Here's a neighbour's house, from Google Maps. As you can see the land is pretty rough and covered in trees. To get the scale, that's a tennis court on the right, and there's a steep cliff to the right of that:
I suppose it's on the map. I have a plan with distances in metres and angles in degrees and minutes between each peg.
The calculation I made was:
25,000 * 5280 / 360 * 10,000As I can now see dividing by 10,000 only gives 4 decimal places, not 5. I can see it clearly now so I have no idea why I got it wrong previously.
Take more water with it?
I have a TomTom XL for in car use but obviously not the same model. The destructions leave much to be desired and specification doesn't feature either. Position outputs in lat/long but not as far as I am aware as I move (at least I haven't noticed that it does). When I get the time I will see how well it agrees with with my 1:25000 digital mapping.
I haven't thought about using my TomTom for hill walking which is what I use my hand held for.
I take my whisky neat but that mistake was made well before I started my nightcap.
Extend a spar horizontally out over the abyss with a plumb bob running from the end of it. if the horizontal distance is too great to measure in one go do it in sections.
That is fine, although it lacks the built in error check of going back to where you started. What you do is establish, using the survey point as a start, one fixed position from which you can see all the others. Because a Total Station measures elevation, azimuth and range with great accuracy you can fix any point visible from your position in three dimensions. Used properly the Total Station will do all the sums and point itself at the position you want in both elevation and azimuth. Getting the position right requires communication with the pole handler who moves around, directed by the TS operator, until the reflector is in the cross hairs and the TS is indicating the right range. The weakness of course is that if you get the TS station position wrong all the survey points are wrong as they are based on measurements from a "known" point.
A meticulous surveyor would use at least two survey positions on the site so each point was surveyed in twice. Both positions should be identical within a mm or so.
That doesn't matter, you don't need to physically visit each point to carry the survey around.
Just the place to build a house! :-). Makes going out for a pint of milk a whole new experience.
As above. Assume for simplicity all pegs can be seen from at least two points. From one point use the TS to establish the peg positions using bearings and range. Move the TS to the other point and repeat from the new reference Each peg should be in the same position.
You are confusing precision with accuracy. All consumer grade (Garmin handheld/TomTom etc) GPS receivers use only a single frequency and something called CA (Coarse/Acquisition code). In the US a system called WAAS (Wide Area Augmentation System) is available to give slightly better accuracy. In Europe EGNOS (European Geostationary Navigation Overlay Service) does the same with receivers capable of using the signal. Neither are available in NZ .
CA cannot give better than a 95% 3m circular error probability. That doesn't mean the GPS position and real position are never the same, it just means you don't know when they are and more usually, they are not.
This is nothing whatsoever to do with what make of receiver you have - the GPS system as whole simply can't do better. Putting a fine scale on a wooden ruler does not make it any more accurate - just more precise. Showing you a grid reference to 10 places of decimals does not mean the reading is accurate to anything like that.
You can try leaving a receiver in one position for hours/days recording where it thinks it is and then try doing some statistical analysis on the scatter plots but you still won't get guaranteed accuracy. you could also try using the PositioNZ nationwide DGPS service for post processing information.
Not enough beer with your supper :-)
I watched them put in three pegs and they did not move the TS while doing those. I agree with one of the pegs for distance. I was about to say that the next peg is 2 metres wrong but I just got hold of a new plan which shows yet another peg on my boundary that is between two neighbouring properties and does not affect my land at all, and that will be the one that the surveyor has just put in. So, the major problem is probably solved, except that the new surveyor didn't like one of my corner pegs but wouldn't say where he thought it should be. It's still there. I'll eventually move my fence to take advantage of the extra 50 square metres of land that I now have.
OK I've done it all now. I used a 3 metre spar 3 times from the top of the cliff down to the building next door which is 16 metres high and I'm sure it's vertical after the millions they spent on it. Then I measured from the back of the building out to the road. The peg appears to be within 50mm of the correct place. I don't think I need to shift any fences. The top peg seems to be 50mm sideways and nobodywill be complaining. The tree that I thought I owned has a third of its roots uphill in the neighbour's property but almost all of the tree leans to my side.
What you need is some proper GPS stuff. You put one on the reference point and it talks to the other GPS units so they can workout the difference to within a cm or two. It takes care of the errors in real time giving positions accurate enough to measure ground movement, build bridges and stuff like that. It will also give height differences.
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Light but strong string tensioned, measure it's length and angle relative to horizontal, the rest is basic trigonometry. The hard bit will be getting an accurate angle measurement, though if the angle is fairly shallow the affects of errors won't be that great.
String 30m long each 5 degree increase in angle reduces the horizontal distance by less than 1m up to 25 degrees. You should be able to measure *much* better than that...
