Geometry problem....

[Rusty]

...or how I wish I'd paid attention in class!

I have a yen for a strawberry bed, but my last venture was a disaster. The bermuda grass swallowed all my strawberries. So this time I thought I'd try one of those strawberry pyramid thingies...until my SO gently reminded me of my "expertise" with the bush hog. And he's right. I'd probably flatten it the first time I mowed.

So then I came up with a new idea: a pyramid constructed out of landscape timbers! (Even I cannot mow them!) What I have in mind is 3 tiers--the first tier constructed of 3' timbers, the second of 2' timbers, and the third of 1-footers.

This is where my lack of geometry skills comes in. I want a structure that is about 6' in diameter in order for it to fit where I want to put it (so the bush hog will fit around it when I mow). To get that, do I need 8 sides or 6? Will the number of sides even make a difference size-wise? For the life of me I cannot figure out how to do the math!

So if any of you budding math geniuses out there have a clue, I'd sure appreciate some help with the math!



Rusty

 

[patandchickens]

The difference between the width of the two is minimal (the difference in the area contained by the figure is a little larger); quite honestly, I htink the simplest and most informative thing to do would be to get yer 3' timbers (or some 3'-long *anythings*) and mock it up on the site.

That way you can see, in person, how much clearance you will have around all parts, and decide which you like best.

Good luck, have fun,

Pat

 

[bid]

Not exactly sure I understand, but 4'x4' would get you close to a 6' diameter- I think-been a long time since geometry class. This is based on having 4 sides.

 

[Rosalind]

This is where my lack of geometry skills comes in. I want a structure that is about 6' in diameter in order for it to fit where I want to put it (so the bush hog will fit around it when I mow). To get that, do I need 8 sides or 6? Will the number of sides even make a difference size-wise? For the life of me I cannot figure out how to do the math!

It does not matter how many sides there are. You can have a 6 foot hexagon or a 6 foot octagon or a 6 foot dodecagon if you'd like. It only depends on how good your miter saw is. You don't have to have an even number, either--you can make a strawberry pentagon or whatever.

For a hexagon, the angle between the sides is 120 degrees. Assuming you get squared off lumber, that means you need to miter one edge to a 60 degree angle (180 - 120 = 60) and then butt one mitered edge + one uncut edge together with wood screws to get the right angle. Or cut all the edges to a 75 degree angle, like so:

__________________________________________
\_________________________________________/

For an octagon, the angle between the sides is 135 degrees. That would entail cutting 78.75 degree angles, which is a little trickier. It depends on how neat and tidy and perfect you want it to be, really...

 

[beavis]

To compute the area of a circle used the formula Api) x (radius)squared

Pi = 3.14159

Radius is half of the diameter.

So the area for a circle with 6 foot diameter would be 3.1459x 3x3 =28.3

 

[patandchickens]

Wait, am I not understanding this properly? I thought the idea was that the sides were going to be 3' long -- exactly, period -- and the question is what-sided figure will have a maximal cross-sectional dimension of 6' ?

Or am I confused?

DH points out that the maximum width ('diameter' if you can call it that for a polygon) of a hexagon with 3' sides is exactly 6'. The same for an octagon with 3' sides will be slightly larger. He just did the math and claims 7'10' which sounds too big to me but I am not going to do the math myself... suffice to say, it will be more than 6'.

Lay it out on the ground and measure, really

Pat

 

[Rosalind]

Wait, are these timbers already pre-cut to lengths of 3', 2', 1'? I thought the idea was just to have three concentric shapes of choice with a maximal distance of 6'?

But if the timbers have not yet been cut, then you cut them to whatever length for the shape you want.

 

[bid]

Take a stake and a piece of string. Pick a midpoint and stretch out the string to 3'. Use a stick or even a can of spray paint and mark out a 6' diameter (just walk in a circle around the stake). Determine how many sides you want and start building it on site. I think the parameters here are it has to be no larger than 6' in diameter.
I think Pat was spot on with the first answer. Mock it up on site. If I had to do the math it would never get built.

I THINK that more sides would give a bit more square footage, but not enough that it would make much of a difference. 6 of one, half dozen of another.

I didn't realize we were going to have math homework. I know I need to work on my spelling too.

 

[vfem]

Here's my idea for my strawberries... a three tiered raised bed. Base is 6x6 then in the middle is 4x4 and the top 2x2. All with untreated 2x4 and then stained and sealed.


http://fromseed.blogspot.com/2009/04/strawberries-in.html

 

[aquarose]

Here's my idea for my strawberries... a three tiered raised bed. Base is 6x6 then in the middle is 4x4 and the top 2x2. All with untreated 2x4 and then stained and sealed.


http://fromseed.blogspot.com/2009/04/strawberries-in.html

I was going to suggest to just make it square like this one.