Long Forgotten Trig and Geometry

[steve-norrell]

I would like to build decorative polygon planters to enclose plastic flower pots and plastic planters. While I located a web site that gives the angles for cutting the sides of various polygons, I need information on how long each side has to be to enclose a circle of specified diameter.

For example, if the flower pot is 18 inches in diameter, how long does each side of a hexagon or octagon have to be to enclose it?

Hopefully someone can help with the math, or better yet, point me to a data source.

Thanks, SN

[Turaj]

If I am not mistaken the formula is as follow. I hope it makes sense!!

L = 2 * [tan (360 / (2 * N))] * R

"N" is the number of sides (Octagon = 8) and "R" is the radius of your pot (18" diameter = 9" radius)

Remember that this is the inside length of your pieces. The outside length would depend on the width of your piece. Just add the width to the radius (in the formula) and you should have the outside measurements.

I am sure there are bunch of tables someplace. Maybe someone else can point us to it.

[dlminehart]

Turaj, please tell me you looked that up somewhere, rather than remembering it or deriving it on the spot! And, quick, what's the 3D counterpart? I.e., to put an cube or dodecahedron around a sphere?

[steve-norrell]

More Help Needed

If I am not mistaken the formula is as follow. I hope it makes sense!!

L = 2 * [tan (360 / (2 * N))] * R

"N" is the number of sides (Octagon = 8) and "R" is the radius of your pot (18" diameter = 9" radius).

My old brain must be missing something. When I apply the formula, I get all negative numbers for hexagons and all positive numbers for octagons. Example: a hexagon to surround a 12 inch pot would have a side that was -76.86 in. long; an octagon for the same size pot would have a side that 6.69 in.

What am I doing wrong.

[LCHIEN]

My old brain must be missing something. When I apply the formula, I get all negative numbers for hexagons and all positive numbers for octagons. Example: a hexagon to surround a 12 inch pot would have a side that was -76.86 in. long; an octagon for the same size pot would have a side that 6.69 in.

What am I doing wrong.

I can't vouch for the equation but a frequent mistake in trig calculations leading to vastly varying answers is swap radians and degrees.
Make sure you are using a calculator with the angle mode for trig set to degrees and not radians.


A quick sanity check:

As N becomes large
the width should approach the circumference/N
e.g. 2 x pi x R / N
and the width should be larger for smaller N.

[crokett]

My professional life has taught me to ask the subject matter expert. I have a buddy who is a civil engineer, so I would just tell him what I want to do and in a day or so he emails back an answer.

And Loring, I'm disappointed. I figured if anybody could answer this, you could.

[jziegler]

Give me until the end of the day, and if I don't get too busy, I may just try to derive something. I bet Loring just didn't have the time. And the civil engineer is probably a good bet.

Jim

Alright, here it is:

side (inside length) = 2 x R / ( tan (90 - 180 / n ) )

where R is the radius, and n is the number of sides. Anyone care to check me?

Further edit:

This isn't right, the original may be. I messed up somewhere along the line. Or, I just did the math wrong (looks ok now)

[Turaj]

Turaj, please tell me you looked that up somewhere, rather than remembering it or deriving it on the spot!

Soryy, but I have never memorized those things!! Had to draw a circle and work with the triangles!
I need more time for 3D and maybe a beer

[Turaj]

.. I get all negative numbers for hexagons and all positive numbers for octagons. Example: a hexagon to surround a 12 inch pot ...

Not sure what is happening but this is what I get:

For hexagon: N=6,
tan (360/(2*6)) = tan (30) = 0.577
L = 2*6*0.577 = 6.98

for an octagon: N=8
tan (360 / (2*8)) = tan (22.5) = 0.414
L = 2 * 6 * 0.414 = 4.97

Am I doing something wrong?

[jziegler]

Those look like the same number that I got, it all looks good. We both got the same answer, just looked at different angles on the same triangle. A quick review of the trig definitions show that our equations are equivalent.

Jim

Not sure what is happening but this is what I get:

For hexagon: N=6,
tan (360/(2*6)) = tan (30) = 0.577
L = 2*6*0.577 = 6.98

for an octagon: N=8
tan (360 / (2*8)) = tan (22.5) = 0.414
L = 2 * 6 * 0.414 = 4.97

Am I doing something wrong?

[steve-norrell]

Needing help with math . . .

Thanks for all the input from folks a lot smarter than I.

Its been more than fifty years since I took Geometry and trig in school so I will have to keep trying.

But then, its really is never to late to learn.

Thanks all!

[bigsteel15]

Autocad does it for me.
Your 12" pot is as follows
Hexagon = 6.9282
Octagon = 4.9706
So same results, but just a shade off on the hexagon.

[steve-norrell]

Success

Today is a good day in the far north. Thanks for all the help. The important clue was radians vs degrees.

Here is a table that gives a few data for different diameters (diameters on the left side) and polygons (number of sides across the top). Hope it is useful to some, like me, who prefer to look up a table than do the math.

[sacherjj]

Not sure what is happening but this is what I get:

For hexagon: N=6,
tan (360/(2*6)) = tan (30) = 0.577
L = 2*6*0.577 = 6.98

for an octagon: N=8
tan (360 / (2*8)) = tan (22.5) = 0.414
L = 2 * 6 * 0.414 = 4.97

Am I doing something wrong?

A Hexagon should have edge lengths the same as radius, because it is made up of 60 degree triangles.

The Radius if the circle is the Hypotenuse of a right trigangle with an opposite side equal to 1/2 edge length. This triangle has an angle of 360/2n (theta).

tan(theta) = l/2r
tan(360/2n) = l/2r
L = 2*r*tan(360/2n)

That formula seems right, but I would expect a side on a Hexagon to be 6"...

Edit: WAIT. I went along with Tan from above, but tangent is Opposite over Adjacent. We want Sine: Opposite of Hypotenuse. SOH-COA-TOA

L = 2*r*sin(360/2n)

That is the right formula. It gives me 6" for a 6" radius circle and a hexagon.

[LCHIEN]

A Hexagon should have edge lengths the same as radius, because it is made up of 60 degree triangles.

The Radius if the circle is the Hypotenuse of a right trigangle with an opposite side equal to 1/2 edge length. This triangle has an angle of 360/2n (theta).

tan(theta) = l/2r
tan(360/2n) = l/2r
L = 2*r*tan(360/2n)

That formula seems right, but I would expect a side on a Hexagon to be 6"...

Edit: WAIT. I went along with Tan from above, but tangent is Opposite over Adjacent. We want Sine: Opposite of Hypotenuse. SOH-COA-TOA

L = 2*r*sin(360/2n)

That is the right formula. It gives me 6" for a 6" radius circle and a hexagon.

You're not correct in this instance.
The round object needs to fit within the hexagon which means that the
radius applies not at the points of the hexagon but at the center of the edges of the hexagon.