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The Great War (1914-1918) Forum

Remembered Today:

N & M P British Trench Map Atlas


whkay

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Please would someone kindly tell me to what unit of measurement do the "cells" equate to when using the measuring tool on the said CD?

Thank you

Mark

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  • 1 month later...

By the word "Cell" I take it to be Map Squares, on the 1/10,000 map it equates to 1,000 yard square.

John

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Thanks John but no, unfortunately I cannot copy an example to here however you are provided with a measure tool where you measure the distance between two chosen points and that distance is shown you in "cells"? There are 367 cells for the lengh of a 500 yard square so it made me wonder what was the unit each single cell equates to?

Mark

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Not quite John..

If you enter this web-site http://www.britishtrenchmaps.com/screenshots.html and scroll down to one of the shown screenshots i.e the Albert one and open it you are presented with a map with numbers in the bottom right hand corner. The two sets shown are not what I'm interested in its the one that appears where the 0.00 is.

Looking back at the map and the toolbox provided at the top, if you press on the sign marked by a + and a ruler (not on the example by the way, it doesn't work) you can then chart the position between two specified points, this gives you a distance in cells between the two points and the 0.00 figure changes to the given distance (in cells). It is this unit of measurement that I'm enquiring about?

I hope this makes sense, difficult to explain without the screenshots, apologies and thanks..

Mark

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Mark

I deduce that the word cells is a misnomer it should be pixels . Using the measuring tool align the crosshairs on the extreme top left hand corner of the map sheet you will find that the Cell Co-ordinates will read 0.00 0.00 and the other box will be 0.00 cels because no measurement has been taken. By using the measuring tool 50 yards equates to 35 pixels plus or minus 0.5 pixels. As far as I am concerned I would not use this for measuring distances, I would stick to the old and tried Romer method.

John

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Thanks for that John, you are of course correct..

Mark

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