5 sided gazebo
Hi,
I need a bit of help! These days I hardly know how to spell trigonometry let alone know how to use it.
I need to know some common figures on how to layout and build a 5 sided gazebo. It is a small structure4 ft per side. The pitch will be 5/12. What figures do I use to cut the hips?
Any other tips you can give me?
Thanks for all, larry haun


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(post #64569, reply #1 of 6)
I'm no architect, and I do not intend to quash anyone's creativity, but 5 sides is going to look a bit odd.
One way:
Draw a circle on graph paper.
Divide 360° by 5= 72°
Use a protractor to mark your circle every 72°.
Connect the dots to form your pentagon.
Now measure the angles with the protractor.
That's your frame with no advanced math.
Or, the math says:
For any Polygon; sum of interior angles = (n – 2)180°
So for Pentagon:
= (5 – 2)180° = (3)180° = 540°
So divide that by 5, and you get 108° corners.
But your question was about the roof:
The way a lot of old timers did it was to form a 'block' or 'box' at the peak to receive the rafters (forgive me if there is a technical term for this thingsomone called it a finial block, but I don't know). This eliminates compound rafter cuts at the peak. Bevel up some blocks to form a small box in the shape of a pentagon for the peak (deep enough to receive rafters), then just make normal cuts on the rafters. Sometimes these 'peak blocks' were enlarged to become the basis for a finial or other decorative element. One on my (hexagonal) porch has a carved pineapple on it.
You could use the same method to form the floor frame, though you would need a supporting post in the center. Might be easier than the alternatives with an odd number of sides.
If you realy wanted to make double cheek cuts at the peak, you can figure the angle of the 5 triangular 'pie pieces' at the peak based upon half of the 108° corners.
The sum of the interior angles of a triangle:
For any polygon: sum of angles = (n – 2)180°
So for a triangle: (3 – 2)180° = (1)180° = 180°
2 of your triangle angles are known to be 54°(108° /2), so the remaining angle (the pie slice) must be 72°. As a reality check: 72°(5)=360°, and there's your pie.
Now you know the value for each angle, and can work from there.
Edited 2/7/2005 3:32 pm ET by csnow
(post #64569, reply #2 of 6)
Hi,
Thanks for your help. A 5 sided structure will look oddyou mean like the pentagon!
Anyway..is there a common figure to lay out a rafter on this structure? I recall that 13 is used to lay out a hip on an octogon instead of 17 on a stardard roof.
Thanks again, larry haun
(post #64569, reply #3 of 6)
"Thanks for your help. A 5 sided structure will look oddyou mean like the pentagon!"
Exactly! Well, not really, I was thinking it would look odd on gazebo scale. Symmetry pleases the eye, so they say...
If you use the peak block idea, the rafter cuts will be simple cuts, not compound, so it is becomes relatively easy to calculate the angles.
Here's one online calculator:
http://www.constructionresource.com/roofframe.php
(post #64569, reply #4 of 6)
Hi,
I have built buildings doing the blocks at the top. That is a good way to do it. I would still like to know what the ratio of rise to run is for a pentagon. Can you help, please?
I think I would like to try building the room by using compound cuts and running the hips to the top. Then I can fill in with a jack or two between. thanks, larry
(post #64569, reply #5 of 6)
Larry  I had to brush up a little on my trig for this one. According to my drawing, I believe tan. 36 deg. = x/12, hence 12 times .7265 = x, or 8.72. Going back to Pythagorean's theorem, 12 sqared plus 8.72 squared = y sqared. Hence, y = the square root of 144 + 76 (which is 220), and the square root of 220 is 14.8
Double checking, the sin of an angle can be found by dividing the length of the side opposite the angle by the length of the hypotenuse. (The side opposite to the angle is the side that is directly across from the center of the angle.) Hence, sin 54 deg. = 12/y, .81=12/y, .81y = 12, y=12/.81, y=14.8
So on an octagon with a 5:12 pitch, the rise:run pitch of the hips should be 5:14.8 (14 & 3/4)
Edited 2/9/2005 12:28 am ET by Huck
"Down these mean streets a man must go who is not himself mean, who is neither tarnished nor afraid...He must be, to use a rather weathered phrase, a man of honor, by instinct, by inevitability, without thought of it, and certainly without saying it."  Raymond Chandler
(post #64569, reply #6 of 6)
Hi,
Thanks for taking the time to help me with the trig. I lost the ability to use that type of math a number of years agolike 40 or so.
The figures you gave me sound right. Anymore, I build Habitat for Humanity houses, do demos at some trade shows, and help friends and family with their projects. The small pentagon gazebo will be an office for a friend who sells rocks to contractors.
www.carpentryforeveryone.com
Thanks again for your help. larry haun