A catenary? Wow! One of my favourite subjects.
This is probably too little too late, and may in any case not yield results which are precise enough, but if the curve of the hose forms a proper catenary which is not too much distorted by its own rigidity, you could do worse than using the catenary equations to get your answers. It may be fun to go through this exercise anyway, just to see how it compares with the results you may meanwhile have obtained by other means.
One of the helpful properties of a catenary is that its gradient at any point is proportional to the distance, measured along the curve, of that point from the curve's "origin", defined as the point where the gradient is zero, i.e. where it's horizontal.
For this you would need to measure the hose's gradient in two places (or more if you like) which are a known distance apart as measured along the curve of the hose. Mark the hose at each point at which you do a gradient measurement, then measure the distance(s) between marks later, when you've gathered the hose in.
So, using your plumbbob you measure the angle which the hose forms with the vertical, convert it to the angle made with the horizontal, then convert that angle into a gradient by taking the arctan.
If you can arrange for the bottom peg gradient to be zero, it will simplify the arithmetic, but if not it won't matter.
If X, Y, and L are the horizontal, vertical, and along-the-hose distances of a point from the origin of the catenary, and G is the gradient at that point, then K is the constant of proportionality which relates gradient to distance: L/K = G
If the point in question is the top peg, and if the bottom peg is at the origin, then L is the distance along the hose between top and bottom pegs. If L=15m, say, and the slope at the top is
45 degrees (G=1), then K would be 15m.One catenary equation tells us that L/K = sinh (X/K) which in this case would tell us that X = K*arcsinh(1) = 13.22m
Another tells us that Y/K = cosh (X/K) - 1 which in this case means that Y/K = cosh(arcsinh(1))-1 and so Y=6.21m
Note that the hypotenuse sqrt(X^2+Y^2) is 14.61m which, as it should be, is shorter than the 15m hose length, because the hose takes a longer path than a straight line.
If the hose is not horizontal at the bottom peg, then you need to calculate two different L values (Ltop and Lbot) from the L difference (i.e. the distance top to bottom along the hose) and from the two gradients (Gtop and Gbot). Suppose that as before Gtop=1 and (Ltop-Lbot)=15m, but this time Gbot and hence Lbot are not zero. Suppose that Gbot=(1/3). This would imply that K = (Ltop-Lbot)/(Gtop-Gbot) = 15m/(2/3) = 22.5m and therefore Lbot=7.5m and Ltop=22.5m.
As before we can use the equations to get Xtop, Xbot, Ytop, Ybot from Ltop, Lbot, and K, and the horizontal and vertical separations between top and bot are then just the differences Xtop-Xbot and Ytop-Ybot respectively. If I haven't made a mistake, they are
12.46m and 8.10m. In this second example the hose is stretched more tightly than in the first, and so the hypotenuse, at 14.86m, is closer to 15m (the length of hose) than in the first example.Ronald Raygun ( snipped-for-privacy@localhost.localdomain) wibbled on Thursday 27 January
2011 14:46:
Fascinating - thanks for that.
"Light but strong" - a favourite of school physics courses! Most string is either cotton (which stretches like anything with varying humidity) or nylon (which is downright elastic).
Still probably better than trying to do catenary equations on a hose with a little water in it somewhere!
Andy
It was "Light inelastic string" when I was young.
I used the hose because it was heavy and not so much affected by crosswinds. Since there's a 16 metre drop from the top of the cliff, it's a bit hard to measure angles. There's also a lot of trees in the way. Here's the ladder I made for the bottom of the cliff:
Neglecting stuff.
GPS is a time based system. It broadcasts from each satellite a really accurate clock. Neglecting for a moment orbital motion.
You know where all the satellites are. The speed of light means that as you are further away from a satellite, you see its clock as further in the past than it 'really' is.
If you can receive 10 signals from satellites, then you know the range to them by comparing the received times with the current time. If there is a delay of .001 second, the distance is 300Km. If it's .002 seconds, it's 600Km, and so on.
You don't however know direction.
So, you have 10 ranges to fixed points. Ideally, these will all meet at one point. Imagine surveying your flat garden - if you have 4 markers, you can cut 4 bamboo sticks that fit between the markers, and any point.
If your measurements are perfect, then they will all meet up in exactly one spot. If they're not, it's quite easy to tell - they don't all meet up, and can't be made to meet up without some errors.
This is the basic idea - with 10 satellites, you have 10 different measurements of distance to known positions. (actually 9, as you don't know time, but...)
If these all meet up to within 10cm, then it's quite reasonable to state that you've got a position accurate to 10cm. If they however are 40m apart - the error is at least 40m.
That's the point. You don't have _any_ distances at all. Not even 9. What you have is _differences_ in distance.
Andy
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