So I'm trying to create 3D collisions for a floor object, and when I jump in-between 2 floor objects, with one being a higher Z axis than the other, as soon as my vertical speed goes even slightly down, I get teleported to be on the floor.
Variable List:
z - Z axis
zspeed - Vertical (Z) speed
grounded - On the ground or not.
Event: Collision with obj_floor
Code:
if z = clamp(z,other.z-5,other.z+1) && zspeed < 0 {
z = other.z;
grounded = true;
}
else { grounded = false; }
if z = other.z {
zspeed = 0;
grounded = true;
}
if z = clamp(z,-5,other.z-2) {
x=xprevious;
y=yprevious;
}
Is there a solution for this? That'd be much appreciated, thank you.
Try something more like this:
if (zspeed > 0 and z > other.z-5 and z < other.z+1) {
z = other.z;
grounded = true;
zspeed = 0;
} else {
grounded = false;
}
Of course there will be other situations outside of this routine where you should should set grounded to false, probably when there is no collision at all.
Basically if the player collides with the platform and is within a certain range of z and is moving downwards (zspeed > 0), set it to z. Otherwise do nothing.
I have never used zspeed, so I assume zspeed of > 0 is moving "downwards".
As for clamp I suspect you have misunderstood what it does. Your use of it suggests you think that it returns true if z is within the min and max. What it actually does is provide you a value of z such that is is within min and max.
So if your value of z is 100000, clamp(z, -5, 5) will return 5. I don't think this is what you want :)
See link:
https://manual.yoyogames.com/GameMaker_Language/GML_Reference/Maths_And_Numbers/Number_Functions/clamp.htm
Related
My goal here is to improve the user experience so that the cursor goes where the user would intuitively expect it to when moving the joystick diagonally, whatever that means.
Consider a joystick that has a different configured speed for each direction.
e.g. Maybe the joystick has a defect where some directions are too sensitive and some aren't sensitive enough, so you're trying to correct for that. Or maybe you're playing an FPS where you rarely need to look up or down, so you lower the Y-sensitivity.
Here are our max speeds for each direction:
var map = {
x: 100,
y: 200,
}
The joystick input gives us a unit vector from 0 to 1.
Right now the joystick is tilted to the right 25% of the way and tilted up 50% of the way.
joystick = (dx: 0.25, dy: -0.50)
Sheepishly, I'm not sure where to go from here.
Edit: I will try #Caderyn's solution:
var speeds = {
x: 100, // max speed of -100 to 100 on x-axis
y: 300, // max speed of -300 to 300 on y-axis
}
var joystick = { dx: 2, dy: -3 }
console.log('joystick normalized:', normalize(joystick))
var scalar = Math.sqrt(joystick.dx*joystick.dx / speeds.x*speeds.x + joystick.dy*joystick.dy / speeds.y*speeds.y)
var scalar2 = Math.sqrt(joystick.dx*joystick.dx + joystick.dy*joystick.dy)
console.log('scalar1' , scalar) // length formula that uses max speeds
console.log('scalar2', scalar2) // regular length formula
// normalize using maxspeeds
var normalize1 = { dx: joystick.dx/scalar, dy: joystick.dy/scalar }
console.log('normalize1', normalize1, length(normalize1))
// regular normalize (no maxpseed lookup)
var normalize2 = { dx: joystick.dx/scalar2, dy: joystick.dy/scalar2 }
console.log('normalize2', normalize2, length(normalize2))
function length({dx, dy}) {
return Math.sqrt(dx*dx + dy*dy)
}
function normalize(vector) {
var {dx,dy} = vector
var len = length(vector)
return {dx: dx/len, dy: dy/len}
}
Am I missing something massive or does this give the same results as regular vector.len() and vector.normalize() that don't try to integrate the maxspeed data at all?
three solutions :
You can simply multiply each component of the input vector by it's respective speed
you can divide the vector itself by sqrt(dx^2/hSpeed^2+dy^2/vSpeed^2)
you can multiply the vector itself by sqrt((dx^2+dy^2)/(dx^2/hSpeed^2+dy^2/vSpeed^2)) or 0 if the input is (0, 0)
the second solution will preserve the vector's direction when the first will tend to pull it in the direction with the greatest max speed. But if the domain of those function is the unit disc, their image will be an ellipse whose radii are the two max speeds
EDIT : the third method does what the second intended to do: if the imput is A, it will return B such that a/b=c/d (the second method was returning C):
Please help me understand how the following code always returns the smallest value in the array. I tried moving position of 3 but it always manages to return it irrespective of the position of it in the array.
let myA = [12,3,8,5]
let myN = 4
function F4(A,N)
{
if(N==1){
return A[0]
}
if(F4(A,N-1) < A[N-1]){
return F4(A,N-1)
}
return A[N-1]
}
console.log(F4(myA,myN))
This is quite tricky to get an intuition for. It's also quite important that you learn the process for tackling this type of problem rather than simply be told the answer.
