I have a short question. I don't know which values I have to put in this function and I can't find any valuable examples on the internet.
This is my function:
I already set up a node and everything else.
node.rotation = SCNVector4Make(x,y,z,w);
What are the values for x, y, z, and w when I want to turn my object with an angle of 45 degrees?
The first value is for "x"
SCNVector4Make(1,0,0,0)
The second is "Y"
SCNVector4Make(0,1,0,0)
The third is "Z"
SCNVector4Make(0,0,1,0)
The fourth "W" is rotation in radians. To rotate your object on the "x" axis 45 degs. It will look like so...
SCNVector4Make(1,0,0,M_PI/4)
M_PI is equal to 180 degs.
from the SCNNode reference:
The four-component rotation vector specifies the direction of the rotation axis in the first three components and the angle of rotation (in radians) in the fourth.
In Swift 4.2 you can use the following values for 45 degrees rotation in SCNVector4Make(x, y, z, w):
X-axis:
node.rotation = SCNVector4Make(1, 0, 0, .pi/4)
Y-axis:
node.rotation = SCNVector4Make(0, 1, 0, .pi/4)
Z-axis:
node.rotation = SCNVector4Make(0, 0, 1, .pi/4)
Remember, w parameter must be in Radians,
so 3.14159 / 4 = 0.78539 radians
(or 180 / 4 = 45 degrees).
Related
I try to explain my problems but perhaps there are to many so I dont know where to start. And I am running out of time :(
I have tested the ability of fungi to alter plastic surfaces after 2 diff timepoints and in two batches. Method of surface investigation was ATR-FT-IR. I now have spectral IR data from 4 different substrates, each exposed to 5 diff fungi for two diff times. Every sample was measured 10 times (very mostly but sadly not always). Logically, I was running control samples (no fungi and no treatment, sample treated but without fungi), also for the two diff batches. SO- for each Substrate, I come up with around 140 columns and 1820 rows. I shrunk the data to respective means and standard deviations with excel and imported it as .xlsx- because .csv absolutely failed and i could figure out why ?! Catastrophe.
> head(pet)
Wavenumbers MEAN_PET_untreated SD_PET_untreated MEAN_c_PET_B1_AL1 SD_PET_B1_AL1 MEAN_c_PET_B1_AL2 SD_c_PET_B1_AL2
1 3997.805 8.021747e-05 0.0003198024 -5.862401e-05 0.0002445300 0.0001309613 0.0004636534
2 3995.877 7.575977e-05 0.0003168603 -4.503153e-05 0.0002384142 0.0001185064 0.0004360579
3 3993.948 7.713719e-05 0.0003169468 -3.218230e-05 0.0002414230 0.0001145128 0.0004352532
4 3992.020 7.847460e-05 0.0003191443 -3.255098e-05 0.0002519945 0.0001258732 0.0004388980
5 3990.091 7.835603e-05 0.0003159916 -4.792059e-05 0.0002617358 0.0001325122 0.0004465352
6 3988.163 7.727790e-05 0.0003063113 -6.286794e-05 0.0002593732 0.0001297744 0.0004532126
My goal was a multiplot, showing averaged spectral data with geom_path and geom_ribbons per fungus, yielding 5 elements per plot (substrate pur, controle t1, controle t2, fungi treat 1, fungi treat 2). The dataset is really large so I had problems to handle it and created these plots manually, so NOT by faceting.
