# Illustrator

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## Can you calculate area of an object to predict balance point?

### Nov 5, 2009 12:31 PM

I'm working on a mobile that has many irregular shapes that need to hang at a precise angle – each has text that needs to be parallel to the ground (the objects will be cut from 1/4-inch aluminum sheet).  Is there a way to calculate area in a 2D Illustrator object, and then find the vertical midpoint that would divide the object into 2 parts of equal area?  An example object is attached.

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Nov 5, 2009 12:52 PM   in reply to Peter Agoos

Untested:

Open the file as a grayscale image in Photoshop. Add a mask and fill the mask with white on the right and black on the left, no gray. Show Historgram. Invert the mask. If the histogram changes, then move the terminator in the mask (transition from white to black) one way or another. Invert mask. Keep this up until the Historgram does not change when the mask is inverted. That’s your balance point.

I would be surprised if the fabricator does not have some way of solving the problem.

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Nov 5, 2009 3:14 PM   in reply to Peter Agoos

Peter,

This is not a matter of simply having the same area on both sides, as I believe the PS Histogram way determines.

Am I right, Scott?

If not, the PS Histogram is definitely the way to go.

What you need is the centre of gravity, and that may be found in either of the following ways:

A) Using the Knife Tool and possibly the Scissors Tool (a bit rough):

A1) Create a (large) copy and cut it out with the Scissors Tool and/or the Knife Tool;

A2) Balance the copy on the egde of the Knife Tool (perpendicular to the line over NEANDERTHAL or in two different directions);

A3) Draw a vertical perpendicular to the line over NEANDERTHAL (through the intersection), calculate the corresponding position on the original size, and place the hole somewhere on the resulting line.

B) Using the Scissors Tool and/or the Knife Tool, the Pin Tool, and the Plumb Tool (a bit more tedious):

B1) Create a (large) copy and cut it out with the Scissors Tool or the Knife Tool;

B2) Pin it upside down at one end of the line over NEANDERTHAL and use the Plumb (line) Tool (hanging from the Pin Tool) to mark the vertical;

B3) Repeat A2) at the other end;

B4) Draw a vertical (perpendicular to the line) through the intersection, calculate the corresponding position on the original size, and place the hole somewhere on the resulting line.

C) With Illy, obviously working on a copy, getting rid of the Clipping Mask and working on the Compound Path (very tedious):

C1) Move the lower left corner of the Compound Path to X = Y = 0;

C2) Determine the simple area, using something like Filter>Telegraphics>Path Area (if you have it) or something similar;

C3) Distort the Compound Path so that the height grows lenarly from 0 to W, possibly by:

C3a) Object>Add Anchor Points until the shape can be well represented by Corner Points;

C3b) Convert all Smooth Anchor Points to Corner Points;

C3c) Distort the Compound Path so that the Y value of each Anchor Point is multiplied by X/W, X being its X value;

C4) Determine the area of the distorted Compound Path and multiply it by W*W (the square of W);

C5) Divide C4) by C2).

D) Let the aluminium cutter do it, possibly more or less as A) or B), or using the Scale Tool.

It would be nice if C3) could be accomplished by the Free Distort Tool, Object>Envelope Distort, or the like, but alas, at least in 10. This calls for a script/action to run through the Anchor Points, maybe something for Kurt, or James.

Kurt?

James?

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Nov 5, 2009 3:21 PM   in reply to Jacob Bugge

Jacob Bugge wrote:

Peter,

This is not a matter of simply having the same area on both sides, as I believe the PS Histogram way determines.

Am I right, Scott?

James?

I don’t think you grok. The Histogram measures the distribution of pixels in the image based on intensity. If the histogram does not change when the image mask is inverted then there are just as many black pixels in the masked area as there are in the unmasked area. Assuming each pixel corresponds to the same mass of Al in the cast piece, this means there’s equal mass on either side of the mask’s terminator.

