How do I a rotate my square radar chart (square->diamond)?

  • Thread starter Thread starter Becs I Know Nathing
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Becs I Know Nathing

Hi There Helpful People

I need to rotate (the whole of my) my square-shaped 'radar' chart 45degrees
so that it is diamond-shaped. At the moment the lines of the axes are
horizontal and vertical and make a cross-shape (not unlike a + shape) and I
need the lines to be on the diagonal (making a X shape).

The chart doesn't have a green dot which I could use to rotate the whole
chart.
What should I do? Is this possible?

I'm using Microsoft Excel 2003, part of the Microsoft Office Professional
Edition.

Many thanks in advance
 
I need to rotate (the whole of my) my square-shaped 'radar' chart 45degrees
so that it is diamond-shaped. At the moment the lines of the axes are
horizontal and vertical and make a cross-shape (not unlike a + shape) and I
need the lines to be on the diagonal (making a X shape).

The chart doesn't have a green dot which I could use to rotate the whole
chart.
Click to expand...

Annoying, isn't it? We've discussed this recently, and I mentioned that
Radar chart should have a control similar to the one provided for Pie
charts, but it doesn't.

I think you're just going to have to turn your table into eight values
instead of four, with the extra four values being an interpolation of
the four original values. So it becomes an "octagonal" radar chart, but
the octagon nature of it is hidden because you're carefully calculating
the intermediate values so that they lie on a straight line. I think
AVERAGE() should work okay.

Or, you could use a graphics program to rotate the picture through 22.5°
 
Thanks for your (speedy) assistance Del. I'll give that a shot and see how I
go.
 
I think copying the chart as a picture and pasting it, followed by rotation
by 45 degrees is the way to go. Eight values (8 spokes) will result in an
octagonal radar chart, with 8 spokes instead of 4. If spokes 1 and 3 have
equal values of a, spoke 2 needs a value of a/sqrt(2) so the three are
collinear, and the relationship for spoke 2's value is more complex if
spokes 1 and 3 have different values.

- Jon
 
This isn’t the most elegant solution, but a little bit of trigonometry is needed for the calculations. Maybe someone can simplify it more. This worked for my four dimensional radar plot that I converted to eight dimensions to rotate it 45 degrees. If you have the original values in cells B2 and B4 and want to calculate what B3 is, use this formula:

=(B2*B4)/(SQRT(B2^2+B4^2)*(SIN((PI()/180)*(135-DEGREES(ASIN(B4/SQRT((B2^2) + (B4^2))))))))



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The above formulae didn't work for me. But this formula does. Same basic idea, i.e. create a 8-dimensional radar chart, where you will use your actual data in every second data point. The points in teh middle will be calculated this way (your 2 original points are in B2 and B4):

=SQRT((2*B2^2+2*B4^2-(SQRT(B2^2+B4^2-2*B2*B4*COS(45)))^2)/4)
 
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