F4<-ggplot(pet)+
geom_errorbar(aes(x = Wavenumbers, y = MEAN_c_PET_B2_AL2, ymin = MEAN_c_PET_B2_AL2 - SD_c_PET_B2_AL2, ymax = MEAN_c_PET_B2_AL2 + SD_c_PET_B2_AL2, group=1), alpha= .1, stat="identity", position = "identity", colour="red")+
geom_path(aes(x = Wavenumbers, y = MEAN_c_PET_B2_AL2), stat="identity", group= 1, colour= "red")+
geom_errorbar(aes(x = Wavenumbers, y = MEAN_c_PET_B2_AL1 ,ymax = MEAN_c_PET_B2_AL1 + SD_c_PET_B2_AL1, ymin = MEAN_c_PET_B2_AL1 - SD_c_PET_B2_AL1, group=1), alpha= .1, stat="identity", position = "identity", colour="purple")+
geom_path(aes(x = Wavenumbers, y = MEAN_c_PET_B2_AL1), stat="identity", group= 1, colour= "purple")+
geom_errorbar(aes(x = Wavenumbers, y = MEAN_PET_untreated, ymax = MEAN_PET_untreated + SD_PET_untreated, ymin = MEAN_PET_untreated - SD_PET_untreated, group=1), alpha= .1, stat="identity", position = "identity", colour="yellow")+
geom_path(aes(x = Wavenumbers, y = MEAN_PET_untreated), stat="identity", group= 1, colour= "yellow")+
geom_errorbar(aes(x = Wavenumbers, y = MEAN_F4_PET_B2_AL1, ymax = MEAN_F4_PET_B2_AL1 + SD_F4_PET_B2_AL1, ymin = MEAN_F4_PET_B2_AL1 - SD_F4_PET_B2_AL1, group=1), alpha= .1, stat="identity", position = "identity", colour="orange")+
geom_path(aes(x = Wavenumbers, y = MEAN_F4_PET_B2_AL1), stat="identity", group= 1, colour= "orange")+
geom_errorbar(aes(x = Wavenumbers, y = MEAN_F4_PET_B2_AL2, ymax = MEAN_F4_PET_B2_AL2 + SD_F4_PET_B2_AL2, ymin = MEAN_F4_PET_B2_AL2 - SD_F4_PET_B2_AL2, group=1), alpha= .1, stat="identity", position = "identity", colour="darkgreen")+
geom_path(aes(x = Wavenumbers, y = MEAN_F4_PET_B2_AL2), stat="identity", group= 1, colour= "darkgreen")+xlab(NULL)+ylab(NULL)+
scale_x_reverse(limits=c(4000 , 500))
So far I summarized the diff ggplots with:
pets<-grid.arrange(F1, F2, F7,F4, F19, ncol = 1, nrow = 5)
ggsave("Multi.pdf", width = 210, height = 297, units = "mm", pets)
This is nearly fine, not elegant and very complicated, but I wont give up at this stage of work as it costed me a whole week. Sadly, I am not really happy with the design, not even to say, I can not use this like it is. Currently, I try to find solutions regarding:
a) Getting rid of empty grid areas left and right to the plotted values. I use scale_x_reverse(limits=c(4000 , 500)), but the range is extended to both sides on the x axis.
b) Creating manually a legend, because even if it would be possible to do this via shared.legend or whatever, it would always yield to many elements. I only want 5 elements with the always repeating, same colors (red=substrate pure, orange= cT_t1, yellow= cT_t2, green= f_t1, purple = f_t2)
c) creating manually a y-labeling (Absorbance), spanning invisible over all plots (vertically)- I tried to label only the 3. plot in the middle, but this leads to a indentation of this plot and the ones above and below appear more left-ragged. If this would be possible, I could use the direct labeling for indicating the respective fungus (e.g. F4).
d) creating a global x labeling- because if I label only the last element, the height of the last plot is reduced by the height of the label.
e) Give it an overall name.
What makes me nervous, too, is that I get an error only for geom_path, telling me that 1 row was removed. But shouldnt this affect also the geom_ribbon? Has it something to do with the fact that I have to call ribbon BEFOR I call geom_path? Otherwise, the lines would have been hidden by the ribbon.
Removed 1 row(s) containing missing values (geom_path).
Also, I am a wondering about the long duration of code execution. 1 element needs 20 seconds, the whole plot 2 minutes to compute. But at least, it is not collapsing like Excel did before- inclusively data loss. Is it normal for such huge datasets? Or could it indicate a very problematic problem?
Ok, finally I hope someone is out there, having had similar work-around-solutions. Because, like I said, I am not willing to spend another week to tidyr or reshape or mutate or whatever.