I just tried it and it’s not helping as much as I expected. That’s because the image is basically black and white, so there are only two values to examine in the graph, which is difficult to get right. So I first flattened the image, then turned Background back into a layer. Now the Histogram is more helpful because it’s not thrown off by transparent pixels. Rather than looking at the graph, I looked at the Mean and Std Dev values. Here they are close enough for jazz.

If I’m right, then you want to drill somewhere along the line in this image…

If you prefer the hole more centred on the cave man’s head, then either edit the shape or move the text. Since you need to move the centre of mass left, you must add mass to the left or remove mass from the right. Shifting the text left should solve it.

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Nov 6, 2009 2:55 AM   in reply to Peter Agoos

I have used the following method to find the balancing point of an irregular pendant, but it takes a bit of trial and error.

1. Shape must be a single closed path. Copy shape into a new document. Leave selected.

2. Open debugging panel (Cmd-Shift-Option-F12) and find path and area as directed by Teri Pettit in this discussion.

3. Make note of the area of original shape.

4. You want to find half this area, so divide this number by 2. Make a note of this number.

5. Draw a rectangle roughly overlapping half of shape along vertical axis.

6. Pathfinder > minus front.

7. In debugging panel, display area for the new shape. Don't worry if area is expressed as a negative number. This reflects path-winding direction.

8. If the new shape is not desired area, undo pathfinder operation, move rectangle and try again. Use vertical guides will help keep track of pathfinder operations.

9. When you find the shape that has half the area of the original shape, you will have found the balancing point of your manufactured object.

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Nov 6, 2009 4:41 AM   in reply to Peter Agoos

Peter,

Here is another possibility, perhaps somewhat easier than C, and quite accurate:

E) Divide a copy of the artwork into vertical parts and calculate the centre of gravity from that (requires some cleanup), here with 10 parts (the whole thing placed with the lower left corner at X = Y = 0):

E1) Create a straight vertical path at 10% width from the left and copy it 8 times;

E2) Select the lot and Pathfinder>Divide;

E3) Cleanup and group subparts into the 10 parts;

E4) For each part Filter>Telegraphics>Patharea or similar, used first on the whole part and then on empty subparts of Compound Paths, in a spreadsheet enter part # (from 1 - 10), total area, area of empty subpart(s), and calculate the resulting area being the former minus twice the latter (Patharea counts the empty subpart twice);

E5) For each part multiply the resulting area by the part # minus 0.5 to get the weighed area;

E6) Calculate the sum of E4 and E5 for all parts, divide the latter by the former, divide by 10, and multiply by total width.

E6) is the distance from the left side to the centre of gravity.

As you can see, it is a bit left of the median of the area as seen in post #4.

I get W = 54.455, H = 88.136, and CG = 28.829 from the left, corresponding to 5.294 (52.94%) in the spreadsheet. I can upload the AI/PDF and ODF files if you wish.

You may try both and see what works.

Sorry for not grokkng, Scott.

The difference between median and centre of gravity is much more clear in this case of identical rectangles, where the median is at the meeting of the two whereas the centre of gravity is quite a bit further to the right.

This is easy to try.

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Nov 6, 2009 5:52 AM   in reply to Peter Agoos

This page describes a Javascript that finds the centroids of selected paths.

It's esentially the same thing as selecting all the anchorPoints of a path and averaging their location. (Object>Path>Average). Of course, the more complicated the shape, the more anchorpoints you need for accuracy--especially along curves. So use Object>Path>AddAnchorPoints or Filter>Distort>Roughen (with Size set to zero) judicously.

The Centroid script should work for simple straight-sided paths, like  those above, and may work for paths like that drawn by Judy with judicously-added points to compensate the "mass" enclosed by curved segments. Compound paths or highly irregular paths would have to be sliced up and treated individually. Probably not practical for the sample drawing.

One could devise a similar script based on this general algorithm:

1. Use CreateObjectMosaic to create same-size black squares of the desired unit (sized as what seems "close enough" for accuracy).

