@blt yes sir o7

sigmah

noice :)

@XEPOH I mean, my satnav is pretty simple, you’re free to make that if you want, but personally I prob won’t,

@Axartar ok

@Solent sort of, my final output is basically just a step away from finding the angle between which way you are headed from the way you want to go. Called the dot product, a value from -1 to 1, where -1 means you’re facing the opposite direction and 1 means you’re heading straight.
.
Like it will work with a PID, but it might make you spin 360* because the dot product is kinda just a rating on how well you’re headed in the right direction, not which direction.
.
If you do want the dot, it’s the bit by all the '█' on the main label

does it putput some kind of heading variable that i could plug into my autopilot?

@RepublicofWrightIsles while the offer is nice, after the end of this upload week I’m going low activity for a while, my studies and my job are picking up in severity, and I don’t really have the time for simple planes tbh

@Axartar thanks

Can you join the New RP RP ( the one I had the election in) army head engineer

thank you very much! 🙏

@Axartar i will hehehehe a lot i wanna learn

@Axartar pheewww this is waaay faster than searching learning online myself hahaha

@griges no sweat, if you need anything gimme a shout, of you need anything

@Axartar thank you very very much!!!
imma try this! ill let you know 😁
credit fo sure 10000%

@griges just type acos(whatever) and it should work with funky trees,
the acos is the inverse of cos, which means its kinda like the opposite of cos,
like how 1/4 x (4x1) = 1
arccos(cosine(1)) = 1.
The formula for finding the angle that uses arccos, uses a bunch of geometry for a proof, I never really had it explained that well, because my teacher was bad at his job. Here is a tutorial of how to use it, im not sure if it shows a proof. The equation normally is cos(theta) = a.b/(|a|*|b|) where the || is a fancy way of writing a^2 + b^2 = c^2, find c

@griges so, you need variables first for where you are headed. im gonna call these
(a,b,c) this is the location of wherever you wanna go. and im not gonna use c, because I only care about horizontal distance atm.
.
The variables of your plane are Longitude, Latitude, Altitude and Heading. imma call these (x,y,z) and theta (theta for angle).
.
to find the vector between you and your destination
distanceX = a-x
distanceY = b-y
.
Also I wall call the length of the vector between you and destination magDist. this is just sqrt(pow(distanceX,2) + pow(distanceY,2)) think a^2 + b^2 = c^2
.
im gonna make sure the length = 1 in equation below
.
for the dot product = sin(Heading)*(distanceX/magDist) + cos(Heading)*(distanceY/magDist)
.
Now you have a dot product, yippee!
.
with the dot product use acos(dot) for an angle.
to make is useful the angle between you and the target is = acos(dot)-heading
.
putting this all into as little lines as I can
acos(sin(Heading)*((a-x)/sqrt(pow(a-x,2) + pow(b-y,2))) +
cos(Heading)*(b-y/sqrt(pow(a-x,2) + pow(b-y,2)))-Heading
.
edit: just fixed smth dumb

yes pls, teach me, i would love to learn

how to add the acos to the ft

@Axartar wait, where was the A and B again? A target and B current?
also why acos

my pos * target?
i devide them to one?
10,15,-10 /10,15,10?
so was it
(10/108/8) + (-10/102/2)

@griges its not scary haha, it looks it but in simple, if you know trig and simple vectors, you can use two formulas and a little neatening to get a result. Im happy to write it out as a big equation if you want

@griges Ok, i think i proof read that fine, the last thing you need is a little formula with the dot product i mentioned earlier. using a formula. and assuming everything was normalised earlier all you need to know is that acos(A.B) = the angle between two vectors, which is fun already.
.
and then im 90% sure because its headings the angle you have to turn to face the right direction is turn_angle = acos(A.B) - Heading

@Axartar yesss I've been trying to implement the SOH CAH TOA 😅
but i figured it wasn't to find angle, its to find distance from known angle
i found the one i want is the inverse trigonometry. i still need to learn that.
.
so how was the math work again?
you right,this sounds scary

😂
so the result of -1,0,1 can it be changed to degrees?
wait, if i example
target x=10 y=15 z= -10
my position is x=2 y=10 z=2
(10-2,15-10,-10-2)
so i got target as (8,5,-12)
say the y is altitude,song long and lat 8 and -12
√(8²+(-12)²) is the distance?
i still dont understand the heading, facing
from that results
,how can i take the heading

@griges basically, you know trigonometry with triangles?. SOH CAH TOA
all of that in this is solely for triangles, the powers are just pythagoras a^2 + b^2 = c^2
.
for this maths im using something called column vectors, which is just a nice way to show vectors, think of (2,3) as 2 along (right), 3 up. help here
.
Also I should say the maths assumes that youre looking down on the world from a birds eye view, which doesnt really mean anything important. just that up means north.
.
There is a few variables we know always, where you are (x,y,z) = (Longitude, Latitude, Altitude), which way you are headed as a heading. headings are the angle away from north you are facing. To use the heading, for your x use cos(Heading), for your y, use sin(Heading)
help here.
.
Also you always know your destination. To find a vector between you and your destination you can use : (destination) x ,y ,z - (your current) x, y, z.
Because vectors are cool and (1,2,3) - (1,2,3) works by x-x, y-y, z-z. which results in (0,0,0) because there is no difference. help here
.
Now, with your triangle knowledge and simple vector knowledge you can use a few important facts. you need to normalise these vectors. This means the hypotenuse of the triangle has a length of 1, always. but it still points in the same direction. you can do this by dividing each side of the triangle by the hypotenuse length. help here
.
next step is the scalar product which sounds scary, but in short we have two vectors, our heading and the vector between us and our destination, which i will call A and B. to find 'how related they are' use the scalar/dot product A.B = x*x + y*y (multiply the x of A and B together and add it to the y of A time y of B).
.
This creates a number between -1 and 1. and basically it tells you a bunch. if its 1. you are looking directly at the island, if its 0 you are looking perpendicular to it. and -1 if you are looking the opposite direction