Bowed Baywindow Roof Framing

[sbebuilders]

Thought I would seperate this bowed baywindow roof framing thread from the Bay Roof to a Point Double Bumpout 45° thread started by Joe Carola. The bowed baywindow can be framed 1000 different ways, but this was my take on it in 1984.


I designed and built a house for a relative in 1984 with this bowed baywindow roof.
The side walls were 33.4” , the overhang was 24”, the pitch was 8/12 with the side
runs being equal to the front bowed window run. I drew it with equal overhang runs,
but I’m not sure if I stacked it with equal overhang runs. Not even sure that I
stacked it with equal common rafter runs. I just looked at the old set of plans I
drew for the bowed baywibdow and I drew it with a 24 point polygon, because I only
had an 8 point polygon drafting template that I was using. This was before cad drawing
programs were an everyday tool.


This is a sketch (on paper) I drew in 1984 for the 5 sided bowed bay window
with a 24 point polygon for window placement calculations . I entered the
sketch calculations into AutoCAD today and ended up with the same dimensions
and angles as I had on paper.

http://www.sbebuilders.com/rfi/image...ow_sketch.html Larger Image of Bowed Baywindow Sketch

http://www.sbebuilders.com/rfi/autoc...-baywindow.dwg Bowed Baywindow Calculations Sketch in AutoCAD DWG Drawing Format


So, Joe B., Joe C., and Joe F. is there a genric formula for the calculating the roof framing angles without calulating each and every angle and dimension?

Attached images are the bowed window calculations sketch and the plan view of the bowed baywindow roof framing.

Sim

[Joe Bartok]

Re: Bowed Baywindow Roof Framing

I should have known this was coming. :) Sim, I'm only going to be online for a little while longer but will take a copy of you drawing to look over on the weekend.

So, Joe B., Joe C., and Joe F. is there a genric formula for the calculating the roof framing angles without calulating each and every angle and dimension?

It's possible to write code for just about anything and I would imagine that one could start of with, say, a regular polygon, solve the corner angle, specify the pitches and calculate only the framing angles and dimensions required.

Several years ago I attempted to make a worksheet that would cover all the bases in terms of pitches and plan angles, roof member dimensions, joint bevel angles and dimensions. LOL ... might just as well have tried to write an AutoCAD program. I was stubborn, though, and persevered for a few years before giving up.

Now I use script that produces all the angles that I know of, print the entire list and figure the dimensions with a calculator, disregarding the angles that aren't needed. But you are likely making a "cheat sheet" to be used by people who don't necessarily understand the trig and are expected only to lay out and cut to the measurements so angles only won't do the trick.

[Joe Bartok]

Re: Bowed Baywindow Roof Framing

When I read your first post I jumped to the conclusion that the formulas were for the purpose of writing a program. If there is a general catch-all formula it might be complicated. The number of sides vary, the triangles are irregular so that means the Law of Sines and Law of Cosines, and throwing in equal overhangs complicates things even more. A program might be a good way to go.

Looks like you will need to script several different irregular plan angle calculations, one for each corner about the line of symmetry where a Hip run intersects an eave. For the drawing you posted, four seperate calculations (or loops? ... determed by the number of sides entered?).

And because of the equal overhangs you will need two iterations of each of the aforementioned calculations. Solve the dimensions and slopes with respect to the insides of the overhangs first, then increase the runs and hence decrease the roof slopes, and re-calc the plan angles, Hips slope angles, (backing angles?, Hip and jack rafter side cut angles?) And finally, the dimensions.

There are a couple of examples of this "two cycle" re-calculation to adjust for the overhangs in the Bay Roof to a Point Double Bumpout 45° thread:
Bay Bumpout equal 135° Corner Angles ... Post #13
Bay Bumpout unequal Corner Angles ... Post #14
The plan angle calcs aren't given in those posts. I did that part using my framing angle program, or you can double check it with your Perl script version or just use the formula.

Just had a thought ... would the corners lie on the perimeter of an ellipse?

[sbebuilders]

Re: Bowed Baywindow Roof Framing

Just had a thought ... would the corners lie on the perimeter of an ellipse?