2. Write a function that averages the center coordinates of all those black squares. That should give you the centroid.

Frankly, I'd just cut the thing and find the hang point on the first sample, then use that as a drill template for the other copies. Depending on the material used, you might also add a weight at the bottom, below the hang point. It's not a simple matter of just dividing the mass in two. You have to have the same righting moment (leverage) on each side. Jacob's two rectangles depict this principle; but his solution above it does not really address it.

If you want to find area of a path, the Label Scripts page, accessible from the same link, has a script for that.

JET

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Nov 6, 2009 5:28 AM   in reply to Scott Falkner

Scott, Judy:

The histogram method will not work, because balance is a question of mass location, not just quantity. Position of the area relative to the fulcrum affects leverage. Balance two bricks on a teeter-totter. They only balance if they are equidistant from the fulcrum. Move one brick closer to the fulcrum and the other brick becomes "heavier". Scott's historgram idea would only tell you if the two bricks are the same size. Raise the arm of the neanderthal and the number of black pixels does not change. But it changes his balance. Judy's method also only tries to ensure the same area on each side. Position of the mass is as important as amount of mass on each side.

JET

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Nov 6, 2009 6:25 AM   in reply to Peter Agoos

I think this method will give you a servicable approximation. Test it on a couple of paper-cutouts and see:

JET

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Nov 6, 2009 10:30 AM   in reply to JETalmage

JETalmage wrote:

Scott, Judy:

The histogram method will not work, because balance is a question of mass location, not just quantity. Position of the area relative to the fulcrum affects leverage.

JET, thank goodness you've weighed in on this topic.

I lost sleep last night because I knew I was wrong in my post. I knew I had to take into consideration the cantilever effect, but couldn't figure out how to calculate it in illustration software. Aaargh!

Peter Agoos, may I make a design suggestion? If you put the 'NEANDERTHAL' lettering on a curved or wavy path, you won't have to worry so much about getting the piece to hang perfectly level.

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Nov 6, 2009 11:55 AM   in reply to JETalmage

James,

I later thought about torque and how that might affect the balance, but I can’t think of a method to compensate for that. I think the only way to get this is through trial and error with the fabricator. Their software might be advanced enough to consider mass and torque to find a balance point. Or they might need to cast/cut some samples and test.

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Nov 6, 2009 11:55 AM   in reply to Peter Agoos

Export the shape to a CAD program. Area calculation and such are simple there. The others have offered some good ideas, though.

Mylenium

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Nov 8, 2009 5:41 AM   in reply to Peter Agoos
I think the only way to get this is through trial and error with the fabricator.

Export the shape to a CAD program. Area calculation and such are simple there.

Path area is calculated in Illustrator; the interface just doesn't provide access to it. Two methods for accessing it have already been suggested: The debugging panel, as mentioned by Judy and the Javascript property as mentioned by me.

As stated, I believe a little simple testing with a couple of paper cutouts will confirm the method suggested in post 9. The assumption upon which it is based is that Illustrator's Center Align functions, when an "anchor" object has not been designated, do not center within the bounds of the selection, but average the centers of the objects (just as the Object>Path>Average... command averages the positions of selected anchorPoints). You can see the suggestion of this with a simple test:

The method, spelled out more explicitly:

1. Scale the black-filled drawing to any practical size you can print. Select it.

2. Copy.

3. Object>Lock>Selection.

4. PasteInFront.

5. Object>Rasterize. Screen resolution should be fine.

6. Filter>Create>ObjectMosaic. Result: Gray. Key in a width (ex: 100). Click the Use Ratio button. (The dialog will calculate the proportional number for height, yielding square rectangles.)

7. Ungroup the results.

8. Select one of the white squares. Select>Same>FillColor. Delete.

9. Select>All.

10. Center Align Horizontally. Center Align Vertically.

11. Attributes Panel: Click the Show Center button.

12. LineTool: Mousedown on the center marker of the selection, ShiftDrag upward. A hangpoint anywhere on that line should balance the cutout perpendicular to its bottom, as oriented on the page.

