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Creating Pointy Shapes with the Star Tool

User’s Guide
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Last update
May 25, 2023 02:42:30 AM GMT

1. Illustrator

User Guide

2. Illustrator basics
   1. Introduction to Illustrator
      1. What’s new in Illustrator
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      3. Illustrator

      system requirements

      4. Illustrator for Apple Silicon
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3. Illustrator on iPad
   1. Introducing Illustrator on iPad
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   4. Selecting and organizing objects
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      5. Cloud documents in Illustrator | Frequently Asked Questions
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5. Adding and editing content
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      10. Correction of contour segments
      11. Making a flower in five easy steps
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   2. 3D objects and materials
      1. Learn more about 3D effects in Illustrator
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   3. Color
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      12. Tab
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      15. Fonts | Frequently Asked Questions and Troubleshooting Tips
      16. Creating a 3D text effect
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   8. Create special effects
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6. Import, export and save
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   2. Creative Cloud Libraries in Illustrator
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7. Printing
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8. Automating tasks
   1. Combining data using the Variables panel
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9. Troubleshooting
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   2. Restoring files after a crash
   3. File problems
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   5. Problems with GPU driver
   6. Problems with Wacom

   devices

   7. Problems with DLL files
   8. Memory problems
   9. Problems with settings file
   10. Problems with fonts
   11. Printer problems
   12. How to share a crash report with Adobe
   13. Illustrator performance improvement

The Star tool lets you create shapes with sharp corners in illustrations.

1. Select tool Star .

2. Drag on the artboard to draw a five-pointed star.

3. If you want to change the number of ends, press the up and down arrow keys while dragging.

4. To draw a star manually, click anywhere on the artboard and enter the options Radius and Points .

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How to draw a star (and not only) in polar coordinates / Sudo Null IT News At best, a solution comes across through the function of the remainder of the division – which is not “pure” from a mathematical point of view, since it does not allow analytical transformations on the function. Apparently, real mathematicians are too busy solving the millennium problems and looking for a simple proof of Fermat’s theorem to pay attention to such banal problems.

Fortunately, in this matter, imagination is more important than knowledge, and to solve this problem you do not need to be a professor of topological sciences – just school-level knowledge is enough.

The formula for an equilateral polygon in polar coordinates looks very simple

and has the following parameters:

— angle;
– number of convex vertices;
– determines how many vertices the sides will lie on one straight line. Negative values ​​are also acceptable for it – it will depend on the sign in which direction the star will bend;
– rigidity – with we get a circle regardless of other parameters, with – a polygon with straight lines, with intermediate values ​​from to – intermediate figures between a circle and a polygon.

With this formula, you can draw a star in two ways:

1)

2) . In this case, two rotations are required instead of one:

The parameter affects the polygon in the following way (here it changes from -1 to 5):

Parameter in the animation:

You can rewrite the original formula in a complex form, and despite the presence of imaginary units in it, the radius value will still remain valid:

At first glance, this may seem pointless, since the formula has become a little more cumbersome – but do not rush to conclusions. Firstly, it lacks the arcsine, which completely changes the mathematical meaning of the formula and allows you to look at the construction of a stellated polygon in a different way. Secondly, compact formulas for special cases can also be obtained from it, for example. Thirdly (and most interestingly), it can be creatively modified and get other, unexpected forms. In order for the appearance of a possible imaginary component in the radius not to cause ambiguity in the calculation, it can be immediately reduced to Cartesian coordinates by multiplying by . Here are examples of some modifications:

As you probably noticed, the rotation of the vector has ceased to be uniform – and precisely because of the appearance of an imaginary component in the radius.

Squares and others

Our formula has a wonderful special case – a square, the formula for which can be written as

or

(choose whichever you like best).

In a slightly more detailed case, intermediate figures between a circle and a square can be defined through a point on the plane

You can also add variability to these figures while maintaining the condition of their passage through a point – directly modulating the parameter itself depending on the angle in such a way that when passing through the diagonals, its multiplier is equal to one. For example, substituting the function for , we get an additional parameter , which can be used to control additional bends. In particular, for the following will be obtained:

In an even more detailed case, it is possible to define not just a square, but a rectangle, and still in polar coordinates:

And even calculate its area (through elliptic integrals):

This will allow you to make profiles with a transition from a circle to a rectangle with a controlled sectional area. Here the area is constant:

And here the area expands exponentially:

Jump to Cartesian coordinates

Any formula in polar coordinates can be expressed through an equation in Cartesian coordinates, and in at least two ways – depending on which the type of gradient on the border of the figure will change. To do this, it is enough to calculate the angle through the arc tangent from the coordinates and bring the formula to a constant through the radius vector by subtracting

or division

The second option is preferable because it gives straight gradients along the sides of the polygon.