Thanx in advance! :)
I would like to rotate only the positive value pixels in my 2d array some degree about the center point. The data represents aerosol concentrations from a plume dispersion model, and the chimney position is the origin of rotation.
I would like to rotate this dispersion pattern given the wind direction.
The concentrations are first calculated for a wind direction along the x-axis and then translated to their rotated position using a 2d linear rotation about the center point of my array (the chimney position) for all points whose concentration is > 0.
The input X,Y to the rotation formula are pixel indexes.
My problem is that the output is aliased since integers become floats. In order to obtain integers, I rounded up or down the output. However, this creates null cells which become increasingly numerous as the angle increases.
Can anyone help me find a solution to my problem? I would like to fix this problem if possible using numpy, or a minimum of packages...
The part of my script that deals with computing the concentrations and rotating the pixel by 50°N is the following. Thank you for your help.
def linear2D_rotation(xcoord,ycoord,azimuth_degrees):
radians = (90 - azimuth_degrees) * (np.pi / 180) # in radians
xcoord_rotated = (xcoord * np.cos(radians)) - (ycoord * np.sin(radians))
ycoord_rotated = (xcoord * np.sin(radians)) + (ycoord * np.cos(radians))
return xcoord_rotated,ycoord_rotated
u_orient = 50 # wind orientation in degres from North
kernel = np.zeros((NpixelY, NpixelX)) # initialize matrix
Yc = int((NpixelY - 1) / 2) # position of central pixel
Xc = int((NpixelX - 1) / 2) # position of central pixel
nk = 0
for Y in list(range(0,NpixelX)):
for X in list(range(0,NpixelY)):
# compute concentrations only in positive x-direction
if (X-Xc)>0:
# nnumber of pixels to origin point (chimney)
dx = ((X-Xc)+1)
dy = ((Y-Yc)+1)
# distance of point to origin (chimney)
DX = dx*pixel_size_X
DY = dy*pixel_size_Y
# compute diffusivity coefficients
Sy, Sz = calcul_diffusivity_coeff(DX, stability_class)
# concentration at ground level below the centerline of the plume
C = (Q / (2 * np.pi * u * Sy * Sz)) * \
np.exp(-(DY / (2 * Sy)) ** 2) * \
(np.exp(-((Z - H) / (2 * Sz)) ** 2) + np.exp(-((Z + H) / (2 * Sz)) ** 2)) # at point away from center line
C = C * 1e9 # convert MBq to Bq
# rotate only if concentration value at pixel is positive
if C > 1e-12:
X_rot, Y_rot = linear2D_rotation(xcoord=dx, ycoord=dy,azimuth_degrees=u_orient)
X2 = int(round(Xc+X_rot))
Y2 = int(round(Yc-Y_rot)) # Y increases downwards
# pixels that fall out of bounds -> ignore
if (X2 > (NpixelX - 1)) or (X2 < 0) or (Y2 > (NpixelY - 1)):
continue
else:
# replace new pixel position in kernel array
kernel[Y2, X2] = C
The original array to be rotated
The rotated array by 40°N showing the data loss
Your problem description is not 100% clear, but here are a few recommendations:
1.) Don't reinvent the wheel. There are standard solutions for things like rotating pixels. Use them! In this case
scipy.ndimage.affine_transform for performing the rotation
a homogeneous coordinate matrix for specifying the rotation
nearest neighbor interpolation (parameter order=0 in code below).
2.) Don't loop where not necessary. The speed you gain by not processing non-positive pixels is nothing against the speed you lose by looping. Compiled functions can ferry around a lot of redundant zeros before hand-written python code catches up with them.
3.) Don't expect a solution that maps pixels one-to-one because it is a fact that there will be points that are no ones nearest neighbor and points that are nearest neighbor to multiple other points. With that in mind, you may want to consider a higher order, smoother interpolation.
Comparing your solution to the standard tools solution we find that the latter
gives a comparable result much faster and without those hole artifacts.
