I replied to this and suggested using a while loop with a single shift
register and an array of 100 points, rotating the array and adding the
latest point to the end of the array to create a buffer of as many points as
you initialise the array to, yet only using one shift register. What is the
problem with this approach?
If you really want it to fade out, rather than simply vanish after n points,
you have to be slightly more sophisticated than the XY graph and use a
picture indicator, but the buffering remains the same.
Thijs Boeree wrote in message
news:3A0FA763.D06FE6C3@amc.uva.nl...
> Hi,
> I allready sent an e-mail to this newsgroup but no one could really help
> me yet.
> I want make a realtime presentation of an XY graph with some sort o
f
> buffer of at least 100 XY positions: a trail, i'm aquiring data of
> eye-movements and i want to display each eye in a XY-graph X =
> horizontal movement and Y is the vertical movement but i don't want just
> one dot on my graph but a trail of 100 points that fades out i do can
> build it with a while loop and a shift register with 100 elements but i
> think there is a better way but don't know which??
>
> Any one!
> Best regards Thijs
>