edge i.e., the angle described by the edge to the panel's flat face. So,114.4/2 = 57.2º, the bevel angle of the meeting corners of the panels or parts. You may find you need to use the complementary angle of 32.8º (90º - 57.2º).

This leaves establishing the mitre gauge setting angle, which is the angle the panel has to be cut at the rising edges of each side from the base line (in the example below). In your case this angle is 57.3º. Basically, that's the angle in the sketch from the base line up each rising corner. As a bit of a footnote, the fact that the required bevel angle and the mitre angle are almost the same, i.e., ~57 degrees is coincidental. Slainte.

PS. I suggest you ignore any number after the decimal point in all the angles given above - spreadsheets can be overly precise, ha, ha. In fact, I suggest you simply try and get within a degree or three of any of the angles I've provided, because I suspect that will be more than close enough for what you're doing. I'm guessing your hopper doesn't need to emulate a finely wrought and delicate piece of furniture, art or sculpture, and near enough will be good enough. This last bit written with humour intended.

View attachment 133069​

Thanks Sgian Dubh (What does the Gaelic mean?). You have it exactly right so that's just what I need.
"a finely wrought and delicate piece of furniture"! You've got to be joking. We're talking windmills. There's nothing that is finely wrought or delicate on a mill unless it's about to fall to pieces. So humour accepted
Martin

 

[Sgian Dubh]

You have it exactly right so that's just what I need.
"a finely wrought and delicate piece of furniture"! You've got to be joking. We're talking windmills. There's nothing that is finely wrought or delicate on a mill unless it's about to fall to pieces. So humour accepted Martin

As Adam said, a hopper type thing in a windmill I strongly suspect is more likely to be butted together at the corners than mitred. Given that, instead of dividing the dihedral angle by 2 to establish the bevel cut to form the mitre, you subtract 90º from the dihedral angle in order to set the bevel cuts required. In this case then, the sum is 114.4º - 90º = 24.4º. You then need to use this 24.4º as the basis for working out how to mark the joint and set any tilt on your saw. This might mean using the complementary angle to 24.4º which is 65.6º calculated 90º - 24.4º = 65.6º.

As before, and as you agree, near enough is more than likely good enough, so you have little need to get overly precious trying to match the 'perfect' calculated angles - it ain't a grand piano, ha, ha. Slainte.

 

[user 43606]

As Adam said, a hopper type thing in a windmill I strongly suspect is more likely to be butted together at the corners than mitred. Given that, instead of dividing the dihedral angle by 2 to establish the bevel cut to form the mitre, you subtract 90º from the dihedral angle in order to set the bevel cuts required. In this case then, the sum is 114.4º - 90º = 24.4º. You then need to use this 24.4º as the basis for working out how to mark the joint and set any tilt on your saw. This might mean using the complementary angle to 24.4º which is 65.6º calculated 90º - 24.4º = 65.6º.

As before, and as you agree, near enough is more than likely good enough, so you have little need to get overly precious trying to match the 'perfect' calculated angles - it ain't a grand piano, ha, ha. Slainte.

Hello can you help me out, I’m looking for a trig formula, it’s for a Compound angle with a butt joint, for a box with sides splaying out at 50 degrees, not the saw set angle, the angle for the edge bevel when marking out on sides with square edges.

 

[Sgian Dubh]

Use the formulae given to calculate tablesaw or compound mitre saw angle settings for any combination of corner and tilt angles. In the drawing below: A = Corner angle formed by the two workpieces. A = 90 for a square corner, 45 for an octagon, 60 for a hexagon and 0 for a straight line. B = Tilt angle of each workpiece away from the base plane. X = Crosscut angle setting. This is the setting for your table saw mitre gauge or SCMS turntable. X = 0 for a square cut. Use the complement of X (90-X) if your saw labels a square cut as 90°. Y = Bevel angle setting. This is the blade tilt setting for your saw. Y = 0 for a square cut. Use the complement of Y (90-Y) if your saw labels a perpendicular cut as 90°. Formulae X = arctan(cos(B) * tan(A/2)) Y = arcsin(sin(B) * sin(A/2))

