How'd I go about this one? I want to tween a value from one to another in x time. While also taking into account that it'd be nice to have an 'ease' at the start and end.
I know, I shouldn't ask really, but I've tried myself, and I'm stuck.
Please assume that to cause a delay, you need to call function wait(time).
One simple approach that might work for you is to interpolate along the unit circle:
To do this, you simply evaluate points along the circle, which ensures a fairly smooth movement, and ease-in as well as ease-out. You can control the speed of the interpolation by changing how quickly you alter the angle.
Assuming you're doing 1-dimensional interpolation (i.e. a simple scalar interpolation, like from 3.5 to 6.9 or whatever), it might be handy to use Y-values from -π/2 to π/2. These are given by the sine function, all you need to do is apply suitable scaling:
angle = -math.pi / 2
start = 3.5
end = 6.9
radius = (end - start) / 2
value = start + radius + radius * math.sin(angle)
I'm not 100% sure if this is legal Lua, didn't test it. If not, it's probably trivial to convert.
You may look at Tweener ActionScript library for inspiration.
For instance, you may borrow necessary equations from here.
If you need further help, please ask.
Related
I am new to LabVIEW and I am trying to read a code written in LabVIEW. The block diagram is this:
This is the program to input x and y functions into the voltage input. It is meant to give an input voltage in different forms (sine, heartshape , etc.) into the fast-steering mirror or galvano mirror x and y axises.
x and y function controls are for inputting a formula for a function, and then we use "evaluation single value" function to input into a daq assistant.
I understand that { 2*(|-Mpi|)/N }*i + -Mpi*pi goes into the x value. However, I dont understand why we use this kind of formula. Why we need to assign a negative value and then do the absolute value of -M*pi. Also, I don`t understand why we need to divide to N and then multiply by i. And finally, why need to add -Mpi again? If you provide any hints about this I would really appreciate it.
This is just a complicated way to write the code/formula. Given what the code looks like (unnecessary wire bends, duplicate loop-input-tunnels, hidden wires, unnecessary coercion dots, failure to use appropriate built-in 'negate' function) not much care has been given in writing it. So while it probably yields the correct results you should not expect it to do so in the most readable way.
To answer you specific questions:
Why we need to assign a negative value and then do the absolute value
We don't. We can just move the negation immediately before the last addition or change that to a subtraction:
{ 2*(|Mpi|)/N }*i - Mpi*pi
And as #yair pointed out: We are not assigning a value here, we are basically flipping the sign of whatever value the user entered.
Why we need to divide to N and then multiply by i
This gives you a fraction between 0 and 1, no matter how many steps you do in your for-loop. Think of N as a sampling rate. I.e. your mirrors will always do the same movement, but a larger N just produces more steps in between.
Why need to add -Mpi again
I would strongly assume this is some kind of quick-and-dirty workaround for a bug that has not been fixed properly. Looking at the code it seems this +Mpi*pi has been added later on in the development process. And while I don't know what the expected values are I would believe that multiplying only one of the summands by Pi is probably wrong.
I am currently trying to solve a complementarity problem with a function that features a downward discontinuity, using the mcpsolve() function of the NLsolve package in Julia. The function is reproduced here for specific parameters, and the numbers below refer to the three panels of the figure.
Unfortunately, the algorithm does not always return the interior solution, even though it exists:
In (1), when starting at 0, the algorithm stays at 0, thinking that the boundary constraint binds,
In (2), when starting at 0, the algorithm stops right before the downward jump, even though the solution lies to the right of this point.
These problems occur regardless of the method used - trust region or Newton's method. Ideally, the algorithm would look for potential solutions in the entire set before stopping.
I was wondering if some of you had worked with similar functions, and had found a clever solution to bypass these issues. Note that
Starting to the right of the solution would not solve these problems, as they would also occur for other parametrization - see (3) this time,
It is not known a priori where in the parameter space the particular cases occur.
