Hip roof calculations — — An Adjuster to

Hip roof calculations - - An Adjuster to

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02/28/2007 12:47 AM

Ok, I have a question on figuring the squares on a simple hip roof.

In a pre-test for certification with a carrier there was a question involving a simple hip roof with a 10′ extention.

They wanted to know what the sf of the roof was and what the sf of the north and west slopes was.

One method is the up and over. You take the length of house x rafter x 2 = total sf. and then take the extention length x the rafter length and add to that = total sf quick and easy.

or 40Long x 20Rafter x 2 = 1600 then add 10Long x 10Rafter x 2 = 200 + 1600 = 1800 sf.

Another method is to break down the roof into 2 trapezoids 2 parallelograms and 2 triangles. (the 3rd triangle will be considered part of the south trapezoid)all the measurements of the individual shapes and add them up. I will call this the individual slope method.

or 40Long + 18Ridge (top) / 2 x 20Rafter = 580 sf for the north slope. The south is the same.

Now the triangle is 36Base x 20Rafter / 2 = 360 then the 10Long x 10Rafter parallelograms for the extentions = 100 sf + 360 = 460 for the west slope. The east is the same.

Now take the 580 + 460 = 1040 for the north and west or 2080 for the total.

2.8 squares difference because the test question was an impossible roof.

With a typical hip, with all slopes having the same pitch. These figures would match perfect and half of the roof would equal to the north and west slopes.

The problem is that the roof that was drawn on the test could not be made with the given dimentions. There was 18′ of ridge on the top of the large hip. The big hip was 40 x 36 with a 20 rafter and the rafter length was the same on all sides. The extention stuck out the south side 10 and had a 10 rafter length. The only way for that to be possible is with a 4′ ridge along the top of the big hip(the difference between the long and short sides of a hip = the top ridge (that is basic geometry provided the rafte length is the same on all slopes). But with the UP AND OVER METHOD you get 18 squares on this roof. If you figure it using (INDIVIDUAL SLOPE method you get 20.8 squares. This is the issue. The answer that was given was the 18 sq. But then if you figure up what is needed to cover only the north and west slopes it was deemed that the answer was 10.4 sq. At this point we were told that we are to figure an entire roof using the up and over method and use geometry to get only the 2 slopes.

That does not make sense to me. How would you answer the test question.

Most of the younger guys say the 20.8 is correct with 10.4 for the 2 slopes only. I am a young guy and I agree with this.

Most of the older guys say it is 18 for the whole roof and that the 10.4 is STILL correct. These guys are my superiors so I do not want to disregard the advice but I feel it is wrong. The class was divided on their opinions but the real question was how to figure the roof on the test. If it was drawn as a perfect roof, either way would work. If it is not a perfect roof. the Individual slope method would be the only accurate one. But the tester may be using the up and over method to figure it and count it wrong on the test. An actual roof would not be an issue. I am only bothered by the fact that the tests may be wrong and I don’t want to miss these questions because of a simple mistake by the one who wrote the test. How should it be figured?

I like to see what your opinions are on this issue. Sorry the message is so long I was just trying to get all the facts in.

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