Joe,

I tried a three point circle, but not an ellipse. I'll have to give the ellipse a try.

I would like to write a script for the bowed bay window roof, but after looking at the bowed bay window sketch for a couple of hours, I couldn't find a starting point. I tried using a 3 point circle to define the bay window points, but realized that I really didn't have enough information to locate the third point of the circle. We know what the bay window face wall run is, but we don't know the length of the bay window wall. If we knew the length of the bay window wall, then we could use a chord of a circle to find the height "D" and locate the third polygon point. But it's still the chicken or egg syndrome. Which came first? Which do need first?

The input variables should be:
Span
Pitch
Number of Windows

Number of windows ... odd numbers only like 3,5,7,9...etc so one of the sides of the bay window is perpendicular to the side walls.

However, the number of windows by it self does not define the central angle of the polygon points of the bay window.

If you use a 10° angle from a 36 sided polygon like Joe Carola suggested, then the projection of the bay windows are completely different from the 24 point polygon I used. So, we might need a input variable of :

Number of polygon points


At least were not writing a script for this curved bowed bay window... yet.

http://www.taunton.com/finehomebuild...33&ac=ts&ra=fp Bow Curved Roof


Sim

[Joe Bartok]

Re: Bowed Baywindow Roof Framing

Sim, not long after logging off Saturday I realized that my notion that the expanded boundary of the roof might lie on an ellipse didn't hold water. Some of the points will lie on a circle but adding an overhang to the side wall throws a monkey wrench in the works. I don't know what kind of a curve (if any!) the points lie on but one way or another it's only complicating the solution.

Here's what I got for the circle. Since no overhang was specified I used an arbitrary value of 20:

Radius on your drawing = 127.9437
Apothem of triangle = 127.9437 × cos 7.5 = 126.84912
Apothem of triangle on overhang boundary = 126.84912 + 20 = 146.84912
Radius of overhang boundary = 146.84912 ÷ cos 7.5 = 148.11628

Given this radius with a bit of basic trigonometry we can find some of the nodes at the intersections of the eaves and Hip runs. Some - but not all of the nodes ... that side wall is messing things up.

A better trick is to use the apothem to locate points on the eave lines. We also know the slopes of these lines ... the tangent of the complementary angle of the apothem in question. Since we have a point and a slope for each of the lines we can determine the equations and hence the intersections of the lines (the nodes).

Given the nodes or intersection points we can use the Pythagorean Theorem to find the lengths in plan and rearranging the dot product formula for vectors allows us to solve the angles.

Bowed Bay Window Diagrams ... these were solved using the method I suggested for the program in Post #3. I took the common runs or altitudes of the triangles in plan view and added 20. The adjusted pitches were entered in my calculator in conjunction with the corner angles (the lines of the overhang boundary are parallel so the corner angle values remain the same). The Javascript recalculated the plan angles, Hip slope angles (and every other angle that might be needed to frame the roof...). I double checked the results with the Pythagorean Theorem, Law of Sines, Law of Cosines, vector math ... (edit: and also made a bristol board polygon. The "3D" images in the link were based on the model).

It took me a hell of a lot longer to make the diagrams than to figure the angles and lengths. In fact while I was at it I did your roof right from scratch. Here is your Bowed Bay Window drawing displayed in terms of the dreaded nodes.

So there are many ways to approach this problem, they all agree with one another, but they all seem to take a bit of button pushing if using a calculator. Of course, if we use Perl or Javascript or whatever the computer won't care about the number and complexity of the operations. I'll keep thinking about this ... maybe a better idea will occur. I am surprised there weren't more posts in this thread over the weekend.

[Joe Bartok]

Re: Bowed Baywindow Roof Framing

I tried using a 3 point circle to define the bay window points, but realized that I really didn't have enough information to locate the third point of the circle. We know what the bay window face wall run is, but we don't know the length of the bay window wall. If we knew the length of the bay window wall, then we could use a chord of a circle to find the height "D" and locate the third polygon point. But it's still the chicken or egg syndrome. Which came first? Which do need first?

...........

However, the number of windows by it self does not define the central angle of the polygon points of the bay window.

