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>> Hi Dave and other McGyverites who have got into this... I forget exactly who, but think it was a German who said that the average respondent "would rather use another man's toothbrush than his ideas", and I'm therefore not unduly shocked that Dave without evidence rejects my proposed sighting method for determining height as having a large expected instrumental error, while favouring his own even more improbable alternative. In a winter contest to do just-possible thought experiments that we guess Jamie will never follow through on anyway, it's really hitting below the belt to get serious and start talking about errors, but I feel I've got to rise to the bait here. Dave, I couldn't follow your pressure formulation as given, but essentially, this seems to be that as you go up the hill and air pressure drops, the volume in the closed end of the manometer will rise proportionally, if temperature stays the same (I think it's Boyle's Law, PV=RT?). This is sure to work as a nice thought experiment but since you are getting finecky, the question is whether any such change would be detectable in practice. Air pressure is halved at about 18,000 feet which should be detectable, but it seems doubtful that Jamie lives that high up; more likely the house is (say) 200 meters above sea level (~600 feet). I looked up Atmospheric Pressure on Wikipedia and if you use the second equation given there for calculating air pressure against elevation (for the lowest zone, starting at sea level), an increase in elevation of 200 m would generate a change in atmospheric pressure of only 2.4 percent. I doubt that a change in volume (= air column height) of 2.4% could even be detected reliably never mind accurately measured with your Tygon tubing method, even if you could avoid water droplets on the wall as you jog up the hill carrying it (Tygon is quite wettable by water, which I think is partly why such instruments often use mercury in glass). If you go back to my suggested sighting method, the two angles are eminently measurable (unlike in your method), and the distance between the two rocks should be easily measurable to better than 1 percent with a calibrated rope, so the main expected error would be in the inaccuracy of reading the two angles from a home-made scale. Because tangents are involved, the actual errors depend strongly on the actual dimensions and angles involved, and I doubt if you can make, fit and read a home-made scale to better than 1 degree. If the house is 200 m up, and the first rock is 1000m out and the second 500 m further, a 1 degree error in setting or reading the scale on the telescope for the two angles would tranlate into a 21-34 percent error in height estimation (worst cases, where the errors run in opposite directions rather than partly cancel out). You could improve this a bit by repeatedly sighting and averaging the angles but I doubt if you could get it down below +/- 10 percent, because of difficulty making and accurately centering the scale. This would not be that bad for a first try. With your method, you too perhaps could improve detectability a bit by running up and down the hill repeatedly, carrying the maonometer, but I think my method of averaging is more restful. Commenting on methods for measuring distance, Lois' method using 3-4-5 Pythagorean triangles that are 'similar' (not 'congruent'): this OK for short distances, as in the example given (factor of 10), but invites the comment that if you have a measuring stick 5 units long already, it would be actually quicker to simply tumble it 6 times to measure out the 30 units required than to set up all the sighting sticks and to make sure that they are vertical. How accurate would it be if you wanted to measure something at, say, 1000 units distant, abandoning the 3-4-5 system but keeping the 4 unit baseline? -- sighting errors over the sticks would be much more significant. The website mentioned by Ray may be interesting historically but doesn't help because ultimately one's paces, thumb or whatever have to be calibrated with a ruler to pull out dimensions that we currently use, like feet or meters. Can't estimate if you can't calibrate. One method not mentioned so far is that years ago, Polaroid cameras introduced a range-sensing method that worked over near to middle distances, that I think employed the reflection of an ultrasonic sound pulse from the desired object, presumably by measuring the time delay. Maybe this is how the hand-held devices used by real estate agents to measure the sizes of rooms work - I'm not sure. Not sure either how my camera autofocusses, either. Final advice to Jamie would be to listen to Peter and rent a surveying instrument, or if he really wants a project, make his own survey system with a laser diode (a few dollars these days), angle scales, tripod, level and and a calibrated measuring pole. Methods using triangulation on maps are practically sensible, but essentially amount to cheating in this McGyver context, because the map has already been calibrated by experts. Steve Quoting David & Alison Webster <dwebster@glinx.com>: > Hi Steve & All, Feb 8, 2007 > Your formulation looks good but I would expect instrument error to > be large & have consequently looked at a second approach; an > improvized differential water manometer. At least that seems like a > reasonable name. Having never used or made anything similar, it would > be best to check this method out before using it to establish runway > elevation for instrument landing purposes. > > Theory: This procedure makes use of the change in air pressure > with elevation. At the initial elevation (house or mean sea level as > convenient) water height in a transparent U-tube (both arms the same) > is recorded just after one end is closed to the air such that air > volume of the closed end is not changed. As rapidly as possible (to > avoid temperature changes) the unit is moved to the other location > (mean sea level or house) and the height of water in the two arms is > recorded. If the initial point is not about half way between the two > final points then there has been a change in water volume or enclosed > air volume due to temperature changes and it may be desirable to > start again. As a rule of thumb, a 5 cm difference from sea level > would represent an elevation difference of about 3.9 metres. > > [DIGRESSION: If I have thought this through correctly, the > difference in water level between the two arms will slightly or > seriously underestimate the difference in air pressure between the > two locations, because change in volume of the air on the closed side > will change air pressure on the closed side. A smaller air pressure > on the closed will be increased and a larger pressure on the closed > side will be decreased by an amount proportional the fractional > change in volume. It might be possible to correct for this it one had > to. ] > > From observed difference (h; cm) in water level between the two > arms (ig