As an illustrative example, consider the following piece of code. Note that the function is smoother, and yet here as well the algorithm cannot find the solution.
function f!(x,fvec)
if x[1] <= 1.8
fvec[1] = 0.1 * (sin(3*x[1]) - 3)
else
fvec[1] = 0.1 * (x[1]^2 - 7)
end
end
NLsolve.mcpsolve(f!,[0.], [Inf], [0.], reformulation = :smooth, autodiff = true)
Once more, setting the initial value to something else than 0 would only postpone the problem. Also, as pointed out by Halirutan, fzero from the Roots package would probably work, but I'd rather use mcpsolve() as the problem is initially a complementarity problem.
Thank you very much in advance for your help.
I'm attempting to gauge the percentage difference between two images.
Having done a lot of reading I seem to have a number of options but I'm not sure what the best method to follow for:
Ease of coding
Performance.
The methods I've seen are:
Non language specific - academic Image comparison - fast algorithm and Mac specific direct pixel access http://www.markj.net/iphone-uiimage-pixel-color/
Does anyone have any advice about what solutions make most sense for the above two cases and have code samples to show how to apply them?
I've had success calculating the difference between two images using the histogram technique mentioned here. redmoskito's answer in the SO question you linked to was actually my inspiration!
The following is an overview of the algorithm I used:
Convert the images to grayscale—compare one channel instead of three.
Divide each image into an n * n grid of "subimages". Then, for subimage pair:
Calculate their colour composition histograms.
Calculate the absolute difference between the two histograms.
The maximum difference found between two subimages is a measure of the two images' difference. Other metrics could also be used (e.g. the average difference betwen subimages).
As tskuzzy noted in his answer, if your ultimate goal is a binary "yes, these two images are (roughly) the same" or "no, they're not", you need some meaningful threshold value. You could produce such a value by passing images into the algorithm and tweaking the threshold based on its output and how similar you think the images are. A form of machine learning, I suppose.
I recently wrote a blog post on this very topic, albeit as part of a larger goal. I also created a simple iPhone app to demonstrate the algorithm. You can find the source on GitHub; perhaps it will help?
It is really difficult to suggest something when you don't tell us more about the images or the variations. Are they shapes? Are they the different objects and you want to know what class of objects? Are they the same object and you want to distinguish the object instance? Are they faces? Are they fingerprints? Are the objects in the same pose? Under the same illumination?
When you say performance, what exactly do you mean? How large are the images? All in all it really depends. With what you've said if it is only ease of coding and performance I would suggest to just find the absolute value of the difference of pixels. That is super easy to code and about as fast as it gets, but really unlikely to work for anything other than the most synthetic examples.
That being said I would like to point you to: DHOG, GLOH, SURF and SIFT.
You can use fairly basic subtraction technique that the lads above suggested. #carlosdc has hit the nail on the head with regard to the type of image this basic technique can be used for. I have attached an example so you can see the results for yourself.
The first shows a image from a simulation at some time t. A second image was subtracted away from the first which was taken some (simulation) time later t + dt. The subtracted image (in black and white for clarity) then shows how the simulation has changed in that time. This was done as described above and is very powerful and easy to code.
Hope this aids you in some way
This is some old nasty FORTRAN, but should give you the basic approach. It is not that difficult at all. Due to the fact that I am doing it on a two colour pallette you would do this operation for R, G and B. That is compute the intensities or values in each cell/pixal, store them in some array. Do the same for the other image, and subtract one array from the other, this will leave you with some coulorfull subtraction image. My advice would be to do as the lads suggest above, compute the magnitude of the sum of the R, G and B componants so you just get one value. Write that to array, do the same for the other image, then subtract. Then create a new range for either R, G or B and map the resulting subtracted array to this, the will enable a much clearer picture as a result.
