# Gambrel math



## AndyGump (Sep 26, 2010)

I prefer pictures.

Andy.

I could get more detailed but I am lazy when I am not paid.:thumbsup:


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## jlmran (Feb 8, 2010)

Andy - I believe your picture is of an octagonal gambrel rather than a 2-hexagonal gambrel. Same result for height (if using full side lengths) but different rafter/span relationships.

Bad photo of a hexagonal gambrel.


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## user1007 (Sep 23, 2009)

Worked for Wolfram Research at on time. Horrible company to work for but Mathematica is a stellar product. You might want to see if one of their online calculators would be helpful. If you are, or have a kid in school? Your college or university probably will let you have a disc and a site license of the whole thing for like $10---even if you are an English lit major or something. The manual, is essentially a book that goes for $65 in bookstores. You could also buy and license your own retail copy for around $1200 as I looked last. 

There are specific add ons for doing things like construction calculations and so forth. They are not overpriced for what they do. Mathematica is worth every penny of what it costs by the way. And I do understand academic pricing and that the site license is expensive. Software people never catch on that it just looks bad to the average consumer to see $10 and $1200 in the same day, for the same product. Adobe is the most clueless I think. And Microsoft? My copy of the full office suite, for personal use but purchased with academic credentials, was under $80. Doesn't it retail for over $400 these days? Sublime to absurd. 

And come on. The argument is for every retail or even personal academic priced piece of product sold, I and others have Chinese 3-year-olds in my spare room duplicating copies? What nonsense. And should I not find it insulting that software companies would assume I would steal and distribute their products? Sorry, I am into stealing trucks carrying pallets of cartoned Clorox!


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## jlmran (Feb 8, 2010)

sdsester said:


> Worked for Wolfram Research at on time. Horrible company to work for but Mathematica is a stellar product. You might want to see if one of their online calculators would be helpful. If you are, or have a kid in school? Your college or university probably will let you have a disc and a site license of the whole thing for like $10---even if you are an English lit major or something. The manual, is essentially a book that goes for $65 in bookstores. You could also buy and license your own retail copy for around $1200 as I looked last.
> 
> There are specific add ons for doing things like construction calculations and so forth. They are not overpriced for what they do. Mathematica is worth every penny of what it costs by the way. And I do understand academic pricing and that the site license is expensive. Software people never catch on that it just looks bad to the average consumer to see $10 and $1200 in the same day, for the same product. Adobe is the most clueless I think. And Microsoft? My copy of the full office suite, for personal use but purchased with academic credentials, was under $80. Doesn't it retail for over $400 these days? Sublime to absurd.
> 
> And come on. The argument is for every retail or even personal academic priced piece of product sold, I and others have Chinese 3-year-olds in my spare room duplicating copies? What nonsense. And should I not find it insulting that software companies would assume I would steal and distribute their products? Sorry, I am into stealing trucks carrying pallets of cartoned Clorox!


Ummm...ok. Don't forget the cost of a computer in your analysis. 

The "ahh-ha" I had in the original post was derived from a plastic Chinese protractor, a 0.7 mechanical pencil, and a scrap piece of paper from an erroneous tax return from the past. Total cost...probably under $10. But then I posted my findings to this forum using an iPhone 4!! Insanity!


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## user1007 (Sep 23, 2009)

jlmran said:


> Ummm...ok. Don't forget the cost of a computer in your analysis.
> 
> The "ahh-ha" I had in the original post was derived from a plastic Chinese protractor, a 0.7 mechanical pencil, and a scrap piece of paper from an erroneous tax return from the past. Total cost...probably under $10. But then I posted my findings to this forum using an iPhone 4!! Insanity!


The guy has a computer or with protractor, mechanical pencil and let's not talk about tax returns right now oK? He tapped the internet. 

Never trust fresh seafood or math calculations being sold from the back of a pickup with Oklahoma plates is what I was taught.

Sent from my Sunbeam 4G capable George Forman grill with waffle iron.


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## jlmran (Feb 8, 2010)

Based on my Chinese protractor and a table of natural sines, the rafter lengths for a full octagon gambrel should equal the span/2.1796. This differs from Andy's picture above (if I converted inches to decimals correctly), and I don't believe it is rounding error. Anybody know why?