Of course, the above assumes all you want to do is solve the stated problem: Find a balancing hangpoint so that the text at the bottom hangs horizontal for a single hanging object. But you also mentioned this is going to be a hanging mobile. Most mobiles (at least interesting ones) are not just one generation deep, but "branch" like an upside-down tree, with additional elements hanging from either side of their parent element. Calculating hangpoints for that, of course, is going to be more involved. But I suspect it could be constructed by an elaboration of the same method. It would be an interesting experiment to try.

JET

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Nov 8, 2009 10:59 AM   in reply to JETalmage

JETalmage wrote:

Most mobiles (at least interesting ones) are not just one generation deep, but "branch" like an upside-down tree, with additional elements hanging from either side of their parent element. Calculating hangpoints for that, of course, is going to be more involved. But I suspect it could be constructed by an elaboration of the same method. It would be an interesting experiment to try.

JET

I disagree. Each item will still hang from its own hangpoint, and only gravity and its own shape will affect how level it is. If our Neanderthal is at one end of a stick and a **** Habilis is at the other end, each will still be suspended from its own point and sit level, or not, without respect to the other form.

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Nov 8, 2009 1:44 PM   in reply to Scott Falkner

Thank you for your solution JET

I usually guess.

To balance multiple items Cadtools by Hotdoor does a great job calculating areas.

Cheers

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Nov 8, 2009 1:48 PM   in reply to Scott Falkner

Scott Falkner wrote:

a **** Habilis is...

Wow. F U C K.

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Nov 8, 2009 2:10 PM   in reply to Scott Falkner
Each item will still hang from its own hangpoint

Yes, each will hang from its own hangpoint. But if any also serves as the "parent" hanger for other "children", then the collective weights of those children (and their descendents) will, of course, change weight distribution on the "parent".

Think of an equilateral triangle hanging from one of its corners. Now thing of additional triangles (or series of triangles) hanging from the other corners.

JET

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Nov 8, 2009 2:35 PM   in reply to JETalmage

JETalmage wrote:

Each item will still hang from its own hangpoint

Yes, each will hang from its own hangpoint. But if any also serves as the "parent" hanger for other "children", then the collective weights of those children (and their descendents) will, of course, change weight distribution on the "parent".

I hadn’t pictured that. I pictured each item at the bottom of its own string, hanging from a more complex arrangement of stick. I still do.

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Jun 6, 2010 5:36 AM   in reply to Peter Agoos

Peter, and others,

I believe it is time for a clarification.

Unfortunately, the very assumption that the centroid of an arbitrary path may be determined by adding Anchor Points, through Object>Path>Add Anchor Points or through Filter>ZigZag (size 0), and then using Object>Path>Average is fundamentally flawed.

Generally, the centroid can only be determined through a calculation based upon the first/static moment of area; and the calculation cannot be based upon the perimeter or specific points upon it, apart from certain special cases with regular shapes and no superfluous Anchor Points, such as triangles and polygons with a certain symmetry.

Basically, a full integration is called for, but an approximation in the form of a summation with a limited number of segments may yield quite accurate results.

In connexion with balance, it is sufficient to determine the horizontal postion of the vertical centreline through the centroid.

As shown in post #6, there is a fundamental difference between median and centreline, and calculation E) showed one rather laborious way of determining the centreline, along with the resulting figures and the balanced Neanderthal.