* =============================================================
SUBROUTINE SUBTRACT(FNAME1,FNAME2,IOS)
* This routine writes a model to files
* =============================================================
* Common :
INCLUDE 'CONST.CMN'
INCLUDE 'IO.CMN'
INCLUDE 'SYNCH.CMN'
INCLUDE 'PGP.CMN'
* Input :
CHARACTER fname1*(sznam),fname2*(sznam)
* Output :
integer IOS
* Variables:
logical glue
character fullname*(szlin)
character dir*(szlin),ftype*(3)
integer i,j,nxy1,nxy2
real si1(2*maxc,2*maxc),si2(2*maxc,2*maxc)
* =================================================================
IOS = 1
nomap=.true.
ftype='map'
dir='./pictures'
! reading first image
if(.not.glue(dir,fname2,ftype,fullname))then
write(*,31) fullname
return
endif
OPEN(unit2,status='old',name=fullname,form='unformatted',err=10,iostat=ios)
read(unit2,err=11)nxy2
read(unit2,err=11)rad,dxy
do i=1,nxy2
do j=1,nxy2
read(unit2,err=11)si2(i,j)
enddo
enddo
CLOSE(unit2)
! reading second image
if(.not.glue(dir,fname1,ftype,fullname))then
write(*,31) fullname
return
endif
OPEN(unit2,status='old',name=fullname,form='unformatted',err=10,iostat=ios)
read(unit2,err=11)nxy1
read(unit2,err=11)rad,dxy
do i=1,nxy1
do j=1,nxy1
read(unit2,err=11)si1(i,j)
enddo
enddo
CLOSE(unit2)
! substracting images
if(nxy1.eq.nxy2)then
nxy=nxy1
do i=1,nxy1
do j=1,nxy1
si(i,j)=si2(i,j)-si1(i,j)
enddo
enddo
else
print *,'SUBSTRACT: Different sizes of image arrays'
IOS=0
return
endif
* normal finishing
IOS=0
nomap=.false.
return
* exceptional finishing
10 write (*,30) fullname
return
11 write (*,32) fullname
return
30 format('Cannot open file ',72A)
31 format('Improper filename ',72A)
32 format('Error reading from file ',72A)
end
! =============================================================
Hope this is of some use. All the best.
Out of the methods described in your first link, the histogram comparison method is by far the simplest to code and the fastest. However key point matching will provide far more accurate results since you want to know a precise number describing the difference between two images.
To implement the histogram method, I would do the following:
Compute the red, green, and blue histograms of each image
Add up the differences between each bucket
If the difference is above a certain threshold, then the percentage is 0%
Otherwise the colors found in the images are similar. So then do a pixel by pixel comparison and convert the difference into a percentage.
I don't know any precise algorithms for finding the key points of an image. However once you find them for each image you can do a pixel by pixel comparison for each of the key points.
I have a projectile that I would like to pass through specific coordinates at the apex of its path. I have been using a superb equation that giogadi outlined here, by plugging in the velocity values it produces into chipmunk's cpBodyApplyImpulse function.
The equation has one drawback that I haven't been able to figure out. It only works when the coordinates that I want to hit have a y value higher than the cannon (where my projectile starts). This means that I can't shoot at a downward angle.
Can anybody help me find a suitable equation that works no matter where the target is in relation to the cannon?
As pointed out above, there isn't any way to make the apex be lower than the height of the cannon (without making gravity work backwards). However, it is possible to make the projectile pass through a point below the cannon; the equations are all here. The equation you need to solve is:
angle = arctan((v^2 [+-]sqrt(v^4 - g*(x^2+2*y*v^2)))/g*x)
where you choose a velocity and plug in the x and y positions of the target - assuming the cannon is at (0,0). The [+-] thing means that you can choose either root. If the argument to the square root function is negative (an imaginary root) you need a larger velocity. So, if you are "in range" you have two possible angles for any particular velocity (other than in the maximum range 45 degree case where the two roots should give the same answer).
I suspect one trajectory will tend to 'look' much more sensible than the other, but that's something to play around with once you have something working. You may want to stick with the apex grazing code for the cases where the target is above the cannon.
