[rrd-users] Re: Monitoring a color lasers 4 toner-levels in the same graph
Alex van den Bogaerdt
alex at ergens.op.het.net
Tue Mar 28 15:15:00 MEST 2006
On Tue, Mar 28, 2006 at 02:02:05PM +0200, Geir Fossum, Gjøvik VGS wrote:
> I'm trying to monitor a HP color laserjet 3500 with 4 different
> tonercartridges with the AREA type in % of tonerlevels.
> Is it possible to have the highest tonerlevel color automatic become the far
> back color and the lovest tonerlevel color i the front ?
> So that i can se all the AREAs simultaneous.
Yes, you can, but it isn't going to be very efficient. I also
estimate that changes (one toner used to be at the highest level but
becomes the next-highest level) are ugly.
Why not consider stacking? The chart would be 400% high, each colour
is visible and it is easy to see which cartridge is almost empty.
If you really want to do this without stacking, you have to generate
several CDEFs, each one looking for the appropriate colour to contain.
You cannot use the available SORT mechanism (available in version 1.2)
because you also want to remember the colour.
I guess it is most easy if you display 16 areas. Four for each colour.
Three of each four will be UNKN (unknown, transparent) and one will be
the real colour.
if C1 is highest then C1a=C1, C1b=UNKN, C1c=UNKN, C1d=UNKN
else if C1 is next highest then C1a=UNKN, C1b=C1, C1c=UNKN, C1d=UNKN
else if C1 is 2nd next then C1a=UNKN, C1b=UNKN, C1c=C1, C1d=UNKN
else C1a=UNKN, C1b=UNKN, C1c=UNKN, C1d=C1
do the same for the other colours.
One of these four will be filled, the other three will be transparent.
This will create the bottom area, the colour most present in your toner.
Then the next best colour is shown, overlapping the previous one. Again
three out of four will be transparent (UNKN) and one will be filled.
and so on for C1c..C4c and C1d..C4d.
The hard part is how to find C1a..C4d. I'm afraid lots and lots
of if-then-else is involved.
Last but not least: which colour do you think is most important?
When all colours are at 100%, one of them is visible and the rest
will be hidden.
Alex van den Bogaerdt
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