I'm not understanding what it is that we don't know. We have a back wall width and slope, therefore a rise. We have a front wall run, a polygon with a known number of sides, and a radius the polygon is based on. This data seems to be more than sufficient ... or am I missing something?

As long as there is enough information to find dimensions of the inner polygon without the overhang we've got it made. Can you post a quick sketch of a hypothetical case, perhaps the "three point circle" quoted above?

[sbebuilders]

Re: Bowed Baywindow Roof Framing

I'm not understanding what it is that we don't know. We have a back wall width and slope, therefore a rise. We have a front wall run, a polygon with a known number of sides, and a radius the polygon is based on. This data seems to be more than sufficient ... or am I missing something?

As long as there is enough information to find dimensions of the inner polygon without the overhang we've got it made. Can you post a quick sketch of a hypothetical case, perhaps the "three point circle" quoted above?

Joe,

The 3 point circle drawing is attached to this post. I think we need to define the input variables as:

Span of Bowed Windows
Projection Of Bowed Windows
Number of Bowed Windows

EDIT: We don't know the number of sides in the polygon. The number of sides of the polygon can range from 16 - 64 depending on the span and projection of the bowed windows.


With the Projection Of Bowed Windows defined we have the apothem of polygon triangle, which should give us the number of sides in the polygon.

Sim

[Joe Bartok]

Re: Bowed Baywindow Roof Framing

This might be a head scratcher and I'm beginning to see where the chicken and egg syndrome comes into play.

Even with a simple three sided projection it seems as if we can't randomly assign values to the width and projection. The angle in the corner, theta, depends on these two dimensions but is also restricted to values which are produced by regular polygons ... 72° for a pentagon, 60° for a hexagon, 45° for an octagon, etc.

We can find the angle by arcsin (Projection ÷ Side of Polygon) but we don't know the length of the side at this stage. And what about a shape with multiple sides where this angle isn't a simple fixed value? Might be able to work around that, though. The width gives us two points and only one more point would be needed to define a circle.

Just some thoughts ...

Edit: Wait a minute. The diagram says center of three point circle. Does this mean we already know where the center is? If so it would make life a lot easier.

[Joe Bartok]

Re: Bowed Baywindow Roof Framing

Another idea. It should be possible to write a program that "hunts" for the best fit polygon. We'll need a front wall width for this. The program would calculate stations, with zero being the center of the front wall, using trial values of the exterior polygon angle.

The exterior angle of a regular polygon with number of sides n is 360°/n. Let's call this angle theta, call the front wall width w. The stations on the X-axis would be:

w/2 + wcos theta + wcos(2 × theta) ... until the front wall width is attained.

Stations on the Y-axis would be:

0 + wsin theta + wsin (2 × theta) ... until the front wall projection is attained.

The program will need conditions so that if the wall width and projection are exceeded it doesn't keep on cyphering on into infinity ...

[Joe Bartok]

Re: Bowed Baywindow Roof Framing

Sim, there might be light at the end of the tunnel.

First of all, though there is a polygon in the geometry for the time being I'm discarding the idea of determining what this polygon actually is. In the real world the width of the back wall and the front wall projection would be any pair of numbers. If a formula existed we might find that a randomly assigned pair of values produces a polygon with 5,382 sides and a radius a kilometer long. To locate points for the wall from its maximum projection we would really be more interested in the central angle or exterior angle and use it in a formula similar to that in my last post.

So here's what I did. I let the maximum projection be at a tangent to the circle. (We really need a chord but at this stage I can't see how to find it). This gives us three points and therefore a radius and central angle.

The angle is incorrect but I used it in the formula to find the lengths of the sides of our hypothetical polygon. Looking at the results will tell us if the angle is too small or large. So I averaged the lengths of the sides, increased the angle, recalculated, studied the results again, averaged the sides and angles, etc ... ad nauseum. Eventually the numbers converged, here's what I ended up with:

From the drawing in your first post:

Span of Bowed Windows = 155.7743
Total Projection of Bowed Windows = 25.3446

Based only on those figures and applying my weird averaging method:

Length of Walls (the sides of the polygon) = 33.39992204
Central Angle = 15.00009623

That's close enough for me. A real mathematician would scoff at how I arrived at my answers, there are logical iterative processes like Newton's method but I can't see how to apply it here. And this was like looking at the answers in the back of a book ... I knew the results beforehand. Tonight I'm going to give this the acid test: randomly selected dimensions for the span and projection and I'm going to add another two sides to make seven triangles.

