I have a couple of questions that may well be answerable here by certain members of my former Corps. I spent the majority of my Sapper life on the admin side of things, so my nitty gritty of matters technical/engineering is mostly in the range between superficial knowledge and non-existant. And, despite many hours of internet research, I still haven't found satisfactory answers to these questions! First, a bit of background I've been pondering what the Germans call a "Schnapsidee" (= 'crazy idea', as distinct from a "Furzidee" = 'brain fart') for a couple or three years or more. It is a true Schnapsidee, I well recall that at the time it occurred to me at least half a bottle of Scotland's finest was flowing through my veins. Discussions with a somewhat technically inclined bloke, my brother-in-law, indicate that the Schnapsidee seems to have merit, but he also cannot answer some of my specific questions or point me towards the answers. Suffice to say, the time is fast approaching to put this Schnapsidee to the test and try to make the gadget a reality all of the other questions have been satisfactorily resolved. To this end, I have now invested in a MIG/MAG welder and other such paraphernalia - fancy that, learning how to weld a year before I become an OAP! Anyway, the plan is that within a few months either the world will be beating a path to my door for licenses to use a gadget that will solve many problems or the local-friendly scrap dealer will have a lucrative day the next time he drives up our strasse! The open questions basically have to do with water and air pressures and how these pressures can be exploited. A minor part of the problem is also getting my head around the Imperial and metric ways of measuring it all! I understand that 1 bar is the equivalent of 14.7 psi. So far so good. I also understand that one of the German/metric definitions of a bar is "the force required to support a column of water 10 metres high". Alles klar! So, if a force of 1 bar is the equivalent of only 14.7 pounds, this then seems to infer that the column of water is itself very thin but how thin? A couple of times I've pratted around with conversion tables and pocket calculators (no answer found on 'tinternet) and always seem to come up with around 1 mm diameter. Could this be correct? Does anyone know the answer? It certainly seems appropriate for the metric system. Let us for the moment assume that the water column is 1 mm in diameter. Would it be true to say that if the diameter was 2 mm, the water column would only rise to 5 meters, and if 4 mm then only to around 2.5 meters? If not, then what other factors need to be taken into consideration when assessing the water column height with a larger column diameter? Do diminishing returns come into play? Finally, calculations for constellation A, the starting point for the gadget, indicate that it will produce combined air and water pressures of around 5 6 bars in constellation B. May I safely assume that these can be used as multiplying factors when calculating the water column heights/diameters leaving constellation B for constellation C? Thus, a force of 5 bars in constellation B would easily raise a water column of say 12 mm in diameter to a height of around 2 metres? Luckily, pressure calculations are not involved in constellations D, E and X they have to do with velocities and rpm! Many thanks in advance for any assistance that can be given.

I couldn't be arsed reading your whole post but the diameter of the column of water makes no difference.(if it's not so small that capillary action comes into play) 10m = 1 atmosphere (approx) whether it's in a 5mm tube or the Atlantic ocean. Go and google u tube manometer. Posted from the ARRSE Mobile app (iOS or Android)

As just said, the diameter of the column makes no difference. What probably wasn't mentioned is that the column isn't open ended, it's closed and contains no air. Any space inside the tube above the water column would be a vacuum. Called a Torrecellian vacuum if I recall. Because a 10m column of water isn't really practical, the lab rat boffins used a column of mercury instead inside a 1 metre tube. Atmospheric pressure of 1 bar would support a column of 760mm with the empty space above being the Torrecellian vacuum. In reallity though you wouldn't even use any column at all, just a pressure gauge. Here lies the next complication for you. At ordinary atmospheric pressure of 1 Bar, the gauge would read 0 (zero). This is known as gauge pressure, and indicates pressures above atmospheric. If your need was for greater accuracy then you would need to either add atmospheric pressure to your reading, or calibrate the gauge to allow for atmospheric pressure. This is then known as Absolute Pressure. Just to translate this into English Imperial units; 1 Bar = 1 Atmosphere = 14.7 psi, although this will vary from day to day with the weather and will decrease as you rise further above sea level, but only by tiny amounts. Your calculations are based on assumptions and before you go any further down that route you would be better advised to have a read up on some basic physics of pressure and volume, etc. Or, just tell us all about your gadget and let us rip it to pieces for you, I mean give it our considered opinion...

The answer to your question is in the measurement...1 bar does not equal 14.7 pounds(approx) 1 bar equals 14.7 psi(pounds per square inch) therefore the diameter of the column is irrelevant because as the head stays constant the volume of water will change, as the diameter(read area, ie square inch) increases or decreases, keeping the pressure constant.This is the problem with dealing in vector and scalar quantities...not every measurement is linear, and many people get caught out.

Or to put it another way, the answer was in your question, i.e. for a volume of water that weighs 14.7 pounds, then the column would have a cross-sectional area of one square inch, so if it was a tube, it would have a diameter of about 1.13 inches. That gives you 14.7 pounds per square inch. Double the volume of water, you double the cross-sectional are of the tube, the height remains the same; make the column big enough to accomodate the Atlantic, as suggested, if the height is 10m, the pressure will be 14.7 psi, and the cross-sectional area will be (weight of water)/14.7.

What he said. Pressure is force per unit area (newtons per sqare metre or pounds force per square inch or something equivalent). It's basic hydrostatics. I did an HND a long time ago, the first term of the fluids lessons were spent on hydrostatics, (U-tube manometers, fluid pressure, etc.). It is 95% common sense (when you know how it works) and a very little GCSE-level maths. Your idea may not have been marketed before now if it's based on a misunderstanding. You'd be best advised to spend a little time poring over a fluid mechanics text book before you spend a lot of your time turning your schnapsidee into metal.