Finding the Area of Arbitrary Polygons With Geometric Algebra

In this short, I show a method that can be used to find the area of any polygon using geometric algebra.

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[derickd]

Wow I've watched your main videos on geometric algebra twice and each time you post one of these I'm more blown away that we don't use this everywhere!

[caspermadlener]

Ah, the shoelace formula. In contexts outside this channel, this would just have been the regular area formulas from Grassmann algebra, but I appreciate the consistent perspective on geometric algebra.

[APaleDot]

Also works in higher dimensions to compute the volume of meshes using tri-, quad-, etc.. vectors but with a different ratio out front.

[mathman]

I have used this many times in the past. The first time was in my undergraduate programming class. It got me an A on my final project for the class in 1983. The project was to determine the area of an n sided polygon just like this video. Then we were programming on a mainframe as the PC wasn’t widely available. If you use it for integration of a function it is basically the trapezoidal rule.

[DeclanMBrennan]

Very neat and beautifully illustrated. One tiny quibble is that the +1 on the last formula is a little big to look like a subscript and I think need a "mod n" after it.

[LLL_elder]

Really neat applications to computational geometry processing too!

[miguelortega]

I believe this is equivalent to the so-called "shoelace method", yes?

[genericcheesewedge]

Very cool, at first I thought there was something missing as it would result in a bivector and not an area, but then I realized the bivector IS the area

[shaharjoselevich]

Even if the origin is not on the same plane as the polygon, forming a kind of pyramid, you get the same resulting bivector! Also, it doesn't even have to be a pyramid and a polygon. If you take any outline of a 2D shape, and any surface with that as it's boundary, then chop it up into little areas and sum up all the bivectors, you will get the area of the 2D shape, regardless of the surface. It's like how one dimension down, adding vectors together, results in just the vector going from start to finish.

[siarya_math]

Note : If you want to use this, the polygon cannot self intersect. Because if it does, the result is you are adding two bivectors that doesn't intersect itself, with opposite orientation. But if we add two bivectors with opposite orientation, we are in a sense, subtracting areas instead of adding areas.

[dzuchun]

I feel like it will work for an origin point on a plane other that one containing a polygon...



the only question I have — what does thing thing represent for an arbitrary 3d-surface? maximum projection area...?

feels like some calculus theorem :think:

[lala]

yup, and you can also divide the polygon into a trapezoid and a triangle if more measurements are given. ^-^

[halyon]

Would be really interesting to see if this expression is manifestly invariant to shift of origin.

[dowmanvarn]

Does the method work for concave polygons as well? I'm concerned that the bivector product could give a negative contribution to the sum, and that would be unphysical in the context of areas.

[anselmschueler]

Does this work with concave shapes or ones with holes?

[jongyonp]

to simplify it, could i put the origin at 1 of the vertices?

i thought of rotating the shape to align with an axis, but that seems more pain than worth. But if 1 edge is already aligned, I'd set the origin to 1 of the vertices at that edge

[NiAlBlack]

I wonder how this relates to the Surveyor's Area Formula. It looks very similar, except that it uses the signed area of the trapezoids formed by each edge with projected onto the x-axis.

[Fezezen]

So does this work with concave polygons too because the origin doesn't need to be witihin the polygon?

[ywenp]

Nice! It should work too for non-convex polygons, right? I would have a hard time to prove it, but in all non convex situations I tried, it results in just the right cancellations of areas to give what we expect.

[jamescook]

We can derive this result without geometric algebra using Green's Theorem. It's a standard result in Calculus III.