Note

Here you also need to remember that at the point (0,0) there is uncertainty due to division by zero – which, however, is easily resolved through the limit (which will be equal in the first case and zero in the second).

The expression can also be simplified to , whose numerator coefficients, when expanded, form an alternating version of the sequence A034839.

The value of the formula from the right side of the equation (in the 2nd case) will change from to if the point falls inside the figure, and from to infinity if not. By choosing different functions to convert it to brightness, you can get different rasterization options. For the exponent (for the first and for the second option) we get

or if saturated

You can use a classic low-pass filter , in which is the filter order that determines the degree of attenuation.

For the first option:

And for the second:

You can also use a piecewise continuous function by explicitly setting the interpolation boundaries.

In addition to rasterization as such, you can also set deformations – for example, compress a chessboard into a circle:

Or even stretch it onto a sphere:

Formula

The classic storytelling style in mathematical texts consists of alternating lemmas/theorems and their proofs, as if the provable statements appeared in the minds of the authors as a revelation from above. And although there is some truth in this, more often the appearance of formulas is preceded by some research work, the description of which can give a greater understanding of their meaning than a formal proof; and the correctness of statements, in turn, can be traced through the correctness of the steps that led to them.

So here – if the article began with a formula in a complex form, then its appearance would be non-obvious and counter-intuitive, and the declared properties would require additional proofs. But in the trigonometric form of recording, the history of its appearance is quite possible to trace.

1) we start with the simplest case – the task of drawing a straight line in polar coordinates. To do this, you need to solve an equation whose solution is obvious.

2) further, the secant argument must be “looped” to provide breaks in the straight line. It is at this stage that other solutions use the “dirty hack” in the form of a remainder from the division. It also uses the sequential taking of the direct and inverse sine functions –

This approach allows you to perform standard mathematical operations on the resulting formula,

for example

you can differentiate it and get a function for a square wave:

Thanks to the same notation, it is possible to simplify the square function in polar coordinates to a more aesthetic form, using the representation of trigonometric functions in complex form. In Wolfram Mathematica, this can be done using the TrigToExp and ExpToTrig functions: 92//TrigToExp//ExpToTrig//Sqrt[#]&//FullSimplify
↓

Thanks to the same record, you can get smooth intermediate figures between a circle and a square with the help of an additional factor due to which the arcsine argument does not reach one -:

And in order for the function to cross a given point, you just need to make an equation and recalculate:

]^2])] /. \[Phi]->Pi/4)==x, k] /. x->k
↓

3) parameters and were simply added in a creative way and their influence was investigated experimentally, in fact.

4) Rectangle is easy to get by going to the parametric view and “stretching” the axes

But after that, the angle will no longer mean, now it is just a parameter that describes the vector through its projections onto the coordinate axes. To go back to polar coordinates, you need to find the length of the vector (via the root of the sum of squares), the angle (via the arc tangent of the ratio), express this angle through and substitute the resulting expression for . 92])]
/. t -> ArcTan[(a Tan[\[Phi]])/b] // Simplify
↓

Simplifying such a formula is already more difficult, and this will require several steps:

1. go to Cartesian coordinates by replacing ;
2. switch to exponential view;
3. simplify;
4. make reverse substitution and ;
5. switch to exponential again;
6. to simplify.

92))] /. \[Phi] -> ArcTan[x, y]
// TrigToExp // Simplify
// # /. {x -> Cos[\[Phi]], y -> Sin[\[Phi]]} &
// TrigToExp // Simplify // FullSimplify
↓

Conclusion

As you can see, even in such a simple and banal thing as a polygon, you can find and come up with something new. And the story does not end there – the area formula for the general case remained unknown, the formula for an arbitrary, and not just a regular polygon remained unknown, and the expansion into power and trigonometric series was left without consideration. It is also likely that a formula of this kind also exists for the 3-dimensional case.

Therefore, if you are told that everything in mathematics has already been invented and only problems remain that are inaccessible to the understanding of an ordinary person, do not believe it. There are many purely practical problems that real mathematicians do not suspect exist, or are not interested in solving them due to the lack of sufficient hype around them, or because they already have a rough idea of ​​\u200b\u200bthe ways to achieve them.