Let us?ÿ assume we have a GPS network of 36 points, how many GPS vectors could we have. Let us start from point 1 where we would have 35 vectors to other points. From point 2 we already have a vector to point 1 so we could have 34 vectors to the remaining points, and so on down to 1 vector from point 35 to 36. What we need to know is the sum of the series from 35 down to 1.
S = n ( averaged value from 1 to n ) = n (1 + n )/2 = 35(1 + 35)/2 = 35(36)/2 = 35(18) 630.
I thought the sum of a series was programmed into the HP calculators but I am at a loss for the symbol for that calculation, or is it so simple they thought I should remember it?
Paul in PA
I'd look at it this way:
sum(35 to 0) = (35+0) + (34+1) + (33+2) ... + (18+17)
There are then 18 term x 35 = 630
But how many are not redundant??ÿ If you are given the first 35 you have all the information, and the rest can be derived from those.
This has little to do with how many measurements are independent - that depends on how many GNSS sessions you do and how many are simultaneous sets.
Do you mean factorial? Symbol is ??!?
Nothing above needs factorials.?ÿ The other way to calculate it is with the combinations formula that you can find on many calculators. Taking 36 things 2 at a time,?ÿ 36C2 = 630.?ÿ The formula behind that calculation involves factorials
n! /[(n-r)!?ÿ r!]
GPS is nothing more than thousands of redundant measurements. My typically OPUS solution is on the order of 3,000 plus measurements. A redundant vector is a set of observations from 1 to 2 at a different time. Having measurements from 1 to 2 and 3, then calculating the distance from 2 to 3 is totally different than having a simultaneous vector from 2 to 3.
I had 4 GPS points on this past Friday in an area with some trees. I had the following vectors and the numbers of satellites in each vector:
1 - 2,?ÿ10 SVs
1 - 3, 12 SVs
1 - 4, 11 SVs
2 - 3, 8 SVs
2 - 4, 8 SVs
3 - 4, 12 SVs
The commonality of satellites in a solution is important, so I excluded the 2-3 and 2-4 vectors and my calculated positions were much tighter. I may begin my traverse this afternoon and I will then know the calculated truth of the position of 2 to 4. BTW, I will begin at point 1 backsighting 4, set a traverse point, then to point 3 and set another traverse point to collect point 4.
Points 1 and 3 are Filed Map monuments, except 1 is a pipe touching an larger iron pin. I set up on the pipe because my notes said I had previously occupied a pipe and I did not notice the iron pin 12 years ago. Points 2 and 4 are mags set in pavement where I had reasonably clear sky.
Paul in PA
There were 1,250 station occupations for the 346 stations included in this project (48 for primary base stations, 1090 for local network stations and 112 for re-observations). The total number of baselines observed was 884. So there is math, then there is GNSS network occupation "reality." Our frequent poster, Mr. Jim Frame, was our prime consultant on this project and the producer of the attached report.?ÿ
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The number of vectors possible given the number of receivers is N*(N-1)/2. so, for 36 receivers the answer is 630. The same would apply to possible vectors given the number of stations. Very Simple
Apologies for the minor hijack, but it's interesting to note that, when r = 2, the combinations formula reduces algebraically?ÿ to n(n-1)/2.