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Stan. Can you help me out?
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Author:  Thechap1 [ Sat May 28, 2016 5:27 pm ]
Post subject:  Stan. Can you help me out?

Stan,

I'm hoping you can help me out?

I'm planning on putting a small area for a fire ring in the back yard and want to make the area into an Octagon shape. It will be made out of 4x4s. I'm hoping to make it 10 to 12 feet at the widest point, if this makes sense.

Considering I will need eight 4x4s what lengths should I get? What degree angle do I cut the 4x4s? And the last question. I plan on using the smaller river type rock inside the Octagon. So how many yards say to fill this area to a depth of two inches or so?

Thanks so much!

Chap

Author:  Thechap1 [ Sat May 28, 2016 5:47 pm ]
Post subject:  Re: Stan. Can you help me out?

JAKE! wrote:
They would be 22.5* ,but just guessing!!
360/8= 45 so that would be divided by 2= 22.5

Approximately 4' long, again just guessing :)


Oh my Jake your killing me brother!! I have no idea what you just said LOL. Pictures please.

Chap

Author:  Hellified [ Sat May 28, 2016 5:59 pm ]
Post subject:  Re: Stan. Can you help me out?

Chap- Just saw your post. Jake can answer it better than I. I will explain the correct angles that Jake provided. 8 sides will divide up a circle into 45 degree segments. But, that angle is divided for each half to miter together...as Jakes 22.5 degrees show.

Place a timber in a miter box and cut 22.5 degree angles on each end. You can play with the length for the desired size of the fire pit.

Author:  Thechap1 [ Sat May 28, 2016 6:38 pm ]
Post subject:  Re: Stan. Can you help me out?

Hellified wrote:
Chap- Just saw your post. Jake can answer it better than I. I will explain the correct angles that Jake provided. 8 sides will divide up a circle into 45 degree segments. But, that angle is divided for each half to miter together...as Jakes 22.5 degrees show.

Place a timber in a miter box and cut 22.5 degree angles on each end. You can play with the length for the desired size of the fire pit.


So Stan and Jake. We are almost there. For the "pad" to be at it's longest part, say ten feet, then how long would each 4x4 be?

Author:  Hellified [ Sat May 28, 2016 7:31 pm ]
Post subject:  Re: Stan. Can you help me out?

Chap- If you want exactly 10 feet across two opposite sides....outside of 4x4's....the long point of each side would be 49 11/16 inches. That is long point to long point.

If you want 10 feet max across opposite corners....then each 4x4 should be cut to 45 15/16 inches. This would then measure 9' -2 7/8" across opposite sides.
Jake is very modest on his math skills. I wish I knew what he has learned!.

Author:  Hellified [ Sat May 28, 2016 7:38 pm ]
Post subject:  Re: Stan. Can you help me out?

Hellified wrote:
Chap- If you want exactly 10 feet across two opposite sides....outside of 4x4's....the long point of each side would be 49 11/16 inches. That is long point to long point.

If you want 10 feet max across opposite corners....then each 4x4 should be cut to 45 15/16 inches. This would then measure 9' -2 7/8" across opposite sides.
Jake is very modest on his math skills. I wish I knew what he has learned!.

Author:  RayNAiken [ Sat May 28, 2016 7:51 pm ]
Post subject:  Re: Stan. Can you help me out?

Octagon calculator

http://www.pagetutor.com/octagon_layout/

Author:  RayNAiken [ Sat May 28, 2016 7:58 pm ]
Post subject:  Re: Stan. Can you help me out?

with either one you're going to have to buy 10ft long 4 x 4s
There would be too much waste with 8 foot ones unless you went with a 9 foot wide octagon pattern.

10 foot wide octagon

Attachment:
10.jpg


12 foot wide octagon

Attachment:
12.jpg

Author:  elwood [ Sat May 28, 2016 8:03 pm ]
Post subject:  Re: Stan. Can you help me out?

Damn! where were you guys when i was in algebra class? ......Oh yeah the internet had not been invented yet. :realcrazy

Author:  RayNAiken [ Sat May 28, 2016 8:20 pm ]
Post subject:  Re: Stan. Can you help me out?

elwood wrote:
Damn! where were you guys when i was in algebra class? ......Oh yeah the internet had not been invented yet. :realcrazy


There are two more methods.
1. Trail & error & trial again.
2. Hunt and Hope.

