yes, im happy to help with upvotes, although i think the trick is to make the entry requirements low, to drum up engagement, but add harsh criteria so its hella difficult to get top marks.
.
As for ideas ive been wanting to host an innovation challenge, where the aim is to have something entirely new, never before seen on the platform where the primary criteria are quality, and originality, because i feel like the usual ww2/ww1/fighter/bomber whatever challenges bottleneck who enters. all yours if you want it
bit of everything, I like building, when something ive built gets validated, im normally more happy and enjoy the build more. I also want to increase my build skill with each build, so yeah
personally build your interior before the cockpit. however make sure to restrain its size roughly. then cover it, smooth it out, also clear glass under opaque glass is 1 way
oh, i helped someone with this a while back, white isnt a colour you can really get. use another VTOL outlet, you can always reduce its engine power to 0 if you dont wanna mess stuff up. Scale the exhaust fat and wide, like a sonic boom, using hex FFFFFF would be white, but you need to remove some red for a whiter white, try something like D6FFFF. Have it running off funky trees, you could use something like
. Variable1 = IAS>343.3?Variable1+0.01:0 Variable2 = clamp01(round(exp(-Variable1)))
@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
@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
Use sets of asterisk like brackets around text to bold (2 asterisk pairs), or italics (1 asterisk pair). Or 3 is both
You can use backticks for the red text (useful to show off funky trees stuff)
Using amounts of hashtags, one hashtags is the largest, then adding more decreases sizes.
![]("image URL") For an image
And use lines of underscores or dashes for a formatted line e.g. dashes:
———————————————————————
Underscores:
@griges yeah, just thought a ufo is probably
One of the few things that could need it, you could try like using a bit of hollow fuselage to like make a shadow over the lights to make them pop more/dark painted fuselage around the lights. But there isn’t really a way to increase brightness past emission 1 (with any effect) to my knowledge
also if u wanna swivel flight cam & or controls, make sure to have the main cockpit/flight comp remain in the same position and rotation, for the sake of physics
@Rb2h yessir
+1@BeastHunter gotta put something out just in case yk
+1yes, im happy to help with upvotes, although i think the trick is to make the entry requirements low, to drum up engagement, but add harsh criteria so its hella difficult to get top marks.
+1.
As for ideas ive been wanting to host an innovation challenge, where the aim is to have something entirely new, never before seen on the platform where the primary criteria are quality, and originality, because i feel like the usual ww2/ww1/fighter/bomber whatever challenges bottleneck who enters. all yours if you want it
tank
+1bit of everything, I like building, when something ive built gets validated, im normally more happy and enjoy the build more. I also want to increase my build skill with each build, so yeah
+1i wouldnt recommend using chatGPT
+1I mean im a fan of an excel spreadsheet, but do you have a functioning prototype?
+1what are they putting in peoples food atm, insanely cool build
+1make a working typewriter
+1insane work.
+1I ain’t doing allat
+1@VictoryLeo I’m glad to hear u like it
+1@LunarEclipseSP ty, its really not that complex though, 90% the rate function icl lol
+1@YarisHatchback yessir, my own custom version
+1@GrizzlitnCFSP all good, glad to see you’re still around
+1I do not like this fact, it makes me feel old. and i thought The states was 21?
+1personally build your interior before the cockpit. however make sure to restrain its size roughly. then cover it, smooth it out, also clear glass under opaque glass is 1 way
+1@keiyronelleavgeek566 finally someone based
+1Insane work for 500 parts
+1RIP.
+1post a build, get upvotes
+1@Rob119 congrats on plat
+1if none of this works you could try another type of engine, but VTOL is prob the best icl
+1oh, i helped someone with this a while back, white isnt a colour you can really get. use another VTOL outlet, you can always reduce its engine power to 0 if you dont wanna mess stuff up. Scale the exhaust fat and wide, like a sonic boom, using hex FFFFFF would be white, but you need to remove some red for a whiter white, try something like D6FFFF. Have it running off funky trees, you could use something like
+1.
Variable1 = IAS>343.3?Variable1+0.01:0Variable2 = clamp01(round(exp(-Variable1)))@blt yes sir o7
+1@ShinyGemsBro I try my best, but my data could get ruined by a light breeze atp
+1What we he banned over, he was a chill guy :/
+1@LunarEclipseSP labels work well
+1@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!
+1@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
@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
@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
@CrestelAeronautics you beat Rb2h, im impressed
+1i forgot
+1[](image url) for an embedded hyperlinkand also
>for like bullet pointsUse sets of asterisk like brackets around text to bold (2 asterisk pairs), or italics (1 asterisk pair). Or 3 is both
You can use backticks for the red text (useful to show off funky trees stuff)
Using amounts of hashtags, one hashtags is the largest, then adding more decreases sizes.
![]("image URL") For an image
And use lines of underscores or dashes for a formatted line e.g. dashes:
———————————————————————
Underscores:
+1
That’s so sick
+1@Theinfinite I will, but im at upload limit rn, so, itll be out like 22:00 GMT
+1Welcome Back
+1@griges I shoulda linked it, you’re right
+1@griges I mean you can try using a lower power impact gun with zero damage into a magnet, idk lol, I used to love impact guns
+1@griges yeah, just thought a ufo is probably
+1One of the few things that could need it, you could try like using a bit of hollow fuselage to like make a shadow over the lights to make them pop more/dark painted fuselage around the lights. But there isn’t really a way to increase brightness past emission 1 (with any effect) to my knowledge
@SuperSuperTheSylph @YarisHatchback. The phantom :)
+1also if u wanna swivel flight cam & or controls, make sure to have the main cockpit/flight comp remain in the same position and rotation, for the sake of physics
+1