# Regular Polygon with alternating side lenghts - a challenge, or a simple thing ?

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This is more a mathematical question, but maybe someone knows a way to to draw this:

I want to draw  a closed regular polygon of 15 edges with alternating sides of 1200mm and 750mm. All I know is these two legths. The question is how to draw it, and  is there a unique radius that matches the outer circle ? If yes, maybe someone knows how to calculate this radius mathematically ? ( I gut stuck in trigonometrical theorems in the end...)

7 hours ago, halfcouple said:

I want to draw  a closed regular polygon of 15 edges with alternating sides of 1200mm and 750mm.

Isn't it going to have an even number of sides (14 in your drawing)? I would figure it out using percentage of the full circle angle of 360 degrees and the geometry created. So a single pair segment would be 360/14 or 25.714 degrees. You could then figure out the 1200mm portion and the 750mm portion of the angle (61.5% and 38.5%), draw one of those angles, bisect it, offset two guides 1200 and 750 apart around a centreline, and where the guides and angle lines intersect draw a segment to find the circle.

Hopefully that makes sense. If I have a little more time later I'll draw it out.

Kevin

Fifteen is the correct number.  There will need to be eight of one length and seven of the other length and you will need to nominate which of the two length there is eight of.

You might be better off posting your query on a geometry/trigonometry maths query website.

Here's a visual representation of my geometric solution -

Kevin

Alternating Polygon.vwx

Kevin, I have to challenge your solution. Your short lines are 751.253. Thats not correct. 🙂

The radius is 2322.5267mm, according to my calculations.

Edited by Stephan Moenninghoff

Here's the file. The formula is based on the knowledge that the angle between all edges is the same.

So the angle = 1/ Edges / (Edges/2 -1) / 360

If you rearrange that you get the angle. From there on you construct your polygon using VWX.

Edited by Stephan Moenninghoff

4 hours ago, Stephan Moenninghoff said:

Kevin, I have to challenge your solution. Your short lines are 751.253. Thats not correct. 🙂

Yes, there is a margin of error. But its likely within construction tolerances and VW's math method.

Edit. my example also has too many segments. I treated each pair as a segment instead of each individual side, so my circle is twice the size needed.

KM

Edited by Kevin McAllister
Error explanation

Just now, Kevin McAllister said:

Yes, there is a margin of error. But its likely within construction tolerances and VW's math method.

KM

Not if you calculate correctly, pal 🙂

When I open your file and measure the alternating segments, they are 866 and 1084, not 750 and 1200?

KM

22 hours ago, Kevin McAllister said:

When I open your file and measure the alternating segments, they are 866 and 1084, not 750 and 1200?

KM

Edited by Stephan Moenninghoff

Hi Guys,

thanks so much for the help ! Think I never got the solution alone. I need to do this with different side lengths so 1200 and 750 are only an example. So  the reproduceable method is to calculate the "double angle" Edges / (Edges/2 -1) / 360 and split it up by the proportion of the side lenghts. Two parallel lines for each side cross the angle at the desired radius.

I will check that for other length combinations.

3 minutes ago, halfcouple said:

Hi Guys,

thanks so much for the help ! Think I never got the solution alone. I need to do this with different side lengths so 1200 and 750 are only an example. So  the reproduceable method is to calculate the "double angle" Edges / (Edges/2 -1) / 360 and split it up by the proportion of the side lenghts. Two parallel lines for each side cross the angle at the desired radius.

I will check that for other length combinations.

It works with any number of edges, as long as there are always two of each. I'll make you a quick Marionette. After my supper.

Edited by Stephan Moenninghoff

Here you go. Good old Marionette. Great for this kind of stuff.

This formula calculates the total of all the interior angles of a polygon -

total angle = (n-2)(180)

where n is the number of sides.

Since all the individual interior angles will be the same in these examples, you can calculate a single interior angle using -

individual angle = total angle/n

where n is again the number of sides.

KM

23 minutes ago, Kevin McAllister said:

This formula calculates the total of all the interior angles of a polygon -

total angle = (n-2)(180)﻿

where n is the number of sides.

Since all the individual interior angles will be the same in these examples, you can calculate a single interior angle using -

individual angle = total angle/n

where n is again the number of sides.

KM

Yep, that's the same. (n-2)(180)/n.

Edited by Stephan Moenninghoff

On 8/3/2019 at 4:19 PM, Kevin McAllister said:

Isn't it going to have an even number of sides (14 in your drawing)?

On 8/4/2019 at 4:32 AM, mike m oz said:

Fifteen is the correct number.  There will need to be eight of one length and seven of the other length and you will need to nominate which of the two length there is eight of.

I don't understand this - if there are eight of one length and seven of the other, then you end up with two of the same length next to each other.

3 hours ago, line-weight said:

I don't understand this - if there are eight of one length and seven of the other, then you end up with two of the same length next to each other.

I don't think that an odd number makes sense here. The offered solution was for an even number of sides.

Line-weight, I didn't think through the problem before posting.  Stephan is correct in that the number of sides needs to be even.

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