- posted on October 17, 2004, 9:17 pm

I'm running a piece of trim 12-inches below ceiling level, all the way around a room. Where the trim will meet in 90* corners, I can miter those no problem. However, the room is on the second floor of a cape, and so I have a sloping ceiling on one side, requiring me to cut the trim to meet both on an inside and outside corner of the slope.

If you take a look at the pictures below, you'll see what I mean. Where the red color and the cream color on the walls meet, that is where the trim will run. The third picture is a profile of the trim I plan on using. I don't know if that matters in calculating angle cuts. Sorry about that picture being fuzzy. I got too close with the digital - even set on macro.

Hope someone can help ! Thanks in advance.

Shawn

http://www.shawn-and-lisa.com/images/Pics/Temp/InsideCorner.JPG

http://www.shawn-and-lisa.com/images/Pics/Temp/OutsideCorner.JPG

http://www.shawn-and-lisa.com/images/Pics/Temp/TrimProfile.JPG

- posted on October 18, 2004, 7:36 pm

This is well beyond my level of expertise... I can only come up with two approaches to this: One is to use two different peices of flat trim, with a wider one on the slanted wall. That's the option I would choose.

The other is to use a transition peice: on the vertical wall, stop just before the corner and miter on an short bit that approaches the corner at right angles, and then go around the corner like a normal outside (or, at the right, inside) corner.

The trouble with this is that each of those transitions will transpose the molding down, from left to right, which means that when you come back around on the other side, the molding won't line up (unless the room is symetrical) in which case that dormered window bay will be higher than the rest of the room.

A third option occurs to me, to wit: find a BIG turned ornamental chunk of wood, cut a 90 degree wedge out of it with a circular saw. (and won't (THAT be fun..) mount those to the wall at the transition points, and cope the molding to interface with those. That would be the prettiest solution, but doing it strikes me as well into the "not a fun job" range.

--Goedjn

- posted on October 18, 2004, 10:24 pm

"willshak" wrote in message

What you say is true when cutting__ _all_ __cuts at 90º. But, this is why you
should have a compound saw. These cuts are nothing more than a typical
compound cut. All angles can be found with a T-Bevel & speed square, or an
angle finder.

For the sake of examples, let's say the slanted wall is 14º out. (Look at first link OP provided) You nail up the trim on the slanted wall, the piece next to glass block would be set for coping of 45º, and the compound would be set at 14º. This makes the piece of trim longer on one end to match the how far out the first piece trails off.

Done it more times than I care to count.

What you say is true when cutting

For the sake of examples, let's say the slanted wall is 14º out. (Look at first link OP provided) You nail up the trim on the slanted wall, the piece next to glass block would be set for coping of 45º, and the compound would be set at 14º. This makes the piece of trim longer on one end to match the how far out the first piece trails off.

Done it more times than I care to count.

- posted on October 18, 2004, 11:31 pm

Cbe wrote:

I was having a hard time describing what I meant, and doubt whether I'll be able to do much better this time around. I was not talking about the miter joint.The miter joint has nothing to do with the problem. The problem remains even with a butt joint. Take one piece of wood and butt it against another piece of the same size, like you were making the corner of a box. Now twist that piece and you will see that when it is twisted, the angle causes its overall height to be shorter than the piece it was butted to.... To hell with the talking... here's a visual. :-) http://www.willshak.com/temp /

I was having a hard time describing what I meant, and doubt whether I'll be able to do much better this time around. I was not talking about the miter joint.The miter joint has nothing to do with the problem. The problem remains even with a butt joint. Take one piece of wood and butt it against another piece of the same size, like you were making the corner of a box. Now twist that piece and you will see that when it is twisted, the angle causes its overall height to be shorter than the piece it was butted to.... To hell with the talking... here's a visual. :-) http://www.willshak.com/temp /

- posted on October 19, 2004, 3:10 pm

You are correct, it can't be made to match no matter how you cut it. By the way, I followed your first explanation just fine. I was mentally trying to come up with a way to use corner blocks at the corners but nothing I could picture looked anything except redneckish.

Harry K

- posted on October 19, 2004, 5:25 pm

A fourth option:

Cut a wedge-shaped backer-block to go behind the molding, such that either ALL of the molding is tilted out to match the sloping ceiling, (Thus creating a flat narrow shelf around the room) or turn that upside down for the sloping part, such that there's a bit of corbeling under the molding for the sloped bit.

Oh, clear as mud. Take a look at www.goedjn.com/notebook/skewmold.gif and look at the diagrams.

--Goedjn

Cut a wedge-shaped backer-block to go behind the molding, such that either ALL of the molding is tilted out to match the sloping ceiling, (Thus creating a flat narrow shelf around the room) or turn that upside down for the sloping part, such that there's a bit of corbeling under the molding for the sloped bit.

Oh, clear as mud. Take a look at www.goedjn.com/notebook/skewmold.gif and look at the diagrams.

--Goedjn

- posted on October 19, 2004, 2:20 am

On Sun, 17 Oct 2004 17:17:24 -0400, "Shawn"

I had kind of a similar problem with some angles on our (irregular) hip roof. I was working on the roof over several years, and went through a lot of iterations, trying to calculate various angles. What I came to was not rocket science, but does involve some trig and geometry.