[sbebuilders]

Re: Bowed Baywindow Roof Framing

Sim, there might be light at the end of the tunnel.

A real mathematician would scoff at how I arrived at my answers, there are logical iterative processes like Newton's method but I can't see how to apply it here.

Joe,

I'll have to look at Newton's method to see if I can use Arc Length and the Apothem. However, I couldn't see any light at the end of the tunnel so I added a input variable to the script.

The input variables are now:
Span
Pitch
Number of Windows
Bowed Bay Window Angle

By adding the bowed bay window angle the script can produce the same results you came up with in http://joe.bartok.googlepages.com/bo...w_diagrams.htm bowed window diagram

However, I'm having problems with the fascia line. See attached images. The side fascia points should be easy to locate, but I can seem to find the right equation that fits all of the possibilities.
Here's a link to the script, it's still in beta mode.

http://www.sbebuilders.com/tools/bowed-baywindow.php bowed baywindow beta script.

Here's the current code I'm using in the script.

$Bowed_Window_Angle = $window_angle;
$Number_Of_Windows = $number_of_windows;
$Bowed_Window_Span = $span;
$Bowed_Window_Overhang = $overhang;

$Bowed_Window_Rad = $Bowed_Window_Angle * $DEGREE_TO_RAD;
$Bowed_Window_Central_Angle_Rad = $Bowed_Window_Rad * $Number_Of_Windows;

$Polygon_Nsides = ($PIE * 2) / $Bowed_Window_Rad;
$Polygon_Cord = $Bowed_Window_Span;

$Radius = ($Polygon_Cord / 2) / sin($Bowed_Window_Central_Angle_Rad / 2);

$Apothem = $Radius * cos($PIE/$Polygon_Nsides);
$Bowed_Window_Wall_Length = $Apothem * tan($Bowed_Window_Rad);

$Overhang_Radius = ($Apothem + $Bowed_Window_Overhang) / cos($Bowed_Window_Rad/2);
$fascia_side_length = get_triangle_SAS($Overhang_Radius,$Bowed_Window_Ra d,$Overhang_Radius);

$fascia_slope_angle = $Bowed_Window_Rad * (($Number_Of_Windows-1)/2);

#$fascia_slope_angle = $RAD180 - ($RAD90 - $Bowed_Window_Rad);
#$fascia_slope_angle = $Bowed_Window_Rad;

#$Fascia_Offset_Y = $Bowed_Window_Overhang * tan($fascia_slope_angle);

$Fascia_Offset_Y = $fascia_side_length * sin($fascia_slope_angle);


Sim

[Joe Bartok]

Re: Bowed Baywindow Roof Framing

Sim, I'll make a copy of your pics and study them tonight. It seems every time there's an interesting thread or post in this forum my computer time runs out. :)

I attempted the seven bay bow window. The 160 span, 30 projection and seven bays were made up out of whole cloth ... as if someone gave me those numbers and said "Build this!". The geometry of my seven bay window is different than yours; I'm not experienced at these so I just drew as seemed best.

It's a hack solution but at least it produced numbers that can be worked with. I didn't bother with the radius and polygon but now that the numbers are known they could be calculated.

However, I'm having problems with the fascia line. See attached images. The side fascia points should be easy to locate, but I can seem to find the right equation that fits all of the possibilities.

To find the fascia line I added to the runs or altitudes of the inner perimeter perimeter (20 was added in my sketches). So now we know the runs to the fascia line and the common slopes can be recalculated. The slopes are entered in pairs with the appropriate corner angle and the plan angles, Hip slopes, etc., are recalculated. This operation must be repeated for each pair of pitches meeting at each of the corner angles.

To summarize, I couldn't find the "right equation" either so I used another hack solution. Somebody is going to come along and say "Yeah, but Tangent-Chord Theorem says ..." and show us the easy way to do this.

