Geometry Question

Randy

By Randy
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wtnhunt Monster Buck

wtnhunt

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If you have to build it with all sides being equal as a hexagon Randy, it is going to be a bit more complicated, but think potash's calculations look right. Honestly I would build a square 6 foot box if I were you, much more simple and will give more room with about the same amount of materials used.

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rhine16 Monster Buck

rhine16

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After i re-did it to make all sides equal, i'm coming out with sides that need to be 2.5 ft. I'm way off from what everyone else is saying, but check my work and see what you think.

The way i set my diagram up is i drew the hexagon and drew my center line(6ft) across where each diagonal meets. From there, i'm just working with one half.

With a half, you can set up with two right triangles and a rectangle. After i tested some stuff, i realized that 2.5 should work. Using 2.5, I went back and dispersed my center line measurements. I found that the bottom side of each triangle should be 1.75:

2.5+ 2x=6

2x=3.5

x=1.75

From there i used pythagoreaum(sp?) theorem on the triangle

a^2+b^2=c^2

1.75^2+b^2=2.5^2

3.0625+b^2=6.25

b^2=3.1875

b=1.78535

Therefore the height from each center line to your flats should be b.

I may be wrong. Just my 2 cents.

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toddyboman Monster Buck

toddyboman

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After i re-did it to make all sides equal, i'm coming out with sides that need to be 2.5 ft. I'm way off from what everyone else is saying, but check my work and see what you think.

The way i set my diagram up is i drew the hexagon and drew my center line(6ft) across where each diagonal meets. From there, i'm just working with one half.

With a half, you can set up with two right triangles and a rectangle. After i tested some stuff, i realized that 2.5 should work. Using 2.5, I went back and dispersed my center line measurements. I found that the bottom side of each triangle should be 1.75:

2.5+ 2x=6

2x=3.5

x=1.75

From there i used pythagoreaum(sp?) theorem on the triangle

a^2+b^2=c^2

1.75^2+b^2=2.5^2

3.0625+b^2=6.25

b^2=3.1875

b=1.78535

Therefore the height from each center line to your flats should be b.

I may be wrong. Just my 2 cents.

Thats WAY TO MUCH work just to build a deer stand....:eek::eek::D

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PotashRLS Monster Buck

PotashRLS

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After i re-did it to make all sides equal, i'm coming out with sides that need to be 2.5 ft. I'm way off from what everyone else is saying, but check my work and see what you think.

The way i set my diagram up is i drew the hexagon and drew my center line(6ft) across where each diagonal meets. From there, i'm just working with one half.

With a half, you can set up with two right triangles and a rectangle. After i tested some stuff, i realized that 2.5 should work. Using 2.5, I went back and dispersed my center line measurements. I found that the bottom side of each triangle should be 1.75:

2.5+ 2x=6

2x=3.5

x=1.75

From there i used pythagoreaum(sp?) theorem on the triangle

a^2+b^2=c^2

1.75^2+b^2=2.5^2

3.0625+b^2=6.25

b^2=3.1875

b=1.78535

Therefore the height from each center line to your flats should be b.

I may be wrong. Just my 2 cents.

Try...... a=1.732, b=2.998 in your triangle. Then your c=3.462'(hypotenuse). In your sketch of the hexagon, corner to corner measurement is 6.924' which gives you 6' perpendicular between parallel sides.

Might be mathematical overkill for a deerstand:rolleyes: Who says you'll never use math after highschool?:eek:

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rhine16 Monster Buck

rhine16

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Try...... a=1.732, b=2.998 in your triangle. Then your c=3.462'(hypotenuse). In your sketch of the hexagon, corner to corner measurement is 6.924' which gives you 6' perpendicular between parallel sides.

Might be mathematical overkill for a deerstand:rolleyes: Who says you'll never use math after highschool?:eek:

Are you doing your dividing line from where one horizontal meets a diagonal to where the other horizontal meets the diagonal?:confused:

I think you may either be doing that or either you're doing an octagon or either i'm just completely lost on how you got 3.462 as your c.

The way i've set it up, c is a known. c is 2.5.

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PotashRLS Monster Buck

PotashRLS

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Are you doing your dividing line from where one horizontal meets a diagonal to where the other horizontal meets the diagonal?:confused:

I think you may either be doing that or either you're doing an octagon or either i'm just completely lost on how you got 3.462 as your c.

The way i've set it up, c is a known. c is 2.5.

I'm a Land Surveyor by trade and just whipped up the 6 sided shape on the Computer in a minute or two. It would be easy to show with a simple diagram.

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texastrophies Monster Buck

texastrophies

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Alright well i would love to see it so i could see where i'm figuring wrong.

Plus it'd probably help Randy.

Just a thought, but, if he wants 6' across then the end pieces have to be over 3' each or they won't meet. The only math I used in that was 3 + 3 = 6 and if you put them at an angle they will have to be longer. :D:D:D

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rhine16 Monster Buck

rhine16

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Alright, this is killing me :mad:

After BuckNRut's post, something just hit me.

In theory, a hexagon is defined as a 6 sided figure. :D

Therefore, BucknRut's, Potash's, and my answer are all correct :eek:.

It all matters on what you want your angles to be. I assumed that you were wanting the angles to be 120 degrees where each horizontal meets diagonal. In my opinion that would definitley make the most sense for a stand. But that is your decision.

BucknRut's 5.2 answer could be made correct, but your angle i described would be like 95 degrees. You might as well have a standard rectangular box then.

Sooo, hopefully this will complete your answer.

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Leo Monster Buck

Leo

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Folks,

A hexagon is made up of six equilateral triangles. Six triangles equal length sides stacked together make one hexagon.

If you are measuring across the flats. Then you are measuring the height of two of those triangles.

The height of the equilateral triangle you are trying to find is 3 which is half of the 6 Randy is looking for.

Since an equilateral triangle has vertice angles of 60 degrees each and the angle the height angle must be half that at 30 degrees. The answer is 3/cos(30) or 3.4641

It simply is no more complicated than that.

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PotashRLS Monster Buck

PotashRLS

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Folks,

A hexagon is made up of six equilateral triangles. Six triangles equal length sides stacked together make one hexagon.

If you are measuring across the flats. Then you are measuring the height of two of those triangles.

The height of the equilateral triangle you are trying to find is 3 which is half of the 6 Randy is looking for.

Since an equilateral triangle has vertice angles of 60 degrees each and the angle the height angle must be half that at 30 degrees. The answer is 3/cos(30) or 3.4641

It simply is no more complicated than that.

That is perfectly correct!!! I bet Randy doesn't even want to build it anymore. Sorry Randy!

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