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Setting Z axis using Waterfall
Posted: Mon Apr 02, 2007 3:13 am
by 9092335
Hello!
I use "waterfall" and I add points using code like this:
for (int i = 0; i <= 40; i++)
{
tChart1.Series[0].Add(i, (autoRand.Next(100)));
}
But I really don't know how to access to Z level. How can I set the new Z level when I add new (x,y) array of points?
Thank you.
Posted: Mon Apr 02, 2007 8:41 am
by 9348258
Hi Tequilla
You can read how "3D series" work in the following
post. It is explained in VCL, but in NET it's the same.
Posted: Mon Apr 02, 2007 9:03 am
by 9092335
9348258 wrote:Hi Tequilla
You can read how "3D series" work in the following
post. It is explained in VCL, but in NET it's the same.
Thank you Edu.
I have read the topic, but I can't find addXYZ method in .NET TeeChart.
There is possibility to add points using "add" method only. I can't add Z value using this method but x,y only.
Posted: Mon Apr 02, 2007 9:40 am
by 9348258
Hi Tequilla
In NET, (AddXY(), AddXYZ() ...) methods don't exist you have to use one of the Add method's overloads...
You have to do something similar as below code:
Code: Select all
Random rnd = new Random();
for (int x = 0; x < 10; x++)
for (int z = 0; z < 4; z++)
waterfall1.Add(x, rnd.Next(), z);
Posted: Mon Apr 02, 2007 9:54 am
by 9092335
Thank you very much!
My problem was I didn't define the type of the chart.
That is why I couldn't input the 3rd parameter in add method as Z value.