Code (without plotting). Please note that I had to transpose and flipud to align the results :
import numpy as np
from scipy import ndimage as sim
from scipy import stats
def mock_data(n, Theta=50, put_neg=True):
y, x = np.ogrid[-20:20:1j*n, -9:3:1j*n, ]
raster = stats.norm.pdf(y)*stats.norm.pdf(x)
if put_neg:
y, x = np.ogrid[-5:5:1j*n, -3:9:1j*n, ]
raster -= stats.norm.pdf(y)*stats.norm.pdf(x)
raster -= (stats.norm.pdf(y)*stats.norm.pdf(x)).T
return {'C': raster * 1e-9, 'Theta': Theta}
def rotmat(Theta, offset=None):
theta = np.radians(Theta)
c, s = np.cos(theta), np.sin(theta)
if offset is None:
return np.array([[c, -s] [s, c]])
R = np.array([[c, -s, 0], [s, c,0], [0,0,1]])
to, fro = np.identity(3), np.identity(3)
offset = np.asanyarray(offset)
to[:2, 2] = offset
fro[:2, 2] = -offset
return to # R # fro
def f_pp(C, Theta):
m, n = C.shape
clipped = np.maximum(0, 1e9 * data['C'])
clipped[:, :n//2] = 0
M = rotmat(Theta, ((m-1)/2, (n-1)/2))
return sim.affine_transform(clipped, M, order = 0)
def linear2D_rotation(xcoord,ycoord,azimuth_degrees):
radians = (90 - azimuth_degrees) * (np.pi / 180) # in radians
xcoord_rotated = (xcoord * np.cos(radians)) - (ycoord * np.sin(radians))
ycoord_rotated = (xcoord * np.sin(radians)) + (ycoord * np.cos(radians))
return xcoord_rotated,ycoord_rotated
def f_OP(C, Theta):
kernel = np.zeros_like(C)
m, n = C.shape
for Y in range(m):
for X in range(n):
if X > n//2:
c = C[Y, X] * 1e9
if c > 1e-12:
dx = X - n//2 + 1
dy = Y - m//2 + 1
X_rot, Y_rot = linear2D_rotation(xcoord=dx, ycoord=dy,azimuth_degrees=Theta)
X2 = int(round(n//2+X_rot))
Y2 = int(round(m//2-Y_rot)) # Y increases downwards
# pixels that fall out of bounds -> ignore
if (X2 > (n - 1)) or (X2 < 0) or (Y2 > (m - 1)):
continue
else:
# replace new pixel position in kernel array
kernel[Y2, X2] = c
return kernel
n = 100
data = mock_data(n, 70)
I have 12 different points and 10 of them are related to the first two; I want to set label for each of this 10 points individually, but sometimes two or more of them have the same coordinate yet I want to show all the label for that coordinate (not on top of each other but readable)
As you can see in the below picture two set of points have the same coordinate and the label of them have overlapping
booleanFunction = np.array(["K","I","H" ,"G", "F", "E" , "D" , "M", "B", "A"])
pointsx = np.empty((rs.shape[1],1))
pointsy = np.empty((rs.shape[1],1))
....
....
....
pl.figure()
pl.hold(True)
pl.plot(X1, Y1, 'ro', X2, Y2, 'y<')
pl.plot(pointsx, pointsy, 'b3')
for i in range (len(pointsx)):
pl.annotate(booleanFunction[i], xy=(pointsx[i], pointsy[i]), xycoords='data', textcoords='data')
I one of my codes to avoid annotation overlap I do something like this:
xoffset = 0.1
switch = -1
for i in range (len(pointsx)):
pl.annotate(booleanFunction[i], xy=(pointsx[i], pointsy[i]),
xytext=(pointsx[i]+switch*xoffset, pointsy[i]),
xycoords='data', textcoords='data')
switch*=-1
This writes your annotated text alternatively shifted left and right xoffset from the point you want to annotate. Of course you can use something similar for the y direction or for both.
I want to convert GPS location (latitude, longitude) into x,y coordinates.
I found many links about this topic and applied it, but it doesn't give me the correct answer!
I am following these steps to test the answer:
(1) firstly, i take two positions and calculate the distance between them using maps.
(2) then convert the two positions into x,y coordinates.