Author:  Flying Wolverine [ Sat May 28, 2016 9:34 pm ]
Post subject:  Re: Stan. Can you help me out?

Lay the 10ft sections of 4x4's over laying end to end on the ground and u will have a visual of the octagon and that of the end cuts - angle cuts you will need.

Lay out the outer pattern and then work inward.......math challenged use the trail and error method........

Good luck! There is more then one way to skin a cat. :)

Author:  MadMuz [ Sat May 28, 2016 9:37 pm ]
Post subject:  Re: Stan. Can you help me out?

My method is:

If at first you don't succeed, :badluck

try again.... :like

then give up and pretend you didn't want to do it anyway :laughing :fuckum :yoda2 :Wolvie

:Confederate

Author:  RayNAiken [ Sat May 28, 2016 10:13 pm ]
Post subject:  Re: Stan. Can you help me out?

Well one could just go with the 8' 4 X 4s.
Cut then in half.
Miter each end 22.5 degrees and put them together.
That would give you an octagon 9 feet- 7-7/8" wide.

Author:  Hellified [ Sun May 29, 2016 1:14 am ]
Post subject:  Re: Stan. Can you help me out?

Ray- Not trying to be a smart asshole here....but we can make it larger. If you take an 8 foot 4x4...and make a 22.5 degree cut across its middle....you can effectively lengthen your 8 foot 4x4 with the overlapping ends being cut across the middle. I have stretched lumber many times this way out of necessity! The tangent of 22.5 degrees times the actual width of a 4 x4...3.5" computes to roughly 1.45". But you have to divide that in half...and allow roughly 1/8" for the saw kerf.

Basically you can get two segments out of an 8 foot 4x4 that measure 48-5/8".

Plugging this " economized stretching of this 4x 4", we now can build an octagon 9'- 9 3/8" across.




.

Author:  Hellified [ Sun May 29, 2016 6:44 am ]
Post subject:  Re: Stan. Can you help me out?

Jake- I just read your request to verify your volume calculation. Since the gravel will only fill inside the 4x4's....that reduces the volume from your 13 cubic ft. estimate to just over 12 cubic feet. But for all practical purposes, both of our answers are correct!

I know you studied much higher levels of math than I will ever understand...such as calculus. The only thing I know about calculus is that is what my dental hygienist scrapes off my teeth every 6 months! Ha

Author:  Hellified [ Sun May 29, 2016 8:12 am ]
Post subject:  Re: Stan. Can you help me out?

Jake- One more red neck comment.. Mathematicians always say pi r square.....FALSE!! Pie r round...I have eaten too many of them.

Author:  Hellified [ Sun May 29, 2016 8:50 am ]
Post subject:  Re: Stan. Can you help me out?

That's offensive! Gabor.....ban Jake!

Author:  MadMuz [ Sun May 29, 2016 10:38 am ]
Post subject:  Re: Stan. Can you help me out?

For me it is '' Pie R gawn" :lick :laughing :yoda2 :Wolvie

Especially apple with cinnamon :beefcake

:Confederate

Author:  MadMuz [ Sun May 29, 2016 10:42 am ]
Post subject:  Re: Stan. Can you help me out?

Hellified wrote:
Ray- Not trying to be a smart asshole here....but we can make it larger. If you take an 8 foot 4x4...and make a 22.5 degree cut across its middle....you can effectively lengthen your 8 foot 4x4 with the overlapping ends being cut across the middle. I have stretched lumber many times this way out of necessity! The tangent of 22.5 degrees times the actual width of a 4 x4...3.5" computes to roughly 1.45". But you have to divide that in half...and allow roughly 1/8" for the saw kerf.

Basically you can get two segments out of an 8 foot 4x4 that measure 48-5/8".

Plugging this " economized stretching of this 4x 4", we now can build an octagon 9'- 9 3/8" across.



Stan, I love it when you talk dirty :laughing :half

Is that Egyptian? Or chinese? :noidea :wol2 :Wolvie

:Confederate

Author:  Hellified [ Sun May 29, 2016 12:20 pm ]
Post subject:  Re: Stan. Can you help me out?

Muz- You like me saying "smart ass".....or does "tangent" get you pumped up? Ha

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