How to configure CircularGauge with higher value of Tickset Minimum than its Maximum?

Gauge for WPF Forum

Posted 9 years ago by Jehiel
Version: 14.1.0620
Avatar

I was trying to plot a circular gauge with a minimum value of 100 and maximum value of 0 and it doesn't look good. Was thinking if I need additional configuration? Or is it already supported? May I request for a sample code that address this scenario?

 

Thanks in advance!

Comments (4)

Posted 9 years ago by Actipro Software Support - Cleveland, OH, USA
Avatar

Hello, I'm not sure that would be possible since the minimum needs to be smaller than the maximum.  Can you describe the scenario you are trying to achieve in more detail?  Something like a range from -100 to 0 should work though.


Actipro Software Support

Posted 9 years ago by Jehiel
Avatar

I was trying to create a program where the user can modify the value of the Minimum and Maximum of a CircularGauge from 0 - 100. That's why I'm testing different possible scenario. Generally, when Minumum value was less than the Maximum value it works fine, but when the user set a higher Minimum value if compared to its Maximum value weird behavior occur.

Posted 9 years ago by Actipro Software Support - Cleveland, OH, USA
Avatar

Perhaps you could do something like watch the two properties for value changes and if one falls outside of the range of the other, back it up as well.  For instance if the minimum is changed to be larger than the maximum, bump up the maximum to the same value.  That way you would retain the integrity of the range.


Actipro Software Support

Posted 9 years ago by Jehiel
Avatar

As a conclusion, it means that circular gauge does not support lower value of max over min value of tickset. To cater this scenario we have to make sure that min is alway lower than its max.

Thanks!

[Modified 9 years ago]

The latest build of this product (v24.1.1) was released 2 months ago, which was after the last post in this thread.

Add Comment

Please log in to a validated account to post comments.