Edit: think I found my mistake. Standby for an update.


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## BigJim (Sep 2, 2008)

jlmran said:


> Based on my Chinese protractor and a table of natural sines, the rafter lengths for a full octagon gambrel should equal the span/2.1796. This differs from Andy's picture above (if I converted inches to decimals correctly), and I don't believe it is rounding error. Anybody know why?
> 
> Edit: think I found my mistake. Standby for an update.


Well.:whistling2:


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## jlmran (Feb 8, 2010)

Andy was correct. For full octagon, rafter length=span/2.6132.

Now I've got to check my hexagon math.


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## jlmran (Feb 8, 2010)

Hexagon number is correct. I think the hexagon looks better.


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## jlmran (Feb 8, 2010)

Just realized that these numbers are not a constant. They only apply if the four roof planes are to have equal rafter lengths. Took me a while to realize that the slopes (pitches) can be maintained while at the same time manipulating their ratio to one another.


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## abracaboom (Dec 27, 2011)

jlmran said:


> Just realized that these numbers are not a constant. They only apply if the four roof planes are to have equal rafter lengths. Took me a while to realize that the slopes (pitches) can be maintained while at the same time manipulating their ratio to one another.


When calculating relationships between lines, it is customary to assign the value of 1 (making it your unit) to the length of the principal line. By assigning, for example, the value of 1 to each of the legs of a right triangle, we come up with the value of the square root of two for its hypothenuse.

Both drawings in this thread start with a right triangle with both legs of unit length (that's why both have "inner" 12-in-12 rafters). When you say that "these numbers are not constant", what you've done is break that one-to-one relationship between the legs of the triangle. 

What I'm trying to say is that the relationship between lines becomes more evident when you use unit lengths for the basic line or lines (in this case one or both legs of a right triangle, not the rafters).


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## abracaboom (Dec 27, 2011)

jlmran said:


> The "ahh-ha" I had in the original post was derived from a plastic Chinese protractor, a 0.7 mechanical pencil, and a scrap piece of paper from an erroneous tax return from the past.


The basic tools for drafting are a straightedge and a compass.


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## jlmran (Feb 8, 2010)

I was simply trying to determine rafter lengths for a gambrel roof. This was the only solution I could determine without the use of anything more than a pocket calculator.


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## abracaboom (Dec 27, 2011)

jlmran said:


> I was simply trying to determine rafter lengths for a gambrel roof. This was the only solution I could determine without the use of anything more than a pocket calculator.


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.


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## jlmran (Feb 8, 2010)

abracaboom said:


> 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?


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## abracaboom (Dec 27, 2011)

jlmran said:


> 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?


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## jlmran (Feb 8, 2010)

abracaboom said:


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


Actually for a hexagon gambrel the number is 2.732.


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## jlmran (Feb 8, 2010)

Rafter length = (1/2 span) ((sqrt 3)-1) 

Doesn't work for all gambrels....only for hexagon based gambrels...correct?


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## ddawg16 (Aug 15, 2011)

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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## abracaboom (Dec 27, 2011)

jlmran said:


> 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.


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## jlmran (Feb 8, 2010)

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.


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## firehawkmph (Dec 12, 2009)

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


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## jlmran (Feb 8, 2010)

firehawkmph said:


> 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.


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## jlmran (Feb 8, 2010)

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.


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## abracaboom (Dec 27, 2011)

jlmran said:


> 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?


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## jlmran (Feb 8, 2010)

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.


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## firehawkmph (Dec 12, 2009)

jlmran said:


> 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


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## abracaboom (Dec 27, 2011)

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).


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## jlmran (Feb 8, 2010)

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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## abracaboom (Dec 27, 2011)

My drawing is a bit too crammed with lines and arcs to make it easy to follow each step from the beginning. 

This link shows, first, how to bisect a line, while at the same time drawing a line perpendicular to another. That was my first step to get my four quadrants. Next it shows how to bisect an angle. That's how I got my 45d angle:
http://en.wikipedia.org/wiki/Bisection

This second link shows how to trisect a right angle, while at the same time drawing an equilateral triangle. That's how I got my 60d angle:
http://www.gap-system.org/~history/HistTopics/Trisecting_an_angle.html

In my drawing, I only drew enough of some circles to get the intersection points I was looking for.


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