(3) then again calculate distance between the two points in the x,y coordinates
and see if it give me the same result in point(1) or not.
one of the solution i found the following, but it doesn't give me correct answer!
latitude = Math.PI * latitude / 180;
longitude = Math.PI * longitude / 180;
// adjust position by radians
latitude -= 1.570795765134; // subtract 90 degrees (in radians)
// and switch z and y
xPos = (app.radius) * Math.sin(latitude) * Math.cos(longitude);
zPos = (app.radius) * Math.sin(latitude) * Math.sin(longitude);
yPos = (app.radius) * Math.cos(latitude);
also i tried this link but still not work with me well!
any help how to convert from(latitude, longitude) to (x,y) ?
Thanks,
No exact solution exists
There is no isometric map from the sphere to the plane. When you convert lat/lon coordinates from the sphere to x/y coordinates in the plane, you cannot hope that all lengths will be preserved by this operation. You have to accept some kind of deformation. Many different map projections do exist, which can achieve different compromises between preservations of lengths, angles and areas. For smallish parts of earth's surface, transverse Mercator is quite common. You might have heard about UTM. But there are many more.
The formulas you quote compute x/y/z, i.e. a point in 3D space. But even there you'd not get correct distances automatically. The shortest distance between two points on the surface of the sphere would go through that sphere, whereas distances on the earth are mostly geodesic lengths following the surface. So they will be longer.
Approximation for small areas
If the part of the surface of the earth which you want to draw is relatively small, then you can use a very simple approximation. You can simply use the horizontal axis x to denote longitude λ, the vertical axis y to denote latitude φ. The ratio between these should not be 1:1, though. Instead you should use cos(φ0) as the aspect ratio, where φ0 denotes a latitude close to the center of your map. Furthermore, to convert from angles (measured in radians) to lengths, you multiply by the radius of the earth (which in this model is assumed to be a sphere).
x = r λ cos(φ0)
y = r φ
This is simple equirectangular projection. In most cases, you'll be able to compute cos(φ0) only once, which makes subsequent computations of large numbers of points really cheap.
I want to share with you how I managed the problem. I've used the equirectangular projection just like #MvG said, but this method gives you X and Y positions related to the globe (or the entire map), this means that you get global positions. In my case, I wanted to convert coordinates in a small area (about 500m square), so I related the projection point to another 2 points, getting the global positions and relating to local (on screen) positions, just like this:
First, I choose 2 points (top-left and bottom-right) around the area where I want to project, just like this picture:
Once I have the global reference area in lat and lng, I do the same for screen positions. The objects containing this data are shown below.
//top-left reference point
var p0 = {
scrX: 23.69, // Minimum X position on screen
scrY: -0.5, // Minimum Y position on screen
lat: -22.814895, // Latitude
lng: -47.072892 // Longitude
}
//bottom-right reference point
var p1 = {
scrX: 276, // Maximum X position on screen
scrY: 178.9, // Maximum Y position on screen
lat: -22.816419, // Latitude
lng: -47.070563 // Longitude
}
var radius = 6371; //Earth Radius in Km
//## Now I can calculate the global X and Y for each reference point ##\\
// This function converts lat and lng coordinates to GLOBAL X and Y positions
function latlngToGlobalXY(lat, lng){
//Calculates x based on cos of average of the latitudes
let x = radius*lng*Math.cos((p0.lat + p1.lat)/2);
//Calculates y based on latitude
let y = radius*lat;
return {x: x, y: y}
}
// Calculate global X and Y for top-left reference point
p0.pos = latlngToGlobalXY(p0.lat, p0.lng);
// Calculate global X and Y for bottom-right reference point
p1.pos = latlngToGlobalXY(p1.lat, p1.lng);
/*
* This gives me the X and Y in relation to map for the 2 reference points.
* Now we have the global AND screen areas and then we can relate both for the projection point.
