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01-03-2012, 09:34 PM   #16
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Quote:
 Originally Posted by abracaboom Each of the two rafters has a rise and a run; the rise of one rafter is the run of the other rafter, and viceversa. All of your dimensions depend on that right triangle with equal legs. It sounded like you were enjoying yourself. If you want to have even more fun, try drawing it using only a straightedge and a compass.
Are you stating that it is possible to calculate the gambrel rafter length with ONLY the Pythagorean theorem?

01-04-2012, 01:14 AM   #17
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Quote:
 Originally Posted by jlmran Are you stating that it is possible to calculate the gambrel rafter length with ONLY the Pythagorean theorem?
If anything, I would have implied it, which I didn't. But now that you mention it, yes, you can calculate the rafter lengths with ONLY the Pythagorean Theorem. It turns out that if you assign a value of one to the total run and the total rise, your rafters equal the square root of three minus one. In other words: rafter length = half the span multiplied by (the square root of three minus 1). If you draw your roof with just a straightedge and a compass, you will see clearly why.

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

01-04-2012, 10:46 AM   #18
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Quote:
 Originally Posted by abracaboom Isn't that a more elegant solution than the span divided by 2.1796?
Actually for a hexagon gambrel the number is 2.732.

 01-04-2012, 11:54 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?
 01-04-2012, 03:30 PM #20 JOATMON     Join Date: Aug 2011 Location: S. California Posts: 10,847 Rewards Points: 854 Blog Entries: 2 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 2-Story Addition Build in Progress Link ... My Garage Build Link and My Jeep Build Link
01-04-2012, 03:48 PM   #21
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Quote:
 Originally Posted by jlmran Rafter length = (1/2 span) ((sqrt 3)-1) Doesn't work for all gambrels....only for hexagon based gambrels...correct?

Only for rafters of equal length at 60 degrees off of the horizontal and vertical lines, like the ones you drew, yes.

In a 90-60-30 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; 01-04-2012 at 04:09 PM.

 01-04-2012, 05:00 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. Attached Images   Last edited by jlmran; 01-04-2012 at 05:03 PM.
 01-04-2012, 05:06 PM #23 Member     Join Date: Dec 2009 Location: Near Cleveland, Ohio Posts: 1,984 Rewards Points: 1,800 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
01-04-2012, 06:33 PM   #24
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Quote:
 Originally Posted by firehawkmph 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
I don't see how that is possible IF the building span is the ONLY known value.

 01-04-2012, 11:09 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; 01-04-2012 at 11:19 PM.
01-05-2012, 03:25 AM   #26
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Quote:
 Originally Posted by jlmran 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.
Ernie's math only works for an octagonal gambrel roof. Note that on the kind of roof you drew, the one for which the ([sqrt 3]-1) formula applies, the rafters don't meet at the circle's perimeter, like Ernie's do.

You're still not using a compass, are you?

Last edited by abracaboom; 01-05-2012 at 03:43 AM.

 01-05-2012, 07:38 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 30-60-90 triangle, with sides of 1-sqrt3-2, 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 ah-ha 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.
01-05-2012, 01:54 PM   #28
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Quote:
 Originally Posted by jlmran I don't see how that is possible IF the building span is the ONLY known value.
Jl,
I was thinking you were working backwards off your drawings you posted. After reading Ernies blog, it all makes sense. Interesting thread.
Mike Hawkins

 01-06-2012, 05:32 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 30-60-90 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 30-60-90 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). Attached Thumbnails   Last edited by abracaboom; 01-06-2012 at 06:41 AM.
01-06-2012, 10:33 PM   #30
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Thanks. I will try to recreate this (and understand all of it) and report back to you. It might be a few days.

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