​

 

[user 43606]

Thanks for your reply and I have a fair idea about this and other practical ways but I’m more interested in the trig formula.
This is an example of a box with the sides splayed out at 60 degrees, these sides have beveled edges top and bottom. I’m interested in the trig formula for the butt joint angle when marking out on square edge material. It’s all part of angle studies and the same formula would apply when marking out a roof Purlin combination joint on a Hip roof, let’s say in a case where the Purlin has combination Butt joint and not a mitered combination mitered joint.

sin é

 

[Jacob]

Thanks for your reply and I have a fair idea about this and other practical ways but I’m more interested in the trig formula.
This is an example of a box with the sides splayed out at 60 degrees, these sides have beveled edges top and bottom. I’m interested in the trig formula for the butt joint angle when marking out on square edge material. It’s all part of angle studies and the same formula would apply when marking out a roof Purlin combination joint on a Hip roof, let’s say in a case where the Purlin has combination Butt joint and not a mitered combination mitered joint.

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Getting into Practical Stereotomy here! It's all about the angles...
Instead of calculating, the alternative is the graphical method as used for roofing geometry the same, which many would find easier.
I'd describe it in detail but its too early in the morning for me and I'd have to do a bit of revision!

 

[user 43606]

Thanks for your reply and I have a fair idea about this and other practical ways but I’m more interested in the trig formula.
This is an example of a box with the sides splayed out at 60 degrees, these sides have beveled edges top and bottom. I’m interested in the trig formula for the butt joint angle when marking out on square edge material. It’s all part of angle studies and the same formula would apply when marking out a roof Purlin combination joint on a Hip roof, let’s say in a case where the Purlin has combination Butt joint and not a mitered combination mitered joint.

sin é

 

[Sgian Dubh]

I'm afraid you've lost me, and unfortunately your photograph doesn't seem to help, either. I'm not sure what it is you're asking for, but I suspect the answer is in the formulae I provided earlier because they both calculate the angles needed for pyramidal type structures, i.e., the dihedral angle and the mitre angle. To mitre the meeting point of the calculated dihedral angle divide the result by two. To work out the required cut for a butt joint subtract ninety degrees from the calculated dihedral angle.

You seem to be almost there with the drawing which you included in one of your posts. It's essentially what Jacob was describing in the use of graphical methods. I've got a similar example. If I resort to graphical methods I turn to CAD for precision, see below. Slainte.

 

[user 43606]

Getting into Practical Stereotomy here! It's all about the angles...
Instead of calculating, the alternative is the graphical method as used for roofing geometry the same, which many would find easier.
I'd describe it in detail but it’s too early in the morning for me and I'd have to do a bit of revision!

I'm afraid you've lost me, and unfortunately your photograph doesn't seem to help, either. I'm not sure what it is you're asking for, but I suspect the answer is in the formulae I provided earlier because they both calculate the angles needed for pyramidal type structures, i.e., the dihedral angle and the mitre angle. To mitre the meeting point of the calculated dihedral angle divide the result by two. To work out the required cut for a butt joint subtract ninety degrees from the calculated dihedral angle.

You seem to be almost there with the drawing which you included in one of your posts. It's essentially what Jacob was describing in the use of graphical methods. I've got a similar example. If I resort to graphical methods I turn to CAD for precision, see below. Slainte.

View attachment 173789​

At least we are on a similar wavelength, it’s not just a matter of subtracting from 90 because the square edge is also tilted and these angles will be marked out with a tee bevel. I’ve never used cad but of course it’s a very handy tool.
sin é

 

[Richard_C]

Looks like we have 2 concurrent active threads with the same question going here, this one hijacked from the original question about windmill hoppers, and the stand alone one about angles for splayed box. Maybe sticking to one will halve the effort (and potential confusion) for those trying to help.