ok i couldnt think of where else to go so can some one here help me find the area of this white area in the picture. or on any odd shape like this with no set number of sides
http://www.angelfire.com/nd2/kylekonline/images/TP-18FM64test.jpg

## All 13 Replies

I'm afraid that none of us can help you out - the image you posted is 404 (not found).

If you fix up the image location, I'm sure one of us can make a suggestion! :-)

here it is it should work jst let me know what you think and if it works or not.

I've tried to open this file and can't do! Are you looking for athe formula to find the area of something? The formula to find the area of a circle is PI * radius squared.

I've tried to open this file and can't do! Are you looking for athe formula to find the area of something? The formula to find the area of a circle is PI * radius squared.

You can't open the attached file? I can ... is there something possibly wrong w/ my database?

Works fine for me.

yeah all i want is a formula is if it is possible. the shape is not quite a circle. I dont know how i would explain it either. but any help would be great. thanks.

If it is meant to be solved as an area, what I would do is set it up as a grid and take it from there (combining shapes to become regular shape and then measuring the are of those), because plain old getting it from the picture is very improbable, unless you have a program to measure the lines, to get the perimeter and multiplying that by itself. Other than that, I have no idea. To have an equation for a shape 'without a name', is different. I would help you, but I'm still in high school, and haven't taken calculus, which I hear is need to take on to these 'without a name' shapes.
If you wanted to find the area of a three dimension Icosahedron, I can help you.

Number of vertices: v
Number of edges: e
Number of faces: f
Edge: a
Radius of circumscribed sphere: R
Radius of inscribed sphere: r
Surface area: S
Volume: V
Dihedral angle between faces: delta (in degrees)
v = 12, e = 30, f = 20
a = (sqrt[50-10 sqrt(5)]/5)R
r = (sqrt[75+30 sqrt(5)]/15)R
R = (sqrt[10+2 sqrt(5)]/4)a
r = (sqrt[42+18 sqrt(5)]/12)a
S = 5 sqrt(3)a2
V = (5[3+sqrt(5)]/12)a3
delta = arccos(-sqrt/3) = 138o 11'
[IMG]http://mathforum.org/dr.math/faq/formulas/images/icosa.gif[/IMG]

so you think i will learn how to do this in calculus. well thats good cause im taking calc this fall. :) but that also sucks cause then i have to put the project off for a while well i will ask some of my buddies that have taken calc thank for the hlep

edit: i guess i should tell everyone why im trying to figure this out. i need to know cause i have been given a project at work to see if i can do with visual basic and the project invovles finding out the area of spaces like this so that they can do other info with the info. there is a program out there that can do this so i figure some how some way i must beable to do it. but the program they have found to do this will cost the company \$50,000.00 :eek: and they dont want to pay that. but hopefully i will figure it out

Do you happen to have an aprox answer as well as the dimensions of the entire pic? Need an aprox to know if i am getting close or not. Thanks

edit: i guess i should tell everyone why im trying to figure this out. i need to know cause i have been given a project at work to see if i can do with visual basic and the project invovles finding out the area of spaces like this so that they can do other info with the info. there is a program out there that can do this so i figure some how some way i must beable to do it. but the program they have found to do this will cost the company \$50,000.00 :eek: and they dont want to pay that. but hopefully i will figure it out

I understand that this is a seriously aged thread, but thought I'd contribute a possible solution anyway.

It makes sense that you'd want to figure out how to process this on your own, but if your employer needs a simple, accurate and inexpensive digital solution for measuring real-size area, then peek over here.

the NIH uses, and makes freely available, a program called NIH IMAGE for Macs - which is also available for Windows from Scion Corp called Scion Image. I have used both for research and find that they measure area accurately to 0.1cm^2.

you can get NIH Image through the National Institues for Health (the mac version) and you can get Scion Image from Scion Corporation (www.scioncorp.com) (the windows version)

I hope that this might apeal to you as a possible solution, and that someone else with a similar requirment might benefit. I have also written several papers on this program's use, email me at maths@iq-data.net if you want some elaboration on their use.

-gkd

If you just want the area in pixels, then I can suggest a simpler method:
if the region you want to measure contains pixels of a similar colour, then you could implement this recursively:

you need firstly:
a list of all the pixels in the image (ie a 2D array), where you store whether they have been visited or not.
a count of the area initialised to 0

Start with a pixel in the image and look at its 8 neighbours. If their colour is similar to the center pixel, and they have not been visited yet, increment the count, and start the procedure again from each of those pixels. In c like pseudocode:

``````void checkPixel(x, y){
for(pixels surrounding x,y){
if(pixelColor(x,y) == pixelColor(x2,y2)){
count++;
checkPixel(x2,y2);
}
}
}``````

Deciding whether a color is similar is the hardest part of this algorithm. The most accurate way to do this is with a gaussian-type function. By recollection, the theoretical optimum is C=exp[(I-Io)/t]^6, where t is the maximum number of pixels diference, and I and Io are the intensities of the comparison pixel and the base pixel respectively. C will range from 0 to 1.

To put this into the above algorithm, you should have the count as a floating point, and incriment it by the value of C, and call checkPixel if C is above 0.5.

t sets the sensitivity of the algorithm, and you will ahve to set this by trial and error.

The effect of using C instead of a cutoff is that pixels that are at the edge, and blurred between the area you are interested in and the rest of the image, are still counted in a userful way.

length times width equals what?

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