Since then, being eager to help as always, Illy has offered to do most of the calculations herself (with the help of Telegraphics>Patharea or similar), so now it is possible to present an improved way, almost exclusively based upon drawing, with just a few simple calculations:

F) Determination of centreline of path with the width W using 10 horizontal segments (the whole thing may be placed with the lower  left corner of the Bounding Box at X = Y = 0):

F1) Create a copy of the artwork in front;

F2) Create a straight vertical path at the left side;

F3) Create 9 copies of 9 each moved by W/10, and Object>Compound Path>Make (only the copies);

F4) Select F1) and F3), Pathfinder>Divide, and delete possible empty paths, to form segments N = 1 - 10;

F5) Select each segment and multiply its H value by (N-0.5)/10 (in other words 0.05, 0.15, 0.25, ... 0.95);

F6) Determine the area of all 10 segments, using Telegraphics>Patharea or similar, subtracting twice the area of possible empty parts of compound paths  if needed;

F7) Determine the area of the original artwork, using Telegraphics>Patharea or similar, subtracting twice the area of possible empty parts of compound paths if needed;

F7) Divide F6) by F7) and multiply by W: this is the distance from the left side to the centreline; move F2) to the front at that distance to form this centreline.

Obviously, this may be repeated at a right angle (or another suitable angle if any) to determine another centreline and thus the actual centroid.

And maybe the whole procedure may be turned into an action/script (maybe even a double action/script for a double procedure).

The following images show rather simple paths, created in normal ways and without superfluous Anchor Points, that have known positions of centroids/centrelines. As it appears, the use of Object>Path>Average may yield extreme errors, whereas the F) way, even with a rather limited number of segments, yields little or no error.

The first path is created from a rectangle, adding/subtracting semicircles along the left side. It is evident that the added and subtracted semicircles outbalance one another so that the centreline is unchanged; the centroid is only raised a little so that it is right above the intersection between the diagonals.

The second path is created from a rectangle, subtracting squares close to the  left side, thus forming a Compound Path. It is evident that the subtracted squares  move the centreline/centroid to the right; a simple calculation shows that the centroid  is 2.86% to the right of the intersection  between the diagonals.

However, as it appears, the greater number of Anchor Points at the left side results in an Object>Path>Average that is quite close to the left side, very far from the centroid, about two thirds from the centroid to the left side,  with an error of about 35% of the total width. And, as it appears, the use of Object>Path>Add Anchor Points or of Filter>Distort>Zigzag hardly makes any change regardless of the chosen numbers because neither changes the uneven distribution.

In contrast, the centreline found using F) is closely coinciding with the centroid, with an error of only 0.13% of the total width and with no error (0.00%) respectively.

It may be noted that even if Object>Path>Add Anchor Points or of Filter>Distort>Zigzag were replaced by some way to ensure an even distribution of Anchor Points along the perimeter there would still be a considerable error because the distrubution of the length of the perimeter itself round the path is uneven.

The third and fourth paths show the importance of avoiding superfluous Anchor Points which may result in considerable errors when using Object>Path>Average even for triangles and symmetrical polygons.

In both cases, a triangle and a rectangle, two versions occur, namely a normal one and one with an additional Anchor Point added at the (lower) left corner, corresponding to a joining of open paths.

As it appears, Object>Path>Average is accurate for the normal versions, but is displaced along the diagonal towards the extra Anchor Point.

In any case, the centreline found using F) is closely coinciding with the centroid, with  an error of only 0.17% of the total width and with no error (0.00%)  respectively.

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Jun 6, 2010 9:40 AM   in reply to Jacob Bugge

Thanks for clearing that up, Jacob.

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Jun 6, 2010 5:02 PM   in reply to Peter Agoos

I often find the best way to do this sort of thing is by pasting a printout of your shape onto a piece of stiffish card, cutting it out as accurately as you can be bothered and then balancing it on the edge of a ruler. In this case the ruler will need to be perpendicular to the lettering. Punch a hole somwhere along the line of the ruler and your Neanderthal chappie will hang straight. I have a feeling that Illie won't help you much. (By the way, shapes of equal area do not necessarily balance on a centre line. Imagine an L-shaped object with legs of equal area; it certainly won't hang straight just by dividing the area in half.)

