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01032012, 08:34 PM  #16  
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Gambrel mathQuote:


01042012, 12:14 AM  #17  
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Isn't that a more elegant solution than the span divided by 2.1796? 

01042012, 09:46 AM  #18  
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01042012, 10:54 AM  #19 
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Gambrel math
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 
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Gambrel math
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....
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01042012, 02:48 PM  #21  
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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 
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Gambrel math
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 
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Gambrel math
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  
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01042012, 10:09 PM  #25 
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Gambrel math
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  
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You're still not using a compass, are you? Last edited by abracaboom; 01052012 at 02:43 AM. 

01052012, 06:38 AM  #27 
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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 
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Gambrel math 
01062012, 04:32 AM  #29 
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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. 
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