Dimensions Of A Graph.

Nov 25, 2008

My macro prduces some text ang a graph on the screen and I can print the result.
Ok so far. But when I run this macro in a different screen resolution, the layout of the page is disturbed: the graph is on the wrong place and the dimensions of the graph are different.

Of course I can drag and resize the graph and include this proces also in the macro but I think this is not the right way. My question is: are in VBA commands which can fix the graph to a location on the sheet/screen and adjust the graph to the requested heighth and width?

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Working With Dimensions

Sep 5, 2007

Is there a way in excel to say if i select Item1 and it is 100mm x 30mm x 100mm, excel will say this will fit into a compartment called A1 and may also fit into a compartment called B6.

I have a list of all the compartments and what the internal dimensions are. So is it just a matter of saying if these dimensions are within those of the compartment then that item will fit in.

I have a list of some what 6000 Items that have dimensions. I may pick 200 or so of these items to fit into special compartments of a machine. Because of the varying sizes i need to know what compartment each item will fit in so i can build a machine big enough and get the right size compartments.

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Jul 3, 2007

I am working with a dynamically declared array. Just wondering if anyone knows of a way to get the dimensions of the array so that I can work with them? e.g. to loop through the array.

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Jun 20, 2014

I am trying to graph dimension measurements that are automatically added as a new worksheet upon completion of a CMM program. Do I need to move all of my data to one work sheet to be able to graph it or is it possible to call the same cell on multiple sheets? Also is there a way to set the graph up so that it updates the information every time a new sheet is added? I have a sample workbook attached that only list one dimension the actual workbook will have many dimensions listed. Sheet 1 is blank and sheets 2-4 contain the information from the CMM.

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Apr 28, 2014

I iterate through a list and store the data in an array.

[Code] ..........

This works fine, but I tried to have 2 dimensions to the array and It no longer works!

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Nov 16, 2009

I have a worksheet with many dimensions and I would like to have a formula or macro to do the following: I need to check 4 different columns that have dimensions and calculate the best combinations in order to fit within the least amount of 96 inch boards.

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Apr 12, 2013

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Aug 13, 2009

I am a woodworker. I have a CAD program that gives me a cut list of all the parts I need. I then take this list into Excel and do some stuff with the data.

all the dimensions are in fractional inches where ' is used to designate feet and " designates inches. I use the find and replace command to get rid of the ' and ". if the dimension is just a fraction like 3/4 Excel thinks this is a date and displays march 4th. I have tried formatting the cells to numbers before i do the find replace with no luck.

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Feb 4, 2010

I am trying to automate the cleaning and compiling of some data, in fact its quite a lot of data, there will be about 200,000 lines over several spreadsheets.

Determine the width of the data on spreadsheet1
Determine the rows with empty cells at the bottom of spreadsheet1
"now that I know the dimensions of this rectangle"
Copy a rectangle (with the same dimensions as the one on sheet1) of data from sheet2
And paste this rectangle into sheet1...

"This is the point of the excercise, I cannot have any gaps in the data (this includes at the end of a spreadsheet) as it needs to be a continous flow from one shreadsheet to another"

This is what I have done so far... I understand if it is laughable, I've only been doing this a couple of days.

Sub HowMuchSpace()

' No. of columns
Dim bWidthOfColumns
Range(Worksheets("Sheet1").Range(":*", ":*").Cells.SpecialCells _(xlCellTypeConstants).Count).Value = bWidthOfColumns

'No. of rows
Dim lNumOfRows
Range("*:*").Value = lNumOfRows

End Sub

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Jul 20, 2009

I have a lot of carton dimensions that are always presented in the same manner and would like to be able to split the individual dimensions into seperate columns.
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Jul 19, 2006

I am working now on macros changing the datas in a pivot table created from a OLAP cube. I have there some hierarchic dimensions in the format of Category.Group.Item etc. When creating a macro with recorder i get something like this...

ActiveSheet.PivotTables("cube").PivotFields("[Dim_Item]").CurrentPageName = _
"[Dim_Item].[All Dim_Item].[CategoryA].[GroupA4].[Item550321]"

Is somewhere out there any tutorial to that [All Dim_Item] thing? ... cause i feel there IS what i need, that with some cunning command i could show an item without knowing his group and category, but i cant get the macro working.