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Mar 15, 2012 5:09 AM   in reply to Peter Agoos

1. Draw a vertical line to divide the image into two pieces roughly equal in area - you can eyeball this.

2. Check the area of each piece and subtract the smaller from the larger (x). See above methods for obtaining area.

3. Measure the  length of the dividing line (t).

4. Divide the area from step 2 (x) by the length of the line (t) and divide the result by two. (x/t/2)

5. Move the dividing line into the part with the greater area the distance obtained in step 4. You balance point will be on this line.

This works well no matter what the shape providing the top and bottom edges are approximately parrallel where the dividing line intersects them for a distance about equal to the number obtained in step 4. On an object 100 mm wide it should not be necessary to move the line more than 5 mm.

If your top and bottom edges are not parrallel, run the procedure again using the vertical line obtaied from the first run as your starting point,  it will then come out to within fractions of whatever measuring unit you're using.

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Mar 15, 2012 5:10 AM   in reply to Peter Agoos

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Mar 15, 2012 7:05 AM   in reply to Jacob Bugge

Yep. I thought I'd seen it before.

The trouble with the whole (hole) thing is that the balance point may be outside the area of the shape.

For example if your object is L-shaped. You won't be able to hang it up properly because the place where the hole should be is outside the shape.

Cut the thing out and balance it. It's the only way to be asolutely sure, especially if you are not brilliant at maths.

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Mar 15, 2012 11:47 PM   in reply to steve fairbairn

If it is possible to draw a line on an object such that the mass is distributed evenly on either side of the line, the object will hang with the line oriented vertically if a hanging hole is positioned reasonably close to one the end of the line (normally the end envisaged as being 'top')  no matter what shape the object - including L shapes; an L shape with a horizontal stroke thicker than the vertical stroke will hang straight.

If it is not possible to draw such a line on an object you have a problem.

ps - check out Alexander Calder's mobiles

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Mar 16, 2012 3:10 AM   in reply to RogerPaine

an L shape with a horizontal stroke thicker than the vertical stroke will hang straight.

How do you work that one out?

Where are you going to put the hole?

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Mar 16, 2012 4:49 AM   in reply to RogerPaine

Roger,

It may work well for an almost symmetrical shape, but if you try it on a shape like the one shown at the bottom of the sixth post, a rather conspicuous rotation is unavoidable. And the L case mentioned by Steve is even worse:

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Mar 16, 2012 7:47 AM   in reply to Peter Agoos

Amazing thread. Some great ideas here. Peter, where does your project stand? Why on earth can't you have more than one mounting point? Why all this work to find the perfect balance point? Surely, there's a hardware (mounting hardware, etc) solution you could use.

---

Admitting I haven't read the whole thread, I want to volunteer two suggestions, one non-digital, and one digital.

(A) My initial thought was, if push comes to shove you can just take the analog approach: print the image on card stock, as large as possible without it becoming flimsy, cut it out (yeah I know . . .), and suspend it by a tack you pierce it with and try and try again until it hangs right.

Though, multiple peircings in the same palce will wear the paper out - Instead, you could make a cardboard cut-out - To do it: print the design large and place it on cardbaord, tape it down so it cannot move, get a pin or thumbtack and repeatedly poke holes along all edges. Remove paper. Now razor cut out the shape outlined in little holes. Use that thick cardboard for balance-testing instead of the thin flimsy paper.

-OR-

(B) Use Photoshop's Average Blur. Import your design to a new flattened white document (which is default). Make it BIG. Crop to the edges, so if you were too align it to the canvas any way, it wouldn't move at all. (this method doesn't address the cantilever effect but your design doesn't have much of that going on

1) Flatten document.

2) Create a single vertical guide where you suspect the balance point to be, just make a quick guess; doesn't matter

3) Select (rectangular marguee) everything on the left side of the vertical guide and do an Average Blur. Also select everything on the right side now (just invert your selection) and Average Blur again.

How close are the two shades of grey?