*/
// This function converts lat and lng coordinates to SCREEN X and Y positions
function latlngToScreenXY(lat, lng){
//Calculate global X and Y for projection point
let pos = latlngToGlobalXY(lat, lng);
//Calculate the percentage of Global X position in relation to total global width
pos.perX = ((pos.x-p0.pos.x)/(p1.pos.x - p0.pos.x));
//Calculate the percentage of Global Y position in relation to total global height
pos.perY = ((pos.y-p0.pos.y)/(p1.pos.y - p0.pos.y));
//Returns the screen position based on reference points
return {
x: p0.scrX + (p1.scrX - p0.scrX)*pos.perX,
y: p0.scrY + (p1.scrY - p0.scrY)*pos.perY
}
}
//# The usage is like this #\\
var pos = latlngToScreenXY(-22.815319, -47.071718);
$point = $("#point-to-project");
$point.css("left", pos.x+"em");
$point.css("top", pos.y+"em");
As you can see, I made this in javascript, but the calculations can be translated to any language.
P.S. I'm applying the converted positions to an HTML element whose id is "point-to-project". To use this piece of code on your project, you shall create this element (styled as position absolute) or change the "usage" block.
Since this page shows up on top of google while i searched for this same problem, I would like to provide a more practical answers. The answer by MVG is correct but rather theoratical.
I have made a track plotting app for the fitbit ionic in javascript. The code below is how I tackled the problem.
//LOCATION PROVIDER
index.js
var gpsFix = false;
var circumferenceAtLat = 0;
function locationSuccess(pos){
if(!gpsFix){
gpsFix = true;
circumferenceAtLat = Math.cos(pos.coords.latitude*0.01745329251)*111305;
}
pos.x:Math.round(pos.coords.longitude*circumferenceAtLat),
pos.y:Math.round(pos.coords.latitude*110919),
plotTrack(pos);
}
plotting.js
plotTrack(position){
let x = Math.round((this.segments[i].start.x - this.bounds.minX)*this.scale);
let y = Math.round(this.bounds.maxY - this.segments[i].start.y)*this.scale; //heights needs to be inverted
//redraw?
let redraw = false;
//x or y bounds?
if(position.x>this.bounds.maxX){
this.bounds.maxX = (position.x-this.bounds.minX)*1.1+this.bounds.minX; //increase by 10%
redraw = true;
}
if(position.x<this.bounds.minX){
this.bounds.minX = this.bounds.maxX-(this.bounds.maxX-position.x)*1.1;
redraw = true;
};
if(position.y>this.bounds.maxY){
this.bounds.maxY = (position.y-this.bounds.minY)*1.1+this.bounds.minY; //increase by 10%
redraw = true;
}
if(position.y<this.bounds.minY){
this.bounds.minY = this.bounds.maxY-(this.bounds.maxY-position.y)*1.1;
redraw = true;
}
if(redraw){
reDraw();
}
}
function reDraw(){
let xScale = device.screen.width / (this.bounds.maxX-this.bounds.minX);
let yScale = device.screen.height / (this.bounds.maxY-this.bounds.minY);
if(xScale<yScale) this.scale = xScale;
else this.scale = yScale;
//Loop trough your object to redraw all of them
}
For completeness I like to add my python adaption of #allexrm code which worked really well. Thanks again!
radius = 6371 #Earth Radius in KM
class referencePoint:
def __init__(self, scrX, scrY, lat, lng):
self.scrX = scrX
self.scrY = scrY
self.lat = lat
self.lng = lng
# Calculate global X and Y for top-left reference point
p0 = referencePoint(0, 0, 52.526470, 13.403215)
# Calculate global X and Y for bottom-right reference point
p1 = referencePoint(2244, 2060, 52.525035, 13.405809)
# This function converts lat and lng coordinates to GLOBAL X and Y positions
def latlngToGlobalXY(lat, lng):
# Calculates x based on cos of average of the latitudes
x = radius*lng*math.cos((p0.lat + p1.lat)/2)
# Calculates y based on latitude
y = radius*lat
return {'x': x, 'y': y}
# This function converts lat and lng coordinates to SCREEN X and Y positions
def latlngToScreenXY(lat, lng):
# Calculate global X and Y for projection point
pos = latlngToGlobalXY(lat, lng)
# Calculate the percentage of Global X position in relation to total global width
perX = ((pos['x']-p0.pos['x'])/(p1.pos['x'] - p0.pos['x']))
# Calculate the percentage of Global Y position in relation to total global height
perY = ((pos['y']-p0.pos['y'])/(p1.pos['y'] - p0.pos['y']))
# Returns the screen position based on reference points
return {
'x': p0.scrX + (p1.scrX - p0.scrX)*perX,
'y': p0.scrY + (p1.scrY - p0.scrY)*perY
}
pos = latlngToScreenXY(52.525607, 13.404572);
pos['x] and pos['y] contain the translated x & y coordinates of the lat & lng (52.525607, 13.404572)
I hope this is helpful for anyone looking like me for the proper solution to the problem of translating lat lng into a local reference coordinate system.
