Compound miter brainteaser

[Josh]

Even though the formula I referenced previously (which can be found as the basis for several compound miter calculators) is easily derived, there was an essential flaw in the construction of the problem (but not in the math, itself). The crux of the problem is that even though the roof pitch is 5/12, the pitch of the rafters is not 5/12 (at least not the ten main rafters which form the wedges which comprise the roof). If we added ten more secondary rafters to the roof frame ran down the middle of each wedge, bisecting the 36 degree angle, those would have a pitch of 5/12.

To put this in a way that is easier to picture, imagine that the secondary rafters are 13 feet long, thus forming a 5-12-13 triangle. The would have a rise of 5 feet, a horizontal run of 12 feet, and an overall length of 13 feet. To keep the eaves of the roof level, the main rafters would, of course, have to drop the same 5 feet total at their end points. However, they would have to be longer than 13 feet. They would in fact be 13/cos(18) feet long. Thus, their pitch would actually be less than 5/12, and that's why the other formula is wrong.

Josh

[dr-whoopie]

somehow after atepting to try and understand

[Enoch Root]

Hah, that's gorgeous... 5/12 is the pitch along a line perpendicular to the direction of the slope, but the rafters are not that. :)

The other formula will work if you adjust your slope to

slope = sqrt(rise^2/(rise^2+run^2))

won't it? (seems to, based on a test using my own proggie)

er

[Morris Dovey]

| Hah, that's gorgeous... 5/12 is the pitch along a line | perpendicular to the direction of the slope, but the rafters are | not that. :) | | The other formula will work if you adjust your slope to | | slope = sqrt(rise^2/(rise^2+run^2)) | | won't it? (seems to, based on a test using my own proggie)

You're obviously having too much fun with this. Just to open things up a bit for all the people who don't have bc (but _do_ have a C compiler):

#include #include #include #include

int main(void) { char input[128]; double sides,rise,run,slant,angle,miter,bevel;

printf("How many sides? "); fflush(stdout); fgets(input,80,stdin); sscanf(input,"%lf\n ",&sides); angle = 2 * M_PI / sides;

printf("What is the slant (degrees or rise/run)? "); fflush(stdout); fgets(input,80,stdin); if (strchr(input,'/')) { sscanf(input,"%lf/%lf",&rise,&run); slant = atan(rise / run); } else { sscanf(input,"%lf",&slant); slant = M_PI * slant / 180; }

miter = atan(cos(slant) * tan(angle / 2)); bevel = asin(sin(slant) * sin(angle / 2));

printf("\nThe miter is %0.2lf degrees\n", miter * 180 / M_PI); printf("The bevel is %0.2lf degrees\n" , bevel * 180 / M_PI);

return EXIT_SUCCESS; }

This program allows entering the "slant" in degrees or as rise/run (like 5/12). It's "quick and dirty" and does no error checking, but should be compilable with even TurboC version 1 (I used 3.0).

-- Morris Dovey DeSoto Solar DeSoto, Iowa USA

[DJ Delorie]

Except for the corner ones, my rafters are. They have to run parallel to the seams in the plywood sheathing.

[DJ Delorie]

If you have neither bc nor a C compiler, my script uses the same math (and allows for rise/run too):

[Morris Dovey]

| "Morris Dovey" writes: || You're obviously having too much fun with this. Just to open things || up a bit for all the people who don't have bc (but _do_ have a C || compiler): | | If you have neither bc nor a C compiler, my script uses the same | math (and allows for rise/run too): | |

script is much appreciated (but not as much your ABPW archive and nowhere near as much as your efforts to make a C compiler freely available to all) - but there are more than 15 miles between my shop and my internet access.

For those with a similar situation, the C source can be found in the collection at

as cbevel.c - sometimes it's a Good Thing to be able to function independently from the internet. :-)

-- Morris Dovey DeSoto Solar DeSoto, Iowa USA

[alexy]

Well, not to be picky, but... There is no flaw in the construction of the problem. The flaw is in the solution, using the roof slope as the slope of the diagonal rafters.

this part, getting the miter angle, is way easier than the dot product solution. Consider 1/2 of one of the roof wedges. Vertically, it is a

12/5/13 right triangle, so the length from the peak to the eve is 13. Horizontally, it is an 18 degree right triangle, with the long leg of

1. Trig tables give us the other leg as 3.9. So the half wedge forming the roof is a right triangle with legs of 13 and 3.9, from which we use trig to get a miter angle of 16.7.

It's the bevel angle that stumps me. I got an answer consistent with the one posted from the calculators, but without more digits of precision, I'm not very confident of mine. And unless one of you computer types want to "program" this solution in Excel , I'm not able to play around with the precision. My solution still causes brain strain when I try to reconstruct it!

[alexy]

Never mind. I created a spreadsheet solution that duplicates the web page cited here, and in so doing got my brain round the bevel angle.

[Relz]

Wouldn't it be 18 degrees?

[Relz]

For what it's worth, I modeled it up in SolidWorks and I got a 16.57 degree miter with a 7.12 degree bevel. So, I think you're pretty close alexy.

