Re: another image problem

• To: mathgroup at smc.vnet.net
• Subject: [mg56996] Re: [mg56975] another image problem
• From: Chris Chiasson <chris.chiasson at gmail.com>
• Date: Thu, 12 May 2005 02:32:37 -0400 (EDT)
• References: <200505110925.FAA24119@smc.vnet.net>
• Reply-to: Chris Chiasson <chris.chiasson at gmail.com>
• Sender: owner-wri-mathgroup at wolfram.com

Yea...

One of the easiest ways is:

Knowing the (x,y) size of the graph in pixels, convert pixel lengths
to x and y scales.

Paste the picture into Mathematica. Make sure the picture is selected,
the border will have dotted lines if that is true. If it isn't
selected, click on it once. Now press down the CTRL key. Follow one of
the lines with the mouse, clicking at regular intervals. Release the
CTRL key. Choose Edit - Copy. Click in an empty notebook area. Choose
Edit - Paste. You will see a list of points. It is up to you to
convert these pixel locations into the function's domain and range.
After that you might interpolate the list of points so that you can
plot or integrate them. The other curve may be treated in the same
manner.

On 5/11/05, sean_incali <sean_incali at yahoo.com> wrote:
> Let's say you have a picture of two overlapping graphs.
>
> You don't have the equations that generated them. but you know the
> graphs is relatively accurate. for example,
>
> http://pg.photos.yahoo.com/ph/sean_incali/detail?.dir=/9264&.dnm=191b.jpg&.src=ph
>
> Now... let's say you wanted to find the area under curve for both
> curves (blue and green) between certain x values...
>
> for example, 425nm <= x <= 475nm , and then 500nm <= x <= 550nm, and
> then 525nm <= x <= 575nm.
>
> is this doable in Mathematica?
>
> thanks In advance for any thoughts.
>
> sean
>
>

--
Chris Chiasson
http://chrischiasson.com/
1 (810) 265-3161

• Prev by Date: Re: Crossing of 3D functions
• Next by Date: Re: ArcTan[1/0] no result, but ArcTan[Infinity] ok. How to resolve?
• Previous by thread: another image problem
• Next by thread: Re: another image problem