


Thread Tools  Display Modes 
01032012, 08:34 PM  #16  
Member
Join Date: Feb 2010
Location: Oklahoma
Posts: 992
Rewards Points: 506

Quote:
Advertisement 

01042012, 12:14 AM  #17  
Retired carpenter
Join Date: Dec 2011
Location: Portland, Oregon
Posts: 303
Rewards Points: 250

Quote:
Isn't that a more elegant solution than the span divided by 2.1796? 

01042012, 09:46 AM  #18  
Member
Join Date: Feb 2010
Location: Oklahoma
Posts: 992
Rewards Points: 506

Quote:




01042012, 10:54 AM  #19 
Member
Join Date: Feb 2010
Location: Oklahoma
Posts: 992
Rewards Points: 506

Rafter length = (1/2 span) ((sqrt 3)1)
Doesn't work for all gambrels....only for hexagon based gambrels...correct? 
01042012, 02:30 PM  #20 
JOATMON

I use autocad....it tells me the length....
but since your on #'s.....I understand now why a 5:12 roof pitch is so popular....easy to calculate the angles....
__________________
Even if you are on the right track, you will still get run over if you just sit there. My 2Story Addition Build in Progress Link ... My Garage Build Link and My Jeep Build Link 
01042012, 02:48 PM  #21  
Retired carpenter
Join Date: Dec 2011
Location: Portland, Oregon
Posts: 303
Rewards Points: 250

Quote:
Only for rafters of equal length at 60 degrees off of the horizontal and vertical lines, like the ones you drew, yes. In a 906030 triangle with a short leg of length one, the hypothenuse's length is two and the long leg's length is square root of three (by the Pythagorean Theorem). That's where the square root of three in the formula comes from. Thus, rafter length equals length of long leg minus length of short leg (that's what I call an elegant solution). There is no "universal ratio" for all possible variations of a gambrel roof. Last edited by abracaboom; 01042012 at 03:09 PM. 

01042012, 04:00 PM  #22 
Member
Join Date: Feb 2010
Location: Oklahoma
Posts: 992
Rewards Points: 506

I'm really dense. I'm still not understanding how Pythagorean theorem only can solve for the rafter lengths...especially given this image. Both roofs have an equal span and an equal height.
Last edited by jlmran; 01042012 at 04:03 PM. 
01042012, 04:06 PM  #23 
Member
Join Date: Dec 2009
Location: Near Cleveland, Ohio
Posts: 2,027
Rewards Points: 1,884

Jlm,
I would think you would just have to break your drawing down to all right triangles, then use the old A sq. + B sq. = C sq. to find the lengths of each hypotenuse and add then add them up. Hope I remembered that right. Mike Hawkins 
01042012, 05:33 PM  #24  
Member
Join Date: Feb 2010
Location: Oklahoma
Posts: 992
Rewards Points: 506

Quote:


01042012, 10:09 PM  #25 
Member
Join Date: Feb 2010
Location: Oklahoma
Posts: 992
Rewards Points: 506

Duh. Seems so easy now, after reading this.
http://erniedipko.blogspot.com/ Thanks to Ernie I feel quite dumb. Thanks to all for your input...I just wasn't getting it. Last edited by jlmran; 01042012 at 10:19 PM. 
01052012, 02:25 AM  #26  
Retired carpenter
Join Date: Dec 2011
Location: Portland, Oregon
Posts: 303
Rewards Points: 250

Quote:
You're still not using a compass, are you? Last edited by abracaboom; 01052012 at 02:43 AM. 

01052012, 06:38 AM  #27 
Member
Join Date: Feb 2010
Location: Oklahoma
Posts: 992
Rewards Points: 506

Correct. It only works for the octagon.
It also took me a while to comprehend the 306090 triangle, with sides of 1sqrt32, but once I FINALLY realized that those values/angles are the basis for the trigonometric functions which I originally used to derive the denominators, then I had another ahha and felt really dumb. I had a trig class in high school in 1989...maybe I knew it once and forgot it. I did use a compass to draw a circle around my protractor derived roofs. I'll need to play some more with it, w/o the use of the Chinese made protractor. 
01052012, 12:54 PM  #28  
Member
Join Date: Dec 2009
Location: Near Cleveland, Ohio
Posts: 2,027
Rewards Points: 1,884

Quote:
I was thinking you were working backwards off your drawings you posted. After reading Ernies blog, it all makes sense. Interesting thread. Mike Hawkins 

01062012, 04:32 AM  #29 
Retired carpenter
Join Date: Dec 2011
Location: Portland, Oregon
Posts: 303
Rewards Points: 250

A picture:
I used points A,B,C and D to draw a line perpendicular to the horizontal passing through A. The circles with centers at A, E and F have a radius of length 1. On the NE quadrant I trisected the right angle NAF to obtain the equilateral triangle AHF to get my 60 degree angle. Point H is the intersection of the circles with centers at A and F. Half of the equilateral triangle is a 306090 triangle (that's why its short leg is half the length of the hypothenuse). On the SW quadrant I showed how to bisect an angle in order to get my 45 degrees using points A,E,I and J. On the NW quadrant, point G is the intersection of the circles with centers at A and E. A line from E to G is at 60 degrees from the horizontal, and I extended that line to its intersection with the vertical line at point M. That gives me a 306090 triangle with a short leg of length 1 (AE), a hypothenuse of length 2 (EM), and a long leg of length sqrt 3 (AM). The 60 degree line EM intersects the 45 degree line AK at point L. Line EL is my lower rafter, and line LN is my upper rafter. Both rafters are the same length. The circle with center at N has a radius equal to the length of the upper rafter and intersects the vertical line at M, meaning that the rafter length is equal to AM minus AN, that is, equal to ([sqrt 3]1). Last edited by abracaboom; 01062012 at 05:41 AM. 
01062012, 09:33 PM  #30 
Member
Join Date: Feb 2010
Location: Oklahoma
Posts: 992
Rewards Points: 506

Thanks. I will try to recreate this (and understand all of it) and report back to you. It might be a few days.
Advertisement 
Thread Tools  
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Building deck on side of Gambrel style roof  questions  sheslostcontrol  Building & Construction  14  05102011 11:15 PM 
Interior walls & gambrel ceiling  Cveroline  Carpentry  2  01272011 09:54 PM 
Insulating/Ventilating Gambrel roof  billybarty  Building & Construction  2  09252009 02:17 PM 
Gambrel Roof Measurements  Hoff77  Building & Construction  2  03152009 03:54 PM 
math for checking for 90 degrees in a corner  coderedcraze  Carpentry  17  12022006 12:40 AM 