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Oct 11, 2007

I am trying to create a spreadsheet containing a table of dimensions that are multiples of up to six loadable variables. I work for a stainless steel tank manufacturer. The body of tanks are formed from coil stock of varying widths. Typically these might be 1200mm, 1500mm, 2000mm and possibly 1800mm as well as the imperial equivalents. This could mean up to eight variables but in reality all are unlikely at any one given time.

We are told the volume that the tank is required to contain and a target preferred diameter. The theoretical height is therefore derived from this information. But now comes the tricky bit. It is desirable, to minimise cost of welding by minimising the number of strakes or wrappers. Can anyone give me a formula or whatever that can be used to do this and hopefully generate a table of close heights up to say 40 meters. Thus allowing an operator to select from say five presented options.

I have a table in a spreadsheet that the three principal widths of 1200mm 1500mm and 2000mm generate. It was created for me a while ago by a gentleman working at the Amsterdam Museum that I’m happy to send for perusal. The trouble is he never told me how he did it and no matter how hard I try, I cannot discover the method or formula. The first column gives the total combined height, the second the number of 1200mm wrappers, the third the number of 1500mm and the last the number of 2000mm. It is that simple

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Dec 10, 2009

In my business I deal with a ton of different dimensions for my products. A lot of times the dimension will be within the cell as follows, "Product A 17x27 Brown", is in a single cell.

Is it even possible to write a code or formula that will find the "17x27" in the description, recognize it as a "dimension" and then run the formula that I need which is

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May 12, 2014

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how can I determine the original dimensions of a picture used in an image control? I can tell that the original resolution is there (by changing the PictureSizeMode to clip), but I can't find any properties that show me how big that image actually is, only ones that return or set the size of the object itself.

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1/3/2000 $600
3/12/2000 $400

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A short example:
Imagine you own 3 different stores and you're selling oranges. So your table looks like this:
http://img179.imageshack.us/my.php?image=orangeshm4.jpg

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the amounts would be (y axis) and the corresponding dates (x axis).

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Davyccexamp1.xls

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Feb 27, 2007

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B: 20 200
C: 30 300
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Feb 16, 2009

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Jun 16, 2009

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Attached File : Sample.xlsx‎

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Sep 28, 2004

Trying to create an N1.85 graph in Excel that has irregular spaced tick marks on the X-axis. Found the following information but no luck with it. Need US measure for this semi-log (10 X N1.85) graph. Also called a semi-expo (Q1.85) graph.

The resulting graph appears to be a log graph in reverse with one scale; the column widths are smaller at the left and become larger as they progress to the right.

Info found:

A 1.85 graph can be constructed manually by establishing a series of 15 values (in the case of the example in D5.2.1) from a base measurement to the exponent of 1.85.

Step 1
Select a base measurement for the desired size of the graph. A base measurement of 1.0 mm will produce a graph to 15 which is approximately 150 mm wide; a base measurement of 1.5 mm will produce a graph approximately 300 mm wide. In the case of a 1 mm base measurement, the x-axis numbers will be the 1-15 series. In the case of a base of 1.5 mm, the numbers will be represented by the series: 1.5, 3.0, 4.5, 6.0 etc. for 15 values.

Step 2
Construct a series of columns to the 1.85 exponent values measured from the zero point. The rows representing the pressure values are linear.

NOTE - A good approximation of the above can be computer-generated by a spreadsheet programme by entering a column width established from the exponential figures by subtracting the preceding value in each case. The column dimensions are displayed in the number of standard characters able to be accommodated in the column width which is slightly inaccurate in linear dimension.

The figures below indicate the values for a graph based on 1.0 mm.

Linear scale Exponential value of linear values = Column width = linear values to 1.85 power exponential value - preceding value

1 1 1
2 3.61 2.61
3 7.63 4.03
4 13.00 5.36
5 19.64 6.64

[Code]....

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I have got all the numbers ready, but I don't know whether or not it is even possible to create a graph like this.

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