@L1nus Basically, you wanna know how your mech walks.
If it walks like a normal human person
-You can find graphs of the gait cycle
-use extension angle against time graphs
-either using a graphing calculator or through other means e.g. Fourier transform
-recreate the graph using regular sinusoidal/other periodic waves
-repeat for each joint, usually hip, knee, ankle.
-copy to other leg, and add a phase factor in the waves to put them about 180* out of sync to each other like normal leg movement.
I have a little documentation on it, and i can link you to my resulting mech if u want
@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
@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
@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
@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
@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 productA.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
@dabestsock landing would be difficult though. for a helicopter, easy. The issue is, you have to know a position point for the start and end of each runway. which is simple, then have it follow that line.
.
this doesnt work with the custom waypoint however, as its just a single point in space
@dabestsock yes and no. assuming no obstacles, its somewhat simple, you could use a PID, of the dot product, which is the ugly piece of maths by each of the '█' in the code
also using a simple gyro to keep roll stable.
.
To avoid obstacles you could just have it fly at like 10000ft or smth tho.
@griges next point? Basically, the distance is the hypotenuse of a triangle made with your current x,y, then using another points x and y, you find a distance.
.
for the direction i use a little vector maths, basically treating the heading of your aircraft as a normal vector, and comparing it to the (normalised) vector between you and the location. I use the scalar product for this.
.
im happy to explain in more detail, just not sure what you mean
@RepublicofWrightIsles i was gonna write out a bunch of stuff, but like in simple, its not that deep, 'true' communism is just homogenous equality etc.
.
I am not defending past atrocities. Ive dug myself a pit here
@BARREND Will do
@L1nus ah. if you find it looks good enough feel free to copy the code from my mahoraga. make sure to have a strong gyro too
@L1nus Basically, you wanna know how your mech walks.
If it walks like a normal human person
I have a little documentation on it, and i can link you to my resulting mech if u want
@Graingy beats me, i had it on my old ipad, late 2016 i think, i just remember building dumb rockets
@MartennLockkenn its all about advertising i think, or being in the right communities/having the right friends
What type of walker are you wanting, like actual physics based walking? Or like ‘fake’ walking where it doesn’t acc walk but uses wheels/jets etc
@Graingy yeah
@Graingy bit under a year without an account
@MartennLockkenn nah not even. I had two builds which reached front page, but i had two which got near none.
@ShinyGemsBro I try my best, but my data could get ruined by a light breeze atp
+1@Graingy and the time without an account lol, either way congrats
@Graingy and somehow you'd get to a mil before i get to plat
What we he banned over, he was a chill guy :/
+1Preach
@LunarEclipseSP labels work well
+1I mean ammo types would be nice, but i want something new, like an EMP, or th british Dragonfire or smth
@XEPOH I mean, my satnav is pretty simple, you’re free to make that if you want, but personally I prob won’t,
@Dissent3R i didn’t know the drama, so it sorta looked like you were talking to yourself, no explanation. Nothing
Schizophrenia simulator
+6@ThomasRoderick jundroo city, not 100% sure what it’s called in the mod menu
+1@Grob0s0VBRa, if you wanted to look at the label tracks, they have, at best, mixed results
+1Happy Bday!
+11:1 walking model of the rizzler
+3@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
@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
@ShinyGemsBro perchance
@griges no sweat, if you need anything gimme a shout, of you need anything
+1@griges just type acos(whatever) and it should work with funky trees,
+1the 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
+1(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-xdistanceY = 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
@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
+1@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.
+1.
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
@griges basically, you know trigonometry with triangles?. SOH CAH TOA
+1all 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
Thought i saw bristol mentioned for a second
@RepublicofWrightIsles fairs, i remember having to search a tutorial for how to build a jet, so no shame at all
@RepublicofWrightIsles ty, im glad i could put some of my maths to use
@dabestsock landing would be difficult though. for a helicopter, easy. The issue is, you have to know a position point for the start and end of each runway. which is simple, then have it follow that line.
.
this doesnt work with the custom waypoint however, as its just a single point in space
@dabestsock yes and no. assuming no obstacles, its somewhat simple, you could use a PID, of the dot product, which is the ugly piece of maths by each of the '█' in the code
also using a simple gyro to keep roll stable.
.
To avoid obstacles you could just have it fly at like 10000ft or smth tho.
@RepublicofWrightIsles back at like 12, thinking this truck was the hype-est thing ever
@RepublicofWrightIsles ah- right, my bad lol
@RepublicofWrightIsles fair, just bear in mind its ancient, i wouldnt wish the levels of cringe upon my worst enemy
@RepublicofWrightIsles my irish friend wasnt best pleased either at the time
@RepublicofWrightIsles tysm, gotta check if youre alright tho, like ur digging deep through my old builds
@RepublicofWrightIsles apparently, its a small website all things considered
@RepublicofWrightIsles ty for all the support dude, its hella kind, dont feel forced tho
@SPsidearm i forgot to remove it from the fish :(
@RepublicofWrightIsles yeah, but dont stray too close to the sun, the guidelines still exist, and ty, its a cool plane lol
@griges next point? Basically, the distance is the hypotenuse of a triangle made with your current x,y, then using another points x and y, you find a distance.
+1.
for the direction i use a little vector maths, basically treating the heading of your aircraft as a normal vector, and comparing it to the (normalised) vector between you and the location. I use the scalar product for this.
.
im happy to explain in more detail, just not sure what you mean
@Graingy @RepublicofWrightIsles this is a planes website its not that deep
@RepublicofWrightIsles i was gonna write out a bunch of stuff, but like in simple, its not that deep, 'true' communism is just homogenous equality etc.
+1.
I am not defending past atrocities. Ive dug myself a pit here
@RepublicofWrightIsles ideology isnt necessarily the same as the execution of the past attempts