Eyebrow dormer class in Portland, Oregon

[David Meiland]

Originally Posted by Lavrans

I'm thinking I'll be the token amateur at this thing.

Lav, I need to be The Guy Who Hasn't A Clue. You have to be something else at least one step above that.

[davenorthup]

Lav - if you need a deal on airfare, I have some US Airways miles that I could sell for a few hundred bucks; less than I suspect you could get a ticket for. Let me know as you get closer to pulling the trigger on this, should lower the prices for you all...

I do not see myself in the NW anytime soon sadly...

[KLyons]

Lavrans,

I'm in! I will send you a PM.

[sbebuilders]

Lavrans,

The last six projects that I bid on are coming back a bust. No one can get funding for their projects. So I'm in. I'll pm you for the details.

Sim

[sbebuilders]

geometery for beer:curve eyebrow roof design

[Lavrans]

Sim- great example of 5 o' clock design.

I put up a page on my website with information on the class.

I'll be getting back to everyone who's expressed interest over the next few days with better contact info, tool list, location, etc. Thank you for your interest and help.

[sbebuilders]

Joe Bartok,

Take a look at this PDF file with tangent circles that define the shape of the eyebrow roof and see if you can find a formula to locate the X-Y point for the radius of the side circle.

Eyebrow Roof Known Variables
1: eyebrow roof height = 24"
2: eyebrow roof width = 136"
3: eyebrow window height = 12"
4: eyebrow window width = 68"
5: window reveal at top = 8"

Thanks,
Sim

[Joe Bartok]

Sim, I'll try to get back later and have a look at the specific numbers in your pdf file. Meanwhile consider this:

We know the point of tangency between the two circles.
We know the intersection of the large circle with the x-axis (at y = zero).
We have an known intersection formed by symmetry about the vertical axis through the center of the large circle.

Therefore we have three known points on the circle.

Javascript Calculator ... Circle Center and Radius given three points on the Circle

There is a link to the theory of calculating the radius further down the page. I used algebra to solve the problem but I think there is a way to find the radius with matrices.

[sbebuilders]

Originally Posted by Joe Bartok

We know the point of tangency between the two circles.
We know the intersection of the large circle with the x-axis (at y = zero).
We have an known intersection formed by symmetry about the vertical axis through the center of the large circle.

Therefore we have three known points on the circle.

Joe,

It's easy to draw geometricaly because of the intersecting apothem line, tangent point and smaller circle radius line segment angle. However, I'm not following you on the known three points on the circle. The way I see it , I only know two points and the radius for the tangent circle is not known.

Thanks for taking a look,
Sim

[Joe Bartok]

Sim, after I posted I downloaded your drawing and had a closer look. I could have responded a bit sooner but had a different idea on how to approach this problem and took the time to make up a graphic.

Interesting … I just posted the same answer to a seemingly different question for a "ski slope" roof over at the BT Forum. Here’s the thread in this forum thread where the math was originally developed:

Shed Rafter into Main Rafter Calculation ... start at Post #6

This is a link to the theory: Circular Arc tangent to Rafter

Use the formulas or open the Excel worksheet at the top of the page. In the first set of fields enter …
Rafter Triangle Rise = 13.87539
Rafter Triangle Run = 21.23616
Length: Difference in Runs = 12.76384
Radius = 96.57785

Or in the second set of fields enter …
Rafter Triangle Rise = 13.87539
Rafter Triangle Run = 21.23616
Chord = 36.72229
Radius = 96.57785

To check the above numbers, in the third set of fields enter …
Rafter Triangle Rise = 13.87539
Rafter Triangle Run = 21.23616
Radius = 96.57784

We now have a radius for the large circle. The difference between my calculated radius and the radius in your drawing is likely due to my using numbers from the pdf which have been rounded off to two decimal places. The sum of the circle radii is:
62.17 + 96.57784 = 158.74784

The center of the large circle from the origin is:
x = 158.74784 sin 33.16° = –86.83172
y = 158.74784 cos 33.16° – 38.17 = 94.72518

Originally Posted by sbebuilders

... the radius for the tangent circle is not known.

Just curious but your pdf drawing does show a radius. How did you get it? AutoCAD?

Also, my eyebrow_radius.gif has been edited (a bit more info added) and a screen capture from your eyebrow_test_window.pdf of the same intersections as shown in my diagram has been uploaded.

[Joe Bartok]

Originally Posted by Joe Bartok

Use the formulas ...

Formulas for different initial givens are listed in both the web page and the Excel spreadsheet. The best one for this situation looks like ...

Radius = ½ Chord ÷ sin Pitch Difference Angle

½ Chord = 18.361097 ... the apothem of the larger tangent circle bisects the chord
Pitch Difference Angle = 33.16° – 22.2° = 10.96°

Radius = 18.361097 ÷ sin 10.96° = 96.574506

Add the two circle radii and solve for x and y as per the last post.