If we take a first view of the code with a few comments and named variables it looks like this:
let myA = [12,3,8,5];
let myN = myA.length;
function F4(A, N) {
// if (once) there is only one element in the array "A", then it must be the minimum, do not recurse
if (N === 1){
return A[0]
}
const valueFromArrayLessLastEl = F4(A,N-1); // Goes 'into' array
const valueOfLastElement = A[N-1];
console.log(valueFromArrayLessLastEl, valueOfLastElement);
// note that the recursion happens before min(a, b) is evaluated so array is evaluated from the start
if (valueFromArrayLessLastEl < valueOfLastElement) {
return valueFromArrayLessLastEl;
}
return valueOfLastElement;
}
console.log(F4(myA, myN))
and produces
12 3 // recursed all the way down
3 8 // stepping back up with result from most inner/lowest recursion
3 5
3
but in order to gain insight it is vital that you approach the problem by considering the simplest cases and expand from there. What happens if we write the code for the cases of N = 1 and N = 2:
// trivially take N=1
function F1(A) {
return A[0];
}
// take N=2
function F2(A) {
const f1Val = F1(A); // N-1 = 1
const lastVal = A[1];
// return the minimum of the first element and the 2nd or last element
if (f1Val < lastVal) {
return f1Val;
}
return lastVal;
}
Please note that the array is not being modified, I speak as though it is because the value of N is decremented on each recursion.
With myA = [12, 3, 8, 5] F1 will always return 12. F2 will compare this value 12 with 3, the nth-1 element's value, and return the minimum.
If you can build on this to work out what F3 would do then you can extrapolate from there.
Play around with this, reordering the values in myA, but crucially look at the output as you increase N from 1 to 4.
As a side note: by moving the recursive call F4(A,N-1) to a local constant I've prevented it being called twice with the same values.
I come from OOP background (C#/Java to be specific) and I really do not understand how R treat the variable from outside the function.
I made this example:
result = list();
result$total = 0;
result$count = 0;
result$something = "abc";
a = 1:10;
b = 10:20;
mapply(function(x, y) {
print(result$something);
# Does not work with either = or <--
result$total <-- result$total + x + y;
result$count <-- result$count + 1;
print(result$count);
}, x = a, y = b);
result$average = result$total / result$count;
print(result$total);
print(result$count);
print(result$average);
Here, clearly result is available to the anonymous function because the program did print "abc" 10 times.
However, the change to its component total and count does not survive. 10 times it prints 1 for the result$count, and the final 3 lines are 0, 0 and NaN.
Why is this happening? What should I do in this case, if I want the function to be able to change the variable value?
Note: in my real case, result is NOT a global variable, but is inside another function, and I will use return (result) from the function.
How can I find a point ( C (x,y,z) ) between 2 points ( A(x,y,z) , B(x,y,z) ) in a thgree.js scene?
I know that with this: mid point I can find the middle point between them, but I don't want the middle point, I want to find the point which is between them and also has distance a from the A point?
in this picture you can see what I mean :
Thank you.
Basically you need to get the direction vector between the two points (D), normalize it, and you'll use it for getting the new point in the way: NewPoint = PointA + D*Length.
You could use length normalized (0..1) or as an absolute value from 0 to length of the direction vector.
Here you can see some examples using both methods:
Using absolute value:
function getPointInBetweenByLen(pointA, pointB, length) {
var dir = pointB.clone().sub(pointA).normalize().multiplyScalar(length);
return pointA.clone().add(dir);
}
And to use with percentage (0..1)
function getPointInBetweenByPerc(pointA, pointB, percentage) {
var dir = pointB.clone().sub(pointA);
var len = dir.length();
dir = dir.normalize().multiplyScalar(len*percentage);
return pointA.clone().add(dir);
}
See it in action: http://jsfiddle.net/8mnqjsge/
Hope it helps.
I know the question is for THREE.JS and I end up looking for something similar in Babylon JS.
Just in case if you are using Babylon JS Vector3 then the formula would translate to:
function getPointInBetweenByPerc(pointA, pointB, percentage) {
var dir = pointB.clone().subtract(pointA);
var length = dir.length();
dir = dir.normalize().scale(length *percentage);
return pointA.clone().add(dir);
}
Hope it help somebody.
This is known as lerp between two points
e.g. in Three:
C = new Three.Vector3()
C.lerpVectors(A, B, a)
also in generic this is just a single lerp (linear interpolation) math (basically (a * t) + b * (1 - t)) on each axis. Lerp can be described as follows:
function lerp (a, b, t) {
return a + t * (b - a)
}
in your case (see above) :
A = {
x: lerp(A.x, B.x, a),
y: lerp(A.y, B.y, a),
z: lerp(A.z, B.z, a)
}
I have a 3-vector, let's say
v = vec3(-4, 2, 3)
I would like to do a max on the absolute values of the components, so the equivalent of:
max(abs(v[0]), max(abs(v[1]), abs(v[2]))) == 4
However, I have a requirement that I need to preserve the sign. So for example:
magic_max(v[0], magic_max(v[1], v[2])) == -4.
It's a trivial problem if I use conditional branching, but I'm trying to do this in as few operations as possible, and avoid branching. Any ideas on where to look? Maybe there's some bit-shifting magic that can be done?
I would determ the max AND the min of all values, and then decide what is abs larger
ma = max(v[0], max(v[1], v[2]));
mi = min(v[0], min(v[1], v[2]));
res = abs(mi) > ma ? mi : ma;
If you want to get the sign, replace the last line with an if
if (abs(mi) > ma) {
sign = -1;
res = mi;
} else {
sign = +1;
res = ma;
}
However, what should happen on (0, 0, 0)? no sign?