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Re: Re: How to increase area of irregular polygon?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg39328] Re: [mg39291] Re: How to increase area of irregular polygon?
*From*: Garry Helzer <gah at math.umd.edu>
*Date*: Tue, 11 Feb 2003 04:42:40 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
Steve,
If you operate on an object with a linear transformation the
area/volume of the image is the original area/volume multiplied by the
determinant of the matrix of the linear transformation.
Assume that you have a list of vectors giving the vertices: p1, . . .
,pn. The center of gravity is the average q=(p1+ . . . +pn)/n (vector
algebra).
1. Translate the figure to the origin by subtracting the centroid from
each pi.
2. Multiply each translated vertex by the number r. This is a
similarity transformation with ratio r and it is linear with matrix
r*Identity. So it multiplies area by r^2 in the 2D case and multiplies
volume by r^3 in the 3D case.
3.Translate back by adding q to each rescaled coordinate.
You now have rescaled the object without moving its center of gravity.
This is the "fix" mentioned below.
On Sunday, February 9, 2003, at 04:51 AM, Steve Gray wrote:
> On Sat, 8 Feb 2003 09:46:12 +0000 (UTC), "John Ng" <johnn at sfu.ca>
> wrote:
>
>> Hi,
>>
>> I wish to know how to increase an irregular polygon's area by, say,
>> 10%.
>> How would I go about doing that? For example, if I have a set of
>> (x,y)
>> coordinates and figured out an area from that and I get an area X,
>> how would
>> I increase the polygon's area by X + 10% ? In other words, if X =
>> 100, I
>> wish to know how to increase X to 110.
>>
>> The area of a polygon is defined as: a = 1/2 *
>> ((x1+x2)(y1-y2)+(x2+x3)(y2-y3)+...+(xn+x1)(yn-y1)). Is this correct?
>>
>> Thanks for your help..
>
> Gray:
> Can't you just multiply all coordinates by Sqrt[1.1]? This
> will move the polygon as well as enlarge it, but that could be fixed.
>
>
>
Garry Helzer
Department of Mathematics
University of Maryland
1303 Math Bldg
College Park, MD 20742-4015
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