u/Zestyclose-Pool-1081

Tenker å ta turen å jakte på litt bringebær. Ønsker gjerne noen tips til gode steder :)

Trying to set up injecting gps data to a pixhawk running arduSub. When i view the EKF i see that it is in const mode and that the horiz abs is Off and red. Which in my mind means something goes wrong with the gps data so it doesn't know/trust where it is horisontally. What could be reasons for this? I am streaming gps data at 10Hz (tried both 5 and 20). The script sets the EKF Origin. The global position is correctly set near the device.

Any help would be greatly apriciated :)

Ba nylig om begrunnelse på en karakter i et teoretisk mattefag. Begrunnelsen jeg fikk var "se løsningsforslag". Jeg kan kanskje forstå det dersom man i løsningsforslaget / sensurveiledning ga noen form for hvis studenten gjør dette og dette får de disse poengene. I dette tilfellet er det bare et allminnlig løsningsforslag. Skal jeg bare ta på gefülen hvilken karakter man får hvis man er så så langt unna løsningsforslaget? Er dette en begunnelse på sensur i det hele tatt? Da det ikke blir gitt noe forklaring på hvilke pretasjoner som gir hvilke karakterer.

I have a pixhawk 6c pro running arduSub. I want to inject gps data to it via mavlink (usb).

I have tried to set the parameters correctly (GPS1_type = 14) and such. Then I (with some help from AI) have written a python script. The script is just a mock script constantly sending one location on loop.

When I view it mission planner. I see that the device recognizes the the gps location it get's from the gps injection (it updates on the map). However when i try to put it into guided mode it says "guided more requires position".

I have tried a setup where i have a gps connected to gps1 and then it all works seamlessly. However i need to inject the gps in the real setup so this solution doesn't work.

What should this program contain to make the gps injection work? Are there any great tutorial or videos on the subject on how to set it up? What are common problems i should check?

Any advice you be greatly appreciated.

Let f[ℤ_p] be a polynomial f(x) = x^2 + x +1 over ℤ_p. Now consider if f is reducible over ℤ_p. Since f is a second order polynomial, being reducible is equivalent to f having a root in ℤ_p. We shall now prove that there exists infinitely such p such that f is reducible over ℤ_p (by PbC).

Assume there exists a finite number of such p. By the well ordering principle there must exist a largest such p, let it be called q. That means that for every prime p bigger than q f has no root in ℤ_p. Now f having root in ℤ_p is the same as at least an element in the Im(f) being composite of p. (∃ a ∈ ℤ_p : f(a) = m*p , m ∈ ℕ) . Consider the image of f, (Im(f)). Since we know that f has no roots in ℤ_p for p > q. We know that for each value f send onto this cannot be a composite of a prime bigger than q. By the fundamental theorem of arithmetic we know that for every natural number it has to have a prime factorization. n = p_1^k_1 * p_2^k_2 * ... p_n ^k_n. By the earlier fact we know that for an element in the image all the prime factors have to be primes on the interval [2, q]. Consider now the element of the image f(q!)

∀ prime, p_i ∈ [2, q], f(q!) = (q!)^2 + q! + 1 ≡ 1 mod p_i, since p! ≡ 0 mod p_i since p_i in q!.

However then f(q!) cannot have any prime factors on the interval [2, q], therefore it must have a prime factors that is bigger than q. Contradiction. Since f(q!) has a prime factor bigger than q, (let's say for the prime r) then f(q!) would be a root in ℤ_r. Which is a contradiction since p was the biggest such prime. Therefore there has to exist infinite p such that f is reducible over ℤ_p.

Now you might be wondering, what does this have to do with primes p≡ 1 mod 3. Well here it comes

We want to find out when f is reducible. That is the same as finding when x^2 + x +1 ≡ 0 mod p. It has solutions iff (2x+1)^2 +3 ≡ 0 mod p (this comes for just algebraically manipulating f)

Let y = 2x+1. Now we are asking the question when does y^2 ≡ -3 mod p. In other words when is -3 a quadratic residue mod p. We can use the Legendre symbol. (-3/p) = (-1/p)*(3/p). Here we use the reciprocity of the primes (assuming 3 is not p but that is not relevant here.) (3/p) = (p/3)* (-1)^( (p-1/2) * (3-1 / 2) ) . (3-1)/2 = 1, (3/p) * (-1)^(p-1/2). Substituting back in we get. (-3/p) = (p/3) * (-1/p) * (-1)^(p-1/2). These ((-1/p), (-1)^(p-1/2)) are the same so they will always either both be -1 or both be +1 so the product is always 1 so we can remove them. (-3/p) = (p/3). We know that 1 is a quadratic residue mod 3 and that 2 is not. And since primes are either 1 mod 3, 2 mod 3, or the number 3 that are our only options. So if (p/3) = -1 (ie no solution) then p≡ 2 mod 3. We have earlier proved that ∃ infinite p such that f is reducible in ℤ_p but that is equivalent to p ≡ 1 mod 3 since p cannot be 2 mod 3. (and there cannot be infinite of p= 3), therefore there must exist infinite primes on the form 3n + 1.

(i am kinda new to the game so this might all be wrong. I am open for all types of criticisms)