The dimensions of the bay window in the attached image are to the inner perimeter ... no overhang. If I get a chance tonight I will make up some side walls, a rise and an overhang and see how the numbers come out.

Where are all the other math gurus? I would have thought this thread would be several pages long by now ...

Here is a link to a larger image of the seven bay bow window, the attachment had to be downsized and it might be tough to read the numbers.

[Joe Bartok]

Re: Bowed Baywindow Roof Framing

Sim, may I ask how you have been coping with these bowed windows?

The input variables are now:
Span
Pitch
Number of Windows
Bowed Bay Window Angle

The biggest headache is that even if I assign an angle it doesn't really help me find a theoretical polygon which "fits". Both the length of side and the angle are variables. Again, I could be missing something here ...

Btw, the polygon in the sketch attached in my previous post would be:

360° ÷ 11.980218° = 30.04953666 ... 30-gon would be close enough.

If we round off the length of the sides to 25 and the angle to 12, then starting at the point labelled origin in my drawing:

½ Span = 25 × [(.5 + cos 12° + cos (2 × 12°) + cos (3 × 12°)] = 80.01775
Projection = 25 × [(sin 12° + sin (2 × 12°) + sin (3 × 12°)] = 30.06084

Close enough. To reiterate, the solution of the sides and angles wasn't based on a 30-gon and a radius (I still haven't calced the radius and have no idea what it is!). The only givens were the 160 span, 30 projection and seven bays.

[sbebuilders]

Re: Bowed Baywindow Roof Framing

I attempted the seven bay bow window. The 160 span, 30 projection and seven bays were made up out of whole cloth ... as if someone gave me those numbers and said "Build this!". The geometry of my seven bay window is different than yours; I'm not experienced at these so I just drew as seemed best.

The dimensions of the bay window in the attached image are to the inner perimeter ... no overhang. If I get a chance tonight I will make up some side walls, a rise and an overhang and see how the numbers come out.

Here is a link to a larger image of the seven bay bow window, the attachment had to be downsized and it might be tough to read the numbers.

Joe,

I used the script to genterate the bowed baywindow DXF file with the same numbers you used in your drawing and sure enough I opened the file in autocad and it dimensioned the same as your results. Which, lead me to belive that the script wasn't displaying the correct answers. I corrected the code in the script and now it displays the same results as your drawing.

The script was displaying:
Bowed Bay Window Wall Length = 25.26513

now it displays
Number of Windows = 7
Bowed Bay Window Angle = 11.98022°
Span of Bowed Bay Windows = 160.00000
Number of Polygon Sides = 30.049536661186
Central Polygon Angle = 83.86153°
Radius of 30.049536661186 Sided Polygon = 119.71888
Atothem of 30.049536661186 Sided Polygon = 119.06521

Bowed Bay Window Wall Length = 24.98695

I changed
Bowed_Window_Wall_Length = (Apothem * tan(Bowed_Window_Rad));
to
Bowed_Window_Wall_Length = (Apothem * tan(Bowed_Window_Rad/2)*2);

How did you formulate your drawing with just the number of bays, span and projection? It seemed almost impossible for me to just use those 3 input varaibles to come up with anything.

Sim

[sbebuilders]

Re: Bowed Baywindow Roof Framing

Sim, may I ask how you have been coping with these bowed windows?

The biggest headache is that even if I assign an angle it doesn't really help me find a theoretical polygon which "fits". Both the length of side and the angle are variables. Again, I could be missing something here ...

I still haven't calced the radius and have no idea what it is!

Joe,

I'm finding this bowed baywindow script the most challenging of all the scripts I've ever written. But I think I’m seeing light at the end of the tunnel.

With the known bay window angle the radius is:

Bowed_Window_Central_Angle_Rad = Bowed_Window_Rad * Number_Of_Windows

Polygon_Nsides = (PIE * 2) / Bowed_Window_Rad
Polygon_Cord = Bowed_Window_Span

Radius = (Polygon_Cord / 2) / sin(Bowed_Window_Central_Angle_Rad / 2)

If I couldn't use the polygon method, I don't think I could draw the bowed baywindow. Still don't have a clue how you drew it without the polygon.

Sim