Calculating area

[Olof]

Hi,

I'm working on a test method for analyzing the quality of a printed image. I have six different measurements for each sample. A good way to present this is to use the radar chart, at least when you have few samples, say if you have two samples it’s easy to see which one is best. The problem is that I have huge amount of samples and I was thinking of calculation the area instead. This makes me able to compare the numbers instead of visually evaluate using the radar chart. Does anyone know how to do this?

Olof

 

[rs2k]

"This makes me able to compare the numbers instead of visually evaluate using the radar chart"

Are you saying that the MAX number is better, or the MIN is better?

Can you post a sample of what numbers you have. and what you are expecting.

Colin.

 

[Olof]

Ok, here's a more detailed description and a background to the problem.

When I do the evaluation I have several things to look at. I evaluate the color depth, the unevenness in the print and finally how different colors bleed into each other. It’s the last thing that is the trickiest one. I start with about 20 different values showing the color bleed for each sample; from these I’ve picked 6 most important values to evaluate.

The color bleed should be as low as possible and the values are somewhere between 0.01 to 0.1 and this makes it difficult to expect a certain result.

 

[fairwinds]

Hi,

=A1*A2*SIN(60*PI()/180)/2

should give you the area of the triangle between 2 data ponts (A1 andA2)
thus the sum of six calculations like this should give you the total area.

 

[Olof]

The formula is not working. Using sin, cos or tan requires that the triangle have one corner that is 90 deg. The general formula of the area triangle is base x height / 4. The tricky part is to find out which length who is the base and this can vary.

One solution that I found was to transform each value into x and y coordinates. Then I can calculate the area using =MDETERM(A1:B2)/2. A1 and B1 is the x coordinate and A2 and B2 is the y coordinate. However, I'm not complete satisfied with this method. Is there an easier way to do this?

 

[fairwinds]

I'm definetely not an expert on trigonometry. I believe however that my formula is correct.

I would very much appreciate someone elses opinion here before admitting that I am wrong.

 

[juicydeuce]

Hi...

Ok.. there are a number of ways to calculate areas of triangle
and one of them is 1/2*b*c*SinA

Where b and c are the length of two sides and A is the
angle between them.

Thus you can use that to calculate the area of a radar chart under these assumptions.

1. Number of fields for each data set is the same (let's call this n)
2. On the radar chart, two adjacent points are connected by a straight line.

Area(A-B) = 1/2*a*b*sin((360/n*pi())/180)

would give the traingular area between data point a and b on a radar chart with n fields.

e.g) Columns A to G contains your data, which means you have 7 sets for
each print. For area of the chart in row 2 would be;

Area = 1/2*(A2*B2+B2*C2+C2*D2+D2*E2+E2*F2+F2*G2+G2*A2)*sin((360/7*pi())/180)

As fairwind suggested, you could carry this out for the rest of the data to get the sum of area. Under the assumption that there are 6 fields for each data, fairwind's formular is correct to calculate the area of triangle.

Juicydeuce

 

[fairwinds]

juicydeuce

Thank you very much for the help!

 

[Olof]

I’m convinced... and a bit embarrassed...
Thanks guys!

 

[Ermmm]

Hello,
This formula does just what I'm looking for although in using it I've come across an answer I can't explain; I'm hoping someone can help me.

I'm using juicy deuce's formula above on a radar chart of 15 points each with a score out of 10. If I give each plot full marks ie 10 out of 10 I draw a circle.
Using the formula above that circle has an area of 305.1 units. Using the formula for the area of a circle A=Pi R squared the area should be 314 units.
Can anyone spot my error?
Thanks


=0.5*(B6*C6+C6*D6+D6*E6+E6*F6+F6*G6+G6*H6+H6*I6+I6*J6+J6*K6+K6*L6+L6*M6+M6*N6+N6*O6+O6*P6+P6*B6)*SIN((360/15*PI()/180))