I'm trying to come up with a mm to ft. equivalent. FGCSVERT (ver. 4.1 5/27/2004) p. 4 maximum loop misclosure (mm) is 8 mm * SQRT(E), where E is the length of the loop in km. This is the same for D, the shortest one-way length of a section in km. I assume I would use D because we start from one BM and close to a second BM without looping back to the beginning.
My math is 8 mm / 1,000,000 mm = x' / 5,280'. Therefore, x = 0.042'. Is that correct?
My allowable misclosure for second order, class II leveling would be 0.042' * distance leveled in miles. We use 0.035' * distance leveled in miles. This makes me wonder if my math is correct.
I also want to confirm all the tolerances we set in our new Leica LS10. We use a maximum sight length of 240.0', which is greater than the 229.7' I calculate from the standards. Our Excel level program flags a sight length greater than 240.0', a minimum ground clearance of line of sight if less than 1.5' (?), a maximum top of rod clearance if the middle wire is higher than the rod, and the delta BS and FS distances if greater or less than 10% between the two.
8mm x (1m/1000mm)x(1 ft/0.3048m)=0.026ft (Confirmed by google).
This for international feet (get accustomed to it), but for an 8mm length there is no practical difference. Another way to think of it - 8mm is 0.008m and 0.008m/0.3048 meters per foot.
You aren't taking the square root of the factor on loop length but need to.
5280/(3937/1200)/1000 = 1.6093km #note 3937/1200 is the conversion from meters to survey feet or vice versa
0.008*SQRT1.6093 = 0.0101mm
0.0101*(3937/1200) = 0.0333ft of misclosure per 5,280ft of level run #note you can't use 0.0333 as a constant, see the following
0.008*SQRT3.2186 = 0.0144mm which = 0.0471ft for 10,560ft (2 miles)
more examples:
0.008*SQRT 1km = 0.008mm misclosure
0.008*SQRT 2km = 0.011mm misclosure
0.008*SQRT 3km = 0.014mm misclosure
0.008*SQRT 4km = 0.016mm misclosure
I know my method to calculate the misclosure isn't the most elegant, but it works for me.
I think 8 mm/sqrt(km) = 0.031 ft/sqrt(miles)
0.008 m /sqrt(km) = (0.008/0.3048) * sqrt(5280 * 0.3048/1000) = 0.31 ft/sqrt(miles)
Note that claiming an order is not just the closure, but a lot of other precautions spelled out in the documents.
In my world we use 12 inches as one foot, 39.37 inches as one meter, and a little dimensional analysis.
Historic boundaries and conservation efforts.
@bill93 0.3048 is correct for internation feet, so yes you are correct if converting to that unit of measure.
I did state that I was converting to survey feet, but I left out the U.S. part of the unit.
The U.S. survey foot is 2ppm longer than the international foot, which in a mile is 0.0106 feet.
Kinda right, but by using 0.008, you've converted your answer from millimeters to meters.
8*sqrt(1.609344) = 10.148794 mm. (1 mile is 1.609344 km, so this gives us the mm misclosure in a mile.)
That's 1.0148794 cm. There are 2.54 cm in an inch (International Foot Measure) and 12 inches in a foot.
1.0148794/2.54/12 = 0.03286 International ft/mile.
In US Survey Feet: 0.033297.
The US Survey Foot calculation uses 39.37 inches per meter instead of 2.54 centimeters per inch for the conversion between metric and imperial measures.
Doing both computations was a nice exercise and instructive for the supposed coming change to Internatonal Feet.
Thanks for confirming my calculation of misclosure when using U.S. survey feet.
My mistake not confirming the misclosure value when using international feet.
Yes, the coming mess of which definition of the foot was used will be fun. State legislatures will define the foot unit for measurements as they apply to the professions within their states, so it will be up to the professional to know how to proceed and document their work.
This discussion could be its own thread.
Thanks again.
Correction: 1.0148794/2.54/12 = 0.03286 International ft/mile should be
1.0148794/2.54/12 = 0.033286 International ft/mile
Correction: 1.0148794/2.54/12 = 0.03286 International ft/mile should be
1.0148794/2.54/12 = 0.033286 International ft/mile
@bill93 you nailed that. Your BS and FS distance overall and within a single set . That’s only part of doing these levels. If field procedures are followed usually meeting the closure is successful. I cannot remember now but for so many degrees of temperature change not applied is like .1mm or so per some thing. Darn brain is not working lol.
Math aside, with respect to the first and second order spec. if you aren't using invar rods you aren't meeting the spec. I've seen many claim to be doing first and second order based on the spec. but not using the specified equipment. You can get accurate enough results for most purposes using equipment specified in third order work. Just don't claim it to meet higher order specifications. Pet peeve