I have a game I am working on that has homing missiles in it. At the moment they just turn towards their target, which produces a rather dumb looking result, with all the missiles following the target around.
I want to create a more deadly flavour of missile that will aim at the where the target "will be" by the time it gets there and I am getting a bit stuck and confused about how to do it.
I am guessing I will need to work out where my target will be at some point in the future (a guess anyway), but I can't get my head around how far ahead to look. It needs to be based on how far the missile is away from the target, but the target it also moving.
My missiles have a constant thrust, combined with a weak ability to turn. The hope is they will be fast and exciting, but steer like a cow (ie, badly, for the non HitchHiker fans out there).
Anyway, seemed like a kind of fun problem for Stack Overflow to help me solve, so any ideas, or suggestions on better or "more fun" missiles would all be gratefully received.
Next up will be AI for dodging them ...
What you are suggesting is called "Command Guidance" but there is an easier, and better way.
The way that real missiles generally do it (Not all are alike) is using a system called Proportional Navigation. This means the missile "turns" in the same direction as the line-of-sight (LOS) between the missile and the target is turning, at a turn rate "proportional" to the LOS rate... This will do what you are asking for as when the LOS rate is zero, you are on collision course.
You can calculate the LOS rate by just comparing the slopes of the line between misile and target from one second to the next. If that slope is not changing, you are on collision course. if it is changing, calculate the change and turn the missile by a proportionate angular rate... you can use any metrics that represent missile and target position.
For example, if you use a proportionality constant of 2, and the LOS is moving to the right at 2 deg/sec, turn the missile to the right at 4 deg/sec. LOS to the left at 6 deg/sec, missile to the left at 12 deg/sec...
In 3-d problem is identical except the "Change in LOS Rate", (and resultant missile turn rate) is itself a vector, i.e., it has not only a magnitude, but a direction (Do I turn the missile left, right or up or down or 30 deg above horizontal to the right, etc??... Imagine, as a missile pilot, where you would "set the wings" to apply the lift...
Radar guided missiles, which "know" the rate of closure. adjust the proportionality constant based on closure (the higher the closure the faster the missile attempts to turn), so that the missile will turn more aggressively in high closure scenarios, (when the time of flight is lower), and less aggressively in low closure (tail chases) when it needs to conserve energy.
Other missiles (like Sidewinders), which do not know the closure, use a constant pre-determined proportionality value). FWIW, Vietnam era AIM-9 sidewinders used a proportionality constant of 4.
I've used this CodeProject article before - it has some really nice animations to explain the math.
"The Mathematics of Targeting and Simulating a Missile: From Calculus to the Quartic Formula":
http://www.codeproject.com/KB/recipes/Missile_Guidance_System.aspx
(also, hidden in the comments at the bottom of that article is a reference to some C++ code that accomplishes the same task from the Unreal wiki)
Take a look at OpenSteer. It has code to solve problems like this. Look at 'steerForSeek' or 'steerForPursuit'.
Have you considered negative feedback on the recent change of bearing over change of time?
Details left as an exercise.
The suggestions is completely serious: if the target does not maneuver this should obtain a near optimal intercept. And it should converge even if the target is actively dodging.
Need more detail?
Solving in a two dimensional space for ease of notation. Take \vec{m} to be the location of the missile and vector \vec{t} To be the location of the target.
The current heading in the direction of motion over last time unit: \vec{h} = \bar{\vec{m}_i - \vec{m}_i-1}}. Let r be the normlized vector between the missile and the target: \vec{r} = \bar{\vec{t} - \vec{m}}. The bearing is b = \vec{r} \dot \vec{h} Compute the bearing at each time tick, and the change thereof, and change heading to minimize that quantity.
The math is harrier in 3d because of the need to find the plane of action at each step, but the process is the same.
You'll want to interpolate the trajectory of both the target and the missile as a function of time. Then look for the times in which the coordinates of the objects are within some acceptable error.