Software or advice on calculating sides of an octagon


Hi fellow woodies,
My latest project involves the planning and construction, of an octagonal (8 sided) summerhouse/workshop on a base measuring 4.5 metres square, with a floor capacity of 4 metres² - 43.056 feet² - 6200.12 inches², and a wall height of (2.14 metres - 7 feet and 0.25 inch - 84.25inches). Try as hard as I can, I've been unable to get my head around the calculations necessary to work out how wide each of the individual 8 sides should be. I visited a book called 'The Woodworker's Complete Shop Reference' by J Churchill. However, my insufficient brain capacity cannot cope with the math, which seems to read as :- Radius=(Length / 2)² / height² / 2 x Height ( / = divided by). I've most likely misinterpreted the math calculation, as this is a weak point with me.
Simple checks using a graphics program seem to suggest a length in the region of 1.57 metres, however, I'm not sure if the prog has also measured the effects of certain viual effects used to make the plans look good when printed (thick outer lines of 5 points and rendering/bevelling and shading etc ...). Can anyone give me an accurate figure for my measurements? Even better, a link to any online programs which could assist with my calcs, as I'm sure I'll need further assistance when it come to the roof.
Kind regards and TIA John (wheelz) Reply address is fictitious, to prevent spam attacks
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<snip>
Hi John,
basic math time (I hope) (bet someone shoots me down)
ok, lets imagine the octagon.
We know that the radius is 2.25 m (cos you told us you have a 4.5m square base.
we also know that the octagon is made of 16 right angle triangles. we know the length of one of the sides, and we know all the internal angles, so we can use trigonometry to give us all the lengths.
In effect 1/2 s (s being side length) is:
2.25 * tan 22.5 = 0.931m
therefore the length of each side, for a perfect octagon is 0.931*2 1.863m (using the scientific calc)
Does this help?
Si
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I got to the same side measure via a different route. But couldn't make sense of how it fit with the rest of his description. Look at your 16 right triangles. They each have an area of .931*2.25/2=1.04m^2, so the total area is 16.8m^2. The original description says 4 m^2, which sounds pretty tiny for the stated purpose. Wonder if it was described as "the same area as a square building 4 meters on a side", and that was misinterpreted as 4 m^2.

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wrote: : : ><snip> : > : >Hi John, : > : >basic math time (I hope) (bet someone shoots me down) : > : >ok, lets imagine the octagon. : > : >We know that the radius is 2.25 m (cos you told us you have a 4.5m square : >base. : > : >we also know that the octagon is made of 16 right angle triangles. we know : >the length of one of the sides, and we know all the internal angles, so we : >can use trigonometry to give us all the lengths. : > : >In effect 1/2 s (s being side length) is: : > : >2.25 * tan 22.5 = 0.931m : > : >therefore the length of each side, for a perfect octagon is 0.931*2 : >1.863m (using the scientific calc) : : I got to the same side measure via a different route. But couldn't : make sense of how it fit with the rest of his description. Look at : your 16 right triangles. They each have an area of : .931*2.25/2=1.04m^2, so the total area is 16.8m^2. The original : description says 4 m^2, which sounds pretty tiny for the stated : purpose. Wonder if it was described as "the same area as a square : building 4 meters on a side", and that was misinterpreted as 4 m^2. : > : >Si
i couldnt decide if 4m square for the interior of the octagon was required, or an octagon that would fir the 4.5 sq m base - so i went for one that would fill the base as much as poss
Si
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Hi fellow woodies,
My latest project involves the planning and construction, of an octagonal (8 sided) summerhouse/workshop on a base measuring 4.5 metres square, with a floor capacity of 4 metres² - 43.056 feet² - 6200.12 inches², and a wall height of (2.14 metres - 7 feet and 0.25 inch - 84.25inches). Try as hard as I can, I've been unable to get my head around the calculations necessary to work out how wide each of the individual 8 sides should be. I visited a book called 'The Woodworker's Complete Shop Reference' by J Churchill. However, my insufficient brain capacity cannot cope with the math, which seems to read as :- Radius=(Length / 2)² / height² / 2 x Height ( / divided by). I've most likely misinterpreted the math calculation, as this is a weak point with me.
Simple checks using a graphics program seem to suggest a length in the region of 1.57 metres, however, I'm not sure if the prog has also measured the effects of certain viual effects used to make the plans look good when printed (thick outer lines of 5 points and rendering/bevelling and shading ***********************
The answer is 1.657 m per side for a perfect octagon. Given that the angled sides are at 45 degrees, and a side is x, that means that a 4 m edge will "see" x + .707x + .707x. So that means 4m = 2.414x, or x = 1.657
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Bob

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I illustrate and describe the math here:
http://www.delorie.com/quake3/octatrap /
(you only need the small section on calculating the octagon, not all the other game-editing steps ;)
I get 1.864 meters too.
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I dunno man....
You can do it graphically by drawing a square in AutoCad, 4 units to a side. Copy the square, rotate by 45 degrees, and drop it back onto the first square. If you measure the sides of the resulting octagon, you get 1.657/side, and a 4m square footprint. Maybe I misread the original question.
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D-oh! 4.5m square, not 4.
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: : D-oh! 4.5m square, not 4.
i think like me you read the original request in many ways -
4.5 sq m base or internal of a 4 m sq or 4 sq m octagon
John - I reckon we've covered all the bases here for you - which one is it?
Si
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Just goes ta show ya, there are at least 3 sides to every octagon...
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If you are going to use Autocad, why the extra steps? Just use the polygon tool.

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I'm so used to drawing odd-shaped mechaincal enclosures, I completely forgot about the polygon tool!
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wrote:

Ummmmm.... how does that make an octagon?
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Regards,
Doug Miller (alphageek at milmac dot com)
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It does if you ignore the corners. I omitted that bit.
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snipped-for-privacy@milmac.com (Doug Miller) wrote:

This is a case where you need to think INSIDE the box!
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