constructing tangent in Illustrator (geometric glyphs)

ghurman
16.May.2008 4.33am
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here’s a common problem when constructing geometric glyphs (glyphs based on basic geometrical shapes):

there’s a given point A on the canvas and an unknown point B on a curve (circle). that’s the case for example when drawing a “6”.

thus, we need to construct a tangent — but how? we can do it approximately, but how to do it precisely, so that the transition between the curve and the tangent flows smoothly but still geometricaly?

AttachmentSize
tangent_unknown_b.gif14.21 KB


Alessandro Segalini
16.May.2008 5.34am
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In your example, Jedan, what you call “point” is actually an area.


bert_vanderveen
16.May.2008 5.45am
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I’d say that would only be possible for instances where the corner A—center of circle—B would be exactly 90 degrees. In the illustration this is not the case. One would have to place B on another spot on the circle. Or lower A.

. . .
Bert Vanderveen BNO


ghurman
16.May.2008 5.55am
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@alessandro
no it’s not an area, i just marked the point to make it bolder.

point B resides on a circle, point A can be anywhere outside the circle/curve.

@bert
the tangent doesn’t have to be a 100% precisely constructed. but it should flow smoothly into the circle/curve. approximation would be OK.


Lars Oppermann
16.May.2008 7.15am
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Since illustrator does not let you draw circles according to non horizontal or vertical diameter you must use a combination of Thales’ and Pythagoras’ Theorems in order to do this geometrically:

* construct a line from the center of your circle (we call this X) to your point B. We’ll call this line a
* in order to construct the Thales circle which will intersect with the original circle at the tangent-point we need the length of a
* create a rectangle with X and B as corners. note the height and width of this rectangle. We’ll call those b and c
* In the example picture, applying Pythagoras to find a = sqrt(b^2 + c^2) = 228.89 pt gives the diameter of the Thales-circle
* Create a circle with a for its height and width and drag its center on the center of the rectangle you constructed on X/B
* The points where this circle intersects with the original circle are the suitable tangent points.


Lars Oppermann
16.May.2008 8.19am
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The naming of a,b and c above is a bit off. Normally one would call the line X/A c and use a^2 + b^2 = c^2... But it works either way it’s just naming ;)


bjharvey
16.May.2008 9.14am
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perhaps this solution will be helpful ...

1) Make a circle using the circle tool
2) add a point (highlighted area)
3) delete the point (select with open arrow tool, delete)
4) click the left point with the pen tool and draw a straight line up
5) expand the outline

or, with the overlap image you posted, you can use
Pathfinder > Divide and then click away the extra pieces using the open arrow tool.


Alessandro Segalini
16.May.2008 9.14am
Alessandro Segalini's picture

“Transition,” “suitable tangent points”, but Jedan, are you connecting a point to a stroke ?


sigurarm
16.May.2008 9.28am
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jharvey’s desciption is quite good. But if I may add a little bit more to it:

1) Make a circle using the circle tool
2) add a point (highlighted area) Use Object > Path > Add Anchor Points. That will give you anchor point exactly at 90°.
3) delete the point (select with open arrow tool, delete)
4) click the left point with the pen tool and draw a straight line up
5) expand the outline

Use Smart Guides. That will help when drawing in 90°


satya
16.May.2008 9.52am
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but how to do it precisely, so that the transition between the curve and the tangent flows smoothly but still geometricaly?

In order to make your curves look more organic/fluid, you will need to break the geometry somewhere.


ghurman
16.May.2008 10.04am
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i used the lars-opperman-method and got the final result.

it looks quite as i wanted it to be. smooth and geometrical. thanks lars!