Relz

[Josh]

I'll bet you made the same mistake in SolidWorks that I alluded to in an earlier post. If you were to start at the peak of the roof and draw

10 lines extending radially outward with a downward pitch of 5/12, you'd come up with angles of 16.57 and 7.12, as you did. However, for a 5/12 slope going straight down the roof (i.e. along a path bisecting two adjacent lines of the ten sided "starfish"), the slope along the ridges would not be 5/12; it would be (5/12)*cos(18). If you were to implement that slope in SolidWorks or Inventor or any other 3D modeling program, you should get the same answers as those given by DJ's algorithm (16.7 miter, and 6.8 bevel).

Josh

[Relz]

I guess I don't understand what you're saying. Could you or someone please explain it to me?

I've modeled two sections of a ten-sided roof with a 5/12 pitch. I've modeled two 1 x 4 "boards" that would join up on the theoretical middle of the rafters. For the boards to come together in both miter and bevel, the CAD software is still telling me that the miter should be 16.57358103 degrees and the bevel should be 7.12327393 degrees.

Sorry, but I've learned to trust my CAD software and either you are all wrong, or more likely, I've put something into my program that's wrong. Also, if anyone has SolidWorks, I am willing to email you my file and maybe you could tell me what I'm doing wrong.

Relz

[alexy]

I'll try different terminology. I thin what you probably modeled has hip rafters at 5:12 rather than common rafters at 5:12.

I don't have solidworks, but have another solid modeler. I'll try it and see what I get.

[alexy]

Actually, before I model, let me explain what I'd do, and see if what you did is equivalent.

Start with a plan view of 1/20 of the gazebo: a right triangle with (for convenience) a long leg of length 12.

Extrude this at 90 degrees, to a height greater than 5, leaving you with a triangular prism.

on the surface defined by the long leg (not hypotenuse) of the first right triangle: draw another right triangle consisting of the top of this surface (12 units long), down one side 5 units, and the hypotenuse.

Extrude cut (or whatever it is called in solidworks) with this triangle to cut away the top of the prism.

What you have left is a solid that if reflected on the short side and repeated 10 times at 36 degrees would be the gazebo (but don't do that). Look at the shape of the "roof" piece to get the miter cut.

Now define a plane perpendicular to the roof, and to the edge of the roof that is the hypotenuse in the plan view. Look at the angle on this plane of the intersection of the roof and the side. This is the bevel cut.

[Josh]

If you can email me a .SAT file I can check it out. Given that you get the exact angles that I get with my dot-product solution above, It seems like you must be defining the hip rafters at 5/12 pitch, which is slightly off from their actual pitch, but then again, maybe we're all wrong ;-)

If you picture the roof comprised of 10 triangular wedges, the two long sides of each triangle would be the hips of the roof, and the short side would be the eave. Now bisect this triangle into two right triangles by drawing a line from the point of the roof to the center of the eve (If the triangle was a Christmas tree, you'd be drawing the trunk). This new line should have a pitch of exactly 5/12 (i.e. should form a 22.62 degree angle from horizontal). Meanwhile, the hip rafters should form an angle of 21.62 degrees from horizontal because they are slightly longer than the common rafter you just drew, but they rise the same distance. In other words, they're slope (rise/run) is less because "rise" is the same, but "run" is longer.

If you measure the angle from your miter joint to horizontal (which should be the same as the pitch of the hip rafters), does it come out to 22.62 degrees or 21.62?

[Relz]

Okay, I think I know where I went wrong. I was modeling the 5/12 pitch on the rafter, not the roof (your tree trunk).

I remodeled the senario as you described and I'm getting a hip rafter angle of 21.62, just as you described. And, now I am getting a bevel degree of

6.83 which is what everyone else was getting.

Thank you Josh and alexy for helping me see where I went wrong. This information will come in handy when I go to build my gazebo. I just may do it ten sided! :-)

Relz

[Josh]

Nice. I put a couple of pictures of my father's gazebo online. They're not very good, but I think I have some better ones at home that I might upload.

[kimnach]

an interesting problem. it is solved by noting that the rafters, when viewed from above, look like an outside corner crown moulding (c.m.) problem.

the pitch of the rafters (5:12 = 22.6degrees) becomes the c.m. angle, while the inclusion angle becomes (180-36)/2 = 72degrees (again, since it's an outside corner crown moulding).

in this specific case the result is: miter = pi/2 - [atan (cos(22.6) / tan (72)] = 16.7deg bevel = atan [cot(pi/2 - 22.6) * sin(pi/2 - 16.7) ] = 6.8degrees

for those who want to visualize what's going on or who aren't necessarily good at trig, check out my tutorial at

-- regards, greg

[entfillet]

Surely this a simple problem. Akin to cutting crown molding, where the number of sides are plugged into the 360 and divided by two to get the miter, and the spring is replicated on the bed of the cutting table, or mathed out to be cut on the flat according to existing tables. The more interesting concern is that the butting joints be undercut, so as to avoid expansion under the inevitable expansion under moisture absorbtion, which would destroy the geometry quickly. This is a carpenter's problem, rather than one for the mathematicians, and a decent carpenter would say to allow for a sixteenth between the joints, along with the underbevel, so that the whole structure has a decent chance of survival. (caveat - it is often assumed that there is no expansion along the lateral line but in this situation it must be alllowed for by creating a relief space and an undercut at the joints)