[sbebuilders]

Originally Posted by Joe Bartok

Formulas for different initial givens are listed in both the web page and the Excel spreadsheet. The best one for this situation looks like ...

Radius = ½ Chord ÷ sin Pitch Difference Angle

½ Chord = 18.361097 ... the apothem of the larger tangent circle bisects the chord
Pitch Difference Angle = 33.16° – 22.2° = 10.96°

Radius = 18.361097 ÷ sin 10.96° = 96.574506

Add the two circle radii and solve for x and y as per the last post.

Joe,

Thanks for the formula. It's right on the money. Yes, I did draw the PDF file in AutoCad to get the radius of the larger tangent circle. It is interesting that the ski slope tangent circle pages and ski slope thread are right inline with what I'm trying to do.

I started in on a cheat sheet for my boys using tangent circles to develop the eyebrow roof and eyebrow window design basd on Billy's eyebrow drawing. I felt by using tangent circles my boys might have an eaiser understanding of how to draw the eyebrow roof. Getting them to draw it out geometrically like Billy's drawing is like pulling teeth with them.

Billy's Eyebrow Drawing

After I developed the cheat sheet using tangent circles and equal radiuses I realized that it only worked for equal radiuses. So I needed to see what I could come up with when the two radiuses are not equal to allow for a standard radius window, rather than the custom eyebrow window using two radiuses to form the outline of the eyebrow window and maintain an equal reveal around the window.

Here's a pretty cool lookng eyebrow roof in Germany

Eyebrow Porch in CA with different size tangent circles allowing for a standard radius top on the door.

Sim

[Joe Bartok]

Originally Posted by sbebuilders

... However, I'm not following you on the known three points on the circle. The way I see it , I only know two points and the radius for the tangent circle is not known.

This is kind of beating a dead horse but here is the three point solution, actually four points if we invoke symmetry. It’s a little less elegant than the "ski slope" roof radius formula but worth looking at. The algebraic solution shown in the attached diagram is based on the Tangent Secant Theorem. The proof is easy to understand and can be found on any geometry text or site on the Internet.

The points to enter in the Javascript Calculator ... Circle Center and Radius given Three Points on the Circle are …

x1 = –68 – 37.65155 = –105.65155
y1 = 0

x2 = –68
y2 = .00000001*

x3 = –34
y3 = 13.87514

Javascript return for Radius = 96.578566

Offset from origin on x-axis …
h = –(96.578566 + 62.17) × sin 33.16° = –86.83212
Javascript return for h = –86.825775

Offset from origin on y-axis …
k = (96.578566 + 62.17) × cos 33.16° – 38.17 = 94.725787
Javascript return for k = 94.725965

*This is an example of Javascript that is technically correct but nevertheless needs to be "patched". The script solves h and k by finding the intersections of the perpendicular bisectors of the chords. Since the secant and chord are parallel to the x-axis the equation of the perpendicular is undefined. Making the entry for y2 == 0 returns k = ∞ … so the input has been cheated to a non-zero value.

[sbebuilders]

Originally Posted by Joe Bartok

This is kind of beating a dead horse but here is the three point solution, actually four points if we invoke symmetry. It’s a little less elegant than the "ski slope" roof radius formula but worth looking at. The algebraic solution shown in the attached diagram is based on the Tangent Secant Theorem. The proof is easy to understand and can be found on any geometry text or site on the Internet.

The points to enter in the Javascript Calculator ... Circle Center and Radius given Three Points on the Circle are …

x1 = –68 – 37.65155 = –105.65155
y1 = 0

x2 = –68
y2 = .00000001*

x3 = –34
y3 = 13.87514

Javascript return for Radius = 96.578566

Offset from origin on x-axis …
h = –(96.578566 + 62.17) × sin 33.16° = –86.83212
Javascript return for h = –86.825775

Offset from origin on y-axis …
k = (96.578566 + 62.17) × cos 33.16° – 38.17 = 94.725787
Javascript return for k = 94.725965

*This is an example of Javascript that is technically correct but nevertheless needs to be "patched". The script solves h and k by finding the intersections of the perpendicular bisectors of the chords. Since the secant and chord are parallel to the x-axis the equation of the perpendicular is undefined. Making the entry for y2 == 0 returns k = ∞ … so the input has been cheated to a non-zero value.

Joe,

Good example of the three point circle.

Here's a roof that has a ski slope and an eyebrow combined.

http://www.boirot-toiture.fr/media-b...lucarne_04.jpg


Sim

[ladnir]

I should post pictures of my dads garage that i helped build it has 3 eyebrows connected together on a sips panel roof with a copper roof. But its done with ellipse windows rather then arch windows.