If not satisifed, proceed to Step #4.

4) Go back in your History Panel to just before you moved the guide last time, in order to try again.

5) Move the vert guide you're using to define your left and right selections until the Average Blurs produce the same exact shade of grey on both sides of the guide.

How close are the two shades of grey?

If satisfied, proceed to Step #6.

If not satisifed, go back to Step #4.

6) Once your greys match, go back in your History Panel to just after you moved the guide last time, which is the correct place for the vertical guide. (Just undo the Average Blurring you did, after placing the guide in the right spot.)

7) Repeat the guide creation and Average Blurring steps, but use a horizontal guide this time and make your selections on top and bottom of it, like you did on left and right sides of your initial vertical guide.

Finally, you'll have a vert and horiz guide identifying the 2D location of your balance point. Too bad this method doesn't account for cantilever weight and who knows what other physics issues, but it ought to get you darn close.

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Mar 16, 2012 7:55 AM   in reply to Mathias17

Your average blur thing won't work unless the shapes are fairly normal.

If they have thin asymmetrical projections or branches sticking out in strange places, these will disappear or blur out before the thicker parts of the shape, thereby upsetting the balance..

The analog method (sticking your shape onto card and clipping it out with scissors) is far the best way.

Balance it on a ruler edge and there's not much guesswork involved.

I have used this method many times and it's petty much foolproof.

That is to say, if a halfwit like me can do it, so can you :-)

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Mar 16, 2012 8:10 AM   in reply to Peter Agoos

@Peter - Oh heck, you did a bunch of them. Lookin' good! Smart method

@Steve - Everything "blurs out" with an Average Blur - it produces a single solid color. It's not really a "blur" per the conventional definition of the word.

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Mar 16, 2012 1:38 PM   in reply to Peter Agoos

It seems that Steve and I could hardly agree more, see the Knife Tool/Scissors Tool solution A) in the third post.

Many years ago, when the European agricultural subsidies became based on acreage, or rather hectareage, it was necessary to be quite sure about the size of each and every irregular field in the hills, and walking the fathom was a bit uncertain, especially where it was impossible to see from one end of the field to the other; both lengths and angles became a bit woolly.

In the end, cutting out the fields from prints of aerial photographs with the Scissors Tool and laying them on the milligram digital scale proved the most accurate way, finally confirmed when the inspector walked every field with GPS equipment, looking rather like a smurf (two years earlier, three other inspectors had inspected a few folds with a total station).

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Mar 16, 2012 2:27 PM   in reply to Peter Agoos

just for the record...this might be helpful...for someone that knows Flash. The WCK is a physics engine I think, it's capable of doing pretty neat stuff...like balancing a Neanderthal figure.

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Mar 20, 2012 1:12 AM   in reply to steve fairbairn

close to the vertical on the top edge of the horizontal - allowing a fairly loose interpretation of L shape

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Mar 20, 2012 1:31 AM   in reply to Jacob Bugge

It does not have to be symmetrical - I've cut some out and tried it - including L shapes - they all work allowing that the centre of gravity is below the hole.

A parallelogram will hang 'straight' even though one corner is significantly higher than the hole - an L shape can be thought of as a parallelogram with a few bits missing, as can any surface the edge of which can be described with a simple closed curve.

If your shape is such that the centre of gravity along the vertical line obtained as described is above the upper edge of the surface it is clearly not possible to hang it straight by any single string method: If it is possible to correctly hang the object by a single string, my method works.

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Mar 20, 2012 2:12 AM   in reply to RogerPaine

The only trouble is that you can't draw irregular shapes of equal predetermined area easily in Illustrator.

Finding the centre of gravity and drawing that vertical line through it involves so much trial and error that it is much quicker and easier to cut out the shape and find its centre of gravity manually.

Been there many times.

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Mar 20, 2012 6:52 PM   in reply to steve fairbairn

I have to disagree - no trial and error here other than the first guess and the result is 99.97% accurate

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