Arc Array - AutoCAD

Arc Array - AutoCAD

SilverlineWoods
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Message 1 of 8

Arc Array - AutoCAD

SilverlineWoods
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Sorry if this is a stupid question since it seems like it should be easy to figure out. How can I center this array (attached) to the arc? The center panels need to be 3' 10" at the top with a 9/16" joint between and equal panel sizes on each end (around 25.8" or so). Thank you in advance!

 

@SilverlineWoods 

Your post title was modified to add the product name and version and to increase findability by @CGBenner  - Original: Arc Array

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Message 2 of 8

Kent1Cooper
Consultant
Consultant

Rather than Array, I would suggest DIVIDE with the pair of lines defined into a Block so you can use it in that command.  You would need to be careful about the orientation of the Block when at zero rotation, its insertion point, etc.  Or, use my DivideMeasurePlus.lsp with its DIV+ command [>here<], which gives you the options of:

1. including the ones at the ends of the path object [regular Divide won't]; and

2. not bothering with the Block, but placing copies of a User selection of the two lines [regular Divide requires either Points or a Block].

And whether you want a joint [both lines] at the ends of the arc, or just one end line there, will affect how you set things up -- you might need to use LENGTHEN on the ends to extend or shorten them by half a joint width.

Kent Cooper, AIA
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Message 3 of 8

SilverlineWoods
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I do appreciate the advise. Not sure I'm skilled enough to make these commands work. Messed with it a little and added the LSP but still couldn't figure it out. Ended up using a work around. Just took the dimensions of the panels I know I need and figured out the arc length. Centered that and did half at a time and was able to get the end panels to match up. Thank you so much for the reply.

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Message 4 of 8

Washingtonn
Collaborator
Collaborator

Here are the steps needed:

1. Obtain the arc length (from the properties palette) and calculate the full panel size and joint width between panels.

2. Calculate the number of full panels to use by dividing arc length by full panel size plus joint width. If a fractional panel remains, then determine the desired minimum partial panel size to be placed at start and end of the array.   You may need to reduce the full panel size by 1 in order to achieve the minimum desired partial panel size.

 

For example, a 120 inch arc length divided by 35.5625 inches (35panel +0.5625 joint) will need 3.374 panels (plus one joint width depending on the extreme end conditions used on the panels). 3 full panels and approximately 3/8 of another panel.  If the partial panel size is to be no less than 1/2 of a full panel, then reduce the panel count by one and divide 1.374 by 2 to get the equal length partial panels to be used at the ends of the array.

 

3. Start your array (PATH) as before, be sure to select the arc as the PATH to follow. The default using the MEASURE method which you will need to change to the DIVIDE method in order to adjust the start and end points of the array.

4. Enter the calculated information for the start offset, end offset and the number of items - which is the number of full panels between the partial panels.  If all is done correctly, the item spacing will be equal to the length of a full panel and one joint spacing. 

 

Message me if you need further clarification or assistance.

 

Washingtonn_0-1762734008789.png

 

 

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Message 5 of 8

j.palmeL29YX
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Accepted solution

@SilverlineWoods wrote:

... a work around. Just took the dimensions of the panels I know I need and figured out the arc length. Centered that and did half at a time and was able to get the end panels to match up.


Assuming that you will frequently need this type of geometry in the future with different dimensions, I would suggest implementing it in a dynamic block.

Attached you find a rather flexible example. The user can change the angles (the length overall and the position) and the radius of the upper arc. Additional in the properties window you can change the length of the panel and the distance (space) between them.

If you're still interested test it ...

 

 

 

Jürgen Palme
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Message 6 of 8

SilverlineWoods
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Enthusiast

I really need to learn these dynamic blocks better. I've been practicing with this one and another solution you had for me on a different post. Thank you, these have been very helpful.

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Message 7 of 8

imadHabash
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Hi,

@SilverlineWoods 

>> I really need to learn these dynamic blocks better.

You can take advantage of the presence of a forum specific to Dynamic Blocks. >> Dynamic Blocks Forum <<

 

Imad Habash

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Message 8 of 8

SilverlineWoods
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Enthusiast

Thanks for the info! Practicing more today so I will be sure to check there if I have any questions.

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