# Octagon dimensions Help

Discussion in 'Landscape Architecture and Design' started by McNeal Lawn, Aug 31, 2003.

1. ### McNeal LawnLawnSite Memberfrom virginia beach vaMessages: 61

I have a customer that wants an octagon bed installed, the bed will have a diameter of 15 feet, constructed with landscape timbers. The problem is I can not calculate the required length of the timbers nor the degree angle for the mitre cut. I am handicapped with limited math knowledge. Can anyone give me a hand in figuring out this problem? Thanks.

2. ### capitalLawnSite Memberfrom Des Moines,IAMessages: 118

Suggest you use the KISS method and will find it goes easier. buy your timbers (8') length, lay them out at job site and then use a chain saw to form up the bed. you should end up around 15.5 ' across. We are currently doing a octogon patio with a circle pattern inside and went to the kiss method after playing with the math numbers. So just paint on the ground your corners and keep adjusting using a string line til u line the corners up . Then lay your timbers and begin cutting your angles at the site. remasure your angles and timber spots after each cut and u should line up correctly.

3. ### mottsterLawnSite Memberfrom South Central KansasMessages: 191

if you use the full 8 foot length tembers you'll end up with a 16 foot bed.
To get the 15 foot diameter (i'm guessing you're meaning side to side and not corner to corner?) you'll need 7.5 foot lengths. cut the timbers to the proper length and then cut the angles from there. Oh...each angle will need to be 22.5 degrees.

4. ### mdvadenLawnSite Bronze Memberfrom Westside OregonMessages: 1,946

Drawing it to scale on 8a.5 x 11 graph paper may help as a visual building block.

We have a small miter saw, so I finish the last part of the cut with a sharp handsaw.

5. ### McNeal LawnLawnSite Memberfrom virginia beach vaMessages: 61

Thanks for the help, all of your suggestions will be used, thanks again

6. ### AGLALawnSite Bronze Memberfrom Cape CodMessages: 1,774

The lengths of each side should be about 5'9" and the parallel sides should be about 13'10" apart.

7. ### jonw9LawnSite Memberfrom Central MichiganMessages: 25

Well, first is it 15' inscribed or circumscribed? Meaning 15' from point to point, or from flat to flat? That will determine length.

AGLA is right on the inscribed bed,
and for circumscribed sides would be 13'10" with the distance between sides being 15'

8. ### Team GopherLawnSite Platinum Memberfrom -Messages: 4,040

Here is a math site that explores the octagon.

9. ### diginaholeLawnSite Memberfrom Port Perry, OntarioMessages: 249

Check your math jonw9, you should get this.

Inscribed: 8 sides @ 5' 9". Distance between parallel sides 13' 10 ". Corner to opposite corner 15' 0".

Circumscribed: 8 sides @ 6' 3" ( actually 6' 2 9/16"). Distance between parallel sides 15' 0". Corner to opposite corner 16' 2 13/16".

10. ### AGLALawnSite Bronze Memberfrom Cape CodMessages: 1,774

I think that the miter cut will be about 67 degrees.