Best
Its better to convert to utm coordinates, and treat that as x and y.
import utm
u = utm.from_latlon(12.917091, 77.573586)
The result will be (779260.623156606, 1429369.8665238516, 43, 'P')
The first two can be treated as x,y coordinates, the 43P is the UTM Zone, which can be ignored for small areas (width upto 668 km).
I want to animate a sprite from point y1 to point y2 with some sort of deceleration. when it reaches point y2, the speed of the object will be 0 so it will completely stop.
I Know the two points, and I know the object's starting speed.
The animation time is not so important to me. I can decide on it if needed.
for example: y1 = 0, y2 = 400, v0 = 250 pixels per second (= starting speed)
I read about easing functions but I didn't understand how do I actually implement it in the
update loop.
here's my update loop code with the place that should somehow implement an easing function.
-(void)onTimerTick{
double currentTime = CFAbsoluteTimeGetCurrent() ;
float timeDelta = self.lastUpdateTime - currentTime;
self.lastUpdateTime = currentTime;
float *pixelsToMove = ???? // here needs to be some formula using v0, timeDelta, y2, y1
sprite.y += pixelsToMove;
}
Timing functions as Bézier curves
An easing timing function is basically a Bézier curve from (0,0) to (1,1) where the horizontal axis is "time" and the vertical axis is "amount of change". Since a Bézier curve mathematically is as
start*(1-t)^3 + c1*t(1-t)^2 + c2*t^2(1-t) + end*t^3
you can insert any time value and get the amount of change that should be applied. Note that both time and change is normalized (in the range of 0 to 1).
Note that the variable t is not the time value, t is how far along the curve you have come. The time value is the x value of the point along the curve.
The curve below is a sample "ease" curve that starts off slow, goes faster and slows down in the end.
If for example a third of the time had passed you would calculate what amount of change that corresponds to be update the value of the animated property as
currentValue = beginValue + amountOfChange*(endValue-beginValue)
Example
Say you are animating the position from (50, 50) to (200, 150) using a curve with control points at (0.6, 0.0) and (0.5, 0.9) and a duration of 4 seconds (the control points are trying to be close to that of the image above).
When 1 second of the animation has passed (25% of total duration) the value along the curve is:
(0.25,y) = (0,0)*(1-t)^3 + (0.6,0)*t(1-t)^2 + (0.5,0.9)*t^2(1-t) + (1,1)*t^3
This means that we can calculate t as:
0.25 = 0.6*t(1-t)^2 + 0.5*t^2(1-t) + t^3
Wolfram Alpha tells me that t = 0.482359
If we the input that t in
y = 0.9*t^2*(1-t) + t^3
we will get the "amount of change" for when 1 second of the duration has passed.
Once again Wolfram Alpha tells me that y = 0.220626 which means that 22% of the value has changed after 25% of the time. This is because the curve starts out slow (you can see in the image that it is mostly flat in the beginning).
So finally: 1 second into the animation the position is
(x, y) = (50, 50) + 0.220626 * (200-50, 150-50)
(x, y) = (50, 50) + 0.220626 * (150, 100)
(x, y) = (50, 50) + (33.0939, 22.0626)
(x, y) = (50+33.0939, 50+22.0626)
(x, y) = (83.0939, 72.0626)
I hope this example helps you understanding how to use timing functions.