- I would just pick an x-y-z reference frame, where x and y are along the walls, and z is vertical.

- A plane can be identified by the vector normal to its surface.

- The intersection of two planes is then just the cross-product of their normals.

- For any two vectors, the normal to the plane containing them is just the cross-product of the vectors.

- The angle between two vectors can be determined by taking the arccos of the normalized dot product of the vectors.

In your case, the intersecting walls would be the planes of interest. You can determine their normal vectors from the pitch, assuming the room is square. The intersection of these planes (the 'vector of intersection') will be the cross-product of their normals.

Once you have that, you can determine the normal to the 'plane of intersection' (ie, the plane where you want to cut the molding) by taking the cross-product of the 'vector of intersection' with the projection of that same vector onto the x-y plane.

Once you have that plane, all that's left is determine the angles of your cuts. In my case, I was cutting flat lumber with a circular saw; so, I wanted two angles: the 'cut angle' and the bevel.

The cut angle is just the angle between the horizontal line of the molding and the 'vector of intersection'. That's the angle I would mark across the wood.

The bevel angle is a little harder to explain; I went around a long time on this. The way I construct it is to recognize that the bevel adjustment rotates perpendicular to the cut line of the saw - that is, in the plane normal to the 'vector of intersection'. Call this plane of rotation the 'bevel plane'. Also, as above, the cut wants to be in the 'plane of intersection'.

Then, the bevel calculation requires two vectors in the 'bevel plane'; namely the intersection of this plane with (1) the 'plane of intersection', and (2) the plane that the saw would cut, if the bevel was set to 0.

(1) is already known. For (2), I picture making the cut with the wood right on one of the walls, with the bevel set to 0. The plane of this imaginary cut would contain both the 'vector of intersection' and the normal to the wall in question; so, the normal would just be the cross-product of these two vectors.

Now, the two vectors required for the bevel calculation are just the intersections of these planes ((1) and (2)) with the 'bevel plane'; these intersections are, again, just the cross-products of the normals to the planes in question. The bevel angle would be the angle between these two vectors.

I know this may be a little thick, and feel free to not read it. If you're comfortable with the math, it might at least a hint of how to proceed.

I've written a program to calculate these things (for a roof, which seems kind of the mirror of your case); but, to be honest, every time I use it, it takes me half a day to figure out what the hell I was thinking. Right now, I can't vouch that it even does what I've outlined above, even assuming that to be correct.

Another plan would be to just take a couple of pieces of flat stock, measure some angles that seem meaningful, and try cutting them and see how they fit. I went through a lot of wood that way.

George

I had kind of a similar problem with some angles on our (irregular) hip roof. I was working on the roof over several years, and went through a lot of iterations, trying to calculate various angles. What I came to was not rocket science, but does involve some trig and geometry.

- I would just pick an x-y-z reference frame, where x and y are along the walls, and z is vertical.

- A plane can be identified by the vector normal to its surface.

- The intersection of two planes is then just the cross-product of their normals.

- For any two vectors, the normal to the plane containing them is just the cross-product of the vectors.

- The angle between two vectors can be determined by taking the arccos of the normalized dot product of the vectors.

In your case, the intersecting walls would be the planes of interest. You can determine their normal vectors from the pitch, assuming the room is square. The intersection of these planes (the 'vector of intersection') will be the cross-product of their normals.

Once you have that, you can determine the normal to the 'plane of intersection' (ie, the plane where you want to cut the molding) by taking the cross-product of the 'vector of intersection' with the projection of that same vector onto the x-y plane.

Once you have that plane, all that's left is determine the angles of your cuts. In my case, I was cutting flat lumber with a circular saw; so, I wanted two angles: the 'cut angle' and the bevel.

The cut angle is just the angle between the horizontal line of the molding and the 'vector of intersection'. That's the angle I would mark across the wood.

The bevel angle is a little harder to explain; I went around a long time on this. The way I construct it is to recognize that the bevel adjustment rotates perpendicular to the cut line of the saw - that is, in the plane normal to the 'vector of intersection'. Call this plane of rotation the 'bevel plane'. Also, as above, the cut wants to be in the 'plane of intersection'.

Then, the bevel calculation requires two vectors in the 'bevel plane'; namely the intersection of this plane with (1) the 'plane of intersection', and (2) the plane that the saw would cut, if the bevel was set to 0.

(1) is already known. For (2), I picture making the cut with the wood right on one of the walls, with the bevel set to 0. The plane of this imaginary cut would contain both the 'vector of intersection' and the normal to the wall in question; so, the normal would just be the cross-product of these two vectors.

Now, the two vectors required for the bevel calculation are just the intersections of these planes ((1) and (2)) with the 'bevel plane'; these intersections are, again, just the cross-products of the normals to the planes in question. The bevel angle would be the angle between these two vectors.

I know this may be a little thick, and feel free to not read it. If you're comfortable with the math, it might at least a hint of how to proceed.

I've written a program to calculate these things (for a roof, which seems kind of the mirror of your case); but, to be honest, every time I use it, it takes me half a day to figure out what the hell I was thinking. Right now, I can't vouch that it even does what I've outlined above, even assuming that to be correct.

Another plan would be to just take a couple of pieces of flat stock, measure some angles that seem meaningful, and try cutting them and see how they fit. I went through a lot of wood that way.

George

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