Hello everybody!
I have a problem. I want to make a graph that looks like the picture from down there. How can i do it. I thought that by moving the grid lines....but is that possible?
Please help me.
Hi,
Take a look at Jon's solution for arbitrary gridlines ...
http://peltiertech.com/Excel/Charts/...Gridlines.html
Yes, it is an answer, but its very complicated and i need to move the vertical lines not the horizontal ones.
My graph has to look exactely like this one.
Something more easy?
Last edited by psystem; 03-07-2007 at 12:22 PM.
Hi,Originally Posted by psystem
hopefully not quite like that, but the attached indicates how to set the horizintal gridlines as logarithmic, which I think is the main part of your question.
hth
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Si fractum non sit, noli id reficere.
Thank you for the solutions, but they do not resolve my problem.
I need to move de grid lines. For ex: if i move 1 closer to 2, the value that is atached to 1 to move closer to 2
F(x)=y (x on Ox axys, y on Oy axys)
The solutions you gave me solves only the moving of the grid without the moving of the values.
I do not understand how your value do not move.
In the example, change the value in cell B1 from 1 to 1.9 and the point moves closer to the '2' gridline.
In cell D1 put .9
in cell B1 put =D$1*C1 and formula fill that to B8
the points all adjust according to the formula?
change D1 to .8
the points all adjust again according to the formula?
Can you explain more clearly how yours doesn't 'move'
---
Si fractum non sit, noli id reficere.
Yae you are right. I solved a part of my problem. Thank you.
But i need you to help me once more. I have 2 more problems: How cean i label only the values 0.001, 0.002, 0.005, 0.01, 0.05, 0.1, 0.5, 1, 2, etc?
And how can i make the border of the chart smaler?
Hi, good to se.Yae you are right
But i need you to help me once more. I have 2 more problems: How cean i label only the values 0.001, 0.002, 0.005, 0.01, 0.05, 0.1, 0.5, 1, 2, etc?
And how can i make the border of the chart smaler?
for Labels, the add-in ChartLabeller that is a free add-in from http://www.appspro.com/Utilities/ChartLabeler.htm which I understand will enable you to label each point.
To reduce the border, select the 'Plot Area' and drag the corner out as far as possible, removing the Legend helps.
hth
---
Si fractum non sit, noli id reficere.
Ok, i'm almost done. I have another one for you.
The graphic is drawn by a function or something like that. I want to know if i can somehow know that function. As you can see in the attachment, on Ox i have some values(diameters) and on the Oy i have percents. I want to find out if i know a percent(ex: 60%) what Ox value is the corespondent. And if it's possible, the vice-versa.
Last edited by psystem; 03-13-2007 at 03:53 AM.
Hi,
Click on the series ...
Add Trendline
select what you want
and display equation on chart ...
I thought of that but is not good. I need something to mach exactely on my graphic(curve).just like in my attachment.
Hi,I thought of that but is not good. I need something to mach exactely on my graphic(curve).
sounds like a job for a Trend type function, is your data of the form that you can feed it through the Trend (I'm not expert in that, but the Help says
TREND(known_y's,known_x's,new_x's,const)
ie
=TREND(B2:B13,A2:A13,A15:A19)
hth
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Si fractum non sit, noli id reficere.
I've tryed but with no luck. Here is my situation: first column i have the diameter and on the second one i have the (cumulative)percents of those diameters.
How can i find out what diameter is the corespondent of 60%?
PS: the graphic is the representation of this situation.
Hi,
looking at your data,
the first .005 (A) was 55 (B = 55)
the second .005 (A = up to .01) was 13 (B = up to 68)
your '60%' lies +5 or -8 within that 55-68 range, possibly around the .009 mark,
however,
I have no idea how to calculate that by formula.
---
Si fractum non sit, noli id reficere.
Bryan made me aware of this discussion, and asked if I could help with f(x)=60 solve for x. Let's see...
1st question: Do you want to do this by linear interpolation or by fitting the data to a curve? 1st instinct is that you would want a curve fitting solution, but I can't be sure.
1st step when curve fitting a set of data is to decide what type of equation you want to use to represent the data. If I did the algebra correctly, y=Aln(x)+B should give a straight line on the semi-log plot you've shown, so maybe some kind of polynomial expansion on that y=A[ln(x)]^2+Bln(x)+C?? Or perhaps you know of some theoretical expression that is supposed to relate particle diameter to percentage.
Once you've decided on a form for the equation, LINEST can be used to find the parameters of the equation. Tushar Mehta has a good explanation of curve fitting here http://www.tushar-mehta.com/publish_...nalysis/16.htm. As long as you choose a linear function (linear as defined by linear algebra, not necessarily "straight line"), the parameters can be obtained using LINEST.
Once you've obtained the best fit curve, then you simply need to solve that equation for x at y=60. In some cases, numerical methods will be required to solve for x.
That probably doesn't help much, but it should get you started.
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