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#1
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![]() Hi to all, I pasted together the below attachment. Can anyone help me with it. I need to know how to start a second graph 55% of the way up the original graphs axis. http://i10.photobucket.com/albums/a1...1980/graph.jpg PS. I cant get the pic bigger but can mail if you leave a reply on the thread. Thanks again! -- Moxy1980 ------------------------------------------------------------------------ Moxy1980's Profile: http://www.excelforum.com/member.php...o&userid=28075 View this thread: http://www.excelforum.com/showthread...hreadid=475830 |
#2
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![]() Moxy The second set of data needs to be set on the second Y axis. Having done that you will need to change the origin of the Y2 axis so that it is 55% less than Y1. It will look better if you make sure that Y1 and Y2 have the same number of intervals and then blank out the Y2 labels Jon -- Jon Quixley ------------------------------------------------------------------------ Jon Quixley's Profile: http://www.excelforum.com/member.php...o&userid=25803 View this thread: http://www.excelforum.com/showthread...hreadid=475830 |
#3
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Alternatively, you can adapt this technique:
http://peltiertech.com/Excel/Charts/StackedCharts.html Instead of four equally spaced zones, you can define any number of arbitrarily spaced zones, and use dummy axes: http://peltiertech.com/Excel/Charts/ArbitraryAxis.html to provide the axis appearance you need. I've used this combination to create large charts (legal size portrait) with five or six zones, each zone having primary and secondary Y axes independent of the other zones, and each zone charting one to five related series. - Jon ------- Jon Peltier, Microsoft Excel MVP Peltier Technical Services Tutorials and Custom Solutions http://PeltierTech.com/ _______ Moxy1980 wrote: Hi to all, I pasted together the below attachment. Can anyone help me with it. I need to know how to start a second graph 55% of the way up the original graphs axis. http://i10.photobucket.com/albums/a1...1980/graph.jpg PS. I cant get the pic bigger but can mail if you leave a reply on the thread. Thanks again! |
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